# Position representation of coherent state and time evolution

1. Mar 16, 2016

### Idoubt

1. The problem statement, all variables and given/known data
I ended up solving the problem as I was typing it up, I am posting what I did anyway as it took so long to type and might be useful to someone else.

I am trying to figure out the position representation of a coherent state and it's time evolution. I should be getting a Guassian wavefunction which is oscillating in position.

2. Relevant equations
Coherent state,
$| \alpha \rangle = \exp (- \frac{|\alpha |^2}{2} ) \sum\limits_{n = 0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} | n \rangle$
Position representation of number state, with $\xi = \sqrt{\frac{\omega}{ \hbar}} x$
$\psi_n (\xi) = \langle x | n \rangle = \left( \frac{\omega}{\pi \hbar}\right)^{1/4} (2^n n!)^{-1/2}\exp (-\frac{\xi^2}{2}) H_n(\xi)$
where $H_n$ are Hermite Polinomials.
Generating function for Hermite Polinomials is,
$\exp (2xt - t^2) = \sum\limits_{n = 0}^{\infty} H_n (x) \frac{t^n}{n!}$
System Hamiltonian is,
$\mathcal{H} = \hbar \omega \left(\frac{1}{2} + a^{\dagger}a \right) = \hbar \omega \left(\frac{1}{2} + \hat{n} \right)$
3. The attempt at a solution

The position representation of a coherent state should be,
$\psi (\alpha) = \langle x | \alpha \rangle = \exp (- \frac{|\alpha |^2}{2} ) \sum\limits_{n = 0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} \langle x | n \rangle \\ = \exp (- \frac{|\alpha |^2}{2} ) \sum\limits_{n = 0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} \left( \frac{\omega}{\pi \hbar}\right)^{1/4} (2^n n!)^{-1/2}\exp (-\frac{\xi^2}{2}) H_n(\xi) \\ = \left( \frac{\omega}{\pi \hbar}\right)^{1/4} \exp (- \frac{|\alpha |^2}{2} ) \exp (-\frac{\xi^2}{2}) \sum\limits_{n = 0}^{\infty} \frac{(\alpha / \sqrt{2})^n}{n!} H_n (\xi)$
Using the generating function of the Hermite polinomials this reduces to,
$\psi (\alpha) = \left( \frac{\omega}{\pi \hbar}\right)^{1/4} \exp (- \frac{|\alpha |^2}{2} ) \exp (-\frac{\xi^2}{2}) \exp (\sqrt{2} \xi \alpha - \frac{\alpha^2}{2})$

Completeing squares gives,
$\psi (\alpha) = \left( \frac{\omega}{\pi \hbar}\right)^{1/4} \exp (- \frac{|\alpha |^2}{2} ) \exp (\frac{\xi^2}{2}) \exp \left( - ( \xi - \frac{\alpha}{\sqrt{2}})^2 \right)$

Now the time evolved state will be,
$| \alpha (t) \rangle = \exp (-i\mathcal{H}t / \hbar) | \alpha \rangle$
$= \exp(- \frac{i\omega t}{2})\exp (-i\omega \hat{n}t) \exp (- \frac{|\alpha |^2}{2} ) \sum\limits_{n = 0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} | n \rangle$
$= \exp(- \frac{i\omega t}{2}) \sum\limits_{n = 0}^{\infty} \frac{(\alpha e^{ -i\omega t})^n}{\sqrt{n!}} | n \rangle$
$= \exp(- \frac{i\omega t}{2}) | \alpha e^{-i \omega t} \rangle$
So the coherent state evolves to another coherent state with a phase $\frac{-i\omega t}{2}$ added overall and $\alpha(t) = \alpha e^{-i \omega t}$

so,
$\psi (\alpha(t)) = \left( \frac{\omega}{\pi \hbar}\right)^{1/4} \exp (- \frac{|\alpha |^2}{2} ) \exp (\frac{\xi^2}{2}) \exp \left( - ( \xi - \frac{\alpha e^{i\omega t}}{\sqrt{2}})^2 \right)$

If I seperate out the real and imaginary part of $\alpha (t)$ as $\alpha_r + i \alpha_i = | \alpha | \cos (\phi - \omega t) + i |\alpha | \sin (\phi - \omega t)$, where $\phi$ is the complex phase of $\alpha$,
$\psi (\alpha(t)) = \left( \frac{\omega}{\pi \hbar}\right)^{1/4} \exp (- \frac{|\alpha |^2}{2} ) \exp (\frac{\xi^2}{2}) \exp \left( - ( (\xi - \frac{\alpha_r}{\sqrt{2}}) - i\frac{\alpha_i}{\sqrt{2}} )^2 \right)$
$= \left( \frac{\omega}{\pi \hbar}\right)^{1/4} \exp (- \frac{|\alpha |^2}{2} ) \exp (\frac{\xi^2}{2}) \exp \left( -(\xi - \frac{\alpha_r}{\sqrt{2}} )^2 + \frac{\alpha_i^2}{2} +2i \frac{\alpha_i}{\sqrt{2}}(\xi - \frac{\alpha_r}{\sqrt{2}} ) \right)$

The probability distribution in x will be,
$| \psi (\alpha (t)) |^2 = \left( \frac{\omega}{\pi \hbar}\right)^{1/2} \exp (-|\alpha |^2) \exp \left( \xi^2 - 2(\xi - \frac{\alpha_r}{\sqrt{2}} )^2 + \alpha_i^2 \right)$
$= \left( \frac{\omega}{\pi \hbar}\right)^{1/2} \exp (-|\alpha |^2) \exp \left( -\xi^2 - \alpha_r^2 + 2\sqrt{2}\xi\alpha_r + \alpha_i^2 \right)$
$= \left( \frac{\omega}{\pi \hbar}\right)^{1/2} \exp (-|\alpha |^2) \exp \left( -(\xi - \sqrt{2}\alpha_r) ^2 + \alpha_r^2 + \alpha_i^2 \right)$
$= \left( \frac{\omega}{\pi \hbar}\right)^{1/2} \exp (-|\alpha |^2) \exp \left( -(\xi - \sqrt{2}\alpha_r) ^2 \right) \exp ( |\alpha|^2 )$

$| \psi_{\alpha}(x) |^2 = \left( \frac{\omega}{\pi \hbar}\right)^{1/2} \exp \left( -(\sqrt{\frac{\omega}{ \hbar}} x - \sqrt{2}|\alpha|\cos (\phi - \omega t)) ^2 \right)$

Which is clearly a normalized Guassian with center at $\sqrt{\frac{2\hbar}{\omega}}| \alpha | \cos (\phi - \omega t)$

2. Mar 21, 2016