# Decomposing a prepared quantum harmonic oscillator state

## 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.

• Probably one of the most complicated questions on SO – Curious Fish Oct 25 '18 at 19:15
• Does $\Psi_\mu (x)= \exp(-\mu \partial_x) ~~\Psi_0(x)$ help you? Did you try simple cases? – Cosmas Zachos Oct 25 '18 at 21:04
• I’ll take a look at $\Psi_\mu(x) = \exp(-\mu \partial_x) \Psi_0(x)$. Could you provide some insight as to where this came from? Was it a relation you recognized? In terms of the simple cases (I’m assuming you mean n=0,1,2,...) I planned on working those out today to see if some recursion relation becomes obvious. – Grant Cates Oct 26 '18 at 5:53
• Taylor expansion of $\Psi_0(x-\mu)$ around x, so in powers of μ. Translation operator. – Cosmas Zachos Oct 26 '18 at 13:02

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 $$\mu$$ : $$|\Psi_\mu\rangle=D(\mu)|0\rangle$$ 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 $$c_n$$

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

• Note that you should use in fact $D\left(\sqrt{\frac{m\omega}{2\hbar}}\mu\right)$. – eranreches Oct 26 '18 at 12:49
• This is exactly what I was looking for! There are a lot of holes in my "foundational" knowledge, because much of what I've learned up to know has been ad-hoc. So, I had to look up a bit on coherent states being displaced vacuum states, but I got there. Thank you for pointing me in the right direction! – Grant Cates Oct 26 '18 at 13:36

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

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