Coherent States of Harmonic Oscillator

[decerto]

Homework Statement

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>

Homework 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>

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 I am 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 I am not sure what <m|n|m'> is equal to, \delta_{nm}\delta_{nm&#039;} maybe?

Im getting e^0 which isn't the right answer as per page 2 of this http://www.phas.ubc.ca/~joanna/phys501/coherent-states.pdf

Any help would be great.

 

[ShayanJ]

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

 

[stevendaryl]

For the second part I am not sure how to even write out the problem. Should the eigenstates in |z&gt; and &lt;z| be |n&gt; or something different

What I think is that it should be is

&lt;z|n|z&gt;=e^{-\frac{|z|^2}{2}}\sum_{m=0}^\infty\frac{z^{\dagger m}}{\sqrt{m!}}e^{-\frac{|z|^2}{2}}\sum_{m&#039;=0}^\infty\frac{z^{m&#039;}}{\sqrt{m&#039;!}}&lt;m|n|m&#039;&gt;

But I am not sure what <m|n|m'> is equal to, \delta_{nm}\delta_{nm&#039;} maybe?

Im getting e^0 which isn't the right answer as per page 2 of this http://www.phas.ubc.ca/~joanna/phys501/coherent-states.pdf

Any help would be great.

You pretty much have it, except that by definition n |m&#039;\rangle = m&#039; |m&#039; \rangle. So \langle m |n| m&#039; \rangle = m&#039; \langle m | m&#039; \rangle.

 

[decerto]

You pretty much have it, except that by definition n |m&#039;\rangle = m&#039; |m&#039; \rangle. So \langle m |n| m&#039; \rangle = m&#039; \langle m | m&#039; \rangle.

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

Nvm got it using the ladder operators

 

[stevendaryl]

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

No, it doesn't give m&#039; as the answer, because m&#039; is summed over. Since \langle m | n |m&#039; \rangle = m&#039; \delta_{m, m&#039;}, you have something like:

e^{-z^2} \sum_{m=0}^\infty \sum_{m&#039;=0}^\infty \frac{z^m z^{m&#039;}}{\sqrt{m! m&#039;!}} m&#039; \delta_{m,m&#039;}

Because of the \delta_{m, m&#039;} the second sum simplifies to just one term, the one where m&#039;=m.