# Integral arising from the proof that the quantum harmonic oscillator satisfies Heisenberg's uncertainty principle

I was trying to verify myself that the stationary states of the quantum harmonic oscillator (QHO) always satisfy Heisenberg's uncertainty relation. In doing so, I came across an integral of the form $$\newcommand{\d}{\mathrm{d}} \mathcal{I} = \int^{+\infty}_{-\infty} \xi^2 e^{-\xi^2} H^2_n\left(\xi\right) \d \xi ,$$ where $$H_n\left(\xi\right)$$ are the physicist's Hermite polynomials, commonly defined by Rodrigues' formula: $$H_n\left(\xi\right) = \left(-1\right)^n e^{\xi^2} \frac{\d^n}{\d \xi^n}e^{-\xi^2} .$$ Although I was able to verify the uncertainty relation without explicitly calculating this integral (I will elaborate on the method later on), I am interested in knowing how to evaluate it.

Details:

The wavefunction of the quantum harmonic oscillator is $$\psi_n = \frac{1}{\sqrt{2^n n!}}\left(\frac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}e^{-\xi^2/2} H_n\left(\xi\right) ,$$ where $$\xi = \sqrt{\frac{m\omega}{\hbar}}x$$ is the dimensionless version of position $$x$$.

I tried to verify Heisenberg's uncertainty relation by brute-forcing my way through the calculations, namely, by calculating $$\left$$, $$\left$$, $$\left$$, and $$\left$$ and plugging them into the LHS of the uncertainty relation: $$\newcommand{\?}[1]{\overset{?}{#1}} \sigma_x\sigma_p = \sqrt{\left - \left^2}\sqrt{\left - \left^2} \?{≥} \frac{\hbar}{2}.$$

Since the integrand is odd in $$\left$$ and $$\left$$, they are both zero.

Now we proceed to calculate $$\left$$ and $$\left$$: \begin{align}\left &= \int^{+\infty}_{-\infty} \psi^*_n\left(x\right) x^2\psi_n\left(x\right) \d x \\ &= \frac{1}{2^n n!\sqrt{\pi}}\frac{\hbar}{m\omega}\int^{+\infty}_{-\infty}\xi^2 e^{-\xi^2} H^2_n\left(\xi\right)\d \xi . \\ \end{align}

\begin{align}\left &= \int^{+\infty}_{-\infty} \psi^*_n\left(x\right) \left(-i\hbar\frac{\d}{\d x}\right)^2\psi_n\left(x\right) \d x \\ &= \frac{\hbar m\omega}{2^n n!\sqrt{\pi}} \int^{+\infty}_{-\infty} H_n\left(\xi\right)e^{-\xi^2}\left[(1-\xi^2)H_n\left(\xi\right)+4nH_{n-1}\left(\xi\right) - 4n(n-1)H_{n-2}\left(\xi\right)\right]\rm d \xi \\ &= \frac{\hbar m\omega}{2^n n!\sqrt{\pi}}\int^{+\infty}_{-\infty}\left(1-\xi^2\right) e^{-\xi^2} H^2_n\left(\xi\right)\d \xi . \\ \end{align}

(The above integral is tedious to simplify, exploiting the orthogonality of the Hermite polynomials is going to save some work)

Using the normalisation integral $$\int^{+\infty}_{-\infty}e^{-\xi^2} H^2_n\left(\xi\right) \d\xi = 2^n n!\sqrt{\pi}$$

to simplify $$\left$$, we can write $$\left$$ and $$\left$$ in terms of $$\,\mathcal{I}$$: $$\left = \frac{\hbar}{m\omega}\left( \frac{\mathcal{I}}{2^n n!\sqrt{\pi}} \right),$$ $$\left = \hbar m\omega\left(1 - \frac{\mathcal{I}}{2^n n!\sqrt{\pi}} \right).$$

Finally, plug them into Heisenberg's relation using $$\sigma_x = \sqrt{\left}$$ and $$\sigma_p = \sqrt{\left}$$, then completing the square: \begin{align} \sigma_x\sigma_p &= \sqrt{\frac{\hbar}{m\omega}\left( \frac{\mathcal{I}}{2^n n!\sqrt{\pi}} \right)}\sqrt{\hbar m\omega\left(1 - \frac{\mathcal{I}}{2^n n!\sqrt{\pi}} \right)} \\ &= \hbar\sqrt{\left(\frac{\mathcal{I}}{2^n n!\sqrt{\pi}}\right)\left(1-\frac{\mathcal{I}}{2^n n!\sqrt{\pi}}\right)} \\ &= \hbar \sqrt{-\left(\frac{\mathcal{I}}{2^n n!\sqrt{\pi}}-\frac{1}{2}\right)^2+\frac{1}{4}} \ge \frac{\hbar}{2}. \end{align}

This proves that the QHO satisfies Heisenberg's uncertainty relation, which was my goal, but I would like to know how to evaluate the integral in question.

• it is a triviality, once corrected: use the recursion relation $H_{n+1}(x)/2 +nH_{n-1}(x) = xH_n(x)$ and orthogonality. May 19, 2023 at 18:00
• From your last chain of equations, it actually follows that $\sigma_x \sigma_p \leq \hbar/2$. Under the square root is subtraction, not addition.
– Gec
May 19, 2023 at 18:39
• Your expression for $\langle p^2 \rangle$ is wrong. May 19, 2023 at 21:39
• The expression for $\langle p^{2} \rangle$ is wrong. It's actually $\langle p^{2} \rangle = \langle \Psi | p^{2} | \Psi\rangle$; this will give you a $\frac{d^{2}}{d x^{2}}$ inside the integral. May 20, 2023 at 5:57
• @ShKol I used $\left(-i\hbar\frac{\rm d}{\rm d x}\right)^2$ to imply applying the momentum operator to $\psi_n$ twice consecutively, getting $\int^{+\infty}_{-\infty} H_n\left(\xi\right)e^{-\xi^2}\left[(1-\xi^2)H_n\left(\xi\right)+4nH_{n-1}\left(\xi\right) - 4n(n-1)H_{n-2}\left(\xi\right)\right]\rm d \xi$ (I added this equation to my question). Is my understanding of $\left<p^2\right>$ problematic? May 20, 2023 at 18:08

You are missing the forest for the trees, and tobogganing into error. Set $$\sqrt{m\omega/\hbar}=1, \implies x=\xi$$.

Recall the basic recursions, $$\bbox[yellow]{ \xi H_n(\xi)= \tfrac{1}{2} H_{n+1}(\xi) +nH_{n-1}(\xi),\\ \partial_\xi H_n= 2n H_{n-1}}~~~,\\ \leadsto \\ \langle x^2\rangle = \frac{1}{2^n n! \sqrt{\pi}}\int\!\!d\xi ~e^{-\xi^2}(\xi H_n)^2\\ = \frac{1}{2^n n! \sqrt{\pi}}\int\!\!d\xi ~e^{-\xi^2}(\tfrac{1}{2} H_{n+1} +nH_{n-1} )^2= n+1/2;\\ \& \\ \langle p^2\rangle = \frac{\hbar^2}{2^n n! \sqrt{\pi}}\int\!\!d\xi ~e^{-\xi^2}(-\xi H_n+\partial_\xi H_n)^2 \\ = \frac{\hbar^2}{2^n n! \sqrt{\pi}}\int\!\!d\xi ~e^{-\xi^2}( -\tfrac{1}{2} H_{n+1} +nH_{n-1})^2= \hbar^2(n+1/2).$$

Consequently, $$\langle x^2\rangle \langle p^2\rangle =\hbar^2 (n+1/2)^2,$$ in comportance with the fine answer by @Gec .

• Why is the integrand of $\left< p^2\right>$ like that? I applied the momentum operator to $\psi_n$ and used the orthogonality condition, but did not get the form you derived. May 24, 2023 at 10:01
• Integrate by parts once to get a perfect square! May 24, 2023 at 10:25
• The expression in the square bracket [ ] in the integrand you provide in your question is spectacularly wrong, as indicated in my early comment... May 24, 2023 at 12:54
• Look at minus the second derivative here! May 24, 2023 at 22:40
• Oh I didn't know these functions had a name and are so well-studied. I'll have a look :) May 25, 2023 at 15:23

Calculating integrals with polynomials is probably not the easiest way to find $$\langle x^2\rangle$$ and $$\langle p^2\rangle$$. In my opinion, it is better to use ladder operators. Or an even simper approach. For the stationary state of an harmonic oscillator, due to the symmetry between kinetic and potential energy terms, we have $$\frac{1}{2m}\langle p^2 \rangle = \frac{m\omega^2}{2} \langle x^2 \rangle = \frac12 E_n = \frac{\hbar\omega}{2}\left(n+\frac12\right).$$ and $$\sqrt{\langle p^2 \rangle}\sqrt{\langle x^2 \rangle} = \hbar\left(n+\frac12\right) \ge \frac{\hbar}2.$$

• The best way, indeed, but the OP wants to evaluate the integral here, not derive the uncertainty. May 20, 2023 at 5:53
• @ShKol Indeed. Well, then these arguments can be used to understand what integrals should be equal to.
– Gec
May 20, 2023 at 6:36