# Coherent States of Harmonic Oscillator

1. Nov 29, 2014

### decerto

1. The problem statement, all variables and given/known data
Given the coherent state of the harmonic oscillator $$|z>=e^{-\frac{|z|^2}{2}}\sum_{n=0}^\infty\frac{z^{n}}{\sqrt{n!}}|n>$$
compute the probability for finding n quanta in the sate $|z>$ and the average excitation number $<z|n|z>$

2. Relevant equations
$|z>=e^{-\frac{|z|^2}{2}}\sum_0^\infty\frac{z^{n}}{\sqrt{n!}}|n>$

Probability=$|<n|z>|^2$

Average excitation number = $<z|n|z>$

3. The attempt at a solution

For the first part I did $|<n|z>|^2=|e^{-\frac{|z|^2}{2}}\sum_{m=0}^\infty\frac{z^{m}}{\sqrt{m!}}<n|m>|^2$
$=|e^{-\frac{|z|^2}{2}}\sum_{m=0}^\infty\frac{z^{m}}{\sqrt{m!}}\delta_{nm}|^2$
$=|e^{-\frac{|z|^2}{2}}\frac{z^{n}}{\sqrt{n!}}|^2$
$=|e^{-|z|^2}\frac{z^{2n}}{n!}|$
$=e^{-|z|^2}\frac{|z|^{2n}}{n!}$

Which is correct but was I right in using m to specify a different energy eigenstate to n to obtain the kroncker delta?

For the second part Im not sure how to even write out the problem. Should the eigenstates in $|z>$ and $<z|$ be $|n>$ or something different

What I think is that it should be is

$<z|n|z>=e^{-\frac{|z|^2}{2}}\sum_{m=0}^\infty\frac{z^{\dagger m}}{\sqrt{m!}}e^{-\frac{|z|^2}{2}}\sum_{m'=0}^\infty\frac{z^{m'}}{\sqrt{m'!}}<m|n|m'>$

But im not sure what <m|n|m'> is equal to, $\delta_{nm}\delta_{nm'}$ maybe?

Im getting $e^0$ which isnt the right answer as per page 2 of this http://www.phas.ubc.ca/~joanna/phys501/coherent-states.pdf

Any help would be great.

2. Nov 29, 2014

### ShayanJ

The first part is correct.
About your question regarding the second part, Pay attention that $n=a^\dagger a$ and $a^\dagger |n>=\sqrt{n+1} |n+1 >$ and $a |n>=\sqrt{n}|n-1>$.

3. Nov 29, 2014

### stevendaryl

Staff Emeritus
You pretty much have it, except that by definition $n |m'\rangle = m' |m' \rangle$. So $\langle m |n| m' \rangle = m' \langle m | m' \rangle$.

4. Nov 29, 2014

### decerto

This gives me m' as the answer but the answer is $|z|^2$?

Nvm got it using the ladder operators

5. Nov 29, 2014

### stevendaryl

Staff Emeritus
No, it doesn't give $m'$ as the answer, because $m'$ is summed over. Since $\langle m | n |m' \rangle = m' \delta_{m, m'}$, you have something like:

$e^{-z^2} \sum_{m=0}^\infty \sum_{m'=0}^\infty \frac{z^m z^{m'}}{\sqrt{m! m'!}} m' \delta_{m,m'}$

Because of the $\delta_{m, m'}$ the second sum simplifies to just one term, the one where $m'=m$.