0
$\begingroup$

I am working on a homework problem from Griffiths QM (Problem 2.11, 3rd Edition). Specifically, I'm working on finding $\left<x^2\right>$ for the ground state and the first excited state of the harmonic oscillator centered at the origin. The (time-independent) wave functions for those two states are given by $$\psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-\frac{m\omega}{2\hbar}x^2}$$ and $$\psi_1(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\sqrt{\frac{2m\omega}{\hbar}}xe^{-\frac{m\omega}{2\hbar}x^2},$$ respectively. Griffiths suggests changing variables with $\xi \equiv \sqrt{\frac{m\omega}{\hbar}}x$ and $\alpha \equiv \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}$, which gives $$\psi_0(\xi) = \alpha e^{-\xi^2/2}$$ and $$\psi_1(\xi) = \sqrt{2}\alpha \xi e^{-\xi^2/2}.$$

To find $\left<x^2\right>$, I thought I'd be smart and find it using the definition of $\xi$ like this: Since $x=\sqrt{\frac{\hbar}{m\omega}}\xi$, then $\left<x^2\right> = \left<\left(\sqrt{\frac{\hbar}{m\omega}}\xi\right)^2\right> = \frac{\hbar}{m\omega}\left<\xi^2\right>$. For the ground state, I'm using $$\left<\xi^2\right> = \int_{-\infty}^{+\infty}\psi_0^*(\xi)\xi^2\psi_0(\xi)d\xi$$ giving $$\left<\xi^2\right> = \frac{\alpha^2\sqrt{\pi}}{2}.$$ Finally, I get $$\left<x^2\right> = \frac{\hbar}{m\omega}\frac{\alpha^2\sqrt{\pi}}{2} = \frac{1}{2}\sqrt{\frac{\hbar}{m\omega}},$$ which I know for a fact is not correct, because, for one thing, the units aren't right (the units I'm getting are m, but I should be getting m$^2$). I've narrowed the problem down to a missing factor of $\sqrt{\frac{\hbar}{m\omega}}$ which comes from converting $dx$ to $d\xi$ in the integral, but I'm wondering, what is wrong with the following equation: $$\left<x^2\right> = \frac{\hbar}{m\omega}\left<\xi^2\right>?$$

$\endgroup$

closed as off-topic by Aaron Stevens, GiorgioP, ZeroTheHero, Kyle Kanos, Jon Custer Apr 22 at 14:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, GiorgioP, ZeroTheHero, Kyle Kanos, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ The issue is your integral that you have for $\langle\xi^2\rangle$. I would suggest starting with the integral for $\langle x^2\rangle$ and then make the variable substitution. You might see the issue then $\endgroup$ – Aaron Stevens Apr 19 at 18:42
2
$\begingroup$

Nothing is wrong with the equation $$\left<x^2\right> = \frac{\hbar}{m\omega}\left<\xi^2\right>$$ it is completely correct.

However, your interpretation of the equation is wrong. If we say $x^2 = \frac{\hbar}{m\omega}\xi^2$ (which is obviously true), then we can immediately go to $\left<x^2\right> = \frac{\hbar}{m\omega}\left<\xi^2\right>$ by simply doing the same thing to both sides of the equation--namely, putting brackets around each side. These brackets must mean the same thing on both sides of the equation for this to be valid, however.

What do brackets mean? They mean take the expectation value. In particular, $\langle A\rangle$ means $\int \psi(x)^*A\psi(x) dx$. It does NOT mean $\int \psi(\xi)^*A\psi(\xi) d\xi$, that is a different thing. So, to go from your correct equation to an actual calculation, you must write

$$\left<x^2\right> = \frac{\hbar}{m\omega}\left<\xi^2\right>=\frac{\hbar}{m\omega}\int\psi(x)^*\xi^2\psi(x) dx$$ and only THEN do a change of variables to write everything in terms of $\xi$.

$\endgroup$

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