- #1
decerto
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Homework Statement
Given the coherent state of the harmonic oscillator [tex]|z>=e^{-\frac{|z|^2}{2}}\sum_{n=0}^\infty\frac{z^{n}}{\sqrt{n!}}|n>[/tex]
compute the probability for finding n quanta in the sate [itex]|z>[/itex] and the average excitation number [itex]<z|n|z>[/itex]
Homework Equations
[itex]|z>=e^{-\frac{|z|^2}{2}}\sum_0^\infty\frac{z^{n}}{\sqrt{n!}}|n>[/itex]
Probability=[itex]|<n|z>|^2[/itex]
Average excitation number = [itex]<z|n|z>[/itex]
The Attempt at a Solution
For the first part I did [itex]|<n|z>|^2=|e^{-\frac{|z|^2}{2}}\sum_{m=0}^\infty\frac{z^{m}}{\sqrt{m!}}<n|m>|^2[/itex]
[itex]=|e^{-\frac{|z|^2}{2}}\sum_{m=0}^\infty\frac{z^{m}}{\sqrt{m!}}\delta_{nm}|^2[/itex]
[itex]=|e^{-\frac{|z|^2}{2}}\frac{z^{n}}{\sqrt{n!}}|^2[/itex]
[itex]=|e^{-|z|^2}\frac{z^{2n}}{n!}|[/itex]
[itex]=e^{-|z|^2}\frac{|z|^{2n}}{n!}[/itex]
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 [itex]|z>[/itex] and [itex]<z|[/itex] be [itex]|n>[/itex] or something different
What I think is that it should be is
[itex]<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'>[/itex]
But I am not sure what <m|n|m'> is equal to, [itex]\delta_{nm}\delta_{nm'}[/itex] maybe?
Im getting [itex]e^0[/itex] 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.