I Gauge invariance of momentum of charged particle

1. Nov 13, 2016

spaghetti3451

I know that, in the presence of a magnetic field, the momentum of a charge particle changes from $p_{i}$ to $\pi_{i}\equiv p_{i}+eA_{i}$, where $e$ is the charge of the particle.

I was wondering if this definition of momentum is gauge-invariant?

How about $\tilde{\pi}_{i}=p_{i}-eA_{i}$?

2. Nov 14, 2016

dextercioby

Well, what do you think about the value of that sign just before the "e"?

3. Nov 14, 2016

The second one is gauge invariant (kinematic momentum). You want the gauge dependent pieces to cancel, one which you get from the transformation of A and the other from the derivative piece (the canonical momentum p). This is just the spatial component of the covariant derivative.

4. Nov 17, 2016

vanhees71

Let's first understand the issue from the point of view of classical mechanics. For simplicity I consider the non-relativistic limit. You start from the Lagrangian for a particle subject to an electromagnetic field described by a scalar potential $\Phi$ and a vector potential $\vec{A}$. The Lagrangian reads
$$L=\frac{m}{2} \dot{\vec{x}}^2-\frac{q}{c} (c \Phi-\dot{\vec{x}} \cdot \vec{A}).$$
The canonical momentum is
$$\vec{\pi}=\frac{\partial L}{\partial \dot{\vec{x}}}=m\dot{\vec{x}} + \frac{q}{c} \vec{A},$$
and this is clearly gauge dependent.

The equation of motion is, of course not gauge dependent since
$$\dot{\vec{\pi}}=m \ddot{\vec{x}} + \frac{q}{c} [\partial_t \vec{A}+(\dot{\vec{x}} \cdot \vec{\nabla}) \vec{A}]=\frac{\partial L}{\partial \vec{x}} = - q \vec{\nabla} \Phi +\frac{q}{c} \vec{\nabla}(\dot{\vec{x}} \cdot \vec{A}).$$
It's easy to see, using
$$\vec{E}=-\vec{\nabla} \Phi-\frac{1}{c} \partial_t \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{A},$$
that everything combines to the gauge-invariant well-known equation of motion
$$m\ddot{\vec{x}}=q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right).$$
$$H=\dot{\vec{x}} \cdot \vec{\pi}-L=\frac{1}{2m} \left (\vec{\pi}-\frac{q}{c} \vec{A} \right)^2+q \Phi.$$
To quantize this in the position representation, we have to realize the canonical (!) momentum as the gradient operator since it's the canonical momentum that generates spatial translations, i.e., we have
$$\hat{\vec{\pi}}=-\hbar \mathrm{i} \vec{\nabla}$$
$$\hat{H}=\frac{1}{2m} \hat{D}^2+q \Phi.$$
Here
$$\hat{D}=-\mathrm{i} \hbar \vec{\nabla}-\frac{q}{c} \vec{A}.$$
Now the time-dependent Schrödinger equation reads
$$\mathrm{i} \hbar \partial_t \psi=\hat{H} \psi.$$
We now check what happens under gauge transformations,
$$\vec{A}'=\vec{A}+\vec{\nabla \chi}, \quad \Phi'=\Phi-\frac{1}{c} \partial_t \chi.$$
Now we can define also a new wave function, which just changes by a phase factor. The right choice is
$$\psi'=\exp \left (\frac{\mathrm{i} q}{\hbar c} \chi \right) \psi.$$
Then we find
$$\hat{D}' \psi'=\exp \left (\frac{\mathrm{i} q}{\hbar c} \chi \right) \hat{D} \psi.$$
So we find that if $\psi$ obeys the time-dependent Schrödinger equation with the electromagnetic potentials $\Phi$ and $\vec{A}$ then $\psi'$ obeys the time-dependent Schrödinger equation with the electromagnetic potentials $\Phi'$ and $\vec{A}'$. Since $\psi$ and $\psi'$ differ just by a phase factor, both represent the same states, since only the ray in Hilbert space determines already the quantum state not the wave function itself. Indeed the only physical content of the wave function is the probability distribution for position, which is $|\psi|^2=|\psi'|^2$, doesn't depend on $\chi$ and thus is gauge invariant.

So the physical outcome of the quantum-theoretical formalism is gauge independent, as it should be.

5. Nov 17, 2016

stevendaryl

Staff Emeritus
I thought that two states $\psi$ and $\psi'$ represent the same state if one is a CONSTANT multiple of the other. In this case, the phase $e^{i \frac{q}{\hbar c} \chi}$ is a function of position and time, so $\psi$ and $\psi'$ aren't constant multiples of each other.

If the definition of "ray" allows for non-constant multiples, then it seems like that would mean that the wave function $\psi(x) = e^{ikx}$ is the same state $\psi'(x) = e^{i k' x}$, where $k \neq k'$. But those aren't the same state, are they? (Yeah, I know they aren't "states" at all in the Hilbert space of square-integrable functions, but they are states for a particle in a box with periodic boundary conditions).

6. Nov 17, 2016

vanhees71

True. I was a bit imprecise in my explanation, but still the point is that multiplication even with a space-time dependent phase factor doesn't change the physics described by this wave function. More precisely, a gauge transformation is a unitary transformation of the operators and wave functions.

Concerning you plane-wave example, you already gave the reason for why this is not a valid argument, because here you consider generalized eigenfunctions of the self-adjoint momentum operator, which are not representing states. If they would, both just describe uniformly distributed position probabilities over the entire infinite position space, but this is not normalizable properly, and thus it's not a wave function that represents a quantum state. For that indeed it must be square-integrable! The purpose of such generalized eigenstates in the continuous spectrum is just the transformation from one to another representation. In your case it's the Fourier transformation between the position and momentum representation,
$$\psi(t,\vec{x})=\frac{1}{(2 \pi \hbar)^{3/2}} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \exp \left (\frac{\mathrm{i} \vec{p} \cdot \vec{x}}{\hbar} \right) \tilde{\psi}(t,\vec{p}),$$
where $\tilde{\psi}(t,\vec{p})$ is the probability distribution for momentum, i.e., in the representation independent Dirac notation
$$\psi(t,\vec{x})=\langle \vec{x}|\psi(t) \rangle, \quad \tilde{\psi}(t,\vec{p})=\langle \vec{p}|\psi(t) \rangle,$$
where I assume we are working in the Schrödinger picture of time evolution.