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I would like to know if there is a convenient identity (and what it is) for

$$\langle n | (a+a^\dagger)^k | m \rangle$$

where $| n \rangle, \, | m \rangle$ are energy eigenstates of a simple harmonic oscillator hamiltonian and $a, \, a^\dagger$ are annihilation and creation operators respectively. $k$ is a natural number. I've done problems for $k = 1, 2, 3$ but it's not clear to me how to generalize.

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    $\begingroup$ It might help if you look at it graphically where you can take $k$ steps up or down and you have to start at $m$ and end at $n$, then just count the number of paths. It's related to Pascal's triangle. $\endgroup$ Commented Feb 13, 2018 at 23:13
  • $\begingroup$ @SeanE.Lake the complication I'm having trouble with is assigning the appropriate products of square roots of eigenvalues $\endgroup$
    – Diffycue
    Commented Feb 13, 2018 at 23:17
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    $\begingroup$ Suggestion 2: try to see if you can figure out what the exapnsion of $(a+a^\dagger)^k$ is in normal order (all $a^\dagger$ to the right of $a$). $\endgroup$ Commented Feb 13, 2018 at 23:28
  • $\begingroup$ @Diffycue my original answer was wrong -- apologies. I updated the derivation to fix it, the result is a bit more complicated. I haven't double checked it but I will when I get a chance. $\endgroup$ Commented Sep 13, 2021 at 12:04

4 Answers 4

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Yes, provided you were willing to perform integrals involving Hermite polynomials in coordinate eigenstates connected to your Fock space.

First, recall $$ a+a^\dagger= \sqrt{2}~ \hat{x}. $$

Then, $$ \psi_m(x)\equiv \left\langle x \mid m \right\rangle = {1 \over \sqrt{2^m m!}}~ \pi^{-1/4} \exp(-x^2 / 2) H_m(x), $$ so inserting a complete set of coordinate eigenstates, $$ \langle n | (a+a^\dagger)^k | m \rangle= 2^{k/2}\langle n | \hat{x}^k | m \rangle = 2^{k/2}\int dx ~\langle n | x\rangle x^k \langle x| m\rangle \\ = 2^{k/2} \frac{1}{\sqrt \pi}\frac{1}{\sqrt{2^{m+n} n! m!}}\int dx ~e^{-x^2} x^k H_n(x) H_m (x). $$ You now have to use the correspondingly messy moment identities for Hermite polynomials, but you may check your low-index specific results for the first few ones, this way.

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Objective:

The objective is to compute the quantity: $$ \langle n|(a+a^{\dagger})^k|m\rangle $$ the Fock states, $|m\rangle$, normalised such that $\langle n|m\rangle=\delta_{n,m}$, and $[a,a^\dagger]=1$.


Preliminary Result:

It is convenient to use coherent state techniques, so in particular we primarily need the relation: \begin{equation} \boxed{ \begin{aligned} I(z,w)&\equiv\langle \bar{z} | (a+a^\dagger)^k |w\rangle\\ & = B_k(w+z,1,0,\dots,0) \,e^{wz} \end{aligned} }\qquad (\star) \end{equation} where $\langle \bar{z}|=\langle 0|e^{za}$ and $|w\rangle=e^{wa^\dagger}|0\rangle$ are coherent states (with $a|w\rangle = w|w\rangle$ and $\langle\bar{z}|a^\dagger=z\langle\bar{z}|$, and $\langle \bar{z}|w\rangle = e^{wz}$), whereas $B_k(w+z,1,0,\dots,0)$ is a complete Bell polynomial. As usual, we normalise $[a,a^\dagger]=1$.

Derivation of ($\star$): Since I received some questions about how to derive $(\star)$ let me point out that I made use of the complete Bell polynomial identity, $B_k(a_1,\dots,a_n) = \partial_t^k \exp(\sum_{s=1}^\infty \frac{1}{s!}a_st^s)|_{t=0}$, and the Baker-Campbell-Hausdorff formula, which reduces to $e^{X+Y}=e^X e^Y e^{-\frac{1}{2}[X,Y]}$, when $[X,Y]$ commutes with $X$ and $Y$; from the latter (since the right-hand side must be symmetric in $X,Y$ as is the left-hand side) it also follows that $e^X e^Y = e^Ye^Xe^{[X,Y]}$. In further detail, making use of these results, \begin{equation} \begin{aligned} I(z,w) &\equiv\langle \bar{z} | (a+a^\dagger)^k |w\rangle\\ &=\partial_t^k\langle \bar{z} | e^{(a+a^\dagger)t} |w\rangle\big|_{t=0}\\ &=\partial_t^k\langle \bar{z} | e^{a^\dagger t} e^{at}e^{\frac{1}{2}[a,a^\dagger]t^2} |w\rangle\big|_{t=0}\\ &=\partial_t^ke^{(z+w)t+\frac{1}{2}t^2}\big|_{t=0} \langle \bar{z} |w\rangle\\ &=B_k(w+z,1,0,\dots,0) \,e^{wz} \end{aligned} \end{equation}


Main Calculation:

Returning to ($\star$), we can extract the quantity of interest from it by differentiating it wrt $z$ and $w$ ($n$ and $m$ times respectively) and then setting $z=w=0$; after including relevant normalisations (assuming $\langle n|m\rangle=\delta_{n,m}$): $$ \boxed{ \begin{aligned} \langle n|(a+a^{\dagger})^k|m\rangle &=\frac{1}{\sqrt{n!m!}}\partial_z^n\partial_w^mI(z,w)\big|_{z,w=0}\\ &=\frac{1}{\sqrt{n!m!}}\partial_z^n\partial_w^m\,B_k(w+z,1,0,\dots,0) \,e^{wz}\big|_{z=w=0}\\ \end{aligned} }\qquad (\star\star) $$

To evaluate the derivatives, I find it convenient to work in terms of cycle index polynomials, $Z_k(a_s)\equiv Z_k(a_1,a_2,\dots,a_k)$, defined by: $$ Z_k(a_1,a_2,\dots,a_k) = \frac{1}{k!}B_k(0!a_1,1!a_2,\dots,(k-1)!a_k). $$ All the properties of cycle index polynomials we need are given in Appendix B of the long paper. In particular, we need the following result: \begin{equation} \begin{aligned} \partial_z^aZ_k(z+w,1,0,\dots,0)&=Z_{k-a}(z+w,1,0,\dots,0), \end{aligned} \end{equation} which follows immediately from the derivative relation in the long paper, and also: $$ Z_{2p}(0,1,0,\dots,0)=\frac{2^{-p}}{p!},\quad Z_{2p+1}(0,1,0,\dots,0) = 0,\qquad(\star\star\star) $$ for some $p=0,1,2,\dots$, which follow immediately, e.g., from the contour integral representation in equation (B.677a) in the long paper.

Plugging these into the above, some elementary manipulations then lead to: \begin{equation} \begin{aligned} \langle n|(a+a^{\dagger})^k|m\rangle &=\frac{k!}{\sqrt{n!m!}}\partial_z^n\partial_w^m\,Z_k(w+z,1,0,\dots,0) \,e^{wz}\big|_{z=w=0}\\ &=\frac{k!}{\sqrt{n!m!}}\partial_z^n\sum_{b=0}^m\binom{m}{b}\big(\partial_w^b Z_k(w+z,1,0,\dots,0)\big)\big(\partial_w^{m-b}e^{wz}\big)\big|_{z=w=0}\\ &=\frac{k!}{\sqrt{n!m!}}\partial_z^n\sum_{b=0}^m\binom{m}{b} Z_{k-b}(z,1,0,\dots,0)z^{m-b}\big|_{z=0}\\ \end{aligned} \end{equation} where we just used the usual Leibniz rule for differentiating a product $m$ times. Similarly, carrying out the remaining derivatives, \begin{equation} \begin{aligned} \langle n|(a+a^{\dagger})^k|m\rangle &=\frac{k!}{\sqrt{n!m!}}\sum_{a=0}^n\binom{n}{a}\sum_{b=0}^m\binom{m}{b} \big(\partial_z^aZ_{k-b}(z,1,0,\dots,0)\big)\big(\partial_z^{n-a}z^{m-b}\big)\big|_{z=0}\\ &=\frac{k!}{\sqrt{n!m!}}\sum_{a=0}^n\binom{n}{a}\sum_{b=0}^m\binom{m}{b} Z_{k-a-b}(0,1,0,\dots,0)(m-b)!\delta_{n-a,m-b}\\ &=\sum_{a=0}^n\frac{k!\sqrt{n!m!}}{(n-a)!a!(m-n+a)!} Z_{k-m+n-2a}(0,1,0,\dots,0)\\ \end{aligned} \end{equation} where in the last line we used the the Kronecker delta to carry out the sum over $b$. From $(\star\star\star)$ and the above relation, it is clear that unless a positive integer, $r$, can be found such that, $$ k=2r+m-n, $$ the quantity $\langle n|(a+a^\dagger)^k|m\rangle$ will vanish. We can then carry out the sum over $a$, taking into account $(\star\star\star)$, to arrive at: $$ \begin{aligned} \langle n|(a+a^{\dagger})^k|m\rangle &=\sum_{a=0}^n\frac{(2r+m-n)!\sqrt{n!m!}}{(n-a)!a!(m-n+a)!} \frac{2^{-r+a}}{(r-a)!}\\ \end{aligned} $$ We can now sum over $a$ (I used Mathematica for this sum), we arrive at the final answer: $$ \boxed{ \langle n|(a+a^{\dagger})^{Q+m-n}|m\rangle = \left\{ \begin{aligned} &\frac{\sqrt{m!n!}}{2^{Q/2}(m-n)!n!(Q/2)!}\,\,\,\,F_{\!\!\!\!\!\!2\,\,\,\,\,1}\big(-n,-\tfrac{Q}{2};1+m-n;2\big),&\textrm{if}\,\, Q\in 2\mathbf{Z}^+\\ &0&\textrm{if}\,\, Q\in2\mathbf{Z}^++1\\ \end{aligned} \right. } $$ where $m\geq n$.


Consistency Check:

As a consistency check, notice that if $m=n+1$ and $Q=0$ this reduces to the expected answer: $$ \langle n|(a+a^{\dagger})|n+1\rangle = \sqrt{n+1}, $$ which can alternatively be computed using the commutator, $[a,a^\dagger]=1$, and the Fock representation of the state, $|n\rangle = \frac{1}{\sqrt{n!}}(a^\dagger)^n|0\rangle$.


References:

D. Skliros & D. Luest, Handle Operators in String Theory, Appendix B

Acknowledgement:

Thanks to @Codename47 for pointing out that there was an error in a previous result.

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    $\begingroup$ Does it matter that you're not using the normalized coherent states? $\endgroup$ Commented Sep 7, 2021 at 15:35
  • $\begingroup$ @ZeroTheHero no it doesn’t matter, because the coherent states are only used as a trick to extract the energy eigenstate result ; the final result is “correctly” normalised as you can check $\endgroup$ Commented Sep 7, 2021 at 15:40
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    $\begingroup$ As it stands, this result appears to be incorrect. In particular, checking for $n=9, m=10, k=1$, it yields $\simeq -2.46$ in implementations in Python and Mathematica while I would expect $\sqrt{5}\simeq 2.24$. It also appears that in the third to last equation, the last denominator should be $(r-a)!$. $\endgroup$ Commented May 30, 2023 at 13:01
  • $\begingroup$ @Codename47 Thanks for pointing this out, I will correct it. By the way, didn't you mean to write $\sqrt{10}$? $\endgroup$ Commented May 31, 2023 at 14:03
  • $\begingroup$ just for the record, as far as I know, what is currently written in my answer is correct $\endgroup$ Commented Jun 3, 2023 at 16:21
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Consider a coherent state $\exp(it(a^{\dagger}+a))| 0 \rangle$, where one can show that $$a\exp(it(a^{\dagger}+a))| 0 \rangle=it\exp(it(a^{\dagger}+a))| 0 \rangle. $$Then it's easy to work out a partial answer $$\langle n|\exp(it(a^{\dagger}+a)|0\rangle=(it)^ne^{-\frac{1}{2}t^2},$$which leads to $$\langle n|(a^{\dagger}+a)^k|0\rangle=(-i)^k\partial_t^k((it)^ne^{-\frac{1}{2}t^2})_{t=0}. $$

Presumably a similar method can yield for the general case but it is certainly more tedious.

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Here's a solution inspired by @Wakabaloola. It doesn't use Bell numbers and is possibly a little more intuitive but I was unable to obtain a sensible expression for the final sum.

We are looking for an elegant computation of $\langle m\vert \hat x^k \vert n\rangle$, where $\vert n\rangle $ is a harmonic oscillator ket.

First: \begin{align} \vert s\rangle&=e^{s\hat a^\dagger}\vert 0\rangle\, ,\\ \hat a\vert s\rangle &=s\vert s\rangle \end{align}

Next, we start with \begin{align} I(z,w)= \langle 0\vert e^{w\hat a}(\hat a+\hat a^\dagger)^k e^{z\hat a^\dagger}\vert 0\rangle \end{align} and compute \begin{align} e^{w\hat a}(\hat a+\hat a^\dagger)e^{-w\hat a}= \hat a+\hat a^\dagger + w\mathbb{I} \end{align} using the usual BCH formula. Moreover: \begin{align} e^{w\hat a}(\hat a+\hat a^\dagger)^2e^{-w\hat a}&= e^{w\hat a}(\hat a+\hat a^\dagger)e^{-w\hat a}e^{w\hat a}(\hat a+\hat a^\dagger)e^{-w\hat a}\, ,\\ &=\left(\hat a+\hat a^\dagger + w\mathbb{I}\right)^2 \end{align} and so by induction \begin{align} e^{w\hat a}(\hat a+\hat a^\dagger)^k&= \left(\hat a+\hat a^\dagger + w\mathbb{I}\right)^k e^{w\hat a}\, . \end{align} Continuing: \begin{align} I(z,w) &= \langle 0\vert\left(\hat a+\hat a^\dagger + w\mathbb{I}\right)^k e^{w\hat a}\vert z\rangle\, ,\\ &=\langle 0\vert\left(\hat a+\hat a^\dagger + w\mathbb{I}\right)^k \vert z\rangle e^{wz}\, ,\\ &=\langle 0\vert\left(\hat a+\hat a^\dagger + w\mathbb{I}\right)^k e^{z\hat a^\dagger}\vert 0\rangle \end{align} since $\hat a\vert z\rangle=z\vert z\rangle.$

Next, we pass $e^{z\hat a^\dagger}$ to the left of $\left(\hat a+\hat a^\dagger + w\mathbb{I}\right)^k$ using the same trick as before: \begin{align} \left(\hat a+\hat a^\dagger + w\mathbb{I}\right)e^{z\hat a^\dagger} &= e^{z\hat a^\dagger} e^{-z\hat a^\dagger}\left(\hat a+\hat a^\dagger + w\mathbb{I}\right)e^{z\hat a^\dagger}\, ,\\ &= e^{z\hat a^\dagger}\left(\hat a+\hat a^\dagger + (z+w)\mathbb{I}\right)\, ,\\ \left(\hat a+\hat a^\dagger + w\mathbb{I}\right)^ke^{z\hat a^\dagger} &= e^{z\hat a^\dagger}\left(\hat a+\hat a^\dagger + (z+w)\mathbb{I}\right)^k\, ,\\ \langle 0\vert\left(\hat a+\hat a^\dagger + w\mathbb{I}\right)^k e^{z\hat a^\dagger}\vert 0\rangle&= \langle 0\vert e^{z\hat a^\dagger}\left(\hat a+\hat a^\dagger + (z+w)\mathbb{I}\right)^k\vert 0\rangle\, ,\\ &=\langle 0\vert\left(\hat a+\hat a^\dagger + (z+w)\mathbb{I}\right)^k\vert 0\rangle \end{align} since $\langle 0\vert\hat a^\dagger=0$. Thus, we now have \begin{align} I(z,w) &= e^{wz} \langle 0\vert(z+\left(\frac{2m\omega}{\hbar}\right)^{1/2} \hat x + w)^k \vert 0\rangle\, ,\\ &= e^{wz} \sum_{p=0}^k \left(\frac{2m\omega}{\hbar}\right)^{p/2} \langle 0\vert \hat x^p \vert 0\rangle (z+w)^{k-p}{k \choose p}\, , \\ &= e^{wz}\sum_{p=0}^k \left(\frac{2m\omega}{\hbar}\right)^{p/2} \langle 0\vert \hat x^p \vert 0\rangle {k \choose p} \sum_q {{k-p}\choose{q}} z^{q}w^{k-p-q} \label{eq:laststep} \end{align} where the binomial expansion is justified since $\hat x$ commutes with $(z+w)\mathbb{I}$. We're getting close.

We want something proportional to $\langle n\vert (\hat a+\hat a^\dagger)^k\vert m\rangle$. We can produce $\langle {n}\vert $ and $\vert {m}\rangle $ by taking $n$ partial derivatives of $I(z,w)$ w/r to $w$ and $m$ partial derivatives of $I(z,w)$ w/r to $z$, respectively: \begin{align} \frac{\partial^{n+m}}{\partial w^n\partial z^m}I(z,w) &=\langle 0\vert\hat a^n e^{w\hat a }(\hat a+\hat a^\dagger)^k (\hat a^\dagger)^me^{z\hat a^\dagger}\vert 0\rangle\, ,\\ &= \sqrt{n!m!} \langle {n}\vert e^{w\hat a }(\hat a+\hat a^\dagger)^k e^{z\hat a^\dagger}\vert {m}\rangle \, ,\\ \sqrt{n!m!} \langle {n}\vert e^{w\hat a }(\hat a+\hat a^\dagger)^k e^{z\hat a^\dagger}\vert {m}\rangle \Bigl\vert_{w=z=0}&= \sqrt{n!m!} \langle {n}\vert (\hat a+\hat a^\dagger)^k \vert{m}\rangle \, . \end{align} Expanding $e^{wz}$, we can now write
\begin{align} \langle {n}\vert (\hat a+\hat a^\dagger)^k \vert {m}\rangle &=\frac{1}{\sqrt{n!m!}} \sum_{p=0}^k \left(\frac{2m\omega}{\hbar}\right)^{p/2} \langle 0\vert x^p\vert 0\rangle\, \, \nonumber \\ &\qquad\qquad\times \partial_z^n\partial_w^m \sum_{q=0}^{k-p} \sum_r \frac{1}{r!} z^{q+r} w^{k-p-q+r}{k-p \choose q}{k \choose p}\big|_{s=w=0} \end{align}

The derivatives are $0$ unless $q+r=n$ and $k-p-q+r=m$, or \begin{align} q=\frac{k-m+n-p}{2}\, ,\qquad r=\frac{m+n+p-k}{2}\, . \end{align} When this is the case we have \begin{align} &\partial_z^n\partial_w^m\, \sum_{q=0}^{k-p} \sum_r \frac{1}{r!} z^{q+r} w^{k-p-q+r}{k-p \choose q}{k \choose p}\big|_{s=w=0}\, ,\\ %%%%%%% &\qquad \qquad =\partial_z^n\partial_w^m\, \frac{z^{n}}{(\frac{m+n+p-k}{2})!} w^{m}{k-p \choose \frac{k-m+n-p}{2} }{k \choose p}\big|_{s=w=0}\, ,\\ &\qquad \qquad = \frac{n!m!}{(\frac{m+n+p-k}{2})!} {k-p \choose \frac{k-m+n-p}{2} }{k \choose p} \end{align} so that \begin{align} \langle n|(\hat a+\hat a^{\dagger})^k|m\rangle& = \frac{n!m!}{\sqrt{n!m!}} \sum_{p=0}^k \left(\frac{2m\omega}{\hbar}\right)^{p/2} \langle 0 \vert x^p \vert 0\rangle \frac{1}{(\frac{m+n+p-k}{2})!} {k-p \choose \frac{k-m+n-p}{2} }{k \choose p} \end{align}

There remains to evaluate $\langle 0\vert x^p \vert 0\rangle$. First note that $\langle 0\vert\hat x^p\vert 0\rangle\ne 0$ only when $p$ is even by parity. Next \begin{align} \langle 0\vert\hat x^p\vert 0\rangle=\sqrt{\frac{m\omega}{\hbar\pi}}\int dx e^{-m\omega x^2/\hbar} x^p\, , \end{align} so let \begin{align} \xi=\sqrt{\frac{m\omega}{\hbar}}x\, ,\qquad dx= \sqrt{\frac{\hbar}{m\omega}} d\xi \end{align} and then \begin{align} \left(\frac{2m\omega}{\hbar}\right)^{p/2} \langle 0\vert \hat x^p\vert 0\rangle&= \frac{1}{\sqrt{\pi}} \left(\frac{2m\omega}{\hbar}\right)^{p/2} \left(\frac{\hbar}{m\omega}\right)^{p/2} \int d\xi e^{-\xi^2} \xi^p = \frac{2^{p/2}}{\sqrt{\pi}} \int d\xi e^{-\xi^2} \xi^p \, ,\\ &= 2^{p/2} \frac{(p-1)!!}{2^{p/2}} = (p-1)!! \end{align}

Putting all this together: \begin{align} \langle {n}\vert\hat x^k\vert{m}\rangle &=\left(\frac{\hbar}{2m\omega}\right)^{k/2} \langle {n}\vert (\hat a+\hat a^\dagger)^k\vert{m}\rangle \, ,\\ &=\left(\frac{\hbar}{m\omega}\right)^{k/2}\frac{\sqrt{n!m!}}{2^{k/2}} \sum_{p=0,2,4\ldots}^k \frac{(p-1)!!}{(\frac{m+n+p-k}{2})!} {k-p \choose \frac{k-m+n-p}{2} }{k \choose p} \end{align}

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  • $\begingroup$ In your final expression the number $k-m+n-p$ in the first binomial coefficient can be negative. How does your result handle this case? Furthermore, your result does not seem to account for the parity like Wakabaloola does, is that by design? $\endgroup$ Commented May 30, 2023 at 10:23
  • $\begingroup$ @Codename47 Thanks for the comment. Not sure I follow about parity: clearly if k+m-n is not even then the matrix element is 0. I also thought I had checked this but so something’s not quite right so give me a bit of time and I will double check again. $\endgroup$ Commented May 30, 2023 at 11:50
  • $\begingroup$ Another small comment is that you evaluate your partial derivatives at the point $|_{s=w=0}$, but I suspect you mean $|_{z=w=0}$? $\endgroup$ Commented May 30, 2023 at 13:07
  • $\begingroup$ if this is urgent I can get to this today else it will be later in the week. $\endgroup$ Commented May 30, 2023 at 13:27
  • $\begingroup$ No urgency, from a few numerical experiments it seems you have the right result if terms where $k-m+n-p<0$ are set to zero, so it is probably just a tweaking of some sum limit. Thank you for your very useful result! $\endgroup$ Commented May 30, 2023 at 13:30

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