Tuesday, 2 March 2010

Gauge Invariance in Quantum Mechanics

This post continues the 'Road to Quantum Field Theory' series.

The Lorentz force law for a non-relativistic particle of charge q, moving with velocity v in both electric (E) and magnetic (B) fields, is given by:
\mathbf{F} = q\mathbf{E} + q\mathbf{v}\times\mathbf{B}
This can be derived from Hamilton's equations using the classical Hamiltonian given below.
H = \frac{1}{2m}\left(\mathbf{p}-q\mathbf{A}\right)^2 + qV
The Schrödinger equation for a charged particle in an EM field is,
\left[\frac{1}{2m}\left(-i\nabla - q\mathbf{A}\right)^2 + qV \right] \psi(\mathbf{x},t) = i\frac{\partial \psi(\mathbf{x},t)}{\partial t}
obtained from the Hamiltonian through the substitution
\mathbf{p} \rightarrow -i\nabla
as is usual for the quantum mechanical momentum operator. We can identify operator combinations:
\mathbf{D} = \nabla - iq\mathbf{A}
D^0 = \frac{\partial}{\partial t} + iqV
which replace the operators
\nabla, \frac{\partial}{\partial t}
when we move from the free-particle Schrödinger equation to the electromagnetic field case.

The solutions for the wavefunctions of the Schrödinger equation describe completely the behaviour of a particle under the influence of the potentials A and V, but these potentials are not unique, as I showed in the previous post in this series. Instead, they can be changed by a gauge transformation:
\mathbf{A}\rightarrow\mathbf{A}^\prime=\mathbf{A}+\nabla\chi
V\rightarrow V^\prime = V - \frac{\partial\chi}{\partial t}
Maxwell's equations for E and B will remain invariant under these transformations (this was the topic of the previous post), but will the physics described by the Schrödinger equation be the same if we make these changes to the potentials there?

The answer is no! The Schrödinger equation is not gauge invariant, since the same wavefunction cannot satisfy both the original and transformed versions! All is not lost, however. The wavefunction itself is not directly observable, whereas E and B are. If we do not require the wavefunction to remain invariant when the potentials undergo a transformation, then we are free to change the wavefunction such that we retain invariance of physical observables. Write the Schrödinger equation in terms of some transformed wavefunction,
\psi\rightarrow\psi^\prime
i.e.
\left[\frac{1}{2m}\left(-i\nabla - q\mathbf{A}^\prime\right)^2 + qV^\prime\right] \psi^\prime(\mathbf{x},t) = i\frac{\partial \psi^\prime (\mathbf{x},t)}{\partial t}

Now, the form in which we have written the equation above is identical to that of the original Schrödinger equation, except that we are writing it in terms of primed quantities (psi', A', V') instead of unprimed (psi, A, V). Both equations describe the same physics, so if we can find such a psi', then the Schrödinger equation is gauge covariant; that is, it maintains the same form under a gauge transformation.

We know the relationship between A, V and A', V', so we can write down the transformation for psi':
\psi^\prime (\mathbf{x},t) = e^{iq\chi(\mathbf{x},t)}\psi(\mathbf{x},t)
where
\chi
is the same space- and time- dependent function appearing in the transformations of A and V. We can verify that this results in the gauge covariance of the Schrödinger equation:
(-i\nabla - q\mathbf{A}^\prime)\psi^\prime=\left[-i\nabla-q\mathbf{A}-q(\nabla\chi)\right]e^{iq\chi}\psi
=q(\nabla\chi)e^{iq\chi}\psi + e^{iq\chi}(-i\nabla\psi) + e^{iq\chi}(-q\mathbf{A}\psi) - q(\nabla\chi)e^{iq\chi}\psi
The first and last terms cancel, leaving:
(-i\nabla - q\mathbf{A}^\prime)\psi^\prime = e^{iq\chi}(-i\nabla - qA)\psi
which can be written as:
(-i\mathbf{D}^\prime \psi^\prime) = e^{iq\chi}(-i\mathbf{D}\psi)


The space-time dependent phase factor feels the action of
\nabla
, but passes through the combined D' operator, converting it to D, so it becomes clear that D'psi' is related to Dpsi in the same way psi' is related to psi! Similarly,
(iD^{0\prime}\psi^\prime = e^{iq\chi}(iD^0\psi)

We can now erite:
\frac{1}{2m}(-i\mathbf{D}^\prime)^2 \psi\prime = e^{iq\chi}\frac{1}{2m}(-i\mathbf{D})^2 \psi = e^{iq\chi}iD^0\psi = iD^{0\prime}\psi^\prime
which demonstrates the correct relationship between psi and psi'. The question remains whether the same physics is described by both psi and psi'. We can see that it is in a number of ways:

1. The probability density is given by
|\psi|^2 = \psi^\dagger \psi
, which is equivalent to
|\psi\prime|^2 = \psi^{\prime\dagger} \psi^\prime = \psi^\dagger e^{-iq\chi}e^{+iq\chi}\psi = \psi^\dagger \psi

2. The probability current,
\psi^\dagger(\nabla\psi) - (\nabla\psi)\psi^\dagger
is not invariant under a gauge transformation, but when we replace
\nabla \rightarrow \mathbf{D} ~~,~~ \frac{\partial}{\partial t} \rightarrow D^0
then,
\psi^\dagger\prime(\mathbf{D}\psi^\prime = \psi^\dagger e^{-iq\chi} e^{iq\chi}(\mathbf{D}\psi) = \psi^\dagger\mathbf{D}\psi
(and similarly for the other term).

Hence, identical physics is described by both the wavefunction psi, and the gauge transformed wavefunction psi'; the gauge invariance of Maxwell's equations presents as a gauge covariance in quantum mechanics, provided that we transform not only the potentials, but the wavefunction also:
\mathbf{A}\rightarrow \mathbf{A}^\prime = \mathbf{A} + \nabla\chi
V\rightarrow V^\prime = V - \frac{\partial \chi}{\partial x}
\psi\rightarrow \psi^\prime = i^{iq\chi}\psi

Finally, we note that the new differential operators,
\mathbf{D} = \nabla - iq\mathbf{A} ~~,~~D^0 = \frac{\partial}{\partial t} + iqV
can be written in a Lorentz covariant form:
D^\mu = \partial^\mu + iqA^\mu

This allows us to write:
-iD^{\prime\mu}\psi^\prime = e^{iq\chi}(-iD^\mu \psi)
and it follows that an equation can involving the operator
\partial^\mu
can be made to be gauge invariant by under the combined transformations:
A^\mu \rightarrow A^{\prime\mu} = A^\mu - \partial^\mu \chi
\psi \rightarrow \psi^\prime = e^{iq\chi}\psi
provided the minimal substitution is also made:
\partial^\mu \rightarrow D^\mu = \partial^\mu + iqA^\mu

This provides a simple mechanism to get the wave equation for a particle in an electromagnetic field from the equation for a free particle - by making the above substitution. This forms the basis of the gauge principle, i.e. the form of an interaction is determined by an insistence on local gauge invariance.

That's enough for now; next time we'll move on to study the Klein-Gordon equation for relativistic spin-zero particles.

2 comments:

  1. Could you post up links to the other parts of the series? It's a bit overwhelming to go over without some context.

    ReplyDelete
  2. Sure. I do this from time to time anyway, but I'll make a new post now with links to all the previous ones.

    ReplyDelete