Finding the Expectation value of harmonic oscillator What is the expectation value for $e^{{\alpha}a^{\dagger}}e^{-\alpha^*{a}}$ of any two states of harmonic oscillator (let say $|n\rangle$ and $|m\rangle$) given below,
$$\langle n|e^{{\alpha}a^{\dagger}}e^{-\alpha^*{a}}|m\rangle$$
The actual problem is finding the above expectation value when $m=n$ and $m\neq{n}$.  For the first case I am able to express the expectation value in terms of $L_n(|\alpha|^2)$ by expanding $e^{{\alpha}a^{\dagger}}$ and $e^{-\alpha^*{a}}$
in terms of finite series, but for the second case $m\neq{n}$, I am stuck here, 
$$\sum_{l=0}^{n}\sum_{k=0}^{m}\frac{(-1)^k(\alpha)^l(\alpha^*)^k}{l!k!}\frac{\sqrt{n!}\sqrt{m!}}{\sqrt{(n-l)!}\sqrt{(m-k)!}}\langle n-l|m-k\rangle$$
Is it possible express this expression in terms of Laguerre, Hermite or any other function? please help me or at least give me some hint.  
I made some progress, I reduced the above expression to $$\sum_{k=0}^{m}\frac{(-1)^k(\alpha)^l(\alpha^*)^k}{l!k!}\frac{\sqrt{(m-k+l)!}\sqrt{m!}}{(m-k)!}$$ now is it possible to express this interms of Laguerre polynomials?
 A: Use the Baker-Campbell-Haussdorf formula to commute the displacement operators,
$$
e^{\alpha a^\dagger} e^{-\alpha^* a} = e^{|\alpha|^2} e^{-\alpha^* a} e^{\alpha a^\dagger} 
$$
and rearrange the matrix element as
$$
\langle n| e^{\alpha a^\dagger} e^{-\alpha^* a} |m \rangle = \frac{e^{|\alpha|^2} }{\sqrt{n! m!}} \langle 0 | a^n e^{-\alpha^* a} e^{\alpha a^\dagger}  (a^\dagger)^m |0 \rangle = \frac{e^{|\alpha|^2} }{\sqrt{n! m!}} \langle 0 | e^{-\alpha^* a} a^n (a^\dagger)^m  e^{\alpha a^\dagger}  |0 \rangle
$$
$$
= (-1)^n \frac{e^{|\alpha|^2} }{\sqrt{n! m!}} \frac{\partial^n}{\partial (\alpha^*)^n}  \frac{\partial^m}{\partial \alpha^m} \langle 0 | e^{-\alpha^* a} e^{\alpha a^\dagger}  |0 \rangle =  (-1)^n \frac{e^{|\alpha|^2} }{\sqrt{n! m!}} \frac{\partial^n}{\partial (\alpha^*)^n}  \frac{\partial^m}{\partial \alpha^m}e^{-|\alpha|^2}  \langle 0 | e^{\alpha a^\dagger}  e^{-\alpha^* a}  |0 \rangle
$$
$$
= (-1)^n \frac{e^{|\alpha|^2} }{\sqrt{n! m!}} \frac{\partial^n}{\partial (\alpha^*)^n}  \frac{\partial^m}{\partial \alpha^m} e^{-|\alpha|^2} = (-1)^{n+m} \frac{e^{|\alpha|^2} }{\sqrt{n! m!}} \frac{\partial^n}{\partial (\alpha^*)^n} \left[ (\alpha^*)^m e^{-|\alpha|^2} \right]
$$
Now try to rearrange the derivative entirely in terms of $|\alpha|^2$:
$$
(-1)^{n+m} \frac{e^{|\alpha|^2} }{\sqrt{n! m!}} \frac{\partial^n}{\partial (\alpha^*)^n} \left[ (\alpha^*)^m e^{-|\alpha|^2} \right] = (-1)^{n+m}  \alpha^{-m} \frac{e^{|\alpha|^2} }{\sqrt{n! m!}} \frac{\partial^n}{\partial (\alpha^*)^n} \left[ (|\alpha|^2)^m e^{-|\alpha|^2} \right] = 
$$
$$
= (-1)^{n+m}  \alpha^{n-m} \frac{e^{|\alpha|^2} }{\sqrt{n! m!}} \frac{\partial^n}{\partial (|\alpha|^2)^n} \left[ (|\alpha|^2)^m e^{-|\alpha|^2} \right] = 
(-1)^{n+m} \sqrt{ \frac{n!}{m!}} (\alpha^*)^{m-n} \left[ \frac{( |\alpha|^2)^{-(m-n)} e^{|\alpha|^2} }{n!} \frac{\partial^n}{\partial (|\alpha|^2)^n} \left[ (|\alpha|^2)^m e^{-|\alpha|^2} \right] \right]
$$
The expression in the big brackets singles out a differential form of Laguerre polynomials, reading
$$
L_n^{(\beta)} (x) = \frac{x^{-\beta} e^x}{n!} \frac{\partial^n}{\partial x^n}\left( x^{n+\beta} e^{-\beta} \right)
$$
So we can eventually identify
$$
\langle n| e^{\alpha a^\dagger} e^{-\alpha^* a} |m \rangle = (-1)^{n+m} \sqrt{ \frac{n!}{m!}} (\alpha^*)^{m-n} L^{(m-n)}_n(|\alpha|^2)
$$
There should be an equivalent expression arising from switching the order of the derivatives in $(-1)^n \frac{e^{|\alpha|^2} }{\sqrt{n! m!}} \frac{\partial^n}{\partial (\alpha^*)^n}  \frac{\partial^m}{\partial \alpha^m} e^{-|\alpha|^2}$, and you could try to symmetrize somehow in $\alpha$, $\alpha^*$ and $n$, $m$.
A: Assuming you have it set up that $\langle n\mid m\rangle=\delta_{nm}$
Then taking the sum:
$$\sum_{l=0}^{n}\sum_{k=0}^{m}\frac{(-1)^k(\alpha)^l(\alpha^*)^k}{l!k!}\frac{\sqrt{n!}\sqrt{m!}}{\sqrt{(n-l)!}\sqrt{(m-k)!}}\langle n-l|m-k\rangle$$
This simplifies to:
$$\sum_{l=0}^{n}\sum_{k=0}^{m}\frac{(-1)^k(\alpha)^l(\alpha^*)^k}{l!k!}\frac{\sqrt{n!}\sqrt{m!}}{\sqrt{(n-l)!}\sqrt{(m-k)!}}\delta_{n-l,m-k}$$
So we have that $n-l=m-k$, lets use this to get rid of the $l$ summation if $n>m$ otherwise, get rid of the $k$ summation. Replace $l=n-m+k$
$$\sum_{l=0}^{n}\sum_{k=0}^{m}\frac{(-1)^k(\alpha)^{n-m+k}(\alpha^*)^k}{(n-m+k)!k!}\frac{\sqrt{n!}\sqrt{m!}}{\sqrt{(m-k)!}\sqrt{(m-k)!}}\delta_{l,n-m+k}$$
$$=\sqrt{n!}\sqrt{m!}(\alpha)^{n-m}\sum_{k=0}^{m}\frac{(-1)^k\left|\alpha\right|^{2k}}{(n-m+k)!k! (m-k)!}$$
$$=\frac{\sqrt{n!}}{\sqrt{m!}}(\alpha)^{n-m}\sum_{k=0}^{m}{m \choose k}\frac{(-1)^k\left|\alpha\right|^{2k}}{(n-m+k)!}$$
In the case $n=m$ you're right in that you can use the formula for the laguerre polynomials
$$L_n(x)=\sum_{k=0}^n {n\choose k} \frac{(-1)^k x^k}{k!}$$
So it becomes $L_n(|\alpha|^2)$
Other than this I'm not sure, one can relabel $k'=m-k$ and use ${m\choose k}={m\choose m-k}$ to get to
$$\frac{\sqrt{n!}}{\sqrt{m!}}(\alpha)^{n-m}|\alpha|^{2m}(-1)^m\sum_{k=0}^{m}{m \choose k}\frac{(-1)^k\left|\alpha\right|^{-2k}}{(n-k)!}$$
Maybe this is useful for you, or someone else might recognise one of these$\ldots$
A: It is at least possible to simplify your second expression.
Note that (if I've not miscalculated)
$$\langle m|e^{{\alpha}a^{\dagger}}e^{-\alpha^*{a}}|n\rangle = \left( \langle n|e^{{-\alpha}a^{\dagger}}e^{\alpha^*{a}}|m\rangle \right)^{*}$$
so we can without loss of generality assume $n > m$.
Now, in your second sum, we can use that
$$\langle n-l|m-k\rangle=\delta_{n-l,m-k} $$
this yields $l=n-m+k$ and we can collapse the sum in $l$ using that replacement (I'm collapsing the sum in l because by assumption, $n-m>0$, therefore that does not yield any strange results). This at least makes the expression considerably simpler. Maybe that helps?
