If there is a hydrogen atom in vacuum,will the momentum of the electron will be constant or not ?Why will it change

Feb 5, 2018

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)

$= \frac{{\hat{p}}_{x}^{2} + {\hat{p}}_{y}^{2} + {\hat{p}}_{z}^{2}}{2 \mu} - \frac{{e}^{2} \cos \theta}{4 \pi {\epsilon}_{0} z}$

$= \frac{{\hat{p}}^{2}}{2 \mu} + \hat{V}$

The linear momentum ought to commute with the kinetic energy operator. It may or may not commute with the potential.

$\left[\hat{H} \text{, } \hat{p}\right]$

$= \left[{\hat{p}}^{2} / \left(2 \mu\right) + \hat{V} \text{, } \hat{p}\right]$

$= \left[{\hat{p}}^{2} / \left(2 \mu\right) \text{, " hatp] + [hatV", } \hat{p}\right]$

= cancel(1/(2mu)[hatp^2", " hatp])^(0) + [hatV", " hatp]

The potential is a function of the angle the electron moves from the $z$ axis. All of the momentum operator commutes except for...

= [- (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.