59
2.6. Comments on the gauge principle in electromagnetism
which may be checked by using the fact that
∫ a
∂
f (t) dt = f (a).
(2.74)
∂a
The notation ψ(f = 0) means just the free-particle solution with f = 0; the
line integral is taken along an arbitrary path ending in the point x. But we
have
∂f
∂f
∂f
df =
dx +
dy +
dz ≡ ∇f · dl.
(2.75)
∂x
∂y
∂z
Hence the integral can be done trivially and the solution becomes
ψ = exp[iq(f (x) − f (−∞))] · ψ(f = 0).
(2.76)
We say that the phase factor introduced by the (in reality, field-free) vector
potential A = ∇f is integrable: the effect of this particular A is merely
to multiply the free-particle solution by an x-dependent phase (apart from
a trivial constant phase). Since this A should give no real electromagnetic
effect, we must hope that such a change in the wavefunction is also somehow
harmless. Indeed Dirac showed (Dirac 1981, pp 92–3) that such a phase
factor corresponds merely to a redefinition of the momentum operator p ˆ. The
essential point is that (in one dimension, say) ˆ
p is defined ultimately by the
commutator (ħ = 1)
[ˆ x, ˆ
p] = i.
(2.77)
Certainly the familiar choice
∂
ˆ
p = −i ∂x
(2.78)
satisfies this commutation relation. But we can also add any function of x
to ˆ
p, and this modified ˆ
p will be still satisfactory since x commutes with
any function of x. More detailed considerations by Dirac showed that this
arbitrary function must actually have the form ∂F/∂x, where F is arbitrary.
Thus
∂
∂F
′
p ˆ = −i
+
(2.79)
∂x
∂x
is an acceptable momentum operator. Consider then the quantum mechanics
defined by the wavefunction ψ(f = 0) and the momentum operator ˆ
p =
−i∂/∂x. Under the unitary transformation (cf (2.76))
iqf (x) ψ(f
ψ(f = 0) → e
= 0)
(2.80)
p ˆ will be transformed to
iqf (x) ˆ
−iqf (x)
p ˆ → e
pe
.
(2.81)
But the right-hand side of this equation is just ˆ
p − q∂f /∂x (problem 2.3),
which is an equally acceptable momentum operator, identifying qf with the
F of Dirac. Thus the case A = ∇f is indeed equivalent to the field-free case.
2.6. Comments on the gauge principle in electromagnetism
which may be checked by using the fact that
∫ a
∂
f (t) dt = f (a).
(2.74)
∂a
The notation ψ(f = 0) means just the free-particle solution with f = 0; the
line integral is taken along an arbitrary path ending in the point x. But we
have
∂f
∂f
∂f
df =
dx +
dy +
dz ≡ ∇f · dl.
(2.75)
∂x
∂y
∂z
Hence the integral can be done trivially and the solution becomes
ψ = exp[iq(f (x) − f (−∞))] · ψ(f = 0).
(2.76)
We say that the phase factor introduced by the (in reality, field-free) vector
potential A = ∇f is integrable: the effect of this particular A is merely
to multiply the free-particle solution by an x-dependent phase (apart from
a trivial constant phase). Since this A should give no real electromagnetic
effect, we must hope that such a change in the wavefunction is also somehow
harmless. Indeed Dirac showed (Dirac 1981, pp 92–3) that such a phase
factor corresponds merely to a redefinition of the momentum operator p ˆ. The
essential point is that (in one dimension, say) ˆ
p is defined ultimately by the
commutator (ħ = 1)
[ˆ x, ˆ
p] = i.
(2.77)
Certainly the familiar choice
∂
ˆ
p = −i ∂x
(2.78)
satisfies this commutation relation. But we can also add any function of x
to ˆ
p, and this modified ˆ
p will be still satisfactory since x commutes with
any function of x. More detailed considerations by Dirac showed that this
arbitrary function must actually have the form ∂F/∂x, where F is arbitrary.
Thus
∂
∂F
′
p ˆ = −i
+
(2.79)
∂x
∂x
is an acceptable momentum operator. Consider then the quantum mechanics
defined by the wavefunction ψ(f = 0) and the momentum operator ˆ
p =
−i∂/∂x. Under the unitary transformation (cf (2.76))
iqf (x) ψ(f
ψ(f = 0) → e
= 0)
(2.80)
p ˆ will be transformed to
iqf (x) ˆ
−iqf (x)
p ˆ → e
pe
.
(2.81)
But the right-hand side of this equation is just ˆ
p − q∂f /∂x (problem 2.3),
which is an equally acceptable momentum operator, identifying qf with the
F of Dirac. Thus the case A = ∇f is indeed equivalent to the field-free case.