# Where did mess up while calculating the expected value of the momentum squared?

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 $$\hbar am.$$

• That $x$ shouldn't be there in the second derivative - you just get a quadratic and a constant term. – Javier Dec 27 '18 at 18:19
• Ah, right, fixed. – Isaac Spivack Dec 27 '18 at 18:50
• 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. – pppqqq Dec 27 '18 at 20:51

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 $$\langle p^{2}\rangle=-\hbar^{2}\left[\left(\frac{2ma}{\hbar}\right)^{2}\int\! \Phi^{*}(x^{2})\Phi\, dx-\left(\frac{2ma}{\hbar}\right)\int\! \Phi^{*}\Phi\,dx\right]=-\hbar^{2}\left[\left(\frac{2ma}{\hbar}\right)^{2}\left(\frac{\hbar}{4ma}\right)-\left(\frac{2ma}{\hbar}\right)\right]=\hbar ma$$