# Finding the Expectation value of harmonic oscillator [closed]

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?

## closed as off-topic by ACuriousMind♦, user36790, CuriousOne, honeste_vivere, GertJul 28 '16 at 14:13

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• if the states are properly normalised then shouldn't $\langle n-l\big| m-k\rangle=\delta_{n-l,m-k}$? – snulty Jul 27 '16 at 10:56
• your reduced expression still has $l$s in them; this should not be the case. I have however not been able to see the Laguerre polynomial for the case $n\neq m$ – Sanya Jul 27 '16 at 13:19
• Is $\langle n\big|n\rangle=1$ or $\sim=n!$? In any case you also need to replace $l=n-m+k$. – snulty Jul 27 '16 at 13:28

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 $n$, $m$.

• I did a quick check using $e^{sX}Ye^{-sX} = Y + s[ X, Y]$ for $X, Y \in \{a, a^\dagger\}$, see again en.wikipedia.org/wiki/Baker–Campbell-Hausdorff_formula, and for $n=1$ and $n=2$ I don't get a $(-1)^n$ factor. – udrv Jul 28 '16 at 15:01
• That is $$\langle n| e^{\alpha a^\dagger} e^{-\alpha^* a} |n \rangle = (n!)^{-1} \langle 0 | a^n e^{\alpha a^\dagger} e^{-\alpha^* a} (a^\dagger)^n |0 \rangle = \\ \langle 0 | e^{\alpha a^\dagger} \left( e^{-\alpha a^\dagger} a e^{\alpha a^\dagger}\right)^n \left( e^{-\alpha^* a} (a^\dagger)e^{\alpha^* a}\right)^n e^{-\alpha^* a} |0 \rangle =\\ \langle 0 | \left( a - \alpha[a^\dagger, a] \right)^n \left( a^\dagger -\alpha^* [a, a^\dagger] \right)^n |0 \rangle = \langle 0 | \left( a + \alpha \right)^n \left( a^\dagger -\alpha^* \right)^n |0 \rangle$$ – udrv Jul 28 '16 at 15:02
• Then for $n=1$ I get $$\langle 1| e^{\alpha a^\dagger} e^{-\alpha^* a} |1 \rangle = \langle 0 | \left( a + \alpha \right) \left( a^\dagger -\alpha^* \right) |0 \rangle = 1 - |\alpha|^2 = L_1(|\alpha|^2)$$ and for $n=2$, $$\langle 2| e^{\alpha a^\dagger} e^{-\alpha^* a} |2 \rangle = (1/2) \langle 0 | \left( a + \alpha \right)^2 \left( a^\dagger -\alpha^* \right)^2 |0 \rangle =$$ The latter eventually reduces to $(1/2) \left( 2 - |\alpha|^2 + |\alpha|^4 \right) = L_2(|\alpha|^2)$. – udrv Jul 28 '16 at 15:02
• sorry, I am also not getting $(-1)^n$, I made a mistake, now I verified and corrected it, your solution is absolutely correct. – Muthu manimaran Jul 28 '16 at 16:49
• Just curious, because of the hold on it, was this question labeled "homework-and-exercises" from the beginning or did you relabel it in the meantime? – udrv Jul 28 '16 at 17:43

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$\ldots$

• Thank you for your effort. I'm just looking for a general expression, It is ok to express it interms of any function or a polynomial, no need to be a Laguerre polynomial. – Muthu manimaran Jul 27 '16 at 14:56
• Apparently it's called a kummer confluent hypergeometric function, for what it's worth: wolframalpha.com/input/… – snulty Jul 27 '16 at 15:13
• @Muthumanimaran and apparently there is a connection to laguerre polynomials: en.wikipedia.org/wiki/… – snulty Jul 27 '16 at 15:17
• @snulty in your last line, there is - I think - an $| \alpha |^{2m}$ too much – Sanya Jul 28 '16 at 9:04

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?

• I beg you pardon they are not $\lambda$ they are $\alpha$, thanks for pointing me out the mistake. – Muthu manimaran Jul 27 '16 at 13:03