- #1

- 1,333

- 75

[tex]

\psi(\mathbf r, t) = \sum_k c_k(t) \phi_k(\mathbf r)

[/tex]

where

[tex]

\hat H_0 \phi_k (\mathbf r) = \epsilon_k \phi_k(\mathbf r)

[/tex]

Plugging the expansion into the time dependent Schroedinger's equation, we can calculate all [itex]c_k[/itex]s. The accepted interpretation of these numbers is that [itex] c_k^* c_k[/itex] is the probability that randomly chosen atom will be in the [itex]k[/itex]-th stationary state.

The problem is, that these numbers are gauge-dependent. How can a probability depend on the choice of gauge?

Proof:

In the initial gauge [itex] \varphi = Kq_p/r, \mathbf A = 0 [/itex]:

[itex] c_k (t) = \langle \phi_k | \psi(t) \rangle = \int \phi_k(\mathbf r) \psi(\mathbf r, t) dV [/itex]

In another gauge [itex] \varphi' = \varphi - \partial_t \chi, \mathbf A ' = \mathbf A + \nabla \chi, \psi' = e ^{-i\frac{q}{\hbar}\chi} \psi[/itex]:

[itex] c_k' (t) = \langle \phi_k | \psi(t)'\rangle = \int \phi_k(\mathbf r) e^{-i\frac{q}{\hbar} \chi(\mathbf r, t)}\psi(\mathbf r, t) dV [/itex]

which can be made equal to any number by suitable choice of [itex]\chi [/itex], for example, time independent [itex] \chi [/itex] rapidly oscillating in spatial coordinates will make it close to zero without change in the field strength (provided it has zero curl, which is not hard to fulfill).

One may object that in the second gauge one should calculate coefficients using new functions for the stationary states, somehow like this:

[itex] c_k' = \langle \phi_k' (\mathbf r, t) | \psi(t)' \rangle [/itex]

where [itex] \phi_k'(\mathbf r, t)[/itex] is a stationary state in the new gauge, but how to find such function? Clearly [itex] \phi_k'(t) = e^{-i\frac{q}{\hbar}\chi} \phi_k [/itex] won't work, because then we'll get

[tex]

\hat H'(t) \phi_k'(t) = [ \epsilon_k - q_e\partial _t \chi (\mathbf r, t)] \phi_k'(t)

[/tex]

with energy eigenvalue dependent on the spatial coordinates as well.

Or is there some other gauge-invariant way to calculate the A, B coefficients?