4
$\begingroup$

A similar question has been asked here before, but that did not contain the particular solution I am after and is now closed. I was wondering if there is a compact analytical formula for matrix elements of the form $$ \langle m|\hat{x}^k|n \rangle, $$ where $|m \rangle$ and $|n \rangle$ are the standard energy eigenstates of a quantum harmonic oscillator. I am aware that elements for a given particular $k$ can be found by using the expression $\hat{x} = \sqrt{\hbar / (2 m \omega)} (\hat{a} + \hat{a}^{\dagger})$ for the position operator in terms of ladder operators and the action of these ladder operators on the number states. However, is there a general closed-form expression for arbitrary $k$? Since, for example, there is an analytical formula for $\langle m|e^{i \hat{x}}|n \rangle$, I thought there might also be one for the matrix element that I am interested in.

More generally, is there an analytical expression for elements of the form $$ \langle m|(\alpha \hat{a}^{\dagger} + \alpha^* \hat{a})^k|n \rangle, $$ where $\alpha$ is a complex number? I have searched for references in textbooks or journal articles containing such expressions, but could not find any.

$\endgroup$
2
$\begingroup$

Just to make you aware of the "bridge" expressions, the real Hermite functions (not Hermite polynomials on which they are based), $\psi_n(x)=\langle x|n\rangle$. You then have $$\langle m| \hat x ^k|n\rangle= \langle m|\int\!\! dx~ x^k ~|x\rangle \langle x |n\rangle = \int\!\! dx~ x^k ~ \psi_m(x) \psi_n(x) . $$ There is a vast literature on such integrals, possibly on this site or its math sibling but it's not clear to me you prefer integrals to the recursions of @Emilio 's answer.

A further trick you could empty in your quest for the more general expression is to consider the exponential generating function of your expression,

$$ \langle m|e^{t(\alpha \hat{a}^{\dagger} + \alpha^* \hat{a})}|n \rangle, $$ whose k t-derivative at t=0 yields your expression. Now, $$ e^{t(\alpha \hat{a}^{\dagger} + \alpha^* \hat{a})}= e^{t^2 \alpha \alpha^* /2} e^{t\alpha \hat a^\dagger} e^{t\alpha^* \hat a} , $$ or the analogous expression for $\hat x$ and $\hat p$...

$\endgroup$
1
$\begingroup$

This is fully determined by the action of the creation and annihilation operators on the number basis, \begin{align} \hat a|n⟩ & = \sqrt{n}|n-1⟩ \\ \hat a^\dagger|n⟩ & = \sqrt{n+1}|n+1⟩. \end{align} Thus, for elements of linear combinations of the two, you have \begin{align} \langle m|\alpha \hat{a}^{\dagger} + \alpha^* \hat{a}|n \rangle & = \alpha\langle m| \hat{a}^{\dagger} |n \rangle + \alpha^*\langle m| \hat{a}|n \rangle \\ & = \alpha\sqrt{n+1}\delta_{m,n+1} + \alpha^* \sqrt{n}\delta_{m,n-1}. \end{align}

For powers of $\hat x$, simply "rinse and repeat". This will probably get quite tedious if you want to calculate this for a high power of $\hat x$, or if you want a systematic formula that covers all of them, but if that's the case then I would suggest you to consider carefully whether you actually need it.

$\endgroup$
4
  • 1
    $\begingroup$ Thank you, but this is what I meant when I said "I am aware that elements for a given particular k can be found...". What I'm after is exactly what you mention in your last paragraph, a systematic formula for general $k$. Maybe there is no neat compact expression, in which case that would answer my question. But since there is one for $\langle m|e^{i \hat{x}}|n \rangle$, for example, I wondered whether there might also be one for my particular case $\endgroup$
    – Quantum
    Jan 7 at 18:45
  • $\begingroup$ @Quantum if you're not interested in the specific form in the question's last paragraph, I would recommend editing the question. $\endgroup$ Jan 7 at 19:22
  • 1
    $\begingroup$ I realised that the exponent was missing in the last paragraph, have now fixed this $\endgroup$
    – Quantum
    Jan 7 at 23:27
  • $\begingroup$ @Quantum I'll leave this up as it doesn't seem to do active harm. It's hard to prove a negative, but on this field I would be pretty confident that if Cosmas isn't aware of it then it doesn't exist. $\endgroup$ Jan 8 at 10:28
1
$\begingroup$

The answer is basically that there are formulas for these matrix elements, but they are too complicated to be useful. However, the expectation $\langle n|x^{2p}|n\rangle$ has a reasonably simple formula.

Generating Function

First of all we can calculate the exponential generating function referred to in the question and in Cosmas Zachos's answer.

The position operator is $ x = \sqrt{\hbar/2m\omega}(a^\dagger + a) $, so setting $ \lambda = t\sqrt{\hbar/2m\omega} $ gives $e^{tx} = e^{\lambda(a^\dagger + a)}$. We can use the Baker–Campbell–Hausdorff formula to normal order this operator, \begin{align} e^{\lambda a^\dagger}e^{\lambda a} = e^{\lambda(a^\dagger + a) + \frac{1}{2}[\lambda a^\dagger, \lambda a]} &\implies e^{\lambda(a^\dagger + a)} = e^{\frac{\lambda^2}{2}}e^{\lambda a^\dagger}e^{\lambda a}\\ &\implies \langle m|e^{tx}|n\rangle = e^{\frac{\lambda^2}{2}}\color{blue}{\langle m|e^{\lambda a^\dagger}}\color{red}{e^{\lambda a}|n\rangle}. \end{align} The annihilation operators satisfy $$ a^k|n\rangle = \left(\sqrt{n-k+1}\ldots\sqrt{n}\right)\,|n-k\rangle = \left(\frac{n!}{(n-k)!}\right)^{1/2}\,|n-k\rangle, $$ so \begin{align} \color{blue}{\langle m|e^{\lambda a^\dagger}} &\color{blue}{= \sum_{l=0}^m \frac{\lambda^l}{l!}\left(\frac{m!}{(m-l)!}\right)^{1/2}\langle m-l|}, & \color{red}{e^{\lambda a}|n\rangle} &\color{red}{= \sum_{k=0}^n \frac{\lambda^k}{k!} \left(\frac{n!}{(n-k)!} \right)^{1/2} |n-k\rangle}. \end{align}

Then

\begin{align} \langle m|e^{tx}|n\rangle &= e^{\frac{\lambda^2}{2}}\color{blue}{\sum_{l=0}^m \frac{\lambda^l}{l!}\left(\frac{m!}{(m-l)!}\right)^{1/2}}\color{red}{\sum_{k=0}^n \frac{\lambda^k}{k!}\left(\frac{n!}{(n-k)!}\right)^{1/2}}\underbrace{\color{blue}{\langle m-l}|\color{red}{n-k\rangle}}_{=\delta_{l,k+m-n}}\\ &= e^{\frac{\lambda^2}{2}}\sum_{k=0}^n \frac{(m!\,n!)^{1/2}\;\lambda^{2k+m-n}}{(k+m-n)!\,(n-k)!\,k!}\tag{$\dagger$}\label{$\dagger$}\\ &= \left(\frac{n!}{m!}\right)^{1/2}\,\lambda^{m-n}e^{\frac{\lambda^2}{2}}\sum_{k=0}^n {m\choose n-k} \frac{\left(\lambda^2\right)^k}{k!}. \end{align}

Using the definition of a generalized Laguerre polynomial,

$$ L_n^{(\alpha)}(x) = \sum_{k=0}^n (-1)^k { n+\alpha \choose n-k } \frac{x^k}{k!}, $$

we find that the generating function is

$$ \langle m|e^{tx}|n\rangle = \left(\frac{n!}{m!}\right)^{1/2}\,\lambda^{m-n}e^{\frac{\lambda^2}{2}}\,L_n^{(m-n)}\left(-\lambda^2\right).$$

If $n = m$, then this reduces to $ \langle n|e^{tx}|n\rangle = e^{\frac{\lambda^2}{2}}L_n\left(-\lambda^2\right)$.

Matrix Elements

The matrix elements $\langle m|x^p|n\rangle$ can be calculated by expanding $\langle m|e^{tx}|n\rangle$ and equating coefficients of $t^p$. Writing $ [f(t)]_{t^p}$ for the coefficient of $t^p$ in the expansion of $f(t)$, ($\ref{$\dagger$}$) gives

\begin{align} \frac{1}{p!}\langle m|x^p|n\rangle &= \left[\color{orange}{e^{\frac{\lambda^2}{2}}}\sum_{k=0}^n \frac{(m!\,n!)^{1/2}\;\lambda^{2k+m-n}}{(k+m-n)!\,(n-k)!\,k!}\right]_{t^p}\\ &= \left( \frac{\hbar}{2m\omega} \right)^{p/2}\left[\color{orange}{\sum_{l=0}^\infty \frac{1}{l!}\left( \frac{\lambda^2}{2} \right)^l} \sum_{k=0}^n \frac{(m!\,n!)^{1/2}\;\lambda^{2k+m-n}}{(k+m-n)!\,(n-k)!\,k!}\right]_{\lambda^p}\\ &= \left( \frac{\hbar}{2m\omega} \right)^{p/2} \sum_{k=0}^N \frac{(m!\,n!)^{1/2}}{2^{(p+n-m)/2-k}\left(\frac{p+n-m}{2}-k\right)!\,(k+m-n)!\,(n-k)!\,k!}\tag{$\star$}\label{$\star$} \end{align} where $ N = \text{min}(n,\frac{p+n-m}{2})$. Mathematica can evaluate this sum for us to give the following expression involving the hypergeometric function:

$$ \langle m|x^p|n\rangle = \left( \frac{\hbar}{m\omega} \right)^{p/2}\cdot \frac{2^{(m-n)/2}p!}{2^p\left( \frac{p+n-m}{2} \right)!} \cdot {m \choose n} \cdot \left( \frac{n!}{m!} \right)^{1/2}{}_2F_1\left( -n,\tfrac{m-n-p}{2};1+m-n;2 \right), $$

(yuck!) which pretty much rules out the possibility that there is a nice closed-form expression.

For the most interesting case $n=m$, we get a fairly nice result (replacing $p\rightarrow 2p$ because expectations of odd moments of $x$ vanish):

$$ \boxed{\langle n| x^{2p} |n\rangle = \left(\frac{\hbar}{m\omega}\right)^p\cdot\frac{(2p)!}{4^p\,p!}\cdot{}_2F_1(-n,-p;1;2).} $$

The equivalent form

$$ \boxed{\langle n| x^{2p} |n\rangle = \left(\frac{\hbar}{m\omega}\right)^p\cdot\frac{(2p)!}{4^p\,p!}\cdot\sum_{k=0}^\text{min(n,p)} {n \choose k}{p \choose k} 2^k,}$$

which follows from ($\ref{$\star$}$), is probably more useful (e.g. for finding asymptotics). It's also possible to use this form to work out the expectation of $x^{2p}$ in the canonical ensemble, which has a neat answer:

$$ \langle x^{2p} \rangle_\beta = \frac{1}{Z}\sum_{n=0}^\infty \langle n|x^{2p}|n\rangle e^{-\beta E_n} = \left( \frac{\hbar}{m\omega} \right)^p\cdot \frac{(2p)!}{4^p\,p!}\cdot \coth^p \left( \frac{\beta \hbar\omega}{2} \right) $$

where $\beta = 1/(k_BT)$ and $Z = \sum e^{-\beta E_n}$ is the partition function.

$\endgroup$
0
$\begingroup$

No and yes. There is a “shortcut” of sorts using the Bargmann map where $\hat a^\dagger \mapsto \xi$ and $\hat a\mapsto \partial_\xi$, which preserves the commutation relations. (Note in above link $\hat a^\dagger$ is denoted by $\hat a^*$). Here $\xi$ is just a dummy variable (they use $z$ in the linked page) with no physical meaning.

Thus \begin{align} \langle n\vert \mapsto \frac{\partial_\xi^n}{\sqrt{n!}}\, ,\qquad \vert n\rangle \mapsto \frac{\xi^n}{\sqrt{n!}} \end{align} and thus your expression can be evaluated as \begin{align} \langle m\vert (\alpha \hat a^\dagger + \alpha^* \hat a)^k \vert n\rangle \mapsto \frac{\partial^m_\xi}{\sqrt{m!}}\left(\alpha\, \xi +\alpha^* \partial_\xi\right)^k \frac{\xi^n}{\sqrt{n!}}\bigl\vert_{\xi=0}\, . \end{align} You still have to work through $k$ power of $\alpha \xi +\alpha^* \partial_\xi$ but if you have access to a symbolic calculator (like Mathematica) then this is actually quite quick.

Even expanding the power by hand is doable if tedious.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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