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, Gert Jul 28 '16 at 14:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Community, CuriousOne, honeste_vivere, Gert
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ if the states are properly normalised then shouldn't $\langle n-l\big| m-k\rangle=\delta_{n-l,m-k}$? $\endgroup$ – snulty Jul 27 '16 at 10:56
  • $\begingroup$ 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 $n\neq m$ $\endgroup$ – Sanya Jul 27 '16 at 13:19
  • 1
    $\begingroup$ Is $\langle n\big|n\rangle=1 $ or $\sim=n!$? In any case you also need to replace $l=n-m+k$. $\endgroup$ – 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 $n$, $m$.

  • $\begingroup$ 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. $\endgroup$ – udrv Jul 28 '16 at 15:01
  • $\begingroup$ That is $$ \langle n| e^{\alpha a^\dagger} e^{-\alpha^* a} |n \rangle = (n!)^{-1} \langle 0 | a^n e^{\alpha a^\dagger} e^{-\alpha^* a} (a^\dagger)^n |0 \rangle = \\ \langle 0 | e^{\alpha a^\dagger} \left( e^{-\alpha a^\dagger} a e^{\alpha a^\dagger}\right)^n \left( e^{-\alpha^* a} (a^\dagger)e^{\alpha^* a}\right)^n e^{-\alpha^* a} |0 \rangle =\\ \langle 0 | \left( a - \alpha[a^\dagger, a] \right)^n \left( a^\dagger -\alpha^* [a, a^\dagger] \right)^n |0 \rangle = \langle 0 | \left( a + \alpha \right)^n \left( a^\dagger -\alpha^* \right)^n |0 \rangle $$ $\endgroup$ – udrv Jul 28 '16 at 15:02
  • $\begingroup$ 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 $(1/2) \left( 2 - |\alpha|^2 + |\alpha|^4 \right) = L_2(|\alpha|^2)$. $\endgroup$ – udrv Jul 28 '16 at 15:02
  • $\begingroup$ sorry, I am also not getting $(-1)^n$, I made a mistake, now I verified and corrected it, your solution is absolutely correct. $\endgroup$ – Muthu manimaran Jul 28 '16 at 16:49
  • $\begingroup$ 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? $\endgroup$ – 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:


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 $l=n-m+k$


$$=\sqrt{n!}\sqrt{m!}(\alpha)^{n-m}\sum_{k=0}^{m}\frac{(-1)^k\left|\alpha\right|^{2k}}{(n-m+k)!k! (m-k)!}$$

$$=\frac{\sqrt{n!}}{\sqrt{m!}}(\alpha)^{n-m}\sum_{k=0}^{m}{m \choose k}\frac{(-1)^k\left|\alpha\right|^{2k}}{(n-m+k)!}$$

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$\ldots$

  • $\begingroup$ 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. $\endgroup$ – Muthu manimaran Jul 27 '16 at 14:56
  • $\begingroup$ Apparently it's called a kummer confluent hypergeometric function, for what it's worth: wolframalpha.com/input/… $\endgroup$ – snulty Jul 27 '16 at 15:13
  • $\begingroup$ @Muthumanimaran and apparently there is a connection to laguerre polynomials: en.wikipedia.org/wiki/… $\endgroup$ – snulty Jul 27 '16 at 15:17
  • $\begingroup$ @snulty in your last line, there is - I think - an $| \alpha |^{2m}$ too much $\endgroup$ – 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?

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.