51
2.4. Gauge invariance (and covariance) in quantum mechanics
D
′ ψ
′ bears to Dψ exactly the same relation as ψ
′ bears to ψ. In just the
same way we find (cf equation (2.30))
(iD
0′ ψ
′ ) = exp(iqχ) · (iD
0 ψ)
(2.38)
where we have used equation (2.32) for V
′ . Once again, D
0′ ψ
′ is simply related
to D
0 ψ. Repeating the operation which led to equation (2.37) we find
1 (−iD
′ )
2 ψ
′
= exp(iqχ) ·
1 (−iD)
2 ψ
2m
2m
= exp(iqχ) · iD
0 ψ
(using equation (2.29))
= iD
0′ ψ
′
(using equation (2.30)).
(2.39)
Equation (2.39) is just (2.33) written in the D notation of equation (2.30),
so we have verified that (2.34) is the correct relationship between ψ
′ and
ψ to ensure consistency between equations (2.29) and (2.33). Precisely this
consistency is summarized by the statement that (2.29) is gauge covariant.
Do ψ and ψ
′ describe the same physics, in fact? The answer is yes, but it
is not quite trivial. It is certainly obvious that the probability densities |ψ|
2
and |ψ
′
|
2 are equal, since in fact ψ and ψ
′ in equation (2.34) are related by
a phase transformation. However, we can be interested in other observables
involving the derivative operators ∇ or ∂/∂t – for example, the current, which
is essentially ψ
∗ (∇ψ) − (∇ψ)
∗ ψ. It is easy to check that this current is
not invariant under (2.34), because the phase χ(x, t) is x-dependent. But
equations (2.37) and (2.38) show us what we must do to construct gaugeinvariant currents: namely, we must replace ∇ by D (and in general also
∂/∂t by D
0 ) since then:
ψ
∗′ (D
′ ψ
′ ) = ψ
∗ exp(−iqχ) · exp(iqχ) · (Dψ) = ψ
∗
Dψ
(2.40)
for example. Thus the identity of the physics described by ψ and ψ
′ is indeed
ensured. Note, incidentally, that the equality between the first and last terms
in (2.40) is indeed a statement of (gauge) invariance.
We summarize these important considerations by the statement that the
gauge invariance of Maxwell equations re-emerges as a covariance in quantum
mechanics provided we make the combined transformation
A → A
′ = A + ∇χ
′
V → V = V − ∂χ/∂t
(2.41)
ψ → ψ
′ = exp(iqχ)ψ
on the potential and on the wavefunction.
The Schr¨ odinger equation is non-relativistic, but the Maxwell equations are
of course fully relativistic. One might therefore suspect that the prescriptions
discovered here are actually true relativistically as well, and this is indeed
Précédent

- 67/979

Suivant