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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##?

|\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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