I'm trying to evaluate


Where $\hat{a}$ and $\hat{a}^\dagger$ are ladder operators and $|n\rangle$ the $n$th Fock state.

For this, I separated the problem in three parts: $k<n$, $k=n$ and $k>n$. In this part I'm a bit confused: Is it possible to evaluate such operations? Some tips on how to start would be really great.

  • $\begingroup$ Do you need to know this state for any $n$ and $k$? $\endgroup$
    – Gec
    Commented Sep 17, 2020 at 18:57
  • $\begingroup$ Yes, if it helps, I just need the projection of them on <0I, but I was curious if it's possible to evaluate analytically $\endgroup$ Commented Sep 17, 2020 at 20:59
  • $\begingroup$ I think I can write a simple formula for $\langle 0|(\hat{a} + \hat{a}^\dagger )^k |n\rangle$. But the analytical formula for the general matrix element $\langle m|(\hat{a} + \hat{a}^\dagger )^k |n\rangle$ is too cumbersome. $\endgroup$
    – Gec
    Commented Sep 18, 2020 at 6:18
  • $\begingroup$ If you could share your progress with that or anything I would be very grateful to see. $\endgroup$ Commented Sep 18, 2020 at 15:03

1 Answer 1


Let's use some math tricks to rewrite $(\hat{a}+\hat{a}^\dagger)^k$ in a normally ordered form. First, recall equalities $$ \frac1{2\pi}\int\limits_{-\pi}^\pi e^{i(k'-k)x} dx = \delta_{k\ k'},\quad \exp(A) = \sum_{n=0}^\infty \frac1{n!} A^n. $$ Then we have $$ (\hat{a}+\hat{a}^\dagger)^k = k! \sum_{n=0}^\infty \delta_{k\ n} \frac1{n!} (\hat{a}+\hat{a}^\dagger)^n = \frac{k!}{2\pi}\int\limits_{-\pi}^\pi e^{-ikx} \sum_{n=0}^\infty \frac1{n!} e^{inx} (\hat{a}+\hat{a}^\dagger)^n\ dx = $$ $$ = \frac{k!}{2\pi}\int\limits_{-\pi}^\pi e^{-ikx}\exp\left(e^{ix} (\hat{a}+\hat{a}^\dagger)\right)\ dx. $$ Due to the commutation relation $[\hat{a},\hat{a}^\dagger] = 1$ we have $$ \exp\left(e^{ix} (\hat{a}+\hat{a}^\dagger)\right) = \exp(e^{ix}\hat{a}^\dagger) \exp(e^{ix}\hat{a}) \exp(e^{2ix}/2) $$ Hence we further obtain $$ (\hat{a}+\hat{a}^\dagger)^k = \frac{k!}{2\pi}\int\limits_{-\pi}^\pi e^{-ikx} \sum_{l,l',m=0}^\infty \frac1{l!l'!m!2^m}^\infty (\hat{a}^\dagger)^l (\hat{a})^{l'} e^{i(l+l'+2m)x}\ dx = $$ $$ =\sum_{l,l',m=0}^\infty \delta_{k\ l+l'+2m} \frac{k!}{l!\ l'!\ m!\ 2^m} (\hat{a}^\dagger)^l (\hat{a})^{l'}\quad (*) $$ In the last expression, operators $\hat{a}^\dagger$ and $\hat{a}$ are normally ordered. Now it is easy to find $\langle 0|(\hat{a}+\hat{a}^\dagger)^k|n\rangle$. Due to $$ \langle 0|(\hat{a}^\dagger)^l(\hat{a})^{l'}|n\rangle = \sqrt{n!}\ \delta_{n\ l'} $$ we get from $(*)$: if $m = (k-n)/2$ is non-negative integer, then $$ \langle 0|(\hat{a}+\hat{a}^\dagger)^k|n\rangle = \frac{k!\sqrt{n!}}{n!\ m!\ 2^m}, $$ else $$ \langle 0|(\hat{a}+\hat{a}^\dagger)^k|n\rangle = 0 $$ I think it is possible to use (*) to find any matrix element $\langle n'|(\hat{a}+\hat{a}^\dagger)^k|n\rangle$.

  • 1
    $\begingroup$ It is really beautiful! I never saw this trick using the exponential and helped me a lot. Really thanks! $\endgroup$ Commented Sep 21, 2020 at 2:29

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