1
$\begingroup$

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 $\hat{p}$.

$\endgroup$
10
  • 4
    $\begingroup$ Applying a derivative twice is tedious? $\endgroup$
    – Kyle Kanos
    Feb 23, 2014 at 3:03
  • $\begingroup$ The integrations $\endgroup$ Feb 23, 2014 at 3:04
  • $\begingroup$ So you can't just use $\int\psi^*\psi\,dx\equiv1$? Are you getting something like $\int\psi^*\cdot f\left(x\right)\psi\,dx$? $\endgroup$
    – Kyle Kanos
    Feb 23, 2014 at 3:05
  • $\begingroup$ Sure normalization is helpful in some cases. I finished all the work but was just looking for something in future cases where you don't even have to go through the inner product and jump straight to the answer through some other value like with the p expectation value being the derivative of the x expectation value. $\endgroup$ Feb 23, 2014 at 3:09
  • 2
    $\begingroup$ This is a bit broad. Certainly there's no general way to just ignore the action of taking an expectation. On the other hand, anything you might be asked to compute for a harmonic oscillator is easy and shouldn't require integrals. $\endgroup$
    – user10851
    Feb 23, 2014 at 3:46

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Hey, thanks for responding! I really love integration under the integral sign. Integrating Gaussians was where I first learn to use it! This is a cool little trick. $\endgroup$ Jun 3, 2014 at 2:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.