# Computing the standard deviation of the momentum operator in quantum mechanics

I have a wave function as follows:

$$\psi = A_o \mathrm e^{-\alpha x^2}$$

And I want to compute the standard deviation of the momentum operator:

$$\Delta p = \sqrt{\langle p^2\rangle-\langle p\rangle^2}$$

I think I might be getting something wrong because when I compute the integrals I get: $$\langle p\rangle=0$$ and $$\langle p^2\rangle$$ negative, which means that $$\Delta p \not \in \mathbb{R}$$, does that make any sense? or the standard deviation of the momentum has to be real?

Note:

$$\int_{0}^{\infty} x^2\mathrm e^{-\alpha x²} dx = \frac{\sqrt{\pi}}{4a^{3/2}}$$

• How are you getting $\langle p^2 \rangle$ negative? The usual intuition about squares being positive is still true in this setting. – jacob1729 Nov 28 '19 at 13:57

How are you calculating the expectation value of $$p^2$$? Remember $$p = -i \hbar \partial_x$$, so $$p^2 = - \hbar^2 \partial^2$$; that minus sign is vital! $$\langle p^2 \rangle$$ is then equal to the integral $$\int_{-\infty}^{\infty} \psi^{\ast}(x) \left(- \partial_x^2 \psi(x) \right) dx$$ which should give a positive result.
Since $$x e^{-2\alpha x^2}$$ is odd and $$\int_{-\infty}^\infty dx\,x e^{-2\alpha x^2}$$ exists, this integral will be $$0$$. By the same argument $$x^2 e^{-2\alpha x^2}$$ is even so it is not possible for this integral to be $$0$$ since the area under the curve of a non-negative function must be non-zero. Note the bounds in the integral, which differ from those you have written in your normalization condition.