Shortcut to find $\hat{p}^2$ expectation value I have been going through several calculations where I am asked to calculate $\langle p^2 \rangle$ and the task is proving to be pretty tedious.  Does anyone know of a shortcut for this?  Such as with $\langle p \rangle$ where:
$$ \langle p \rangle = m\frac{d\langle x\rangle}{dt} $$
I have seen a few specific examples where it can be done knowing the energy eigenvalues and the potential...
$$ \frac{\langle p^2 \rangle}{2m} + \langle V \rangle = E_n $$
But I was hoping for something more fruitful than this.
I guess another way would be to utilize the Hermicity of $\hat{p}$.
 A: I haven't found a really good shortcut, but the following can make the integration much simpler in some cases. The time independent Schrodinger Equation:
$$ \frac{\hat{p}^2}{2m}\Psi+V\Psi=E\Psi $$
$$ \frac{\hat{p}^2}{2m}\Psi=(E-V)\Psi  $$
$$ \hat{p}^2\Psi=2m(E-V)\Psi  $$
So....
$$ \langle p^2\rangle = \int\Psi^*\hat{p^2}\Psi dx = \int\Psi^*[2m(E-V)\Psi]dx $$
I thought that this was a nice little trick.  Hopefully someone can get use out of it.
A: For a specific class of problems, the expectation value of $p^2$ can in fact be calculated much more easily than by brute integration. Essentially, for the ground state of the harmonic oscillator and related states (more technically, gaussian states) one can use a differentiate-inside-the-integral trick to do this.
For definiteness, consider a simple harmonic oscillator and set $\hbar=m=1$, but leave the frequency $\omega$ free. Because of the units, the characteristic length scale is $1/\sqrt\omega$, and the ground state is
$$
\psi(x)=\left(\frac\omega\pi\right)^{1/4}e^{-\frac12\omega x^2}.
$$
As you well note, taking the expected value of the hamiltonian leads to one simplification,
$$
\frac12\langle p^2\rangle+\frac12\omega^2\langle x^2\rangle=\frac12\omega.
$$
This leaves you with the task of calculating the expectation value of the potential, which no longer includes derivatives but still looks like more pain than one really wants on a first go:
$$
\langle x^2\rangle=\sqrt{\frac\omega\pi}\int_{-\infty}^\infty x^2e^{-\omega x^2}\,\text dx.
$$
The trick here is to use the fact that $\psi$ is normalized, or in other words to start from the relatively easy integral
$$
\int_{-\infty}^\infty e^{-\omega x^2}\,\text dx=\sqrt{\frac\pi\omega},
$$
and to differentiate both sides with respect to $\omega$:
$$
-\int_{-\infty}^\infty x^2e^{-\omega x^2}\,\text dx=-\frac{1}{2\omega}\sqrt{\frac\pi\omega}.
$$
This gives an easy calculation of
$$
\langle x^2\rangle=\frac{1}{2\omega},
$$
and with that the equivalent
$$
\langle p^2\rangle=\frac{\omega}{2}.
$$

Now, this trick is of course relatively specialized. You can apply it to anything that has a simple gaussian as a probability density, like coherent states, more or less directly. You can also use this specific case as a base and derive from it, using algebraic tricks, non-gaussian states like the rest of the oscillator eigenstates. These are general enough that it's worth keeping, relatively sharp, in some accessible drawer of your desk, which is why I'm posting it despite its limitations.
However, it is pretty limited to these cases. In general, there is nothing for it but to integrate. If your state is an eigenstate of some hamiltonian then you can get away with transferring the derivatives to an expectation value of the potential, but then you still have to buckle up and integrate. Keep your pencils sharp and G&R handy.
