The last thing I discussed in the last post was about the energy eigenstates of the continuous field. The ground state $|0\rangle$ classically corresponds to there being no displacement in the chain at any spatial index $x$ and quantum mechanically corresponds to each oscillator for each normal mode index $k$ being in its ground state, while the first excited state $|k\rangle = a^{\dagger} (k)|0\rangle$ for a given $k$ classically corresponds to a traveling plane wave normal mode of wavevector $k$ and quantum mechanically corresponds to only the oscillator at the given normal mode index $k$ being in its first excited state (and all others being in their ground states). The excited state $|k\rangle$ has energy $E = \hbar v|k|$ above the ground state and overall momentum $p = \hbar k$ above the ground state. This post will discuss what the second and higher excited states are. Follow the jump to see more.
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Showing posts with label physics. Show all posts
Showing posts with label physics. Show all posts
Tuesday, 27 August 2013
Monday, 26 August 2013
Operators and States of the Continuous Quantum Field
Posted on 19:43 by Unknown
In my last post about intuiting and visualizing quantum field theory, I discussed the diagonalization of the Hamiltonian and overall momentum and how they become operators. In this post I'm going to discuss more the meanings of the operators and associated quantum states of this field. Follow the jump to see more.
Read more »Saturday, 24 August 2013
Diagonalizing and Quantizing the Continuous Field Hamiltonian
Posted on 19:34 by Unknown
In my previous post I discussed the intuition behind the classical acoustic field in one dimension. Now I'm going to talk about diagonalizing the Hamiltonian and making the step into quantum field theory. Follow the jump to see what it's like.
Read more »Friday, 23 August 2013
Classical Discrete and Continuum Fields
Posted on 09:45 by Unknown
I've been reading various documents about quantum field theory over the last several weeks, specifically about the canonical quantization of quantum fields. In doing so, I've come to realize that quantum mechanics has a lot of crazy math and even crazier physical interpretations, and I just took that for granted, but now those things are coming back to haunt me in quantum field theory. It is very hard for me to wrap my head around, and I feel like I could use a lot more help in visualizing and intuiting what certain concepts in canonical quantization mean. This will be the first of a few posts which are outlets for me to gather my thoughts and put them out there for you all to see and correct; this one will be about classical fields.
I feel like the easiest quantum field system to study is the phonon. It is a spin-0 bosonic system, so it can be described by a scalar field. Furthermore, said field can be restricted to one dimension, which simplifies the math even further. This means that taking the continuum limit becomes a bit easier than in three dimensions. Follow the jump to see how it goes.
Read more »I feel like the easiest quantum field system to study is the phonon. It is a spin-0 bosonic system, so it can be described by a scalar field. Furthermore, said field can be restricted to one dimension, which simplifies the math even further. This means that taking the continuum limit becomes a bit easier than in three dimensions. Follow the jump to see how it goes.
Friday, 16 August 2013
Reflection: 2013 Summer UROP
Posted on 13:33 by Unknown
Wow. This summer has been incredibly busy, productive, and fun all at once. I can't believe it's already over!
So what did I do this summer?
My primary concern this summer was my UROP. I have been able to bring it very close to an end point; I wasn't able to finish it up completely, but I guess that was an unrealistic expectation because that's just not how science works. It doesn't wrap up cleanly; it's an ongoing process. I learned a whole lot more about Scheme and MEEP in the process, though, which was great.
On a related note, another UROP project fell by the wayside (as I wrote about earlier this summer) once I realized it was based on flawed calculations. To be honest, I'm not really sure if I want to pick up that project again and try to bring it to some sort of conclusion or if that's really worth my time.
My secondary concern was preparing for graduate school. I took the GRE this past Tuesday, and I am happy to say that went quite well. I have also been studying for the Physics GRE, along with making my list of graduate schools/programs/professors that I want to further investigate and send applications.
My tertiary concern was making another video for the MIT-K12 project. That went off successfully as well.
Apart from that, not being around my usual set of friends for the summer had a silver lining. While I would have certainly liked to have been able to hang out with them more, I was able to become a lot closer to a few people who usually live on my floor during the semester and hang out with them a lot more. Compared to the end of last semester, where I would basically just say "hi" to them but not a whole lot more, I now intend to hang out with them significantly more during this coming semester.
What didn't I do? These things didn't happen because I didn't have the time or energy to carry them out.
I wasn't able to edit and publish all the videos I took of 8.033 lectures from 2 years ago. In fact, I couldn't really look at those at all.
I wasn't able to do much work for OCW as I had planned.
(Actually, that's mostly it.)
I'm excited for the coming semester. My classes all look quite exciting, and I'm still deciding what I want to do regarding my UROP once my current project can truly said to be concluded. That said, I feel a bit sad that this has been my last summer at MIT, and it is already over. After that, I only have 9 more months at this place. I hope I can make those 9 months really special. Before that, though, I'll be going on a vacation with my family for a few days and then spending the remaining 1.5 weeks of August at home. Yay!
So what did I do this summer?
My primary concern this summer was my UROP. I have been able to bring it very close to an end point; I wasn't able to finish it up completely, but I guess that was an unrealistic expectation because that's just not how science works. It doesn't wrap up cleanly; it's an ongoing process. I learned a whole lot more about Scheme and MEEP in the process, though, which was great.
On a related note, another UROP project fell by the wayside (as I wrote about earlier this summer) once I realized it was based on flawed calculations. To be honest, I'm not really sure if I want to pick up that project again and try to bring it to some sort of conclusion or if that's really worth my time.
My secondary concern was preparing for graduate school. I took the GRE this past Tuesday, and I am happy to say that went quite well. I have also been studying for the Physics GRE, along with making my list of graduate schools/programs/professors that I want to further investigate and send applications.
My tertiary concern was making another video for the MIT-K12 project. That went off successfully as well.
Apart from that, not being around my usual set of friends for the summer had a silver lining. While I would have certainly liked to have been able to hang out with them more, I was able to become a lot closer to a few people who usually live on my floor during the semester and hang out with them a lot more. Compared to the end of last semester, where I would basically just say "hi" to them but not a whole lot more, I now intend to hang out with them significantly more during this coming semester.
What didn't I do? These things didn't happen because I didn't have the time or energy to carry them out.
I wasn't able to edit and publish all the videos I took of 8.033 lectures from 2 years ago. In fact, I couldn't really look at those at all.
I wasn't able to do much work for OCW as I had planned.
(Actually, that's mostly it.)
I'm excited for the coming semester. My classes all look quite exciting, and I'm still deciding what I want to do regarding my UROP once my current project can truly said to be concluded. That said, I feel a bit sad that this has been my last summer at MIT, and it is already over. After that, I only have 9 more months at this place. I hope I can make those 9 months really special. Before that, though, I'll be going on a vacation with my family for a few days and then spending the remaining 1.5 weeks of August at home. Yay!
Posted in break, college, crystal, internship, MIT, photonic, physics, quantum mechanics, Reflection, science, UROP
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Wednesday, 17 July 2013
Skepticism in My Photonics UROP
Posted on 14:47 by Unknown
I've mentioned here on several occasions that I've been doing a UROP regarding nanophotonics/photonic crystals. Specifically, my first project was in determining whether particular types of photonic structures might enhance absorptivity of light, which would help solar cells that convert that light into electricity. The goal is that the absorptivity enhancement (versus no texturing over the solar cell) should be over as broad of a frequency band as possible, because it is difficult to manufacture many different kinds of photonic structures just to satisfy performance demands over many narrow frequency bands. I worked on this project for about 4 months, because that was the time that my postdoctoral UROP supervisor was around (he moved after that). I was able to get some really nice-looking results in that time, and I figured there wasn't much more I needed to do to wrap it up, so I felt comfortable generally moving on to a new project (which I have been working on since 2012 February or so). The enhancement results looked great compared to existing designs, so I thought we might be on to something here. Recently, things started gearing up for a publication submission.
Today, it all came crashing down. Why? Another postdoctoral UROP supervisor (who I have worked with since last year primarily on my more current project but recently joined in to help progress of the older project, which is the subject of this post) asked me some hard questions about what I was really doing. Because of this, I realized that one of the parameter choices in my calculations that were giving such nice results was fatally flawed. When I fixed that issue, the results I was getting suddenly looked significantly less compelling; with that, any dreams of publication were dashed.
Why did this happen? It boils down to me not being skeptical enough about what was going on. A large part of this has to do with the fact that because this was my first UROP, I didn't have a great idea of what was going on in terms of details. And because that time I spent was only about 4 months and was followed immediately by a new project, I didn't spend much more time on that project after that. Ultimately I got complacent in more ways than one. Because the code I was using was based on existing code for similar calculations, I figured it must have been written to work even with the modifications I was making. I also figured that because I had been getting consistently good results from what I had done over those 4 months, I just needed to worry about those results on the surface and not the fundamentals operation of the code. Those two assumptions combined such that even though I had seen the results of not being careful in my second UROP project and had become much more careful about checking that code as a result, I didn't think I needed to apply the same level of care in checking the code used for the first project.
After realizing the implications of this, I did a few more calculations in a significantly more mopey mood. But then I thought about this and I realize that I shouldn't feel so bad about this. Why is that? Here are a few reasons in no particular order.
1. I've made similar mistakes before, and I've really come to learn from them. One example of something that I thought was going great but turned out badly has to do with email. People who know me may have heard this story, and people who have known me for a while may have actually been there to see me do this, but I won't share the story now; I believe it is sufficient to say that I am now a lot more careful when sending emails especially to large groups of people. A reverse example actually comes from my second UROP project: for a while I was making a mistake in my code that was giving garbage, but after many months of trying various fixes, one particular fix solved all the other issues. Since then I have been a lot more careful about checking my code for that project (though I guess I was confident enough about the code used in the first project that I thought such a high level of care might be unnecessary).
2. I didn't think I would be in the position of having my work for both projects on paper until very recently. Now I can go back to thinking that in any case my second project work would be more likely to go on paper (especially as I know that I have taken a lot more care in checking my code for that project).
3. Several months ago, when my second project was stalling, I was asking myself why it wasn't going as smoothly or quickly as my first project. Now I know that the first project should have in fact gone as slowly as the second project for the work to become as solid and carefully checked. The other part of this issue is that I have been working on the second project continuously, so I have been able to make continuous adjustments to the code and work progressively higher levels of care in checking the code in a smooth manner. Because I essentially stopped working on the first project after those first four months, if I adopt more careful code-checking now, it'll feel more like I'm starting over from scratch, which makes the process feel a lot more frustrating.
4. With all this, I feel like I have already learned a lot more from this lesson than I would have if everything was fine and dandy and this work did get submitted for publication.
5. If nothing else, I hold out hope that I may be able to salvage some good results with the fixes I have made to the code of the first project.
There are two morals to this story. The first is that I shouldn't just check the code I run; I should check it in an actively skeptical manner, always questioning each and every line. The second is that C++ is way more painful to read and (to a lesser extent) write than Scheme is for the kinds of calculations I run.
Today, it all came crashing down. Why? Another postdoctoral UROP supervisor (who I have worked with since last year primarily on my more current project but recently joined in to help progress of the older project, which is the subject of this post) asked me some hard questions about what I was really doing. Because of this, I realized that one of the parameter choices in my calculations that were giving such nice results was fatally flawed. When I fixed that issue, the results I was getting suddenly looked significantly less compelling; with that, any dreams of publication were dashed.
Why did this happen? It boils down to me not being skeptical enough about what was going on. A large part of this has to do with the fact that because this was my first UROP, I didn't have a great idea of what was going on in terms of details. And because that time I spent was only about 4 months and was followed immediately by a new project, I didn't spend much more time on that project after that. Ultimately I got complacent in more ways than one. Because the code I was using was based on existing code for similar calculations, I figured it must have been written to work even with the modifications I was making. I also figured that because I had been getting consistently good results from what I had done over those 4 months, I just needed to worry about those results on the surface and not the fundamentals operation of the code. Those two assumptions combined such that even though I had seen the results of not being careful in my second UROP project and had become much more careful about checking that code as a result, I didn't think I needed to apply the same level of care in checking the code used for the first project.
After realizing the implications of this, I did a few more calculations in a significantly more mopey mood. But then I thought about this and I realize that I shouldn't feel so bad about this. Why is that? Here are a few reasons in no particular order.
1. I've made similar mistakes before, and I've really come to learn from them. One example of something that I thought was going great but turned out badly has to do with email. People who know me may have heard this story, and people who have known me for a while may have actually been there to see me do this, but I won't share the story now; I believe it is sufficient to say that I am now a lot more careful when sending emails especially to large groups of people. A reverse example actually comes from my second UROP project: for a while I was making a mistake in my code that was giving garbage, but after many months of trying various fixes, one particular fix solved all the other issues. Since then I have been a lot more careful about checking my code for that project (though I guess I was confident enough about the code used in the first project that I thought such a high level of care might be unnecessary).
2. I didn't think I would be in the position of having my work for both projects on paper until very recently. Now I can go back to thinking that in any case my second project work would be more likely to go on paper (especially as I know that I have taken a lot more care in checking my code for that project).
3. Several months ago, when my second project was stalling, I was asking myself why it wasn't going as smoothly or quickly as my first project. Now I know that the first project should have in fact gone as slowly as the second project for the work to become as solid and carefully checked. The other part of this issue is that I have been working on the second project continuously, so I have been able to make continuous adjustments to the code and work progressively higher levels of care in checking the code in a smooth manner. Because I essentially stopped working on the first project after those first four months, if I adopt more careful code-checking now, it'll feel more like I'm starting over from scratch, which makes the process feel a lot more frustrating.
4. With all this, I feel like I have already learned a lot more from this lesson than I would have if everything was fine and dandy and this work did get submitted for publication.
5. If nothing else, I hold out hope that I may be able to salvage some good results with the fixes I have made to the code of the first project.
There are two morals to this story. The first is that I shouldn't just check the code I run; I should check it in an actively skeptical manner, always questioning each and every line. The second is that C++ is way more painful to read and (to a lesser extent) write than Scheme is for the kinds of calculations I run.
Monday, 3 June 2013
FOLLOW-UP: Rebutting the New York Times Review of "A Universe From Nothing"
Posted on 18:37 by Unknown
Somehow a few months ago, my family discovered the existence of this blog, and the first post they read was this one, for which this post is a follow-up. They then bought the aforementioned book for me so that I could read it and perhaps understand the criticisms laid out in the New York Times article that I rebutted.
Well, I read the book. It was a fairly interesting read. Having taken the class 8.033 — Relativity, I would say that everything up to around the middle of the book is stuff I've seen before in the context of general relativity. After that comes some stuff that is new to me, like the ideas behind inflation, virtual particles, and how multiverses can be explained within the framework of quantum field theory. They were all new and fairly strange ideas, but I accepted them fairly easily because it was clear how they fit into the framework of quantum field theory. After finishing the book, I went back to read the book review as well as my rebuttal of it, and I have to say that in many ways, the book review looks even slightly more ridiculous than before, and I'm actually quite happy with the assessment I laid out about a year ago.
Well, I read the book. It was a fairly interesting read. Having taken the class 8.033 — Relativity, I would say that everything up to around the middle of the book is stuff I've seen before in the context of general relativity. After that comes some stuff that is new to me, like the ideas behind inflation, virtual particles, and how multiverses can be explained within the framework of quantum field theory. They were all new and fairly strange ideas, but I accepted them fairly easily because it was clear how they fit into the framework of quantum field theory. After finishing the book, I went back to read the book review as well as my rebuttal of it, and I have to say that in many ways, the book review looks even slightly more ridiculous than before, and I'm actually quite happy with the assessment I laid out about a year ago.
Posted in Book Review, FOLLOW-UP, philosophy, physics, quantum electrodynamics, quantum mechanics, religion, science
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Monday, 6 May 2013
Expected Utility in Quantum States
Posted on 07:12 by Unknown
Last semester, 14.04 — Intermediate Microeconomic Theory covered choice theory under uncertainty; at the same time, I was taking 8.05 — Quantum Physics II, where we had talked about 2-state systems and the formalism of quantum states being complex vectors in a Hilbert space, and as choice under uncertainty discusses how consumers make choices based on states of the world, I thought it would be cool to extend it to quantum states, but I wasn't sure how to do that at that time. Now that 14.03 — Microeconomic Theory and Public Policy is talking about choice under uncertainty as well this semester, and now that I have had some time for the stuff about 2-state systems to simmer in my head, I think I have a slightly better idea of how to think about extended the states of the world to include the superpositions as allowed by quantum mechanics.
One of the simplest examples would be the 2-state system. In the class I am taking now, a typical 2-state system would be a fair coin which can either take on values of heads or tails (each with probability $\frac{1}{2}$). What might this look like in quantum mechanics? We could replace the coin with the spin $S_z$ of an electron. A state of the electron corresponding to being measured as $|\uparrow_z \rangle$ or $|\downarrow_z \rangle$ each with equal probability could be labeled as $|\psi \rangle = \frac{1}{\sqrt{2}} \left(|\uparrow_z \rangle + |\downarrow_z \rangle\right)$. This indeed gives equal probabilities of being measured as spin-up or spin-down in the $z$-direction. The problem is that this state is exactly $|\psi \rangle = |\uparrow_x \rangle$, so the probability of the state being measured in the $x$-direction as spin-up is unity. This leads to issues in trying to reconcile the interpretation of payoffs for different states of the world; this particular state would pay $w(|\uparrow_z \rangle)$ or $w(|\downarrow_z \rangle)$ with equal probabilities until $S_z$ is measured, but would pay $w(|\uparrow_x \rangle)$ with certainty as $S_x$ has essentially already been measured, collapsing a previously unknown state into this one. So there seems to be an issue with trying to stuff the interpretation of probabilistic measurement of a state of the world into the idea of superposing quantum states, as certain measurable states of the world do not commute ($[S_i, S_j] = i\hbar \epsilon_{ijk} S_k$ for instance). So what can be done now?
Recall that each state of the world $j$ has an associated probability $\mathbb{P}_j$. Yet, once a state is measured, those probabilities are meaningless, because a state becomes observed or not observed; it cannot be "observed in fraction $\mathbb{P}_j$" or anything like that. Taking the frequentist view, these probabilities are only meaningful in the large ensemble limit; after a large number of observations, the fraction of those observations corresponding to the state $j$ converges to $\mathbb{P}_j$. This is exactly the same rationale behind the density matrix, which describes the state of an ensemble of systems which are usually not all prepared in the same quantum state, but in the large number limit, the fraction of those prepared in the state $|\psi_j \rangle$ is $\mathbb{P}_j$ such that $\sum_j \mathbb{P}_j = 1$. It is then defined as \[ \rho = \sum_j \mathbb{P}_j |\psi_j \rangle \langle \psi_j | . \] Note that in general $\langle \psi_j | \psi_{j'} \rangle \neq \delta_{jj'}$ because the states do not in general need to be orthogonal. Furthermore, different state preparations can have the same density matrix, in which case the "different" state preparations are actually physically indistinguishable if a quantum mechanical measurement is made.
So what does this have to do with the states of the world? If the example of a fair coin flip is again translated into measuring each eigenvalue of $S_z$ with equal probability, then the density matrix becomes $\rho = \frac{1}{2} \left(|\uparrow_z \rangle \langle \uparrow_z | + |\downarrow_z \rangle \langle \downarrow_z |\right)$. Now it is much more plausible to define a wealth $w(|\psi_j \rangle)$ dependent on the state of the world.
For example, let us now consider an unfair "coin" represented by the density matrix $\rho = \frac{1}{3} \left(|\uparrow_z \rangle \langle \uparrow_z | + 2|\downarrow_z \rangle \langle \downarrow_z |\right)$. What is the probability of measuring this system in the state $|\uparrow_x \rangle$? That would be $\langle \uparrow_x | \rho | \uparrow_x \rangle = \frac{1}{2}$. Similarly, $\langle \downarrow_x | \rho | \downarrow_x \rangle = \frac{1}{2}$. It is interesting to note that equal probabilities are achieved for spin-up and spin-down in the $x$-direction (and $y$-direction as well) for this system; the difference is that now there are off-diagonal terms in the density matrix describing the degree of interference between the states of $S_x$ in the two populations of spins.
What can we do with this? Payoffs now need to be defined over every plausible noncommuting observable in the Hilbert space (because commuting observables yield the same state, and the payoff is really defined for the state). Thankfully this is fairly simple for spins, as the noncommuting observables in question are the components of $\vec{S}$.
Let us return to the previous unfair "coin". Classically (so it would be a coin), the result of the coin flip would be the end of the story. Let us suppose that the consumer had a risk-neutral utility $V(w) = w$. Let us also suppose that there was a game which payed off 20 if heads and 0 if tails and cost 10 to enter. The expected payoff, accounting for the cost of entering, would be $\mathbb{E}(w) = -\frac{10}{3}$, so the consumer would prefer to not play.
Classically that would be the whole story. Quantum mechanically, though, there is more to the story. If the owner of the game decided to measure $S_z$, then the result would be the same as the classical result. If the owner decided instead to measure $S_x$ or $S_y$, then $\mathbb{E}(w) = 0$ so the consumer would be indifferent between playing and not playing, which is certainly a different outcome from preferring not to play. In fact, if the consumer does not know for sure what the owner decides to measure, then probabilities could be assigned to the measurement choice itself, and these could then be used to find the expectation value of payoff or utility over all possible choices, where each of those choices will have an expected payoff or utility as well.
One thing to look further at is how to extend this to include continuous ensembles. The example in class was how a state of the world might be the outdoor temperature measured in some range. The quantum mechanical equivalent might be having a continuum of systems, each infinitesimal one having a position inside a given range of allowed positions; the only issue with this is that matter is not continuous, so I don't think it is possible to have a density matrix in the form of $\rho = \int |\psi(s) \rangle \langle \psi(s)| \, ds$ for some continuous index $s$. However, that can certainly be further investigated, and maybe I'll do that next time. The point is that states of the world in analyzing expected utility can easily be generalized to include quantum states through density matrices, and the question of which observable to measure in a noncommuting set brings out some interesting behavior not seen in classical states.
One of the simplest examples would be the 2-state system. In the class I am taking now, a typical 2-state system would be a fair coin which can either take on values of heads or tails (each with probability $\frac{1}{2}$). What might this look like in quantum mechanics? We could replace the coin with the spin $S_z$ of an electron. A state of the electron corresponding to being measured as $|\uparrow_z \rangle$ or $|\downarrow_z \rangle$ each with equal probability could be labeled as $|\psi \rangle = \frac{1}{\sqrt{2}} \left(|\uparrow_z \rangle + |\downarrow_z \rangle\right)$. This indeed gives equal probabilities of being measured as spin-up or spin-down in the $z$-direction. The problem is that this state is exactly $|\psi \rangle = |\uparrow_x \rangle$, so the probability of the state being measured in the $x$-direction as spin-up is unity. This leads to issues in trying to reconcile the interpretation of payoffs for different states of the world; this particular state would pay $w(|\uparrow_z \rangle)$ or $w(|\downarrow_z \rangle)$ with equal probabilities until $S_z$ is measured, but would pay $w(|\uparrow_x \rangle)$ with certainty as $S_x$ has essentially already been measured, collapsing a previously unknown state into this one. So there seems to be an issue with trying to stuff the interpretation of probabilistic measurement of a state of the world into the idea of superposing quantum states, as certain measurable states of the world do not commute ($[S_i, S_j] = i\hbar \epsilon_{ijk} S_k$ for instance). So what can be done now?
Recall that each state of the world $j$ has an associated probability $\mathbb{P}_j$. Yet, once a state is measured, those probabilities are meaningless, because a state becomes observed or not observed; it cannot be "observed in fraction $\mathbb{P}_j$" or anything like that. Taking the frequentist view, these probabilities are only meaningful in the large ensemble limit; after a large number of observations, the fraction of those observations corresponding to the state $j$ converges to $\mathbb{P}_j$. This is exactly the same rationale behind the density matrix, which describes the state of an ensemble of systems which are usually not all prepared in the same quantum state, but in the large number limit, the fraction of those prepared in the state $|\psi_j \rangle$ is $\mathbb{P}_j$ such that $\sum_j \mathbb{P}_j = 1$. It is then defined as \[ \rho = \sum_j \mathbb{P}_j |\psi_j \rangle \langle \psi_j | . \] Note that in general $\langle \psi_j | \psi_{j'} \rangle \neq \delta_{jj'}$ because the states do not in general need to be orthogonal. Furthermore, different state preparations can have the same density matrix, in which case the "different" state preparations are actually physically indistinguishable if a quantum mechanical measurement is made.
So what does this have to do with the states of the world? If the example of a fair coin flip is again translated into measuring each eigenvalue of $S_z$ with equal probability, then the density matrix becomes $\rho = \frac{1}{2} \left(|\uparrow_z \rangle \langle \uparrow_z | + |\downarrow_z \rangle \langle \downarrow_z |\right)$. Now it is much more plausible to define a wealth $w(|\psi_j \rangle)$ dependent on the state of the world.
For example, let us now consider an unfair "coin" represented by the density matrix $\rho = \frac{1}{3} \left(|\uparrow_z \rangle \langle \uparrow_z | + 2|\downarrow_z \rangle \langle \downarrow_z |\right)$. What is the probability of measuring this system in the state $|\uparrow_x \rangle$? That would be $\langle \uparrow_x | \rho | \uparrow_x \rangle = \frac{1}{2}$. Similarly, $\langle \downarrow_x | \rho | \downarrow_x \rangle = \frac{1}{2}$. It is interesting to note that equal probabilities are achieved for spin-up and spin-down in the $x$-direction (and $y$-direction as well) for this system; the difference is that now there are off-diagonal terms in the density matrix describing the degree of interference between the states of $S_x$ in the two populations of spins.
What can we do with this? Payoffs now need to be defined over every plausible noncommuting observable in the Hilbert space (because commuting observables yield the same state, and the payoff is really defined for the state). Thankfully this is fairly simple for spins, as the noncommuting observables in question are the components of $\vec{S}$.
Let us return to the previous unfair "coin". Classically (so it would be a coin), the result of the coin flip would be the end of the story. Let us suppose that the consumer had a risk-neutral utility $V(w) = w$. Let us also suppose that there was a game which payed off 20 if heads and 0 if tails and cost 10 to enter. The expected payoff, accounting for the cost of entering, would be $\mathbb{E}(w) = -\frac{10}{3}$, so the consumer would prefer to not play.
Classically that would be the whole story. Quantum mechanically, though, there is more to the story. If the owner of the game decided to measure $S_z$, then the result would be the same as the classical result. If the owner decided instead to measure $S_x$ or $S_y$, then $\mathbb{E}(w) = 0$ so the consumer would be indifferent between playing and not playing, which is certainly a different outcome from preferring not to play. In fact, if the consumer does not know for sure what the owner decides to measure, then probabilities could be assigned to the measurement choice itself, and these could then be used to find the expectation value of payoff or utility over all possible choices, where each of those choices will have an expected payoff or utility as well.
One thing to look further at is how to extend this to include continuous ensembles. The example in class was how a state of the world might be the outdoor temperature measured in some range. The quantum mechanical equivalent might be having a continuum of systems, each infinitesimal one having a position inside a given range of allowed positions; the only issue with this is that matter is not continuous, so I don't think it is possible to have a density matrix in the form of $\rho = \int |\psi(s) \rangle \langle \psi(s)| \, ds$ for some continuous index $s$. However, that can certainly be further investigated, and maybe I'll do that next time. The point is that states of the world in analyzing expected utility can easily be generalized to include quantum states through density matrices, and the question of which observable to measure in a noncommuting set brings out some interesting behavior not seen in classical states.
Friday, 26 April 2013
Thoughts on Typesetting
Posted on 17:12 by Unknown
In 2012 IAP, I taught myself how to use LaTeX by typesetting the 8.033 — Relativity lecture notes. I also did this so that the lecturer I had that semester and the lecturer for the following semester would both have these notes at their disposal; for the record, the former is on sabbatical this academic year, while the latter did indeed use it when he taught the class this past fall. I needed to teach myself LaTeX because I was going to be taking 8.13, which I did this past fall, and that requires LaTeX use for papers (and recommends Beamer for presentations as well). That said, recently I was hanging out with some friends and a couple of them suggested that LaTeX isn't really necessary as far as producing scientific papers goes, because Microsoft Office Word 2013 has an advanced enough equation editor that it can essentially replace LaTeX, especially as it now recognizes basic LaTeX syntax. At first, given how much I had used LaTeX (and also given some of my past negative feelings toward Microsoft), I felt a little defensive. But then I realized I should give the comparison a fairer shot, so I decided to see if I could try to replicate one of my PRL-formatted LaTeX-typeset papers in LibreOffice Writer. Follow the jump to see my findings.
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Posted in class, compatibility, dream, installation, LibreOffice, mathematics, microsoft office, physics, science, word
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Thursday, 25 April 2013
Magnetic Field Tensor and Continuum Optical Modes
Posted on 05:57 by Unknown
There is actually a third thing in this post, but I'm not going to list that in the title. Also, the two things in the title are separate and unrelated. Follow the jump to see it all.
Monday, 15 April 2013
Harmonic Oscillator from Fields not Potentials
Posted on 12:57 by Unknown
I finally started writing my paper for 8.06 yesterday. Before that, though, I had asked a couple questions about the topic to my UROP supervisor, whose primary area of expertise is actually in QED and Casimir problems. I was asking him why the book The Quantum Vacuum by Peter Milonni uses the magnetic potential $\vec{A}$ instead of the electromagnetic fields $\vec{E}$ and $\vec{B}$ to expand in Fourier modes and derive the harmonic oscillator Hamiltonian. He said that is just a matter of choice, and in fact the same derivation can be done using the fields rather than the potential; moreover, he encouraged me to try this out for myself, and I did just that. Lo and behold, the right answer popped out by modifying the derivation in that book to use the fields and the restrictions of the Maxwell equations instead of the potential and the Coulomb gauge choice; follow the jump to see what it looks like. I'm going to basically write out the derivation in the book and show how at each point I modify it.
Posted in class, college, MIT, physics, qed, quantum electrodynamics, quantum mechanics
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Tuesday, 9 April 2013
Charge Conservation and Legendre Transformations
Posted on 11:20 by Unknown
As a follow-up (sort of, but not exactly) to my previous post on the matter, I would like to post a few updates and new questions, using Einstein summation throughout for convenience. The first has to do with why $p^{\mu} = \int T^{(0, \mu)} d^3 x$ is a Lorentz-contravariant vector. Apparently Noether's theorem says that if some Noether current $J^{\mu}$ generates a symmetry and satisfies \[ \partial_{\mu} J^{\mu} = 0 \] then the quantity \[ q = \int J^{(0)} d^3 x \] called the Noether charge is Lorentz-invariant and conserved. The first is not easy to show, but apparently some E&M textbooks do it for the example of electric charge. The second is fairly easy to show: using the condition that $\int \partial_{j} J^{j} d^3 x = \int_{\partial \mathbb{R}^3} J^{j} d\mathcal{S}_{j} = 0$ (from the divergence theorem applied to all of Euclidean space) in conjunction with $\partial_{\mu} J^{\mu} = 0$, the result $\dot{q} = 0$ follows.
As an example, let us consider the generator of rotations and Lorentz boosts for a general energy distribution: that is the 3-index angular momentum tensor \[ M^{\mu \nu \sigma} = x^{\mu} T^{\nu \sigma} - x^{\nu} T^{\mu \sigma} .\] Given that $\partial_{\nu} T^{\mu \nu} = 0$ then $\partial_{\sigma} M^{\mu \nu \sigma} = (\partial_{\sigma} x^{\mu})T^{\nu \sigma} + x^{\mu} \partial_{\sigma} T^{\nu \sigma} - (\partial_{\sigma} x^{\nu})T^{\mu \sigma} - x^{\nu} \partial_{\sigma} T^{\mu \sigma}$ $= \delta_{\sigma}^{\; \mu} T^{\nu \sigma} - \delta_{\sigma}^{\; \nu} T^{\mu \sigma} = T^{\nu \mu} - T^{\mu \nu} = 0$. Therefore the 3-index angular momentum is a proper Noether current. Its corresponding conserved charge is the 2-index angular momentum integrated over spatial directions: \[ L^{\mu \nu} = \int M^{(\mu \nu, 0)} d^3 x \] (except for perhaps a sign because $M^{\mu \nu \sigma}$ is antisymmetric in its indices) and it should be easy now to show that $\dot{L}^{\mu \nu} = 0$, which is cool. The only remaining question I have is whether it is more correct to say $L^{\mu \nu} = x^{\mu} p^{\nu} - x^{\nu} p^{\mu}$ where $p^{\mu} = \int T^{(\mu, 0)} d^3 x$ as before or if the better definition is the one integrating $M^{(\mu \nu, 0)}$ over space.
Now I have an even bigger question looming ahead of me though. The Noether current generating spacetime translational symmetry is exactly the stress-energy tensor derivable as the Legendre transformation of the Lagrangian. The term involving the conjugate momenta is easy, but the term involving the Lagrangian is confusing. For a scalar field $\phi$ (and for a vector field this is easily replaced with $A^{\sigma}$), what I have seen is \[ T^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)} \partial^{\nu} \phi - \mathcal{L} B^{\mu \nu} .\] The problem is that the tensor $B^{\mu \nu}$ seems to depend on either the field $\phi$ used or on the notation consistently used. Sometime $B^{\mu \nu} = \delta^{\mu \nu}$, while other times $B^{\mu \nu} = \eta^{\mu \nu}$. I'm not really sure which it is supposed to be, as sometimes for scalar fields $B = \delta$ is used, while for the electromagnetic field $B = \eta$ is used, and sometimes the notation isn't even that consistent. The issue is that either one would properly specify a Lorentz-contravariant 2-index tensor, but only one of them actually defines the translational symmetry Noether current properly. Which one is it? The issue appears to be akin to the problem of two grammatically correct sentences where one carries meaning and makes sense while the other makes no sense at all (e.g. "colorless green ideas sleep furiously").
As an example, let us consider the generator of rotations and Lorentz boosts for a general energy distribution: that is the 3-index angular momentum tensor \[ M^{\mu \nu \sigma} = x^{\mu} T^{\nu \sigma} - x^{\nu} T^{\mu \sigma} .\] Given that $\partial_{\nu} T^{\mu \nu} = 0$ then $\partial_{\sigma} M^{\mu \nu \sigma} = (\partial_{\sigma} x^{\mu})T^{\nu \sigma} + x^{\mu} \partial_{\sigma} T^{\nu \sigma} - (\partial_{\sigma} x^{\nu})T^{\mu \sigma} - x^{\nu} \partial_{\sigma} T^{\mu \sigma}$ $= \delta_{\sigma}^{\; \mu} T^{\nu \sigma} - \delta_{\sigma}^{\; \nu} T^{\mu \sigma} = T^{\nu \mu} - T^{\mu \nu} = 0$. Therefore the 3-index angular momentum is a proper Noether current. Its corresponding conserved charge is the 2-index angular momentum integrated over spatial directions: \[ L^{\mu \nu} = \int M^{(\mu \nu, 0)} d^3 x \] (except for perhaps a sign because $M^{\mu \nu \sigma}$ is antisymmetric in its indices) and it should be easy now to show that $\dot{L}^{\mu \nu} = 0$, which is cool. The only remaining question I have is whether it is more correct to say $L^{\mu \nu} = x^{\mu} p^{\nu} - x^{\nu} p^{\mu}$ where $p^{\mu} = \int T^{(\mu, 0)} d^3 x$ as before or if the better definition is the one integrating $M^{(\mu \nu, 0)}$ over space.
Now I have an even bigger question looming ahead of me though. The Noether current generating spacetime translational symmetry is exactly the stress-energy tensor derivable as the Legendre transformation of the Lagrangian. The term involving the conjugate momenta is easy, but the term involving the Lagrangian is confusing. For a scalar field $\phi$ (and for a vector field this is easily replaced with $A^{\sigma}$), what I have seen is \[ T^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)} \partial^{\nu} \phi - \mathcal{L} B^{\mu \nu} .\] The problem is that the tensor $B^{\mu \nu}$ seems to depend on either the field $\phi$ used or on the notation consistently used. Sometime $B^{\mu \nu} = \delta^{\mu \nu}$, while other times $B^{\mu \nu} = \eta^{\mu \nu}$. I'm not really sure which it is supposed to be, as sometimes for scalar fields $B = \delta$ is used, while for the electromagnetic field $B = \eta$ is used, and sometimes the notation isn't even that consistent. The issue is that either one would properly specify a Lorentz-contravariant 2-index tensor, but only one of them actually defines the translational symmetry Noether current properly. Which one is it? The issue appears to be akin to the problem of two grammatically correct sentences where one carries meaning and makes sense while the other makes no sense at all (e.g. "colorless green ideas sleep furiously").
Posted in class, college, MIT, physics, qed, quantum electrodynamics, quantum mechanics
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Monday, 8 April 2013
Long-Term Review: Chakra 2013.02 "Benz"
Posted on 14:19 by Unknown
I did this long-term review on my normal UROP desktop computer with the 64-bit edition of the OS. Follow the jump to see how it fared. Also do note that there are more days logged because I intend to use it for about 60-80 full hours of work, which is the equivalent of 7-10 full days in the summer, though now I am working on a part-time basis as classes have started.
Read more »Wednesday, 27 March 2013
Hamiltonian Density and the Stress-Energy Tensor
Posted on 13:44 by Unknown
As an update to a previous post about my adventures in QED-land for 8.06, I emailed my recitation leader about whether my intuition about the meaning of the Fourier components of the electromagnetic potential solving the wave equation (and being quantized to the ladder operators) was correct. He said it basically is correct, although there are a few things that, while I kept in mind at that time, I still need to keep in mind throughout. The first is that the canonical quantization procedure uses the potential $\vec{A}$ as the coordinate-like quantity and finds the conjugate momentum to this field to be proportional to the electric field $\vec{E}$, with the magnetic field nowhere to be found directly in the Hamiltonian. The second is that there is a different harmonic oscillator for each mode, and the number eigenstates do not represent the energy of a given photon but instead represent the number of photons present with an energy corresponding to that mode. Hence, while coherent states do indeed represent points in the phase space of $(\vec{A}, \vec{E})$, the main point is that the photon number can fluctuate, and while classical behavior is recovered for large numbers $n$ of photons as the fluctuations of the number are $\sqrt{n}$ by Poisson statistics, the interesting physics happens for low $n$ eigenstates or superpositions thereof in which $a$ and $a^{\dagger}$ play the same role as in the usual quantum harmonic oscillator. Furthermore, the third issue is that only a particular mode $\vec{k}$ and position $\vec{x}$ can be considered, because the electromagnetic potential has a value for each of those quantities, so unless those are held constant, the picture of phase space $(\vec{A}, \vec{E})$ becomes infinite-dimensional. Related to this, the fourth and fifth issues are, respectively, that $\vec{A}$ is used as the field and $\vec{E}$ as its conjugate momentum rather than using $\vec{E}$ and $\vec{B}$ because the latter two fields are coupled to each other by the Maxwell equations so they form an overcomplete set of degrees of freedom (or something like that), whereas using $\vec{A}$ as the field and finding its conjugate momentum in conjunction with a particular gauge choice (usually the Coulomb gauge $\nabla \cdot \vec{A} = 0$) yields the correct number of degrees of freedom. These explanations seem convincing enough to me, so I will leave those there for the moment.
Another major issue that I brought up with him for which he didn't give me a complete answer was the issue that the conjugate momentum to $\vec{A}$ was being found through \[ \Pi_j = \frac{\partial \mathcal{L}}{\partial (\partial_t A_j)} \] given the Lagrangian density $\mathcal{L} = \frac{1}{8\pi} \left(\vec{E}^2 - \vec{B}^2 \right)$ and the field relations $\vec{E} = -\frac{1}{c}\partial_t \vec{A}$ & $\vec{B} = \nabla \times \vec{A}$. This didn't seem manifestly Lorentz-covariant to me, because in the class 8.033 — Relativity, I had learned that the conjugate momentum to the electromagnetic potential $A^{\mu}$ in the above Lagrangian density would be the 2-index tensor \[ \Pi^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu})} .\] This would make a difference in finding the Hamiltonian density \[ \mathcal{H} = \sum_{\mu} \Pi^{\mu} \partial_t A_{\mu} - \mathcal{L} = \frac{1}{8\pi} \left(\vec{E}^2 + \vec{B}^2 \right). \] I thought that the Hamiltonian density would need to be a Lorentz-invariant scalar just like the Lagrangian density. As it turns out, this is not the case, because the Hamiltonian density represents the energy which explicitly picks out the temporal direction as special, so time derivatives are OK in finding the momentum conjugate to the potential; because the Lagrangian and Hamiltonian densities looks so similar, it looks like both could be Lorentz-invariant scalar functions, but deceptively, only the former is so. At this point, I figured that because the Hamiltonian and (not field conjugate, but physical) momentum looked so similar, they could arise from the same covariant vector. However, there is no "natural" 1-index vector with which to multiply the Lagrangian density to get some sort of covariant vector generalization of the Hamiltonian density, though there is a 2-index tensor, and that is the metric. I figured here that the Hamiltonian and momentum for the electromagnetic field could be related to the stress-energy tensor, which gives the energy and momentum densities and fluxes. After a while of searching online for answers, I was quite pleased to find my intuition to be essentially spot-on: indeed the conjugate momentum should be a tensor as given above, the Legendre transformation can then be done in a covariant manner, and it does in fact turn out that the result is just the stress-energy tensor \[ T^{\mu \nu} = \sum_{\mu, \xi} \Pi^{\mu \xi} \partial^{\nu} A_{\xi} - \mathcal{L}\eta^{\mu \nu} \] (UPDATE: the index positions have been corrected) for the electromagnetic field. Indeed, the time-time component is exactly the energy/Hamiltonian density $\mathcal{H} = T_{(0, 0)}$, and the Hamiltonian $H = \sum_{\vec{k}} \hbar\omega(\vec{k}) \cdot (\alpha^{\star} (\vec{k}) \alpha(\vec{k}) + \alpha(\vec{k}) \alpha^{\star} (\vec{k})) = \int T_{(0, 0)} d^3 x$. As it turns out, the momentum $\vec{p} = \sum_{\vec{k}} \hbar\vec{k} \cdot (\alpha^{\star} (\vec{k}) \alpha(\vec{k}) + \alpha(\vec{k}) \alpha^{\star} (\vec{k}))$ doesn't look similar just by coincidence: $p_j = \int T_{(0, j)} d^3 x$. The only remaining point of confusion is that it seems like the Hamiltonian and momentum should together form a Lorentz-covariant vector $p_{\mu} = (H, p_j)$, yet if the stress-energy tensor respects Lorentz-covariance, then integrating over the volume element $d^3 x$ won't respect transformations in a Lorentz-covariant manner. I guess because the individual components of the stress-energy tensor transform under a Lorentz boost and the volume element does as well, then maybe the vector $p_{\mu}$ as given above will respect Lorentz-covariance. (UPDATE: another issue I was having but forgot to write before clicking "Publish" was the fact that only the $T_{(0, \nu)}$ components are being considered. I wonder if there is some natural 1-index Lorentz-convariant vector $b_{\nu}$ to contract with $T_{\mu \nu}$ so that the result is a 1-index vector which in a given frame has a temporal component given by the Hamiltonian density and spatial components given by the momentum density.) Overall, I think it is interesting that this particular hang-up was over a point in classical field theory and special relativity and had nothing to do with the quantization of the fields; in any case, I think I have gotten over the major hang-ups about this and can proceed reading through what I need to read for the 8.06 paper.
Another major issue that I brought up with him for which he didn't give me a complete answer was the issue that the conjugate momentum to $\vec{A}$ was being found through \[ \Pi_j = \frac{\partial \mathcal{L}}{\partial (\partial_t A_j)} \] given the Lagrangian density $\mathcal{L} = \frac{1}{8\pi} \left(\vec{E}^2 - \vec{B}^2 \right)$ and the field relations $\vec{E} = -\frac{1}{c}\partial_t \vec{A}$ & $\vec{B} = \nabla \times \vec{A}$. This didn't seem manifestly Lorentz-covariant to me, because in the class 8.033 — Relativity, I had learned that the conjugate momentum to the electromagnetic potential $A^{\mu}$ in the above Lagrangian density would be the 2-index tensor \[ \Pi^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\nu})} .\] This would make a difference in finding the Hamiltonian density \[ \mathcal{H} = \sum_{\mu} \Pi^{\mu} \partial_t A_{\mu} - \mathcal{L} = \frac{1}{8\pi} \left(\vec{E}^2 + \vec{B}^2 \right). \] I thought that the Hamiltonian density would need to be a Lorentz-invariant scalar just like the Lagrangian density. As it turns out, this is not the case, because the Hamiltonian density represents the energy which explicitly picks out the temporal direction as special, so time derivatives are OK in finding the momentum conjugate to the potential; because the Lagrangian and Hamiltonian densities looks so similar, it looks like both could be Lorentz-invariant scalar functions, but deceptively, only the former is so. At this point, I figured that because the Hamiltonian and (not field conjugate, but physical) momentum looked so similar, they could arise from the same covariant vector. However, there is no "natural" 1-index vector with which to multiply the Lagrangian density to get some sort of covariant vector generalization of the Hamiltonian density, though there is a 2-index tensor, and that is the metric. I figured here that the Hamiltonian and momentum for the electromagnetic field could be related to the stress-energy tensor, which gives the energy and momentum densities and fluxes. After a while of searching online for answers, I was quite pleased to find my intuition to be essentially spot-on: indeed the conjugate momentum should be a tensor as given above, the Legendre transformation can then be done in a covariant manner, and it does in fact turn out that the result is just the stress-energy tensor \[ T^{\mu \nu} = \sum_{\mu, \xi} \Pi^{\mu \xi} \partial^{\nu} A_{\xi} - \mathcal{L}\eta^{\mu \nu} \] (UPDATE: the index positions have been corrected) for the electromagnetic field. Indeed, the time-time component is exactly the energy/Hamiltonian density $\mathcal{H} = T_{(0, 0)}$, and the Hamiltonian $H = \sum_{\vec{k}} \hbar\omega(\vec{k}) \cdot (\alpha^{\star} (\vec{k}) \alpha(\vec{k}) + \alpha(\vec{k}) \alpha^{\star} (\vec{k})) = \int T_{(0, 0)} d^3 x$. As it turns out, the momentum $\vec{p} = \sum_{\vec{k}} \hbar\vec{k} \cdot (\alpha^{\star} (\vec{k}) \alpha(\vec{k}) + \alpha(\vec{k}) \alpha^{\star} (\vec{k}))$ doesn't look similar just by coincidence: $p_j = \int T_{(0, j)} d^3 x$. The only remaining point of confusion is that it seems like the Hamiltonian and momentum should together form a Lorentz-covariant vector $p_{\mu} = (H, p_j)$, yet if the stress-energy tensor respects Lorentz-covariance, then integrating over the volume element $d^3 x$ won't respect transformations in a Lorentz-covariant manner. I guess because the individual components of the stress-energy tensor transform under a Lorentz boost and the volume element does as well, then maybe the vector $p_{\mu}$ as given above will respect Lorentz-covariance. (UPDATE: another issue I was having but forgot to write before clicking "Publish" was the fact that only the $T_{(0, \nu)}$ components are being considered. I wonder if there is some natural 1-index Lorentz-convariant vector $b_{\nu}$ to contract with $T_{\mu \nu}$ so that the result is a 1-index vector which in a given frame has a temporal component given by the Hamiltonian density and spatial components given by the momentum density.) Overall, I think it is interesting that this particular hang-up was over a point in classical field theory and special relativity and had nothing to do with the quantization of the fields; in any case, I think I have gotten over the major hang-ups about this and can proceed reading through what I need to read for the 8.06 paper.
Posted in class, college, MIT, physics, qed, quantum electrodynamics, quantum mechanics
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Tuesday, 26 March 2013
Schrödinger and Biot-Savart
Posted on 10:47 by Unknown
There were two things that I would like to post here today. The first is something I have been mulling over for a while. The second is something that I thought about more recently.
Time evolution in nonrelativistic quantum mechanics occurs according to the [time-dependent] Schrödinger equation \[ H|\Psi\rangle = i\hbar \frac{\partial}{\partial t} |\Psi\rangle .\] While this at first may seem intractable, the trick is that typically the Hamiltonian is not time-dependent, so a candidate solution could be $|\Psi\rangle = \phi(t)|E\rangle$. Plugging this back in yields time evolution that occurs through the phase $\phi(t) = e^{-\frac{iEt}{\hbar}}$ applied to energy eigenstates that solve \[ H|E\rangle = E \cdot |E\rangle \] and this equation is often called the "time-independent Schrödinger equation". When I was taking 8.04 — Quantum Physics I, I agreed with my professor who called this a misnomer, in that the Schrödinger equation is supposed to only describe time evolution, so what is being called "time-independent" is more properly just an energy eigenvalue equation. That said, I was thinking that the "time-independent Schrödinger equation" is really just like a Fourier transform of the Schrödinger equation from time to frequency (related to energy by $E = \hbar\omega$), so the former could be an OK nomenclature because it is just a change of basis. However, there are two things to note: the Schrödinger equation is basis-independent, whereas the "time-independent Schrödinger equation" is expressed only in the basis of energy eigenstates, and time is not an observable quantity (i.e. Hermitian operator) but is a parameter, so the change of basis/Fourier transform argument doesn't work in quite the same way that it does for position versus momentum. Hence, I've come to the conclusion that it is better to call the "time-independent Schrödinger equation" as the energy eigenvalue equation.
Switching gears, I was thinking about how the Biot-Savart law is derived. My AP Physics C teacher told me that the Ampère law is derived from the Biot-Savart law. However, this is patently not true, because the Biot-Savart law only works for charges moving at a constant velocity, whereas the Ampère law is true for magnetic fields created by any currents or any changing electric fields. In 8.022 — Physics II, I did see a derivation of the Biot-Savart law from the Ampère law, showing that the latter is indeed more fundamental than the former, but it involved the magnetic potential and a lot more work. I wanted to see if that derivation still made sense to me, but then I realized that because magnetism essentially springs from the combination of electricity and special relativity and because the Biot-Savart law relies on the approximation of the charges moving at a constant velocity, it should be possible to derive the Biot-Savart law from the Coulomb law and special relativity. Indeed, it is possible. Consider a charge $q$ whose electric field is \[ \vec{E} = \frac{q}{r^2} \vec{e}_r \] in its rest frame. Note that the Coulomb law is exact in the rest frame of a charge. Now consider a frame moving with respect to the charge at a velocity $-\vec{v}$, so that observers in the frame see the charge move at a velocity $\vec{v}$. Considering only the component of the magnetic field perpendicular to the relative motion, noting that there is no magnetic field in the rest frame of the charge yields, and considering the low-speed limit (which is the range of validity of the Biot-Savart law) $\left|\frac{\vec{v}}{c}\right| \ll 1$ so that $\gamma \approx 1$ yields $\vec{B} \approx -\frac{\vec{v}}{c} \times \vec{E}$. Plugging in $-\vec{v}$ (the specified velocity of the new frame relative to the charge) for $\vec{v}$ (the general expression for the relative velocity) and plugging in the Coulomb expression for $\vec{E}$ yields the Biot-Savart law \[ \vec{B} = \frac{q\vec{v} \times \vec{e}_r}{cr^2}. \] One thing to be emphasized again is that the Coulomb law is exact in the rest frame of the charge, while the Biot-Savart law is always an approximation because a moving charge will have an electric field that deviates from the Coulomb expression; the fact that the Biot-Savart law is a low-speed inertial approximation is why I feel comfortable doing the derivation this way.
Time evolution in nonrelativistic quantum mechanics occurs according to the [time-dependent] Schrödinger equation \[ H|\Psi\rangle = i\hbar \frac{\partial}{\partial t} |\Psi\rangle .\] While this at first may seem intractable, the trick is that typically the Hamiltonian is not time-dependent, so a candidate solution could be $|\Psi\rangle = \phi(t)|E\rangle$. Plugging this back in yields time evolution that occurs through the phase $\phi(t) = e^{-\frac{iEt}{\hbar}}$ applied to energy eigenstates that solve \[ H|E\rangle = E \cdot |E\rangle \] and this equation is often called the "time-independent Schrödinger equation". When I was taking 8.04 — Quantum Physics I, I agreed with my professor who called this a misnomer, in that the Schrödinger equation is supposed to only describe time evolution, so what is being called "time-independent" is more properly just an energy eigenvalue equation. That said, I was thinking that the "time-independent Schrödinger equation" is really just like a Fourier transform of the Schrödinger equation from time to frequency (related to energy by $E = \hbar\omega$), so the former could be an OK nomenclature because it is just a change of basis. However, there are two things to note: the Schrödinger equation is basis-independent, whereas the "time-independent Schrödinger equation" is expressed only in the basis of energy eigenstates, and time is not an observable quantity (i.e. Hermitian operator) but is a parameter, so the change of basis/Fourier transform argument doesn't work in quite the same way that it does for position versus momentum. Hence, I've come to the conclusion that it is better to call the "time-independent Schrödinger equation" as the energy eigenvalue equation.
Switching gears, I was thinking about how the Biot-Savart law is derived. My AP Physics C teacher told me that the Ampère law is derived from the Biot-Savart law. However, this is patently not true, because the Biot-Savart law only works for charges moving at a constant velocity, whereas the Ampère law is true for magnetic fields created by any currents or any changing electric fields. In 8.022 — Physics II, I did see a derivation of the Biot-Savart law from the Ampère law, showing that the latter is indeed more fundamental than the former, but it involved the magnetic potential and a lot more work. I wanted to see if that derivation still made sense to me, but then I realized that because magnetism essentially springs from the combination of electricity and special relativity and because the Biot-Savart law relies on the approximation of the charges moving at a constant velocity, it should be possible to derive the Biot-Savart law from the Coulomb law and special relativity. Indeed, it is possible. Consider a charge $q$ whose electric field is \[ \vec{E} = \frac{q}{r^2} \vec{e}_r \] in its rest frame. Note that the Coulomb law is exact in the rest frame of a charge. Now consider a frame moving with respect to the charge at a velocity $-\vec{v}$, so that observers in the frame see the charge move at a velocity $\vec{v}$. Considering only the component of the magnetic field perpendicular to the relative motion, noting that there is no magnetic field in the rest frame of the charge yields, and considering the low-speed limit (which is the range of validity of the Biot-Savart law) $\left|\frac{\vec{v}}{c}\right| \ll 1$ so that $\gamma \approx 1$ yields $\vec{B} \approx -\frac{\vec{v}}{c} \times \vec{E}$. Plugging in $-\vec{v}$ (the specified velocity of the new frame relative to the charge) for $\vec{v}$ (the general expression for the relative velocity) and plugging in the Coulomb expression for $\vec{E}$ yields the Biot-Savart law \[ \vec{B} = \frac{q\vec{v} \times \vec{e}_r}{cr^2}. \] One thing to be emphasized again is that the Coulomb law is exact in the rest frame of the charge, while the Biot-Savart law is always an approximation because a moving charge will have an electric field that deviates from the Coulomb expression; the fact that the Biot-Savart law is a low-speed inertial approximation is why I feel comfortable doing the derivation this way.
Posted in AP, class, college, electricity, MIT, physics, quantum electrodynamics, quantum mechanics, school
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Thursday, 21 March 2013
Time and Temperature are Complex
Posted on 13:52 by Unknown
In a post from a few days ago, I briefly mentioned the notion of imaginary time with regard to angular momentum. I'd like to go into that a little further in this post.
In 3 spatial dimensions, the flat (Euclidean) metric is $\eta_{ij} = \delta_{ij}$, which is quite convenient, as lengths are given by $(\Delta s)^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$ which is just the usual Pythagorean theorem. When a temporal dimension is added, as in special relativity, the coordinates are now $x^{\mu} = (ct, x_{j})$, and the Euclidean metric becomes the Minkowski metric $\eta_{\mu \nu} = \mathrm{diag}(-1, 1, 1, 1)$ so that $\eta_{tt} = -1$, $\eta_{(t, j)} = 0$, and $\eta_{ij} = \delta_{ij}$. This means that spacetime intervals become $(\Delta s)^2 = -(c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$, which is the normal Pythagorean theorem only if $\Delta t = 0$. In general, time coordinate differences contribute negatively to the spacetime interval. In addition, Lorentz transformations are given by a hyperbolic rotation by a [hyperbolic] angle $\alpha$ equal to the rapidity given by $\frac{v}{c} = \tanh(\alpha)$. This doesn't look quite the same as normal Euclidean geometry. However, a transformation to imaginary time, called a Wick rotation, can be done by setting $\tau = it$, so $x^{\mu} = (ic\tau, x_{j})$, $\eta_{\mu \nu} = \delta_{\mu \nu}$, $(\Delta s)^2 = (c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$ as in the usual Pythagorean theorem, and the Lorentz transformation is given by a real rotation by an angle $\theta = i\alpha$ (though I may have gotten some of these signs wrong so forgive me) where $\alpha$ is now imaginary. Now, the connection to the component $L_{(0, j)}$ of the angular momentum tensor should be more clear.
I first encountered this in the class 8.033 — Relativity, where I was able to explore this curiosity on a problem set. That question and the accompanying discussion seemed to say that while this is a cool thing to try doing once, it isn't really useful, especially because it does not hold true in general relativity with more general metrics $g_{\mu \nu} \neq \eta_{\mu \nu}$ except in very special cases. However, as it turns out, imaginary time does play a role in quantum mechanics, even without the help of relativity.
Schrödinger time evolution occurs through the unitary transformation $u = e^{-\frac{itH}{\hbar}}$ satisfying $uu^{\dagger} = u^{\dagger} u = 1$. This means that the probability that an initial state $|\psi\rangle$ ends after time $t$ in the same state is given by the amplitude (whose square is the probability [density]) $\mathfrak{p}(t) = \langle\psi|e^{-\frac{itH}{\hbar}}|\psi\rangle$. Meanwhile, assuming the states $|\psi\rangle$ form a complete and orthonormal basis (though I don't know if this assumption is truly necessary), the partition function $Z = \mathrm{trace}\left(e^{-\frac{H}{k_B T}}\right)$, which can be expanded in the basis $|\psi\rangle$ as $Z = \sum_{\psi} \langle\psi|e^{-\frac{H}{k_B T}}|\psi\rangle$. This, however, is just as well rewritten as $Z = \sum_{\psi} \mathfrak{p}\left(t = -\frac{i\hbar}{k_B T}\right)$. Hence, quantum and statistical mechanical information can be gotten from the same amplitudes using the substitution $t = -\frac{i\hbar}{k_B T}$, which essentially calls temperature a reciprocal imaginary time. This is not really meant to show anything more deep or profound about the connection between time and temperature; it is really more of a trick stemming from the fact that the same Hamiltonian can be used to solve problem in quantum mechanics or equilibrium statistical mechanics.
As an aside, it turns out that temperature, even when measured in an absolute scale, can be negative. There are plenty of papers of this online, but suffice it to say that this comes from a more general statistical definition of temperature. Rather than defining it (as it commonly is) as the average kinetic energy of particles, it is better to define it as a measure of the probability distribution that a particle will have a given energy. Usually, particles tend to be in lower energy states more than in higher energy states, and as a consequence, the temperature is positive. However, it is possible (and has been done repeatedly) under certain circumstances to cleverly force the system in a way that causes particles to be in higher energy states with higher probability than in lower energy states, and this is exactly the negative temperature. More formally, $\frac{1}{T} = \frac{\partial S}{\partial E}$ where $E$ is the energy and $S$ is the entropy of the system, which is a measure of how many different states the system can possibly have for a given energy. For positive temperature, if two objects of different temperatures are brought into contact, energy will flow from the hotter one to the colder, cooling the former and heating the latter until equal temperatures are achieved. For negative temperature, though, if an object with negative temperature is brought in contact with an object that has positive temperature, each object tends to increase its own entropy. Like most normal objects, the latter does this by absorbing energy, but by the definition of temperature, the former does this by releasing energy, meaning the former will spontaneously heat the latter. Hence, negative temperature is hotter than positive temperature; this is a quirk of the definition of reciprocal temperature, so really what is happening is that absolute zero on the positive side is still the coldest possible temperature, absolute zero on the negative side is now the hottest temperature, and $\pm \infty$ is in the middle.
This was really just me writing down stuff that I had been thinking about a couple of months ago. I hope this helps someone, and I also await the day when TV newscasters say "complex time brought to you by..." instead of "time and temperature brought to you by...".
In 3 spatial dimensions, the flat (Euclidean) metric is $\eta_{ij} = \delta_{ij}$, which is quite convenient, as lengths are given by $(\Delta s)^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$ which is just the usual Pythagorean theorem. When a temporal dimension is added, as in special relativity, the coordinates are now $x^{\mu} = (ct, x_{j})$, and the Euclidean metric becomes the Minkowski metric $\eta_{\mu \nu} = \mathrm{diag}(-1, 1, 1, 1)$ so that $\eta_{tt} = -1$, $\eta_{(t, j)} = 0$, and $\eta_{ij} = \delta_{ij}$. This means that spacetime intervals become $(\Delta s)^2 = -(c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$, which is the normal Pythagorean theorem only if $\Delta t = 0$. In general, time coordinate differences contribute negatively to the spacetime interval. In addition, Lorentz transformations are given by a hyperbolic rotation by a [hyperbolic] angle $\alpha$ equal to the rapidity given by $\frac{v}{c} = \tanh(\alpha)$. This doesn't look quite the same as normal Euclidean geometry. However, a transformation to imaginary time, called a Wick rotation, can be done by setting $\tau = it$, so $x^{\mu} = (ic\tau, x_{j})$, $\eta_{\mu \nu} = \delta_{\mu \nu}$, $(\Delta s)^2 = (c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$ as in the usual Pythagorean theorem, and the Lorentz transformation is given by a real rotation by an angle $\theta = i\alpha$ (though I may have gotten some of these signs wrong so forgive me) where $\alpha$ is now imaginary. Now, the connection to the component $L_{(0, j)}$ of the angular momentum tensor should be more clear.
I first encountered this in the class 8.033 — Relativity, where I was able to explore this curiosity on a problem set. That question and the accompanying discussion seemed to say that while this is a cool thing to try doing once, it isn't really useful, especially because it does not hold true in general relativity with more general metrics $g_{\mu \nu} \neq \eta_{\mu \nu}$ except in very special cases. However, as it turns out, imaginary time does play a role in quantum mechanics, even without the help of relativity.
Schrödinger time evolution occurs through the unitary transformation $u = e^{-\frac{itH}{\hbar}}$ satisfying $uu^{\dagger} = u^{\dagger} u = 1$. This means that the probability that an initial state $|\psi\rangle$ ends after time $t$ in the same state is given by the amplitude (whose square is the probability [density]) $\mathfrak{p}(t) = \langle\psi|e^{-\frac{itH}{\hbar}}|\psi\rangle$. Meanwhile, assuming the states $|\psi\rangle$ form a complete and orthonormal basis (though I don't know if this assumption is truly necessary), the partition function $Z = \mathrm{trace}\left(e^{-\frac{H}{k_B T}}\right)$, which can be expanded in the basis $|\psi\rangle$ as $Z = \sum_{\psi} \langle\psi|e^{-\frac{H}{k_B T}}|\psi\rangle$. This, however, is just as well rewritten as $Z = \sum_{\psi} \mathfrak{p}\left(t = -\frac{i\hbar}{k_B T}\right)$. Hence, quantum and statistical mechanical information can be gotten from the same amplitudes using the substitution $t = -\frac{i\hbar}{k_B T}$, which essentially calls temperature a reciprocal imaginary time. This is not really meant to show anything more deep or profound about the connection between time and temperature; it is really more of a trick stemming from the fact that the same Hamiltonian can be used to solve problem in quantum mechanics or equilibrium statistical mechanics.
As an aside, it turns out that temperature, even when measured in an absolute scale, can be negative. There are plenty of papers of this online, but suffice it to say that this comes from a more general statistical definition of temperature. Rather than defining it (as it commonly is) as the average kinetic energy of particles, it is better to define it as a measure of the probability distribution that a particle will have a given energy. Usually, particles tend to be in lower energy states more than in higher energy states, and as a consequence, the temperature is positive. However, it is possible (and has been done repeatedly) under certain circumstances to cleverly force the system in a way that causes particles to be in higher energy states with higher probability than in lower energy states, and this is exactly the negative temperature. More formally, $\frac{1}{T} = \frac{\partial S}{\partial E}$ where $E$ is the energy and $S$ is the entropy of the system, which is a measure of how many different states the system can possibly have for a given energy. For positive temperature, if two objects of different temperatures are brought into contact, energy will flow from the hotter one to the colder, cooling the former and heating the latter until equal temperatures are achieved. For negative temperature, though, if an object with negative temperature is brought in contact with an object that has positive temperature, each object tends to increase its own entropy. Like most normal objects, the latter does this by absorbing energy, but by the definition of temperature, the former does this by releasing energy, meaning the former will spontaneously heat the latter. Hence, negative temperature is hotter than positive temperature; this is a quirk of the definition of reciprocal temperature, so really what is happening is that absolute zero on the positive side is still the coldest possible temperature, absolute zero on the negative side is now the hottest temperature, and $\pm \infty$ is in the middle.
This was really just me writing down stuff that I had been thinking about a couple of months ago. I hope this helps someone, and I also await the day when TV newscasters say "complex time brought to you by..." instead of "time and temperature brought to you by...".
Posted in class, college, mathematics, MIT, physics, qed, quantum electrodynamics, quantum mechanics, statistical mechanics
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Wednesday, 20 March 2013
Nonzero Electromagnetic Fields in a Cavity
Posted on 08:03 by Unknown
The class 8.06 — Quantum Physics III requires a final paper, written essentially like a review article of a certain area of physics that uses quantum mechanics and that is written for the level of 8.06 (and not much higher). At the same time, I have also been looking into other possible UROP projects because while I am quite happy with my photonic crystals UROP and would be pleased to continue with it, that project is the only one I have done at MIT thus far, and I would like to try at least one more thing before I graduate. My advisor suggested that I not do something already done to death like the Feynman path integrals in the 8.06 paper but instead to do something that could act as a springboard in my UROP search. One of the UROP projects I have been investigating has to do with Casimir forces, but I pretty much don't know anything about that, QED, or [more generally] QFT. Given that other students have successfully written 8.06 papers about Casimir forces, I figured this would be the perfect way to teach myself what I might need to know to be able to start on a UROP project in that area. Most helpful thus far has been my recitation leader, who is a graduate student working in the same group that I have been looking into for UROP projects; he has been able to show me some of the basic tools in Casimir physics and point me in the right direction for more information. Finally, note that there will probably be more posts about this in the near future, as I'll be using this to jot down my thoughts and make them more coherent (no pun intended) for future reference.
Anyway, I've been able to read some more papers on the subject, including Casimir's original paper on it as well as Lifshitz's paper going a little further with it. One of the things that confused me in those papers (and in my recitation leader's explanation, which was basically the same thing) was the following. The explanation ends with the notion that quantum electrodynamic fluctuations in a space with a given dielectric constant, say in a vacuum surrounded by two metal plates, will cause those metal plates to attract or repel in a manner dependent on their separation. This depends on the separation being comparable to the wavelength of the electromagnetic field (or something like that), because at much larger distances, the power of normal blackbody radiation (which ironically still requires quantum mechanics to be explained) does not depend on the separation of the two objects, nor does it really depend on their geometries, but only on their temperatures. The explanation of the Casimir effect starts with the notion of an electromagnetic field confined between two infinite perfectly conducting parallel plates, so the fields form standing waves like the wavefunctions of a quantum particle in an infinite square well. This is all fine and dandy...except that this presumes that there is an electromagnetic field. This confused me: why should one assume the existence of an electromagnetic field, and why couldn't it be possible to assume that there really is no field between the plates?
Then I remembered what the deal is with quantization of the electromagnetic field and photon states from 8.05 — Quantum Physics II. The derivation from that class still seems quite fascinating to me, so I'm going to repost it here. You don't need to know QED or QFT, but you do need to be familiar with Dirac notation and at least a little comfortable with the quantization of the simple harmonic oscillator.
Let us first get the classical picture straight. Consider an electromagnetic field inside a cavity of volume $\mathcal{V}$. Let us only consider the lowest-energy mode, which is when $k_x = k_y = 0$ so only $k_z > 0$, stemming from the appropriate application of boundary conditions. The energy density of the system can be given as \[H = \frac{1}{8\pi} \left(\vec{E}^2 + \vec{B}^2 \right)\] and the fields that solve the dynamic Maxwell equations \[\nabla \times \vec{E} = -\frac{1}{c} \frac{\partial \vec{B}}{\partial t}\] \[\nabla \times \vec{B} = \frac{1}{c} \frac{\partial \vec{E}}{\partial t}\] as well as the source-free Maxwell equations \[\nabla \cdot \vec{E} = \nabla \cdot \vec{B} = 0\] can be written as \[\vec{E} = \sqrt{\frac{8\pi}{\mathcal{V}}} \omega Q(t) \sin(kz) \vec{e}_x\] \[\vec{B} = \sqrt{\frac{8\pi}{\mathcal{V}}} P(t) \cos(kz) \vec{e}_y\] where $\vec{k} = k_z \vec{e}_z = k\vec{e}_z$ and $\omega = c|\vec{k}|$. The prefactor comes from normalization, the spatial dependence and direction come from boundary conditions, and the time dependence is somewhat arbitrary. I think this is because the spatial conditions are unaffected by time dependence if they are separable, and the Maxwell equations are linear so if a periodic function like a sinusoid or complex exponential in time satisfies Maxwell time evolution, so does any arbitrary superposition (Fourier series) thereof. That said, I'm not entirely sure about that point. Also note that $P$ and $Q$ are not entirely arbitrary, because they are restricted by the Maxwell equations. Plugging the fields into those equations yields conditions on $P$ and $Q$ given by \[\dot{Q} = P\] \[\dot{P} = -\omega^2 Q\] which looks suspiciously like simple harmonic motion. Indeed, plugging these electromagnetic field components into the Hamiltonian [density] yields \[H = \frac{1}{2} \left(P^2 + \omega^2 Q^2 \right)\] which is the equation for a simple harmonic oscillator with $m = 1$; this is because the electromagnetic field has no mass, so there is no characteristic mass term to stick into the equation. Note that these quantities have a canonical Poisson bracket $\{Q, P\} = 1$, so $Q$ can be identified as a position and $P$ can be identified as a momentum, though they are actually neither of those things but are simply mathematical conveniences to simplify expressions involving the fields; this will become useful shortly.
Quantizing this yields turns the canonical Poisson bracket relation into the canonical commutation relation $[Q, P] = i\hbar$. This also implies that $[E_a, B_b] \neq 0$, which is huge: this means that states of the photon cannot have definite values for both the electric and magnetic fields simultaneously, just as a quantum mechanical particle state cannot have both a definite position and momentum. Now the fields themselves are operators that depend on space and time as parameters, while the states are now vectors in a Hilbert space defined for a given mode $\vec{k}$, which has been chosen in this case as $\vec{k} = k\vec{e}_z$ for some allowed value of $k$. The raising and lowering operators $a$ and $a^{\dagger}$ can be defined in the usual way but with the substitutions $m \rightarrow 1$, $x \rightarrow Q$, and $p \rightarrow P$. The Hamiltonian then becomes $H = \hbar\omega \cdot \left(a^{\dagger} a + \frac{1}{2} \right)$, where again $\omega = c|\vec{k}|$ for the given mode $\vec{k}$. This means that eigenstates of the Hamiltonian are the usual $|n\rangle$, where $n$ specifies the number of photons which have mode $\vec{k}$ and therefore frequency $\omega$; this is in contrast to the single particle harmonic oscillator eigenstate $|n\rangle$ which specifies that there is only one particle and it has energy $E_n = \hbar \omega \cdot \left(n + \frac{1}{2} \right)$. This makes sense on two counts: for one, photons are bosons, so multiple photons should be able to occupy the same mode, and for another, each photon carries energy $\hbar\omega$, so adding a photon to a mode should increase the energy of the system by a unit of the energy of that mode, and indeed it does. Also note that these number eigenstates are not eigenstates of either the electric or the magnetic fields, just as normal particle harmonic oscillator eigenstates are not eigenstates of either position or momentum. (As an aside, the reason why lasers are called coherent is because they are composed of light in coherent states of a given mode satisfying $a|\alpha\rangle = \alpha \cdot |\alpha\rangle$ where $\alpha \in \mathbb{C}$. These, as opposed to energy/number eigenstates, are physically realizable.)
So what does this have to do with quantum fluctuations in a cavity? Well, if you notice, just as with the usual quantum harmonic oscillator, this Hamiltonian has a ground state energy above the minimum of the potential given by $\frac{1}{2} \hbar\omega$ for a given mode; this corresponds to having no photons in that mode. Hence, even an electrodynamic vacuum has a nonzero ground state energy. Equally important is the fact that while the mean fields $\langle 0|\vec{E}|0\rangle = \langle 0|\vec{B}|0\rangle = \vec{0}$, the field fluctuations $\langle 0|\vec{E}^2|0\rangle \neq 0$ and $\langle 0|\vec{B}^2|0 \rangle \neq 0$; thus, the electromagnetic fields fluctuate with some nonzero variance even in the absence of photons. This relieves the confusion I was having earlier about why any analysis of the Casimir effect assumes the presence of an electromagnetic field in a cavity by way of nonzero fluctuations even when no photons are present. Just to tie up the loose ends, because the Casimir effect is introduced as having the electromagnetic field in a cavity, the allowed modes are standing waves with wavevectors given by $\vec{k} = k_x \vec{e}_x + k_y \vec{e}_y + \frac{\pi n_z}{l} \vec{e}_z$ where $n_z \in \mathbb{Z}$, assuming that the cavity bounds the fields along $\vec{e}_z$ but the other directions are left unspecified. This means that each different value of $\vec{k}$ specifies a different harmonic oscillator, and each of those different harmonic oscillators is in the ground state in the absence of photons. You'll be hearing more about this in the near future, but for now, thinking through this helped me clear up my basic misunderstandings, and I hope anyone else who was having the same misunderstandings feels more comfortable with this now.
Anyway, I've been able to read some more papers on the subject, including Casimir's original paper on it as well as Lifshitz's paper going a little further with it. One of the things that confused me in those papers (and in my recitation leader's explanation, which was basically the same thing) was the following. The explanation ends with the notion that quantum electrodynamic fluctuations in a space with a given dielectric constant, say in a vacuum surrounded by two metal plates, will cause those metal plates to attract or repel in a manner dependent on their separation. This depends on the separation being comparable to the wavelength of the electromagnetic field (or something like that), because at much larger distances, the power of normal blackbody radiation (which ironically still requires quantum mechanics to be explained) does not depend on the separation of the two objects, nor does it really depend on their geometries, but only on their temperatures. The explanation of the Casimir effect starts with the notion of an electromagnetic field confined between two infinite perfectly conducting parallel plates, so the fields form standing waves like the wavefunctions of a quantum particle in an infinite square well. This is all fine and dandy...except that this presumes that there is an electromagnetic field. This confused me: why should one assume the existence of an electromagnetic field, and why couldn't it be possible to assume that there really is no field between the plates?
Then I remembered what the deal is with quantization of the electromagnetic field and photon states from 8.05 — Quantum Physics II. The derivation from that class still seems quite fascinating to me, so I'm going to repost it here. You don't need to know QED or QFT, but you do need to be familiar with Dirac notation and at least a little comfortable with the quantization of the simple harmonic oscillator.
Let us first get the classical picture straight. Consider an electromagnetic field inside a cavity of volume $\mathcal{V}$. Let us only consider the lowest-energy mode, which is when $k_x = k_y = 0$ so only $k_z > 0$, stemming from the appropriate application of boundary conditions. The energy density of the system can be given as \[H = \frac{1}{8\pi} \left(\vec{E}^2 + \vec{B}^2 \right)\] and the fields that solve the dynamic Maxwell equations \[\nabla \times \vec{E} = -\frac{1}{c} \frac{\partial \vec{B}}{\partial t}\] \[\nabla \times \vec{B} = \frac{1}{c} \frac{\partial \vec{E}}{\partial t}\] as well as the source-free Maxwell equations \[\nabla \cdot \vec{E} = \nabla \cdot \vec{B} = 0\] can be written as \[\vec{E} = \sqrt{\frac{8\pi}{\mathcal{V}}} \omega Q(t) \sin(kz) \vec{e}_x\] \[\vec{B} = \sqrt{\frac{8\pi}{\mathcal{V}}} P(t) \cos(kz) \vec{e}_y\] where $\vec{k} = k_z \vec{e}_z = k\vec{e}_z$ and $\omega = c|\vec{k}|$. The prefactor comes from normalization, the spatial dependence and direction come from boundary conditions, and the time dependence is somewhat arbitrary. I think this is because the spatial conditions are unaffected by time dependence if they are separable, and the Maxwell equations are linear so if a periodic function like a sinusoid or complex exponential in time satisfies Maxwell time evolution, so does any arbitrary superposition (Fourier series) thereof. That said, I'm not entirely sure about that point. Also note that $P$ and $Q$ are not entirely arbitrary, because they are restricted by the Maxwell equations. Plugging the fields into those equations yields conditions on $P$ and $Q$ given by \[\dot{Q} = P\] \[\dot{P} = -\omega^2 Q\] which looks suspiciously like simple harmonic motion. Indeed, plugging these electromagnetic field components into the Hamiltonian [density] yields \[H = \frac{1}{2} \left(P^2 + \omega^2 Q^2 \right)\] which is the equation for a simple harmonic oscillator with $m = 1$; this is because the electromagnetic field has no mass, so there is no characteristic mass term to stick into the equation. Note that these quantities have a canonical Poisson bracket $\{Q, P\} = 1$, so $Q$ can be identified as a position and $P$ can be identified as a momentum, though they are actually neither of those things but are simply mathematical conveniences to simplify expressions involving the fields; this will become useful shortly.
Quantizing this yields turns the canonical Poisson bracket relation into the canonical commutation relation $[Q, P] = i\hbar$. This also implies that $[E_a, B_b] \neq 0$, which is huge: this means that states of the photon cannot have definite values for both the electric and magnetic fields simultaneously, just as a quantum mechanical particle state cannot have both a definite position and momentum. Now the fields themselves are operators that depend on space and time as parameters, while the states are now vectors in a Hilbert space defined for a given mode $\vec{k}$, which has been chosen in this case as $\vec{k} = k\vec{e}_z$ for some allowed value of $k$. The raising and lowering operators $a$ and $a^{\dagger}$ can be defined in the usual way but with the substitutions $m \rightarrow 1$, $x \rightarrow Q$, and $p \rightarrow P$. The Hamiltonian then becomes $H = \hbar\omega \cdot \left(a^{\dagger} a + \frac{1}{2} \right)$, where again $\omega = c|\vec{k}|$ for the given mode $\vec{k}$. This means that eigenstates of the Hamiltonian are the usual $|n\rangle$, where $n$ specifies the number of photons which have mode $\vec{k}$ and therefore frequency $\omega$; this is in contrast to the single particle harmonic oscillator eigenstate $|n\rangle$ which specifies that there is only one particle and it has energy $E_n = \hbar \omega \cdot \left(n + \frac{1}{2} \right)$. This makes sense on two counts: for one, photons are bosons, so multiple photons should be able to occupy the same mode, and for another, each photon carries energy $\hbar\omega$, so adding a photon to a mode should increase the energy of the system by a unit of the energy of that mode, and indeed it does. Also note that these number eigenstates are not eigenstates of either the electric or the magnetic fields, just as normal particle harmonic oscillator eigenstates are not eigenstates of either position or momentum. (As an aside, the reason why lasers are called coherent is because they are composed of light in coherent states of a given mode satisfying $a|\alpha\rangle = \alpha \cdot |\alpha\rangle$ where $\alpha \in \mathbb{C}$. These, as opposed to energy/number eigenstates, are physically realizable.)
So what does this have to do with quantum fluctuations in a cavity? Well, if you notice, just as with the usual quantum harmonic oscillator, this Hamiltonian has a ground state energy above the minimum of the potential given by $\frac{1}{2} \hbar\omega$ for a given mode; this corresponds to having no photons in that mode. Hence, even an electrodynamic vacuum has a nonzero ground state energy. Equally important is the fact that while the mean fields $\langle 0|\vec{E}|0\rangle = \langle 0|\vec{B}|0\rangle = \vec{0}$, the field fluctuations $\langle 0|\vec{E}^2|0\rangle \neq 0$ and $\langle 0|\vec{B}^2|0 \rangle \neq 0$; thus, the electromagnetic fields fluctuate with some nonzero variance even in the absence of photons. This relieves the confusion I was having earlier about why any analysis of the Casimir effect assumes the presence of an electromagnetic field in a cavity by way of nonzero fluctuations even when no photons are present. Just to tie up the loose ends, because the Casimir effect is introduced as having the electromagnetic field in a cavity, the allowed modes are standing waves with wavevectors given by $\vec{k} = k_x \vec{e}_x + k_y \vec{e}_y + \frac{\pi n_z}{l} \vec{e}_z$ where $n_z \in \mathbb{Z}$, assuming that the cavity bounds the fields along $\vec{e}_z$ but the other directions are left unspecified. This means that each different value of $\vec{k}$ specifies a different harmonic oscillator, and each of those different harmonic oscillators is in the ground state in the absence of photons. You'll be hearing more about this in the near future, but for now, thinking through this helped me clear up my basic misunderstandings, and I hope anyone else who was having the same misunderstandings feels more comfortable with this now.
Posted in class, college, electricity, MIT, physics, qed, quantum electrodynamics, quantum mechanics, statistical mechanics, UROP
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Monday, 18 March 2013
A Less-Seen View of Angular Momentum
Posted on 07:16 by Unknown
Many people learn in basic physics classes that angular momentum is a scalar quantity that describes the magnitude and direction of rotation, such that its rate of change is equal to the sum of all torques $\tau = \dot{L}$, akin to Newton's equation of motion $\vec{F} = \dot{\vec{p}}$. People who take more advanced physics classes, such as 8.012 — Physics I, learn that in fact angular momentum and torque are vectors; in the case of fixed-axis rotation, the moment of inertia (the rotational equivalent to mass) is a scalar so $\vec{L} = I\vec{\omega}$ means that angular momentum points in the same direction as angular velocity. By contrast, in general rigid body motion, the moment of inertia becomes anisotropic and becomes a tensor, so \[\vec{L} = \stackrel{\leftrightarrow}{I} \cdot \vec{\omega}\] implies that angular momentum is no longer parallel to angular velocity, but instead the components are related (using Einstein summation for convenience) by \[L_i = I_{ij} \omega_{j}.\] This becomes important in the analysis of situations like gyroscopes and torque-induced precession, torque-free precession, and nutation.
There is one problem though: there is nothing particularly vector-like about angular momentum. It is constructed as a vector essentially for mathematical convenience. The definition $\vec{L} = \vec{x} \times \vec{p}$ only works in 3 dimensions. Why is this? Let's look at the definition of the cross product components: in 3 dimensions, the permutation tensor has 3 indices, so contracting it with 2 vectors produces a third vector $\vec{c} = \vec{a} \times \vec{b}$ such that $c_i = \varepsilon_{ijk} a_{j} b_{k}$. One trick that is commonly taught to make the cross product easier is to turn the first vector into a matrix and then perform matrix multiplication with the column representation of the second vector to get the column representation of the resulting vector: the details of this rule are hard to remember, but the source is simple, as it is just $a_{ij} = \varepsilon_{ijk} a_{k}$. Now let us see what happens to angular velocity and angular momentum using this definition. Angular velocity was previously defined as a vector through $\vec{v} = \vec{\omega} \times \vec{x}$. We know that $\vec{x}$ and $\vec{v}$ are true vectors, while $\vec{\omega}$ is a pseudovector (defined by it flipping direction when the coordinate system undergoes reflection), so $\vec{\omega}$ is vector to be made into a tensor. Using the previous definition that in 3 dimensions $\omega_{ij} = \varepsilon_{ijk} \omega_{k}$, then \[v_i = \omega_{ij} x_{j}\] now defines the angular velocity tensor. Similarly, angular momentum is a pseudovector, so it can be made into a tensor through $L_{ij} = \varepsilon_{ijk} L_{k}$. Substituting this into the equation relating angular momenta and angular velocities yields \[L_{ij} = I_{ik} \omega_{kj}\] meaning the matrix representation of the angular momentum tensor is now the matrix multiplication of the matrices representing the moment of inertia and angular velocity tensors.
This has another consequence: the meaning of the components of the angular velocity and angular momentum become much more clear. Previously, $L_{j}$ was the generator of rotation in the plane perpendicular to the $j$-axis, and $\omega_{j}$ described the rate of this rotation: for instance, $L_z$ and $\omega_z$ relate to rotation in the $xy$-plane. This is somewhat counterintuitive. On the other hand, the tensor definitions $L_{ij}$ and $\omega_{ij}$ deal with rotations in the $ij$-plane: for example, $L_{xy}$ generates and $\omega_{xy}$ describes rotations in the $xy$-plane, which seem much more intuitive. Also, with this, $L_{ij} = x_{i} p_{j} - p_{i} x_{j}$ becomes a definition (though there may be a numerical coefficient that I am missing, so forgive me).
The nice thing about this formulation of angular velocities and momenta as tensor quantities is that this is generalizable to 4 dimensions, be it 4 spatial dimensions or 3 spatial and 1 temporal dimension (as in relativity). $L_{\mu \nu} = x_{\mu} p_{\nu} - p_{\mu} x_{\nu}$ now defines the generator of rotation in the $\mu\nu$-plane. Similarly, $\omega_{\mu \nu}$ defined in $L_{\mu \nu} = I_{\mu}^{\; \xi} \omega^{\xi}_{\; \nu}$ describes the rate of rotation in that plane. The reason why these cannot be vectors any more is that the permutation tensor gains an additional index, so contracting it with two vectors yields a tensor with 2 indices; this means that the cross product as laid out in 3 dimensions does not work in any other number of dimensions (except, interestingly enough, for 7, and that is because a 7-dimensional Cartesian vector space can be described through the algebra of octonions which does have a cross product, just as 2-dimensional vectors can be described by complex numbers and 3-dimensional vectors can be described by quaternions).
This has further nice consequence for special relativity. The Lorentz transformation as given in $x^{\mu'} = \Lambda^{\mu'}_{\; \mu} x^{\mu}$ is a hyperbolic rotation through an angle $\alpha$, equal to the rapidity defined as $\alpha = \tanh(\beta)$. A hyperbolic rotation is basically just a normal rotation through an imaginary angle. This can actually be seen by transforming to coordinates with imaginary time (called a Wick rotation, which may come back up in a post in the near future): $x^{\mu} = (ct, x^{j}) \rightarrow (ict, x^{j})$, allowing the metric to change as $\eta_{\mu \nu} = \mathrm{diag}(-1, 1, 1, 1) \rightarrow \delta_{\mu \nu}$. This changes the rapidity to just be a real angle, and the Lorentz transformation becomes a real rotation. Because only the temporal coordinate has been made imaginary while the spatial coordinates have been left untouched, because the Lorentz transformation is now a real rotation, and because angular momentum generates real rotations, then it can be said that the angular momentum components $L_{(0, j)}$ generate Lorentz boosts along the $j$-axis. This fact remains true even if the temporal coordinate is not made imaginary and the metric remains with an opposite sign for the temporal component, though the math of Lorentz boost generation becomes a little more tricky. That said, typically the conservation of angular momentum implies symmetry of the system under rotation, thanks to the Noether theorem. Naïvely, this would imply that conservation of $L_{(0, j)}$ is associated with symmetry under the Lorentz transformation. The truth is a little more complicated (but not by too much), as my advisor and I found from a few Internet searches. Basically, in nonrelativistic mechanics, just as momentum is the generator of spatial translation, position is the generator of (Galilean) momentum boosting: this can be seen in the quantum mechanical representation of momentum in the position basis $\hat{p} = -i\hbar \frac{\partial}{\partial x}$, and the analogous representation of position in the momentum basis $\hat{x} = i\hbar \frac{\partial}{\partial p}$. If the system is invariant under translation, then the momentum is conserved and the system is inertial, whereas if the system is invariant under boosting, then the position is conserved and the system is fixed at a given point in space. In relativity, the analogue to a Galilean momentum boost is exactly the Lorentz transformation, so conservation of $L_{(0, j)}$ corresponds to the system being fixed at its initial spacetime coordinate; this is OK even in relativity because spacetime coordinates are invariant geometric objects, even if their components transform covariantly.
There are a few remaining issues with this analysis. One is that rotations in 3 dimensions are just sums of pairs of rotations in planes, and rotations in 4 dimensions are just sums of pairs of rotations in 3 dimensions. This relates in some way (that I am not really sure of) to symmetries under special orthogonal/unitary transformations in those dimensions. In dimensions higher than 4, things get a lot more hairy, and I'm not sure if any of this continues to hold. Also, one remaining issue is that in special relativity, because the speed of light is fixed and finite, rigid bodies cease to exist except as an approximation, so the description of such dynamics using a moment of inertia tensor generalized to special relativity may not work anymore (though the description of angular momentum as a tensor should still work anyway). Finally, note that the generalization of particle momentum $p_{\mu}$ to a distribution of energy lies in the stress-energy tensor $T_{\mu \nu}$, so the angular momentum of such a distribution becomes a tensor with 3 indices that looks something like (though maybe not exactly like) $L_{\mu \nu \xi} = x_{\mu} T_{\nu \xi} - x_{\nu} T_{\mu \xi}$. In addition, stress-energy tensors with relativistic angular momenta may change the metric itself, so that would need to be accounted for through the Einstein field equations. Anyway, I just wanted to further explore the formulations and generalizations of angular momentum, and I hope this helped in that regard.
There is one problem though: there is nothing particularly vector-like about angular momentum. It is constructed as a vector essentially for mathematical convenience. The definition $\vec{L} = \vec{x} \times \vec{p}$ only works in 3 dimensions. Why is this? Let's look at the definition of the cross product components: in 3 dimensions, the permutation tensor has 3 indices, so contracting it with 2 vectors produces a third vector $\vec{c} = \vec{a} \times \vec{b}$ such that $c_i = \varepsilon_{ijk} a_{j} b_{k}$. One trick that is commonly taught to make the cross product easier is to turn the first vector into a matrix and then perform matrix multiplication with the column representation of the second vector to get the column representation of the resulting vector: the details of this rule are hard to remember, but the source is simple, as it is just $a_{ij} = \varepsilon_{ijk} a_{k}$. Now let us see what happens to angular velocity and angular momentum using this definition. Angular velocity was previously defined as a vector through $\vec{v} = \vec{\omega} \times \vec{x}$. We know that $\vec{x}$ and $\vec{v}$ are true vectors, while $\vec{\omega}$ is a pseudovector (defined by it flipping direction when the coordinate system undergoes reflection), so $\vec{\omega}$ is vector to be made into a tensor. Using the previous definition that in 3 dimensions $\omega_{ij} = \varepsilon_{ijk} \omega_{k}$, then \[v_i = \omega_{ij} x_{j}\] now defines the angular velocity tensor. Similarly, angular momentum is a pseudovector, so it can be made into a tensor through $L_{ij} = \varepsilon_{ijk} L_{k}$. Substituting this into the equation relating angular momenta and angular velocities yields \[L_{ij} = I_{ik} \omega_{kj}\] meaning the matrix representation of the angular momentum tensor is now the matrix multiplication of the matrices representing the moment of inertia and angular velocity tensors.
This has another consequence: the meaning of the components of the angular velocity and angular momentum become much more clear. Previously, $L_{j}$ was the generator of rotation in the plane perpendicular to the $j$-axis, and $\omega_{j}$ described the rate of this rotation: for instance, $L_z$ and $\omega_z$ relate to rotation in the $xy$-plane. This is somewhat counterintuitive. On the other hand, the tensor definitions $L_{ij}$ and $\omega_{ij}$ deal with rotations in the $ij$-plane: for example, $L_{xy}$ generates and $\omega_{xy}$ describes rotations in the $xy$-plane, which seem much more intuitive. Also, with this, $L_{ij} = x_{i} p_{j} - p_{i} x_{j}$ becomes a definition (though there may be a numerical coefficient that I am missing, so forgive me).
The nice thing about this formulation of angular velocities and momenta as tensor quantities is that this is generalizable to 4 dimensions, be it 4 spatial dimensions or 3 spatial and 1 temporal dimension (as in relativity). $L_{\mu \nu} = x_{\mu} p_{\nu} - p_{\mu} x_{\nu}$ now defines the generator of rotation in the $\mu\nu$-plane. Similarly, $\omega_{\mu \nu}$ defined in $L_{\mu \nu} = I_{\mu}^{\; \xi} \omega^{\xi}_{\; \nu}$ describes the rate of rotation in that plane. The reason why these cannot be vectors any more is that the permutation tensor gains an additional index, so contracting it with two vectors yields a tensor with 2 indices; this means that the cross product as laid out in 3 dimensions does not work in any other number of dimensions (except, interestingly enough, for 7, and that is because a 7-dimensional Cartesian vector space can be described through the algebra of octonions which does have a cross product, just as 2-dimensional vectors can be described by complex numbers and 3-dimensional vectors can be described by quaternions).
This has further nice consequence for special relativity. The Lorentz transformation as given in $x^{\mu'} = \Lambda^{\mu'}_{\; \mu} x^{\mu}$ is a hyperbolic rotation through an angle $\alpha$, equal to the rapidity defined as $\alpha = \tanh(\beta)$. A hyperbolic rotation is basically just a normal rotation through an imaginary angle. This can actually be seen by transforming to coordinates with imaginary time (called a Wick rotation, which may come back up in a post in the near future): $x^{\mu} = (ct, x^{j}) \rightarrow (ict, x^{j})$, allowing the metric to change as $\eta_{\mu \nu} = \mathrm{diag}(-1, 1, 1, 1) \rightarrow \delta_{\mu \nu}$. This changes the rapidity to just be a real angle, and the Lorentz transformation becomes a real rotation. Because only the temporal coordinate has been made imaginary while the spatial coordinates have been left untouched, because the Lorentz transformation is now a real rotation, and because angular momentum generates real rotations, then it can be said that the angular momentum components $L_{(0, j)}$ generate Lorentz boosts along the $j$-axis. This fact remains true even if the temporal coordinate is not made imaginary and the metric remains with an opposite sign for the temporal component, though the math of Lorentz boost generation becomes a little more tricky. That said, typically the conservation of angular momentum implies symmetry of the system under rotation, thanks to the Noether theorem. Naïvely, this would imply that conservation of $L_{(0, j)}$ is associated with symmetry under the Lorentz transformation. The truth is a little more complicated (but not by too much), as my advisor and I found from a few Internet searches. Basically, in nonrelativistic mechanics, just as momentum is the generator of spatial translation, position is the generator of (Galilean) momentum boosting: this can be seen in the quantum mechanical representation of momentum in the position basis $\hat{p} = -i\hbar \frac{\partial}{\partial x}$, and the analogous representation of position in the momentum basis $\hat{x} = i\hbar \frac{\partial}{\partial p}$. If the system is invariant under translation, then the momentum is conserved and the system is inertial, whereas if the system is invariant under boosting, then the position is conserved and the system is fixed at a given point in space. In relativity, the analogue to a Galilean momentum boost is exactly the Lorentz transformation, so conservation of $L_{(0, j)}$ corresponds to the system being fixed at its initial spacetime coordinate; this is OK even in relativity because spacetime coordinates are invariant geometric objects, even if their components transform covariantly.
There are a few remaining issues with this analysis. One is that rotations in 3 dimensions are just sums of pairs of rotations in planes, and rotations in 4 dimensions are just sums of pairs of rotations in 3 dimensions. This relates in some way (that I am not really sure of) to symmetries under special orthogonal/unitary transformations in those dimensions. In dimensions higher than 4, things get a lot more hairy, and I'm not sure if any of this continues to hold. Also, one remaining issue is that in special relativity, because the speed of light is fixed and finite, rigid bodies cease to exist except as an approximation, so the description of such dynamics using a moment of inertia tensor generalized to special relativity may not work anymore (though the description of angular momentum as a tensor should still work anyway). Finally, note that the generalization of particle momentum $p_{\mu}$ to a distribution of energy lies in the stress-energy tensor $T_{\mu \nu}$, so the angular momentum of such a distribution becomes a tensor with 3 indices that looks something like (though maybe not exactly like) $L_{\mu \nu \xi} = x_{\mu} T_{\nu \xi} - x_{\nu} T_{\mu \xi}$. In addition, stress-energy tensors with relativistic angular momenta may change the metric itself, so that would need to be accounted for through the Einstein field equations. Anyway, I just wanted to further explore the formulations and generalizations of angular momentum, and I hope this helped in that regard.
Wednesday, 6 March 2013
More on 2012 Fall
Posted on 06:00 by Unknown
Last semester, I was taking 8.05, 8.13, 8.231, and 14.04, along with continuing my UROP. I was busy and stressed basically all the time. Now I think I know why: it turns out that the classes I was taking were much closer to graduate classes in material, yet they came with all the trappings of an undergraduate class, like exams (that were not intentionally easy). Let me explain a little more.
8.05 — Quantum Physics II is where the linear algebra formalism and bra-ket notation of quantum mechanics are introduced and thoroughly investigated. Topics of the class include analysis of wavefunctions in 1-dimensional potentials, vectors in Hilbert spaces, matrix representations of operators, 2-state systems, applications to spin, NMR, continuous Hilbert spaces (e.g. position), the harmonic oscillator, coherent & squeezed states as well as the representation of photon states and the electromagnetic field operators forming a harmonic oscillator, angular momentum, addition of angular momenta, and Clebsch-Gordan coefficients. OK, so considering that most of these things are expected knowledge for the GRE in physics, this is probably more like a standard undergraduate quantum mechanics curriculum rather than a graduate-level curriculum. That said, apparently this perfectly substitutes for the graduate-level quantum theory class, because I know of a lot of people who go right from 8.05 to the graduate relativistic quantum field theory class.
8.13 — Experimental Physics I is generally a standard undergraduate physics laboratory class (although it is considered standard in the sense that its innovations have spread far and wide). The care and detail in performing experiments, analyzing data, making presentations, and writing papers seem like fairly obvious previews of graduate life as an experimental physicist.
8.231 — Physics of Solids I might be the first class on this list that actually could be considered a graduate-level class for undergraduates, also because the TAs for that class have said that it is basically a perfect substitute for the graduate class 8.511 — Theory of Solids I, allowing people who did well in 8.231 to take the graduate class 8.512 & mdash; Theory of Solids II immediately after that. 8.231 emphasized that it is not a survey course but intends to go deep into the physics of solids. I would say that it in fact did both: it was both fairly broad and incredibly deep. Even though the only prerequisite is 8.044 — Statistical Physics I with the corequisite being 8.05, 8.231 really requires intimate familiarity with the material of 8.06 — Quantum Physics III, which is what I am taking this semester. 8.06 introduces in fairly simple terms things like the free electron gas (which is also a review from 8.044), the tight-binding model, electrons in an electromagnetic field, the de Haas-van Alphen effect, and the integer quantum Hall effect, and it will probably talk about perturbation theory and the nearly-free electron gas. 8.231 requires a good level of comfort with these topics, as it goes into much more depth with all of these, as well as the basic descriptions of crystals and lattices, reciprocal space and diffraction, intermolecular forces, phonons, band theory, semiconductor theory and doping, a little bit of the fractional quantum Hall effect (which is much more complicated than its integer counterpart), a little bit of topological insulator theory, and a little demonstration on superfluidity and superconductivity.
14.04 &; Intermediate Microeconomic Theory is the other class I can confidently say is much closer to a graduate class than an undergraduate class, because I talked to the professor yesterday and he said exactly this. He said that typical undergraduate intermediate microeconomic theory classes are more like 14.03 &; Microeconomic Theory and Public Policy (which I am taking now), where the constrained optimization problems are fairly mechanical, and there may be discussion on the side of applications to real-world problems. By contrast, 14.04 last semester focused on the fundamentals of abstract choice theory with a lot more elegant mathematical formalism, the application of those first principles to derive all of consumer and producer choice theories, partial and general equilibrium, risky choice theory, subjective risky choice theory and its connections to Arrow-Debreu securities and general equilibrium, oligopoly and game theory, asymmetric information, and other welfare problems. The professor was saying that by contrast to a typical such class elsewhere, 14.04 here is much closer to a graduate microeconomic theory/decision theory class, and the professor wanted to achieve that level of abstract conceptualization while not going too far for an undergraduate audience.
At this point, I'm hoping that the experiences from last semester pay off this semester. It looks like that has been working so far!
8.05 — Quantum Physics II is where the linear algebra formalism and bra-ket notation of quantum mechanics are introduced and thoroughly investigated. Topics of the class include analysis of wavefunctions in 1-dimensional potentials, vectors in Hilbert spaces, matrix representations of operators, 2-state systems, applications to spin, NMR, continuous Hilbert spaces (e.g. position), the harmonic oscillator, coherent & squeezed states as well as the representation of photon states and the electromagnetic field operators forming a harmonic oscillator, angular momentum, addition of angular momenta, and Clebsch-Gordan coefficients. OK, so considering that most of these things are expected knowledge for the GRE in physics, this is probably more like a standard undergraduate quantum mechanics curriculum rather than a graduate-level curriculum. That said, apparently this perfectly substitutes for the graduate-level quantum theory class, because I know of a lot of people who go right from 8.05 to the graduate relativistic quantum field theory class.
8.13 — Experimental Physics I is generally a standard undergraduate physics laboratory class (although it is considered standard in the sense that its innovations have spread far and wide). The care and detail in performing experiments, analyzing data, making presentations, and writing papers seem like fairly obvious previews of graduate life as an experimental physicist.
8.231 — Physics of Solids I might be the first class on this list that actually could be considered a graduate-level class for undergraduates, also because the TAs for that class have said that it is basically a perfect substitute for the graduate class 8.511 — Theory of Solids I, allowing people who did well in 8.231 to take the graduate class 8.512 & mdash; Theory of Solids II immediately after that. 8.231 emphasized that it is not a survey course but intends to go deep into the physics of solids. I would say that it in fact did both: it was both fairly broad and incredibly deep. Even though the only prerequisite is 8.044 — Statistical Physics I with the corequisite being 8.05, 8.231 really requires intimate familiarity with the material of 8.06 — Quantum Physics III, which is what I am taking this semester. 8.06 introduces in fairly simple terms things like the free electron gas (which is also a review from 8.044), the tight-binding model, electrons in an electromagnetic field, the de Haas-van Alphen effect, and the integer quantum Hall effect, and it will probably talk about perturbation theory and the nearly-free electron gas. 8.231 requires a good level of comfort with these topics, as it goes into much more depth with all of these, as well as the basic descriptions of crystals and lattices, reciprocal space and diffraction, intermolecular forces, phonons, band theory, semiconductor theory and doping, a little bit of the fractional quantum Hall effect (which is much more complicated than its integer counterpart), a little bit of topological insulator theory, and a little demonstration on superfluidity and superconductivity.
14.04 &; Intermediate Microeconomic Theory is the other class I can confidently say is much closer to a graduate class than an undergraduate class, because I talked to the professor yesterday and he said exactly this. He said that typical undergraduate intermediate microeconomic theory classes are more like 14.03 &; Microeconomic Theory and Public Policy (which I am taking now), where the constrained optimization problems are fairly mechanical, and there may be discussion on the side of applications to real-world problems. By contrast, 14.04 last semester focused on the fundamentals of abstract choice theory with a lot more elegant mathematical formalism, the application of those first principles to derive all of consumer and producer choice theories, partial and general equilibrium, risky choice theory, subjective risky choice theory and its connections to Arrow-Debreu securities and general equilibrium, oligopoly and game theory, asymmetric information, and other welfare problems. The professor was saying that by contrast to a typical such class elsewhere, 14.04 here is much closer to a graduate microeconomic theory/decision theory class, and the professor wanted to achieve that level of abstract conceptualization while not going too far for an undergraduate audience.
At this point, I'm hoping that the experiences from last semester pay off this semester. It looks like that has been working so far!
Thursday, 21 February 2013
A Less-Seen View of Complex Numbers
Posted on 19:30 by Unknown
This post is a little different from the last one, only because it's more about mathematics than physics. It's based on thoughts I have been having about complex numbers and how they relate to 2-dimensional vectors. Follow the jump to see more.
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