- #1
taishizhiqiu
- 63
- 4
Look at the following derivation:
##
p=\frac{im}{\hbar}[H,r]
##
if ##H|\psi\rangle=E|\psi\rangle##, then
##
\langle \psi|p|\psi \rangle = \frac{im}{\hbar}\langle \psi|Hr-rH|\psi \rangle = \frac{im}{\hbar}\langle \psi|r|\psi \rangle(E-E)=0
##
What's wrong with my derivation or it is true that average momentum of energy eigenstates is always zero?
##
p=\frac{im}{\hbar}[H,r]
##
if ##H|\psi\rangle=E|\psi\rangle##, then
##
\langle \psi|p|\psi \rangle = \frac{im}{\hbar}\langle \psi|Hr-rH|\psi \rangle = \frac{im}{\hbar}\langle \psi|r|\psi \rangle(E-E)=0
##
What's wrong with my derivation or it is true that average momentum of energy eigenstates is always zero?