Let's take a simple example: a single charged particle moves in an external, uniform magnetic field \( \vec{B} = (0,0,B) \). As you know, there are multiple possible choices of the vector potential \( \vec{A} \) which will give this same field. For example, let's suppose that
\[ \begin{aligned} \vec{A} = (-yB, 0, 0). \end{aligned} \]
The Hamiltonian for our particle can be written out explicitly as
\[ \begin{aligned} \hat{H} = \frac{1}{2m} \left[ \left( \hat{p}_x + \frac{e}{c} B \hat{y} \right)^2 + \hat{p}_y^2 + \hat{p}_z^2 \right]. \end{aligned} \]
Both \( \hat{p}_x \) and \( \hat{p}_z \) commute with the Hamiltonian, so we can label our energy eigenstates simultaneously with those momenta, \( \ket{E} \rightarrow \ket{E,p_x,p_z} \). Acting on such an eigenstate, the Hamiltonian takes the form
\[ \begin{aligned} \hat{H} = \frac{\hat{p}_y^2}{2m} + \frac{{p}_z^2}{2m} + \frac{1}{2m} \left( {p}_x + \frac{e}{c} B \hat{y} \right)^2. \end{aligned} \]
As far as the \( \hat{y} \) and \( \hat{p}_y \) variables are concerned, this is nothing more than a simple harmonic oscillator:
\[ \begin{aligned} \hat{H} = \frac{\hat{p}_y^2}{2m} + \frac{1}{2} m \omega_c^2 (\hat{y} - y_0)^2 + \frac{p_z^2}{2m} \end{aligned} \]
where \( \omega_c \) happens to be the cyclotron frequency \( eB/mc \). The energy eigenstates are thus
\[ \begin{aligned} E_n(p_z) = \hbar \omega_c \left( n+\frac{1}{2} \right) + \frac{p_z^2}{2m}, \end{aligned} \]
i.e. harmonic oscillator states plus a free-particle contribution in the \( z \)-direction. This is the quantum version of orbiting around the magnetic field lines.
Now, as I said there are many possible vector potential choices here; another example is the choice
\[ \begin{aligned} \vec{A} = (-yB/2, xB/2, 0). \end{aligned} \]
Now the Hamiltonian is
\[ \begin{aligned} \hat{H} = \frac{1}{2m} \left[ \left( \hat{p}_x + \frac{eB}{2c} \hat{y} \right)^2 + \left( \hat{p}_y - \frac{eB}{2c} \hat{x} \right)^2 + \hat{p}_z^2 \right] \end{aligned} \]
This is somewhat more complicated to deal with than the previous example. We can rewrite things using the kinematical momenta:
\[ \begin{aligned} \hat{\Pi}_x = \hat{p}_x + \frac{eB}{2c} \hat{y} \\ \hat{\Pi}_y = \hat{p}_y - \frac{eB}{2c} \hat{x}. \end{aligned} \]
We know their commutation relation already:
\[ \begin{aligned} [\hat{\Pi}_x, \hat{\Pi}_y] = \frac{i\hbar e}{c} B_z = i \hbar m \omega_c. \end{aligned} \]
This looks almost like the relationship between coordinate and momentum; if we define the variable
\[ \begin{aligned} \hat{Q} = -\hat{\Pi}_y / \hbar \omega_c \end{aligned} \]
then \( [\hat{Q}, \hat{\Pi}_x] = i \hbar \), and the Hamiltonian can be rewritten as
\[ \begin{aligned} \hat{H} = \frac{\hat{p}_z^2}{2m} + \frac{\hat{\Pi}_x^2}{2m} + \frac{1}{2} m \omega_c^2 \hat{Q}^2. \end{aligned} \]
Once again, we find free-particle motion in the \( z \)-direction, added to a harmonic oscillator at the cyclotron frequency. So clearly, some important things (like the energy spectrum) are invariant under redefinition of \( \vec{A} \).
However, notice that for the first gauge choice, \( \hat{p}_x \) was a constant of the motion, but in the second example it is not! The expectation value \( \ev{\hat{p}_x} \) will evidently not be equal in our two scenarios. The short answer to this confusion is that the canonical momentum \( \hat{p} \) is not, in fact, a gauge-invariant quantity; only the kinematical momentum \( \hat{\Pi} \) is actually physical.
This immediately begs the question, why is \( \vec{p} - e\vec{A} / c \) the "real" operator here? Moreover, why does dealing with the electromagnetic field force us to redefine our fundamental operators, when changes to a scalar potential were so trivial?
The key is in the difference between a global phase transformation, and one that depends on position. In general both the scalar and vector potential can be position-dependent, but suppose we have a gauge transformation of the form
\[ \begin{aligned} \phi \rightarrow \phi,\\ \vec{A} \rightarrow \vec{A} + \nabla \Lambda(\vec{x}). \end{aligned} \]
The object \( \nabla \Lambda(\vec{x}) \) which modifies the vector potential is a function, which does not have to be uniform in space. To have invariance under this transformation, we are in fact asking for invariance under local gauge transformations, i.e. under a phase shift which can vary in space,
\[ \begin{aligned} \psi(x,t) \rightarrow \psi'(x,t) = e^{i \theta(x)} \psi(x,t). \end{aligned} \]
(Note: following a discussion in class, I should clarify this point. As far as I know, it's not obvious that this particular transformation rule for \( \psi(x) \) follows from the gauge transformation as written. Instead, consider the gauge transformation as motivation for the fact that something \( \vec{x} \)-dependent has to happen to \( \psi(x) \). We'll work backwards and show that \( e^{i\theta(x)} \) does correspond to exactly the vector potential gauge transformation written above.)
If we ignore the dynamics of the wavefunction for a moment, then such a transformation looks like it isn't a problem; transition probabilities between two states are given by
\[ \begin{aligned} |\sprod{\phi'}{\psi'}|^2 = |\bra{\phi} e^{-i\theta(x)} e^{i\theta(x)} \ket{\psi}|^2 = |\sprod{\phi}{\psi}|^2. \end{aligned} \]
However, the Schrödinger equation is a different story. To leave the physics invariant, we must demand that if we perform a gauge transfomation at time \( t_0 \), then at a later time \( t \) all physical observables will be unchanged. After a rotation, the Schrödinger equation becomes
\[ \begin{aligned} i \hbar e^{i\theta({x})} \frac{\partial \psi}{\partial t} = \left[ \frac{\hat{p}^2}{2m} + V(\hat{x}) \right] e^{i \theta({x})} \psi(x,t). \end{aligned} \]
The potential energy term will let us simply pull the \( e^{i\theta(x)} \) through, so that part of the Schrödinger equation looks the same as the original equation for the untransformed wavefunction. The momentum term, however, gives us something extra:
\[ \begin{aligned} \hat{p} e^{i \theta({x})} \psi(x,t) = -i\hbar \frac{\partial}{\partial x} \left( e^{i\theta(x)} \psi(x,t) \right) \\ = e^{i \theta(x)} \left( -i\hbar \frac{\partial \psi}{\partial x} + \hbar \frac{\partial \theta}{\partial x} \psi(x,t) \right), \end{aligned} \]
and applying \( \hat{p} \) again will just give us more \( \theta \)-dependent terms, which will alter the Schrödinger equation: we can't cancel the phase factor \( e^{i\theta(x)} \).
The fundamental problem here is most easily seen in position space, where momentum acts as a derivative. I'm going to be dealing with lots of functions of \( \hat{x} \), so I'll suppress the hats for now. Remember that the derivative can be written in terms of the difference between two infinitesmally-separated points:
\[ \begin{aligned} \frac{\partial \psi}{\partial x} = \lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon} \left[ \psi(x+\epsilon) - \psi(x) \right]. \end{aligned} \]
This makes it obvious that we can't expect our original Hamiltonian to be invariant under local transformations; we're asking for symmetry under an arbitrary phase rotation \( e^{i\theta(x)} \) at every point in space, and here we have an expression containing two terms which transform with different phases.
The only way out of this puzzle is to introduce a new operator of some kind, which will also transform in some way under the local gauge rotation and compensate for the phase difference between nearby points. The simplest thing we can do is define an object called the comparator, which transforms in the following way:
\[ \begin{aligned} {U}(x_1, x_2) \rightarrow e^{i \theta(x_1)} {U}(x_1, x_2) e^{-i \theta(x_2)}. \end{aligned} \]
This ensures that the object \( {U}(x_1, x_2) \psi(x_2) \) has the same phase as \( \psi(x_1) \). (If you've studied general relativity, the comparator is an example of a parallel transporter, an object that tells us how to shift from one point to another in order to make comparisons.) To make this symmetric under exchange of \( x_1 \) with \( x_2 \), we require \( \hat{U} \) to be unitary. We can now construct a new sort of derivative, the covariant derivative, which will rotate uniformly with the gauge transformation:
\[ \begin{aligned} D_x \psi(x) \equiv \lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon} \left[ \psi(x+\epsilon) - {U}(x+\epsilon, x) \psi(x) \right]. \end{aligned} \]
Since we have a unitary operator with a small parameter \( \epsilon \), we can write it as a series expansion in terms of another operator:
\[ \begin{aligned} \hat{U}(x+\epsilon, x) = 1 + i\frac{e}{\hbar c} \epsilon \hat{A}_x(x) + \mathcal{O}(\epsilon^2) \end{aligned} \]
The function \( {A} \), which appears in the expansion of the comparator, is sometimes called a connection. The constant \( e \) is arbitrary, but you can of course already see that \( e \) will be the electric charge, and \( A_x \) is the vector potential (in the \( x \) direction.)
To see how the connection operator transforms under a local phase rotation, we substitute this infinitesmal version of \( {U} \) back into the definition above:
\[ \begin{aligned} {U}(x+\epsilon, x) \rightarrow e^{i \theta(x+\epsilon)} {U}(x+\epsilon,x) e^{-i\theta(x)} \\ = e^{i \theta(x+\epsilon)} \left( 1 + \frac{ie}{\hbar c} \epsilon {A}_x(x) + ... \right) e^{-i \theta(x)}. \end{aligned} \]
Once again, we're interested in the limit as \( \epsilon \rightarrow 0 \). We can Taylor expand out the exponential on the left,
\[ \begin{aligned} e^{i \theta(x+\epsilon)} = e^{i \theta(x)} + \epsilon \theta'(x) e^{i \theta(x)} + ... \end{aligned} \]
which leads to
\[ \begin{aligned} {U}(x+\epsilon,x) \rightarrow 1 + \frac{ie}{\hbar c} \epsilon {A}_x(x) + \epsilon \theta'(x) + \mathcal{O}(\epsilon^2). \end{aligned} \]
Since the first two terms are the original comparator \( {U} \), we conclude that the transformation must give us the last term, i.e. we must have
\[ \begin{aligned} A_x \rightarrow A_x + \frac{\hbar c}{e} \theta'(x). \end{aligned} \]
Once again, you know that this is the gauge transformation rule of the vector potential; in three dimensions we have the gradient of the phase. Since we know that the vector potential is supposed to transform as
\[ \begin{aligned} \vec{A} \rightarrow \vec{A} + \nabla \Lambda(x), \end{aligned} \]
we can identify the relationship between the function \( \Lambda \) and the phase \( \theta \) by which the wavefunction transforms as
\[ \begin{aligned} \theta(x) = \frac{e\Lambda(x)}{\hbar c}. \end{aligned} \]
Finally, in terms of the connection, we can rewrite the covariant derivative without the limit:
\[ \begin{aligned} D_x \psi(x) = \left[ \frac{\partial}{\partial x} - i\frac{e}{\hbar c}A_x(x) \right] \psi(x) \end{aligned} \]
or
\[ \begin{aligned} -i\hbar D_x \psi(x) = \left[ -i \hbar \frac{\partial}{\partial x} - \frac{e}{c} {A}_x(\hat{x}) \right] \ = \hat{p}_x - \frac{e}{c} {A}_x(\hat{x}) = \hat{\Pi}_x. \end{aligned} \]
Thus, the kinematical momentum can be thought of as the momentum associated with the covariant derivative, i.e.
\[ \begin{aligned} \hat{\Pi}_i = -i\hbar D_i. \end{aligned} \]
The kinematical momentum is gauge invariant by construction, but we can also see it by explicit calculation: if we require the Schrödinger equation to be invariant under gauge transformations
\[ \begin{aligned} \ket{\psi} \rightarrow \tilde{\ket{\psi}} = e^{i \theta(x)} \ket{\psi}, \end{aligned} \]
then we find
\[ \begin{aligned} i \hbar \frac{\partial}{\partial t} \tilde{\ket{\psi}} = \hat{H} \tilde{\ket{\psi}} \\ i \hbar \frac{\partial}{\partial t} e^{i \theta(x)} \ket{\psi} = \hat{H} e^{i \theta(x)} \ket{\psi} \\ i \hbar \frac{\partial}{\partial t} \ket{\psi} = \hat{H} \ket{\psi} + [\hat{H}, e^{i \theta(x)}] \ket{\psi} \end{aligned} \]
(note that the operator \( \hat{H} \) did not transform: we're taking a "Schrödinger-picture-like" approach to gauge transformations, where only the state vectors are rotated.) So the gauge invariance of the Schrödinger equation requires that this commutator vanish, i.e. we must have
\[ \begin{aligned} \hat{H} e^{i \theta(x)} = e^{i \theta(x)} \hat{H} \Rightarrow \hat{H} = e^{-i \theta(x)} \hat{H} e^{i \theta(x)}. \end{aligned} \]
Obviously, the potential part of the Hamiltonian satisfies this relation, since functions of \( \hat{x} \) always commute. For the momentum, we see that
\[ \begin{aligned} e^{-i \theta(x)} \hat{\vec{p}} e^{i \theta(x)} = e^{-i \theta(x)} \left[ \hat{\vec{p}}, e^{i \theta(x)} \right] + \hat{\vec{p}} \\ = - e^{-i e \Lambda(x) / \hbar c} (-i \hbar \nabla) e^{i e \Lambda(x) / \hbar c} + \hat{\vec{p}} \\ = \hat{\vec{p}} + \frac{e}{c} \nabla \Lambda, \end{aligned} \]
where I've used the identity we derived awhile back that commutating \( \hat{p} \) with a function of \( \hat{x} \) gives us the derivative of the function. Thus, the combination \( \hat{\vec{\Pi}} = \hat{\vec{p}} - e \hat{\vec{A}} / c \) will transform into itself and give us a vanishing commutator; the kinematical momentum \( \hat{\vec{\Pi}} \) is gauge invariant, and hence must be the operator which appears in the Hamiltonian, rather than \( \hat{\vec{p}} \) alone. We have been forced to introduce a new operator, the vector potential \( \vec{A} \), to construct a gauge-invariant Hamiltonian.
This is a profound insight, although it might at this point seem like a somewhat circular argument: the gauge symmetry of the vector potential requires a symmetry under a spatially-dependent field redefinition, which in turn requires the existence of a vector potential, and so on. But the existence of a local gauge invariance in the quantum theory is the more fundamental observation; the presence of such a simple-looking symmetry leads inevitably to the existence of a vector potential \( \vec{A} \), transforming in the right way so that the only gauge-invariant combinations we can construct are the electromagnetic fields, \( \vec{E} \) and \( \vec{B} \).
Stated in a different way, the electromagnetic interaction is a direct consequence of the existence of a local gauge symmetry. In fact, electromagnetism is an example of a "gauge force", and what's more, all of the forces we know the quantum description for (i.e. not gravity) arise fundamentally from gauge symmetries. (You might ask how we can define a more general gauge symmetry, since the local dependence on phase looks like the most general possibility. The nuclear forces come from non-Abelian gauge symmetries, in which the comparator depends not just on the distance between two points, but on the path which we take to get from \( A \) to \( B \). This is starting to push into the realm of quantum field theory; we won't study non-Abelian interactions in this course.)