Undergrad Physical interpretation of this coherent state

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SUMMARY

The discussion focuses on the physical interpretation of coherent states in quantum mechanics, specifically the coherent state defined as ##|\alpha\rangle \equiv e^{\alpha A^{\dagger} - \bar{\alpha} A} |0\rangle##. It is established that ##|\alpha\rangle## is an eigenvector of the lowering operator ##A## with eigenvalue ##\alpha##. The inner product of two coherent states is derived as ##|\langle \alpha | \beta \rangle|^2 = e^{|\alpha - \beta|^2}##, indicating that coherent states span the Hilbert space. The mean value of momentum is shown to be proportional to the imaginary part of ##\alpha##, linking the quantum description to classical motion through Ehrenfest's theorem.

PREREQUISITES
  • Understanding of quantum harmonic oscillators and their operators, specifically ##A## and ##A^{\dagger}##.
  • Familiarity with coherent states and their mathematical representation in quantum mechanics.
  • Knowledge of inner product calculations in Hilbert spaces.
  • Basic grasp of Ehrenfest's theorem and its implications in quantum-classical correspondence.
NEXT STEPS
  • Study the derivation and implications of coherent states in quantum optics.
  • Learn about the mathematical properties of the raising and lowering operators in quantum mechanics.
  • Explore the applications of Ehrenfest's theorem in connecting quantum mechanics to classical physics.
  • Investigate the position and momentum representations of coherent states in more detail.
USEFUL FOR

Quantum physicists, students of quantum mechanics, and researchers interested in the relationship between quantum states and classical systems will benefit from this discussion.

ergospherical
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Given the usual raising & lowering operators ##A^{\dagger}## & ##A## for a quantum harmonic oscillator, consider a coherent state ##|\alpha\rangle \equiv e^{\alpha A^{\dagger} - \bar{\alpha} A} |0\rangle##. I first check that ##|\alpha\rangle## is an eigenvector of ##A##. I already proved that if ##X##, ##Y## commute with ##[X,Y]## then ## e^{X+Y} = e^{X} e^{Y} e^{-\frac{1}{2}[X,Y]}##, which is applicable here because both ##A^{\dagger}## & ##A## clearly commute with ##[A^{\dagger}, A] = 1##, therefore\begin{align*}
|\alpha \rangle &= e^{\alpha A^{\dagger}} e^{-\bar{\alpha} A} e^{-\frac{1}{2}[\alpha A^{\dagger},-\bar{\alpha} A]} |0\rangle = e^{\frac{1}{2}|\alpha|^2} e^{\alpha A^{\dagger}} |0\rangle
\end{align*}where I used that ##e^{-\bar{\alpha} A}|0 \rangle = (1 - \bar{\alpha} A + \dots)|0\rangle = |0 \rangle##. Upon application of ##A##,\begin{align*}
A|\alpha\rangle = e^{\frac{1}{2}|\alpha|^2} (A e^{\alpha A^{\dagger}}) |0 \rangle = e^{\frac{1}{2}|\alpha|^2} ([A , e^{\alpha A^{\dagger}}] + e^{\alpha A^{\dagger}} A) |0 \rangle &= e^{\frac{1}{2}|\alpha|^2} (\alpha e^{\alpha A^{\dagger}} + e^{\alpha A^{\dagger}} A) |0 \rangle \\
&= \alpha e^{\frac{1}{2}|\alpha|^2} e^{\alpha A^{\dagger}} |0\rangle \\
&= \alpha |\alpha \rangle
\end{align*}which means that ##|\alpha \rangle## is of eigenvalue ##\alpha##. I make use of a similar operator identity ##e^X e^Y = e^Y e^X e^{[X,Y]}## to calculate the inner product of two general states:\begin{align*}
\langle \alpha | \beta \rangle = e^{\frac{1}{2} (|\alpha|^2 + |\beta|^2)}\langle 0 | e^{\bar{\beta} A} e^{\alpha A^{\dagger}}| 0 \rangle &= e^{\frac{1}{2} (|\alpha|^2 + |\beta|^2)}\langle 0 | e^{\alpha A^{\dagger}} e^{\bar{\beta} A} e^{[\bar{\beta} A, \alpha A^{\dagger}]} | 0 \rangle \\
&= e^{\frac{1}{2}(|\alpha|^2 + |\beta|^2)}e^{-2\alpha \bar{\beta}} \langle 0 | e^{\alpha A^{\dagger}} e^{\bar{\beta} A} | 0 \rangle \\
&= e^{\frac{1}{2}(|\alpha|^2 -2\alpha \bar{\beta} + |\beta|^2))}
\end{align*}where I used the fact that ##\langle 0 | e^{\alpha A^{\dagger}} \leftrightarrow e^{\bar{\alpha} A} |0\rangle##. It is therefore also the case that ##|\langle \alpha | \beta \rangle|^2 = e^{|\alpha - \beta|^2}##, that the set ##\{ |\alpha \rangle \}_{\alpha \in \mathbf{C}}## spans the space and that one can select a basis from a suitable subset of ##\{ |\alpha \rangle \}_{\alpha \in \mathbf{C}}##.

To consider the physical interpretation of ##|\alpha(t) \rangle## for a general complex ##\alpha \in \mathbf{C}##, it is advised to calculate ##\langle \alpha | P | \alpha \rangle##,\begin{align*}
\langle \alpha | P | \alpha \rangle = \frac{i}{\sqrt{2}} \langle \alpha | (A^{\dagger} - A) | \alpha \rangle &= \frac{i}{\sqrt{2}} \langle \alpha | (\bar{\alpha} - \alpha) |\alpha \rangle \\
&= \frac{i}{\sqrt{2}} (\bar{\alpha} - \alpha) \\
&= \sqrt{2} \mathrm{Im}(\alpha)
\end{align*}How am I supposed to interpret that the mean value of the momentum is proportional to the imaginary part of ##\alpha##? Also, how would I use this result to describe, qualitatively, how the position and momentum space wavefunctions evolve (I already worked out that ##|\alpha(t) \rangle = e^{-i\omega t/2} | e^{-i\omega t} \alpha \rangle##?
 
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Thanks for the link. Is there a way to see that the position and momentum basis wavefunctions oscillate like in a classical system without further calculation apart from the OP? I only ask because my problem sheet suggests that it is not necessary to work anything else out explicitly.
 
You can refer to Ehrenfest's theorem. Since the harmonic oscillator has a linear set of equations of motion for ##x## and ##p## the expectation values fulfill precisely the same equations of motion as the classical harmonic oscillator. A coherent state is one of minimal uncertainty product, ##\Delta x \Delta p=\hbar/2##, and thus are closest to a classical description of the system, especially for large ##|\alpha|##.
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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