Closed expression for expected values of $\hat{p}\,\,^{2j}$ for the vacuum state

Question

I am wondering if there is a closed expression for the expected value $\left<0\lvert \hat{p}\,\,^{2j}\lvert 0\right>$ with $j\in\mathbb{N}$, where $\left|0\right>$ is the vacuum state of the harmonic oscillator and $\hat{p}=\frac{\hat{a}-\hat{a}\,\,^{\dagger}}{\sqrt{2}i}$, being $\hat{a}\,\,^{\dagger}$ the creation operator. Thanks.

Answer 1

I am wondering if there is a closed expression for the expected value $\left<0\lvert \hat{p}\,\,^{2j}\lvert 0\right>$ with $j\in\mathbb{N}$, where $\left|0\right>$ is the vacuum state of the harmonic oscillator

Yes. Consider the function $$ f(\alpha) = \langle 0|e^{\alpha \hat p}|0\rangle\;.\tag{A} $$

We can use such a function to generate the expectation values of all the powers of $\hat p$, since: $$ \left.\frac{d^n f}{d\alpha^n}\right|_{\alpha=0} = \langle 0|\hat p^n|0\rangle\;.\tag{B} $$

But, also, by using the Baker Campbell Hausdorff formula, we can find a closed form expression for $f(\alpha)$: $$ f(\alpha) = e^{\alpha^2/4}\;.\tag{C} $$

Use Eq. (C) on the LHS of Eq. (B) to find an expression for the expectation value of whatever power of $\hat p$ you like.

Answer 2

The quickest way to compute this overlap is probably by inserting an identity resolved in terms of the $p$-eigenstates: $$ \mathbf{I} = \int |p\rangle\langle p| dp . $$ Then you get $$ \langle 0|p^{2n}|0\rangle = \int \langle 0|p^{2n}|p_0\rangle\langle p_0|0\rangle dp_0 = \int \langle 0|p_0\rangle p_0^{2n}\langle p_0|0\rangle dp_0 = \int \psi_0^2(p_0)p_0^{2n} dp_0 , $$ where $$ \psi_0(p_0)= \langle p_0|0\rangle = \pi^{-1/4}\exp(-p_0^2/2) . $$ So you end up with $$ \langle 0|p^{2n}|0\rangle = \int p_0^{2n} \pi^{-1/2} \exp(-p_0^2) dp_0 . $$ The result represents the even moments of the Gaussian wavefunction of the vacuum state in the $p$-basis. One can compute such moments with the aid of a generating function $$ \langle 0|p^{2n}|0\rangle = \left. \left(-\partial_{\xi}\right)^n \int \pi^{-1/2}\exp(-\xi p_0^2) dp_0 \right|_{\xi=1} = \left. \left(-\partial_{\xi}\right)^n \frac{1}{\sqrt{\xi}} \right|_{\xi=1} = \frac{\Gamma(n+1/2)}{\sqrt{\pi}n!} . $$