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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Tuesday, 27 August 2013
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