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

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.

• I'd guess it will be something similar to this. IIRC, there is also another question here regarding a similar issue, but I cannot find it right now... Commented May 14 at 1:05

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.

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 $$\langle 0|p^{2n}|0\rangle = \left. \left(-\partial_{\xi}\right)^n \int \pi^{-1/2}\exp(-\xi p_0^2) dp_0 \right|_{\xi=1} = \left. \left(-\partial_{\xi}\right)^n \frac{1}{\sqrt{\xi}} \right|_{\xi=1} = \frac{\Gamma(n+1/2)}{\sqrt{\pi}n!} .$$

• Your last line can be simplified: It is proportional to the $𝑛$-th derivative of a Gaussian integral with respect to an artificial parameter (Feynman integration trick)--or more simply the even moments of the Gaussian distribution (up to normalization), and should evaluate to something like a double factorial! I'd suggest to add this, since it really makes the problem "closed". Commented May 14 at 6:17