Particles in the Continuous Quantum Field

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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Operators and States of the Continuous Quantum Field

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.

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Diagonalizing and Quantizing the Continuous Field Hamiltonian

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.

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Classical Discrete and Continuum Fields

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.

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Harmonic Oscillator from Fields not Potentials

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.

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Charge Conservation and Legendre Transformations

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").

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Hamiltonian Density and the Stress-Energy Tensor

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.

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Time and Temperature are Complex

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...".

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Nonzero Electromagnetic Fields in a Cavity

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.

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