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I have to calculate the integral

$$\frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}H_n(x)e^{-x^2+kx}H_l(x)\;\mathrm{d}x$$

where $H_n(x)$ is the $n^{th}$ Hermite polynomial and prove that it equals

$$\sqrt{\frac{m_<!}{m_>!}}\left(\frac{k}{\sqrt{2}}\right)^{|n-l|}L_{m_<}^{|n-l|}\left(-\frac{k^2}{2}\right)\exp\left(\frac{k^2}{4}\right)$$

where $m_<$ and $m_>$ denote the smaller and the larger respectively of the two indices $n$ and $l$ and where $L_n^m$ are the associated Laguerre polynomials.

The last term is $\exp(k^2/4)$, hence I suppose that I begin with

$$\frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}H_n(x)e^{-x^2+kx-\frac{k^2}{4}}e^{\frac{k^2}{4}}H_l(x)\;\mathrm{d}x$$ $$\frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}e^{\frac{k^2}{4}}\int_{-\infty}^{+\infty}H_n(x)e^{-(x-\frac{k}{2})^2}H_l(x)\;\mathrm{d}x$$

but here I'm stuck... Thanks for your help!

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Cross-posted to physics.stackexchange.com/q/54063/2451 –  Qmechanic Feb 15 '13 at 17:57

2 Answers 2

Since the phys.se question was closed, I'm posting my answer here.

One way to do this is by induction, first on $n$ and then on $l$. The base case is easy, since $H_0(x)$ is constant, and the integral is a simple gaussian; the integral for $n=1$ and $l=0$ is also easy. Then fix $l=1$ and assume the formula for arbitrary $n$. Then the formula can be proven for $n+1$ by using the recurrence relation for $H_{n+1}$, $$H_{n+1}(x)=2xH_n(x)-2nH_n(x),$$ changing the $2x$ factor for a derivative with respect to $k$, and applying a recurrence relation for the Laguerre polynomial on the right-hand side. That will prove the general case under $l=1$. Then using a similar induction procedure for $1\leq l\leq n$ will prove the full statement.

I know it's ugly, but it should work.

The other possibility is to do what everyone else does: reduce it to the matrix element $\langle m|\hat{D}(\alpha)|n\rangle$ and then blindly cite* Cahill and Glauber (Ordered expansions in boson amplitude operators. Phys. Rev. 177 no. 5 (1969), pp. 1857-1881, Appendix B.). What they do, if my thesis notes are to be trusted, is compare the matrix element $$\langle m|\hat{D}(\beta)|\alpha\rangle=\langle m|e^{\frac12 (\beta\alpha^\ast-\beta^\ast\alpha)}|\alpha+\beta\rangle=\frac{1}{\sqrt{m!}}(\beta+\alpha)^m e^{-\frac12|\beta|^2-\frac12|\alpha|^2-\beta^\ast\alpha} $$ to the generating function of the Laguerre polynomials, $$ (1+y)^m e^{-xy}=\sum_{n=0}^\infty L_n^{(m-n)}(x) y^n $$ (which is valid for all $y\in\mathbb{C}$; take $y=\beta/\alpha$ up to conjugates) and from there to the original one expanding the coherent state $|\alpha\rangle$ in a number state expansion, comparing coefficients of $\alpha^n$.

(Note also that you will have to do a rotation to complex $k$. This is because your integral is of the form $\langle m|e^{k\hat{x}}|n\rangle$ and for real $k$ the operator $e^{k\hat{x}}$ is not unitary. Doing that also brings your desired result into the much nicer form $L_{m_<}^{|n-l|}(k^2/2)e^{-\frac14 k^2}$, which oscillates for small $k$ and then decays. Changing $k$ for $ik$ is valid because both sides of your target equality are entire functions of $k\in\mathbb{C}$, and proving them equal in one axis is enough by analytic continuation.)

If you ask me, this is just as ugly. But I'd tell you to do both ways since you'll learn a lot from each. If you give up, the magic google keyword is "displaced number states".


*Fun fact: papers that need this matrix element usually also cite Cahill and Glauber's other paper (page 1883, same journal, same volume), which does not relate to it. Beware of citing blindly!

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Thank you so much ! –  mwoua Feb 16 '13 at 14:56
    
Oh ! Don't you have an idea for this ? physics.stackexchange.com/q/54060/20994 –  mwoua Feb 16 '13 at 15:01
up vote 0 down vote accepted

Finally I found how to do it. I post it, if someone is interested.

\begin{align} D_{ln}(\varkappa) &= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}H_n(\tilde{x})e^{-\tilde{x}^2+\varkappa \tilde{x}}H_l(\tilde{x})\;\mathrm{d}\tilde{x} \notag\\ &= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}H_n(\tilde{x})e^{-\tilde{x}^2+\varkappa \tilde{x}-\varkappa^2/4}e^{\varkappa^2/4}H_l(\tilde{x})\;\mathrm{d}\tilde{x} \notag\\ &= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}e^{\varkappa^2/4}\underbrace{\int_{-\infty}^{+\infty}H_n(\tilde{x})e^{-(\tilde{x}-\varkappa/2)^2}H_l(\tilde{x})\;\mathrm{d}\tilde{x}}_I \end{align}

If we pose $x = \tilde{x}-\frac{\varkappa}{2}$ in this expression, the integral $I$ becomes

\begin{equation*} I = \int_{-\infty}^{+\infty}H_n(x+\varkappa/2)e^{-x^2}H_l(x+\varkappa/2)\;\mathrm{d}x \end{equation*}

We know that

\begin{equation*} H_n(x+a) = \sum_{p=0}^n \frac{n!}{(n-p)!p!}(2a)^{n-p}H_p(x) \end{equation*}

Hence, the integral $I$ becomes

\begin{align*} I &= \int_{-\infty}^{+\infty} \sum_{p=0}^n \frac{n!}{(n-p)!p!}\varkappa^{n-p}H_p(x) e^{-x^2} \sum_{q=0}^l \frac{l!}{(l-q)!q!}\varkappa^{l-q}H_q(x)\;\mathrm{d}x \\ &= \sum_{p=0}^n\sum_{q=0}^l \frac{n!}{(n-p)!p!}\varkappa^{n-p}\frac{l!}{(l-q)!q!}\varkappa^{l-q}\int_{-\infty}^{+\infty}H_p(x) e^{-x^2}H_q(x)\;\mathrm{d}x \\ \end{align*}

The Hermite polynomials are orthogonal in the range $(-\infty,\infty)$ with respect to the weighting function $e^{-x^2}$ and satisfy

\begin{alignat*}{2} &&&\int_{-\infty}^{+\infty}H_p(x) e^{-x^2}H_q(x)\;\mathrm{d}x = \sqrt{\pi}2^pp!\;\delta_{pq} \\ &\Rightarrow\quad&& I = \sum_{p=0}^n\sum_{q=0}^l \frac{n!}{(n-p)!p!}\frac{l!}{(l-q)!q!}\varkappa^{n+l-p-q}\cdot \sqrt{\pi}2^pp!\;\delta_{pq} \end{alignat*}

As this integral is nil if we have not $p=q$, we can replace the two sums by only one that goes from 0 to $\min(n,l)$. Let us say that $n<l$. Hence, the full expression for the $D$-matrix is

\begin{align} D_{ln}(\varkappa) &= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}e^{\varkappa^2/4} \sum_{p=0}^n \frac{n!}{(n-p)!p!}\frac{l!}{(l-p)!p!}2^pp!\sqrt{\pi}\;\varkappa^{n+l-2p} \notag\\ &= \frac{\varkappa^{n+l}}{\sqrt{2^nn!}\sqrt{2^ll!}}e^{\varkappa^2/4} \sum_{p=0}^n \frac{n!}{(n-p)!p!}\frac{l!}{(l-p)!}2^p\;\varkappa^{-2p} \notag\\ &= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{n+l}e^{\varkappa^2/4} \sum_{p=0}^n \frac{l!}{(n-p)!(l-p)!p!}\left(\frac{\varkappa^2}{2}\right)^{-p} \notag\\ &= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{p=0}^n \frac{l!}{(n-p)!(l-p)!p!}\left(\frac{\varkappa^2}{2}\right)^{n-p} \end{align}

Associated Laguerre polynomials $L_a^b(x)$ are given by

\begin{equation*} L_a^b(x) = \sum_{k=0}^{a}(-1)^k \frac{(a+b)!}{(a-k)!(b+k)!k!}x^k \end{equation*}

It suggests us to transform the expression of the $D$-matrix by posing $k=n-p$. Hence, we have

\begin{align} D_{ln}(\varkappa) &= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{k=n}^0 \frac{(l)!}{(n-(n-k))!(l-(n-k))!(n-k)!}\left(\frac{\varkappa^2}{2}\right)^{n-(n-k)} \notag\\ &= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{k=0}^n \frac{l!}{k!(l-n+k)!(n-k)!}\left(\frac{\varkappa^2}{2}\right)^{k} \notag\\ &= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{k=0}^n (-1)^k\frac{([l-n]+n)!}{(n-k)!([l-n]+k)!k!}\left(-\frac{\varkappa^2}{2}\right)^{k} \notag\\ &= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4}L_n^{l-n}\left(-\frac{\varkappa^2}{2}\right) \end{align}

It should be remembered that we had supposed that $n<l$. But that could be otherwise. In order to be general, $n_<$ and $n_>$ will be defined as $n_<=\min{(n,l)}$ and $n_>=\max{(n,l)}$ and $l-n=|l-n|$. We then have

\begin{equation} D_{ln}(\varkappa) = \sqrt{\frac{n_<!}{n_>!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{|l-n|}L_{n_<}^{|l-n|}\left(-\frac{\varkappa^2}{2}\right)e^{\varkappa^2/4} \end{equation}

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