Given the wave function $$\psi(x)=A\exp\left[-a \left(\frac{mx^{2}}{\hbar}+it\right)\right]$$ I would like to calculate $\sigma_{p}$.

\begin{align}\langle p\rangle &=\int \psi^{\star}\left(\frac{\hbar}{i}\frac{\partial}{\partial x}\right)\psi dx\\ &=2imaA^{2}\int xe^{-kx^{2}}dx\\ &=0 \end{align} since the integrand is odd. (I let $k=\frac{2am}{\hbar}$)

Similarly, \begin{align} \langle p^2\rangle &=\int \psi^{\star}\left(\frac{\hbar}{i}\frac{\partial}{\partial x}\right)^{2}\psi dx\\ &=-2A^{2}ma\hbar \left[\int kx^{2}e^{-kx^{2}}dx-\int e^{-kx^{2}}dx\right]\\ &=-2A^{2}ma\hbar\left[\frac{1}{2} \sqrt{\frac{\pi}{k}}-\frac{1}{2} \sqrt{\frac{\pi}{k}}\right]\\ &=0 \end{align}

But doesn't this imply $\sigma_{p}=\sqrt{\langle \widehat p^{2}\rangle -\langle \widehat p\rangle ^{2}}=0$ ?

I think that I must have made a mathematical error, because this result as it stands would violate the uncertainty principle as I understand it: $\sigma_{x}\sigma_{p}\geq\frac{\hbar}{2}$

But I don't see anything incorrect. Does $\sigma_{p}=0$ violate the uncertainty principle?


closed as off-topic by Kyle Kanos, Brandon Enright, John Rennie, JamalS, Emilio Pisanty Jan 26 '15 at 15:14

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    $\begingroup$ The integrals you have look fine, but you've evaluated one of them with a factor out. $\endgroup$ – Holographer Jan 25 '15 at 19:40
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    $\begingroup$ You also need to find the normalization constant, $A$, in order to get the value of $\sigma_p$. $\endgroup$ – Ihle Jan 25 '15 at 22:23

You have made an error in your calculation of ⟨p2⟩. You've evaluated two integrals, the second of which is off by a factor of two.

  • $\begingroup$ Ah: $\int ^{\infty}_{-\infty} e^{-kx^{2}}dx = \sqrt{\dfrac{\pi}{k}}$ $\endgroup$ – user44816 Jan 25 '15 at 23:16

There is an error in my calculation.

In fact: $\int ^{\infty}_{-\infty} e^{-kx^{2}}dx = \sqrt{\dfrac{\pi}{k}}$

So $$\langle \widehat p^{2} \rangle=-2A^{2}ma\hbar [\frac{-1}{2}\sqrt{\frac{\pi}{k}}]=ma\hbar$$ given that $A^{2}=\sqrt{\frac{2am}{\pi\hbar}}$ as calculated using the normalization condition.

This, can be used to calculate $$\sigma_{p}=\sqrt{ma\hbar}$$ It can similarly be shown that $$\langle x^{2}\rangle=\frac{\hbar}{4ma}$$ and $$\langle x \rangle= 0$$so $$\sigma_{x}=\sqrt{\frac{\hbar}{4ma}}$$ and finally: $$\sigma_{p}\sigma_{x}=\sqrt{\frac{\hbar^{2}ma}{4ma}}=\frac{\hbar}{2}\geq\frac{\hbar}{2}$$which agrees with the uncertainty principle.