Representation of the displacement-operator in number basis

Question

According to the original paper of Glauber and Cahill

Ordered Expansions in Boson Amplitude Operators. K. E. Cahill and R. J. Glauber. Phys. Rev. 177 no. 5, 1857-1881 (1969).

the displacement operator's matrix elements with respect to the number-basis read:

$$ ⟨m|D(\alpha)|n⟩=\sqrt{\frac{n!}{m!}}\cdot\alpha^{m-n}\cdot e^{-|\alpha|^2/2}\cdot L_n^{(m-n)}\left(|\alpha|^2\right)\quad\text{when }m\geq n, $$

where $L_n^{(k)}$ are the associated Laguerre polynomials. According to their definition the associated Laguerre polynomials are defined for $k\geq0$, or $k>-1$ if we look at their scipy implementation. However, it seems to me as if we come into troubles if we want to find a matrix representation with respect to the number basis, because then we would need all matrix elements, especially those with $n>m$.

What am I missing? How do the matrix elements for $n>m$ look like?

Thank you in advance for your help!

Answer 1

The displacement operator satisfies the identity $$ \hat{D}^{\dagger}(\alpha) = \hat{D}(-\alpha). $$ Therefore, when $m<n$, \begin{align*} ⟨m|\hat{D}(\alpha)|n⟩ &=\left(⟨n|\hat{D}^{\dagger}(\alpha)|m⟩\right)^* =\left(⟨n|\hat{D}(-\alpha)|m⟩\right)^*\\ &=\sqrt{\frac{m!}{n!}}(-\alpha^*)^{n-m}e^{-\tfrac12|-\alpha|^2}L_m^{(n-m)}(|-\alpha|^2)\\ &=\sqrt{\frac{m!}{n!}}(-\alpha^*)^{n-m}e^{-\tfrac12|\alpha|^2}L_m^{(n-m)}(|\alpha|^2). \end{align*}

Answer 2

Remember the definition $D(\alpha)=\exp(\alpha a^\dagger - \alpha^* a)$. Using BCH we can rewrite this in two ways: $$D(\alpha) = e^{-|\alpha|^2/2}e^{\alpha a^\dagger}e^{-\alpha^* a} = e^{|\alpha|^2/2}e^{-\alpha^* a}e^{\alpha a^\dagger}.\tag1\label1$$ The matrix elements of the displacement operator in the number basis read $$\langle n|D(\alpha)|m\rangle = \frac{1}{\sqrt{n! m!}}\langle a^n D(\alpha) a^{\dagger m}\rangle,$$ where $\langle \cdot\rangle$ denotes the vacuum expectation value. We can go two ways for this calculation, using the two different ways to write $D(\alpha)$ given in \eqref{1}.

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Using $D(\alpha) = e^{-|\alpha|^2/2}e^{\alpha a^\dagger}e^{-\alpha^* a}$

We have $$\langle a^n e^{\alpha a^\dagger}e^{-\alpha^* a} a^{\dagger m}\rangle = \sum_{i=0}^m\frac{(-\alpha^*)^i}{i!}\langle a^n e^{\alpha a^\dagger} a^i a^{\dagger m}\rangle = \sum_{i=0}^m\sum_{j=0}^\infty \frac{(-\alpha^*)^i\alpha^j}{i!j!}\langle a^n a^{\dagger j} a^i a^{\dagger m}\rangle.\label{sum1}\tag2$$ The expectation value $\langle a^n a^{\dagger j} a^i a^{\dagger m}\rangle$ is only nonzero when $n+i=j+m$. This allows to collapse the sum over $j$ replacing $j=n+i-m$, with the additional constraint that we must have $j\ge0$, which thus translates into $i\ge m-n$. But $i$ must also be positive, while $m-n$ might not be. We thus conclude that the appropriate constraint on $i$ is $\max(0,m-n)\le i\le m$. In summary, \eqref{sum1} becomes $$\langle a^n e^{\alpha a^\dagger}e^{-\alpha^* a} a^{\dagger m}\rangle = \sum_{i=\max(0,m-n)}^m \frac{(-\alpha^*)^{i}\alpha^{n+i-m}}{i!(n+i-m)!} \langle a^n a^{\dagger(n+i-m)}a^i a^{\dagger m}\rangle$$

We then see that $$\langle a^n a^{\dagger(n+i-m)}a^i a^{\dagger m}\rangle = \sqrt{m!\frac{m!}{(m-i)!}\frac{((m-i)+(n+i-m))!}{(m-i)!}n!} = \frac{m!n!}{(m-i)!},$$ where we used the identities $$a^{\dagger j}|\ell\rangle = \sqrt{\frac{(\ell+j)!}{\ell!}}|\ell+j\rangle, \qquad a^j |\ell\rangle = \sqrt{\frac{\ell!}{(\ell-j)!}}|\ell-j\rangle \,\,\,\text{(for $\ell\ge j$)}.$$ We conclude that $$\langle a^n e^{\alpha a^\dagger}e^{-\alpha^* a} a^{\dagger m}\rangle = \sum_{i=\max(0,m-n)}^m (-\mu)^i \alpha^{n-m}\binom{m}{i}\binom{n}{m-i}(m-i)!,$$ where $\mu\equiv|\alpha|^2$, and $$\langle n|D(\alpha)|m\rangle = \frac{e^{-\mu/2}}{\sqrt{n!m!}} \sum_{i=\max(0,m-n)}^m (-\mu)^i \alpha^{n-m}\binom{m}{i}\binom{n}{m-i}(m-i)!$$ The summation can actually be simplified using the range $\sum_{i=0}^m$, as the corresponding terms vanish whenever $i< m-n$ anyway, due to the $\binom{n}{m-i}$ factor.