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

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.

• You object to integrating by parts? Commented Sep 20, 2022 at 21:02

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.

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: $$\int i\hbar\frac{\partial}{\partial x}\underbrace{|\psi|^2}_{=\psi^*\psi}\mathrm{d}x =\int\psi\left(i\hbar\frac{\partial\psi^*}{\partial x}\right)\mathrm{d}x +\int\psi^*\left(i\hbar\frac{\partial\psi}{\partial x}\right)\mathrm{d}x \stackrel{\text{eq.}(2),(1)}=\langle\widehat{p}\rangle-\langle\widehat{p}\rangle=0.$$

• How does the argument of vanishing wave function at infinity work for a plane wave? Commented Sep 20, 2022 at 21:23
• @LucasBaldo 1. The plane waves are not elements of $L^2(\mathbb R)$ and 2. the argument that a generic wave function of that space vanishes at infinity is wrong; there are wave functions in that space which do not fulfill this requirement. Despite that, roughly speaking, you want to choose the domain of $P$ s.t. it is, among other things, hermitian/symmetric. See e.g. this classic paper; example 2 and its 'resolution' in the appendix. As far as I remember, there are also some questions regarding this here on SE. Commented Sep 21, 2022 at 5:39
• Technically $L^2$ alone does not get you vanishing at infinity, although the counterexamples are rather physically unnatural.
– Ian
Commented Sep 21, 2022 at 18:26

(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)