# Properties of Expectation Values Under Variable Substitution [closed]

I am working on a homework problem from Griffiths QM (Problem 2.11, 3rd Edition). Specifically, I'm working on finding $$\left$$ 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 $$\left$$, 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 $$\left = \left<\left(\sqrt{\frac{\hbar}{m\omega}}\xi\right)^2\right> = \frac{\hbar}{m\omega}\left<\xi^2\right>$$. 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 $$\left = \frac{\hbar}{m\omega}\frac{\alpha^2\sqrt{\pi}}{2} = \frac{1}{2}\sqrt{\frac{\hbar}{m\omega}},$$ 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: $$\left = \frac{\hbar}{m\omega}\left<\xi^2\right>?$$

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

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• 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 – Aaron Stevens Apr 19 at 18:42

Nothing is wrong with the equation $$\left = \frac{\hbar}{m\omega}\left<\xi^2\right>$$ 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 $$\left = \frac{\hbar}{m\omega}\left<\xi^2\right>$$ 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
$$\left = \frac{\hbar}{m\omega}\left<\xi^2\right>=\frac{\hbar}{m\omega}\int\psi(x)^*\xi^2\psi(x) dx$$ and only THEN do a change of variables to write everything in terms of $$\xi$$.