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Prove that the expectation value of momentum is always a real number. 📂Quantum Mechanics

Prove that the expectation value of momentum is always a real number.

Summary

The expectation value of the momentum operator p\langle p \rangle is always a real number.

Explanation

In fact, not only the momentum operator but all Hermitian operators have eigenvalues that are always real.

Proof

The expectation value of momentum is as follows.

p=ψ(ix)ψdx \displaystyle \langle p \rangle = \int \psi^{\ast} \left( \frac{\hbar}{i}\frac{\partial}{\partial x} \right) \psi dx

Additionally, the complex conjugate of the expectation value of momentum is as follows. p=ψ(ix)ψdx \displaystyle \langle p \rangle ^{\ast}= \int \psi \left( \frac{\hbar}{-i}\frac{\partial}{\partial x} \right) \psi^{\ast} dx By subtracting the two values and showing it is zero, the proof is complete.

pp=i(ψψx+ψψx)dx=ix(ψψ)dx=i[ψψ]+=0 \begin{align*} \langle p \rangle -\langle p \rangle ^{\ast} &= \frac{\hbar}{i} \int \left( \psi^{\ast} \frac{\partial \psi}{\partial x}+\psi\frac{\partial \psi^{\ast}}{\partial x} \right) dx \\ &= \frac{\hbar}{i} \int \frac{\partial}{\partial x} \left( \psi^{\ast} \psi \right) dx \\ &= \frac{\hbar}{i} \left[ \psi^{\ast}\psi \right] ^{+\infty}_{-\infty} \\ &= 0 \end{align*}

The final equality holds because the wave function must satisfy ψ(±)=0\psi (\pm \infty) = 0.