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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:

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

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

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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.

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