When we calculate the expectation value of the momentum operator, we use
$$\langle p\rangle
= \int \psi^*\left(−i\hbar\frac{\partial}{\partial x}\right)\psi\, \mathrm dx\tag{1}.$$
I'm wondering if we can get $\langle p\rangle$ by
$$\langle p\rangle
= \int \left(i\hbar\frac{\partial}{\partial x}\psi^*\right)\psi\,\mathrm dx\tag{2},$$
due to the fact that the momentum operator is
Hermitian and therefore $\langle p\rangle=\langle\psi|p\psi⟩=⟨p\psi|\psi⟩$. Does (2) work to get $\langle p\rangle$?
Actually, I've seen a similar form of (1) $=$ (2) at the proof of hermiticity of momentum operator, but when it comes to obtaining $⟨p⟩$, people just use (1). Is there any reason for using only (1)? I guess the only reason not to use (2) might be the risk of calculating (2) as $$\langle p\rangle
= \int \left(i\hbar\frac{\partial}{\partial x}\right)|\psi|^2\mathrm dx$$
by mistake, which calculates $\psi^*\psi$ first, producing the wrong answer.
Best Answer
Yes, these two equations are equivalent as long as the wave function goes to zero at the boundaries.
For example, if the boundary of integration is $\pm\infty$, we see that the two above-quoted equations are related by a single integration by parts as long as the boundary term is zero: $$ \int \psi^*\left(−i\hbar\frac{\partial}{\partial x}\right)\psi dx =\int \left[\frac{\partial}{\partial x}\left(−i\hbar\psi^*\psi\right) +\left(i\hbar\frac{\partial }{\partial x}\psi^*\right)\psi\right]dx $$ $$ =\left.-i\hbar|\psi|^2\right|^{\infty}_{-\infty}+\int\left(i\hbar\frac{\partial }{\partial x}\psi^*\right)\psi dx $$ $$ = 0 + \int\left(i\hbar\frac{\partial }{\partial x}\psi^*\right)\psi dx $$
This is the boundary term, which is usually zero, as discussed above. If the boundary term is not zero then your two expressions for $\langle p \rangle$ (your Eq. (1) and Eq. (2)) are not the same.