Technically, if we want to prove an operator $\hat{A}$ to be an hermitian,we should prove $ \left< \psi \hat{A}|\psi \right> =\left< \psi |\hat{A}\psi \right>$. It works well in Cartesian coordinate when proving the momentum operator $\hat{p}$.

However when dealing with $\hat{p}_r\equiv-i\hbar\left(\frac{\partial}{\partial r}+\frac{1}{r}\right)$ in Spherical coordinate.

In Spherical coordinate system, we have:

$$ \langle\psi \hat{A} \mid \psi\rangle=\int_{0}^{\infty} \int_{0}^{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\hat{A} \psi)^{\dagger} \psi r^{2} \sin \theta \mathrm{d} \varphi \mathrm{d} \theta \mathrm{d} r=4 \pi \int_{0}^{\infty}(\hat{A} \psi)^{\dagger} \psi r^{2} \mathrm{~d} r $$

Imply $\hat{p}_r$ , then we have: $$ \begin{aligned} & 4 \pi \int_{0}^{\infty}\left(-i \hbar\left(\frac{\partial}{\partial r}+\frac{1}{r}\right) \psi\right)^{\dagger} \psi r^{2} \mathrm{~d} r \\ =& 4 \pi i \hbar \int_{0}^{\infty}\left(\frac{1}{r} \frac{\partial}{\partial r}(r \psi)\right)^{\dagger} \psi r^{2} \mathrm{~d} r \\ =& 4 \pi i \hbar \int_{0}^{\infty}\left(\frac{\partial}{\partial r}(r \psi)\right)^{\dagger}(\psi r) \mathrm{d} r \\ =& 4 \pi i h\left(\left.(r \psi)^{\dagger}(\psi r)\right|_{0} ^{\infty}-\int_{0}^{\infty}(r \psi)^{\dagger} \frac{\partial}{\partial r}(\psi r) \mathrm{d} r\right) \end{aligned} $$

When proving $\hat{p}$ to be an hermitian, the first term now vanish because there's no extra $r$ in the equation and the wave function equals 0 when it goes to $\pm \infty$.But now, this step cannot be used here.

So how to prove it?


1 Answer 1


That term still vanishes. Since the integral $\int_0^\infty |\psi|^2 r^2\, \mathrm dr$ is finite, $|\psi|^2 r^2$ must vanish at infinity.

Note that there are two small inaccuracies in your derivation so far, but correcting them will not change the calculations significantly. First, in order to show that $A$ is Hermitian, it is not enough to show $\langle A\psi, \psi\rangle = \langle \psi, A\psi \rangle$. You need $$ \langle A\psi_1, \psi_2 \rangle = \langle \psi_1, A\psi_2 \rangle \tag{1} $$ for any $\psi_1$ and $\psi_2$. Second, in the step where you evaluate the $\mathrm d\varphi \mathrm d\theta$ integrals, you assume that $\psi$ is spherically symmetric (i.e., that it does not depend on $\varphi$ or $\theta$). However, you need to show (1) for all $\psi_1$, $\psi_2$ and not only spherically symmetric ones.


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