If there is a hydrogen atom in vacuum,will the momentum of the electron will be constant or not ?Why will it change
1 Answer
Whenever the operator commutes with the Hamiltonian, and is not also a function of time, its observable is conserved over time.
That is, in Heisenberg's equation of motion,
#d/(dt) << Psi | hatp | Psi >> = i/ℏ << Psi | [hatH", " hatp] | Psi >> + << Psi | (dhatp)/(dt) | Psi >>#
Both right-hand terms would go to zero, respectively.
I have already written the Hamiltonian before. I suppose you don't need me to explain what it means.
#hatH = -ℏ^2/(2mu) nabla^2 - e^2/(4piepsilon_0r)#
#= (hatp_x^2 + hatp_y^2 + hatp_z^2)/(2mu) - (e^2costheta)/(4piepsilon_0z)#
#= (hatp^2)/(2mu) + hatV#
The linear momentum ought to commute with the kinetic energy operator. It may or may not commute with the potential.
#[hatH", " hatp]#
#= [hatp^2/(2mu) + hatV", " hatp]#
#= [hatp^2/(2mu)", " hatp] + [hatV", " hatp]#
#= cancel(1/(2mu)[hatp^2", " hatp])^(0) + [hatV", " hatp]#
The potential is a function of the angle the electron moves from the
#= [- (e^2costheta)/(4piepsilon_0z), -iℏ (del)/(delz)]#
#= e^2/(4piepsilon_0)[(costheta)/(z), iℏ (del)/(delz)]#
#= (iℏe^2)/(4piepsilon_0)[costheta/z, (del)/(delz)] ne 0#
We have just shown that
#i/ℏ << Psi | [hatH", " hatp] | Psi >> ne 0#
Thus,
#d/(dt) << Psi | hatp | Psi >> = cancel(i/ℏ << Psi | [hatH", " hatp] | Psi >>)^(ne 0) + cancel(<< Psi | (dhatp)/(dt) | Psi >>)^(0) ne 0#
So linear momentum is not a constant of the motion in the hydrogen atom, wherever it is in the universe.
