Momentum operator's position when we calculate the expectation value of momentum

Question

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 $\langle p\rangle$?

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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.

Answer 1

When we calculate the expectation value of the momentum operator, we use $$⟨p⟩ = \int ψ^*\left(−iℏ\frac{∂}{∂x}\right)ψ\ dx\tag{1}.$$ I'm wondering if we can get $⟨p⟩$ by $$⟨p⟩ = \int \left(iℏ\frac{∂}{∂x}ψ^*\right)ψ\ dx\tag{2},$$

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: $$ \int \psi^*\left(−i\hbar\frac{\partial}{\partial x}\right)\psi dx =\int \left[\frac{\partial}{\partial x}\left(−i\hbar\psi^*\psi\right) +\left(i\hbar\frac{\partial }{\partial x}\psi^*\right)\psi\right]dx $$ $$ =\left.-i\hbar|\psi|^2\right|^{\infty}_{-\infty}+\int\left(i\hbar\frac{\partial }{\partial x}\psi^*\right)\psi dx $$ $$ = 0 + \int\left(i\hbar\frac{\partial }{\partial x}\psi^*\right)\psi dx $$

I guess the only reason not to use (2) might be the risk of calculating (2) as $$⟨p⟩ = \int \left(iℏ\frac{∂}{∂x}\right)|ψ|^2\ dx$$ by mistake, which calculates $ψ^*ψ$ first, producing the wrong answer.

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.

Answer 2

The momentum operator is hermitian ($\widehat{p}^\dagger=\widehat{p}$), therefore the expectation value is a real number, which is of course important for physical interpretation: $$\langle\widehat{p}\rangle^* =\langle\psi|\widehat{p}|\psi\rangle^* =\langle\psi|\widehat{p}^\dagger|\psi\rangle =\langle\psi|\widehat{p}|\psi\rangle =\langle\widehat{p}\rangle.$$ By writing this in integral form, if you take equation (1) for the expectation value of the momentum operator, you simply get equation (2): $$\langle\widehat{p}\rangle =\langle\widehat{p}\rangle^* \stackrel{\text{eq.}(1)}=\left(\int\psi^*\left(-i\hbar\frac{\partial\psi}{\partial x}\right)\mathrm{d}x\right)^* =\int\psi\left(i\hbar\frac{\partial\psi^*}{\partial x}\right)\mathrm{d}x,$$ which therefore does indeed work. Why equation (1) is used more often than equation (2) is probably just a form of convention. I don't quite see, how you can risk calculating the expectation operator like in your third equation, which is the integral across a derivative and therefore vanishes as the wavefunction has to vanish with increasing distance ($\lim_{|x|\rightarrow\infty}\psi(x)=0$) due to being normed. You can also subtract equation (1) from equation (2) to get this result: $$\int i\hbar\frac{\partial}{\partial x}\underbrace{|\psi|^2}_{=\psi^*\psi}\mathrm{d}x =\int\psi\left(i\hbar\frac{\partial\psi^*}{\partial x}\right)\mathrm{d}x +\int\psi^*\left(i\hbar\frac{\partial\psi}{\partial x}\right)\mathrm{d}x \stackrel{\text{eq.}(2),(1)}=\langle\widehat{p}\rangle-\langle\widehat{p}\rangle=0.$$

Answer 3

(2) does work to get $\langle P \rangle$ but acting $\hat{P}$ on the bra rather than the ket introduces an extra minus sign which might cause trouble. So it is just easier to use (1)