Decomposing a prepared quantum harmonic oscillator state

Question

Problem

I'm attempting to decompose a system prepared in state $|\Psi_\mu\rangle$, defined by

$$\Psi_\mu(x) = \left( \frac{m\omega}{\pi \hbar} \right)^{1/4} \exp \left( -\frac{m\omega (x-\mu)^2}{2\hbar} \right)$$

where $H_n$ is the nth Hermite polynomial

$$H_n(x) = (-1)^n e^{x^2} \frac{d^n}{dx^n}e^{-x^2}= \frac{d^n}{dt^n}e^{2xt-t^2}\Big\vert_{t=0} \, ,$$

into eigenstates of the quantum harmonic oscillator Hamiltonian.

My Approach

I know that any arbitrary state, such as $\Psi_\mu(x)$, can be expressed as a superposition of eigenstates of the Hamiltonian $\psi_n(x)$, because they form a complete, orthonormal basis

$$\Psi_\mu(x) = \sum\limits_{n=0}^{\infty}c_n \psi_n(x)$$

with

$$\psi_n(x) = \sqrt{\frac{1}{2^n n!}} \left( \frac{m \omega}{\pi \hbar} \right)^{1/4} H_n \left( \sqrt{\frac{m\omega}{\hbar}} x \right) \exp \left( -\frac{m\omega x^2}{2\hbar} \right) $$

with the weights $c_n$ given by

$$c_n = \int dx \psi_n^*(x) \Psi_\mu(x) \, .$$

So, if I can solve this integral for index $n$, I have decomposed $\Psi_\mu(x)$.

My problem

I'm not certain how to evaluate the integral for $c_n$

\begin{align} c_n =& \sqrt{\frac{1}{2^n n!}} \left( \frac{m \omega}{\pi \hbar} \right)^{1/2}\\ & \int dx H_n \left( \sqrt{\frac{m\omega}{\hbar}}x \right) \exp \left( -\frac{m\omega x^2}{2\hbar} \right) \exp \left( -\frac{m\omega (x-\mu)^2}{2\hbar} \right) \, . \end{align}

Having to integrate around the nth derivative of x in $H_n(x)$ is throwing me off. I feel like I'm either unaware of some existing machinery for integrating Hermite polynomials (although my search has mainly turned up results dealing with orthnormality) or there is another, more elegant approach.

Answer 1

First, set $u=\sqrt{\frac{m\omega}{\hbar}}x$ and $u_{0}=\sqrt{\frac{m\omega}{\hbar}}\mu$ such that

$$c_{n}=\frac{1}{\sqrt{2^{n}n!\pi}}\int{\rm d}uH_{n}(u)e^{-\frac{1}{2}u^{2}}e^{-\frac{1}{2}(u-u_{0})^{2}}$$

Now let's try to use the generating function for the Hermite polynomials

$$e^{2ut-t^{2}}=\sum_{n=0}^{\infty}H_{n}(u)\frac{t^{n}}{n!}$$

Multiply the above by $e^{-\frac{1}{2}u^{2}-\frac{1}{4}u_{0}^{2}}$ and choose $t=\frac{u_{0}}{2}$ to get

$$e^{-\frac{1}{2}(u-u_{0})^{2}}=e^{-\frac{1}{2}u^{2}+uu_{0}-\frac{1}{2}u_{0}^{2}}=e^{-\frac{1}{2}u^{2}-\frac{1}{4}u_{0}^{2}}\sum_{n=0}^{\infty}H_{n}(u)\frac{u_{0}^{n}}{2^{n}n!}$$

so this matches the expression inside the integral. Now we can put it above to get

$$c_{n}=\frac{1}{\sqrt{2^{n}n!\pi}}\int{\rm d}uH_{n}(u)e^{-\frac{1}{2}u^{2}}e^{-\frac{1}{2}u^{2}-\frac{1}{4}u_{0}^{2}}\sum_{k=0}^{\infty}H_{k}(u)\frac{u_{0}^{k}}{2^{k}k!}=$$

$$=\frac{1}{\sqrt{2^{n}n!\pi}}e^{-\frac{1}{4}u_{0}^{2}}\sum_{k=0}^{\infty}\frac{u_{0}^{k}}{2^{k}k!}\int{\rm d}uH_{n}(u)H_{k}(u)e^{-u^{2}}$$

Now use the orthogonality relation $\int{\rm d}uH_{n}(u)H_{k}(u)e^{-u^{2}}=\sqrt{\pi}2^{n}n!\delta_{nk}$ to achieve

$$c_{n}=\frac{1}{\sqrt{2^{n}n!\pi}}e^{-\frac{1}{4}u_{0}^{2}}\sum_{k=0}^{\infty}\frac{u_{0}^{k}}{2^{k}k!}\sqrt{\pi}2^{n}n!\delta_{nk}=\color{blue}{\frac{u_{0}^{n}}{\sqrt{2^{n}n!}}e^{-\frac{1}{4}u_{0}^{2}}}$$

Answer 2

I think that a more simple solution can be obtained by noticing that your state is given by the displacement of the ground state by the factor $\mu$ [1]: $$|\Psi_\mu\rangle=D(\mu)|0\rangle$$ where $D(\mu)$ is the displacement operator. This is justified by the fact that your state is described by the Wigner function of the ground state, displaced on the $x$ axis. Therefore, $|\Psi_\mu\rangle$ is a coherent state described by the Fock expansion: $$|\Psi_\mu\rangle=e^{-\frac{\mu^2}{2}}\sum_n\frac{\mu^n}{\sqrt{n!}}|n\rangle$$ Which gives your coefficient $c_n$

---

[1] Maybe a renormalization of the constant $\mu$ is needed. I'll check this.