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I have the correct answer except with a negative sign.

The wave function is given as, $$\Phi=A\exp\left[-a\left(\frac{mx^2}{\hbar} + it\right)\right]$$

By squaring the momentum quantity, I found the expectation value of momentum squared to be $\langle-\hbar^2\frac {\partial^2}{\partial x^2}\rangle$.

I then computed the second derivative of $\Phi$ and found it to be $$\frac{\partial^2\psi}{\partial x^2}=A\exp\left[-a\left(\frac{mx^2}{\hbar} + it\right)\right]\cdot\left[4\left(\frac {am}\hbar\right)^2x^2-\left(\frac {am}\hbar\right)\right].$$

The expectation value can therefore be written as $$-\hbar^2 4\left(\frac {am}\hbar\right)^2(\int \Phi^*(x^2)\Phi\mathop{}\!\mathrm dx -\frac {am}\hbar\int \Phi^*\Phi \mathop{}\!\mathrm dx)$$

$\int \Phi^*(x^2)\Phi\mathop{}\!\mathrm dx$ is just the expectation value for $x^2$, and the other integral is just 1 (since the wave function is normalized).

I previously found the expectation value $x^2$ to be $\frac \hbar{4ma}$.

The expectation value of momentum squared should then simplify to $$-\hbar^2 4\left(\frac {am}\hbar\right)^2\cdot(\frac \hbar{4ma}-\frac {am}\hbar)=-\hbar am +4a^3m^3$$ The given answer is $$\hbar am.$$

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    $\begingroup$ That $x$ shouldn't be there in the second derivative - you just get a quadratic and a constant term. $\endgroup$ – Javier Dec 27 '18 at 18:19
  • $\begingroup$ Ah, right, fixed. $\endgroup$ – Isaac Spivack Dec 27 '18 at 18:50
  • $\begingroup$ Hi @Isaac and welcome on this site. You already have an answer below, but I would suggest you also to check that your final answer ($\langle p^2 \rangle= -\hbar a m +4 (am)^3$) is dimensionally inconsistent. That should allow you to easily go back and trace the error in your calculation. $\endgroup$ – pppqqq Dec 27 '18 at 20:51
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First of all, your second derivative is wrong it should be $$\frac{d^{2}\Phi}{dx^{2}}=\Phi\left[\left(\frac{2ma}{\hbar}\right)^{2}x^{2}-\left(\frac{2ma}{\hbar}\right)\right]$$

Second, you wrote wrong the expression for the expectation value $$\langle p^{2}\rangle=-\hbar^{2}\left[\left(\frac{2ma}{\hbar}\right)^{2}\int\! \Phi^{*}(x^{2})\Phi\, dx-\left(\frac{2ma}{\hbar}\right)\int\! \Phi^{*}\Phi\,dx\right]=-\hbar^{2}\left[\left(\frac{2ma}{\hbar}\right)^{2}\left(\frac{\hbar}{4ma}\right)-\left(\frac{2ma}{\hbar}\right)\right]=\hbar ma$$

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