46.3 Gauss’s law and time-independent gauge transformations

46.3 Gauss’s law and time-independent gauge transformations

Concept links · terms present in this machine draft; source roles are unverified: Lie algebra · group action · U(1)

Returning to the U(1) case, a problem with the temporal gauge is that Gauss’s law (equation 46.4) is not necessarily satisfied. At the same time, the group ${\mathcal{G}}_0\subset \mathcal{G}$ of time-independent gauge transformation will act non-trivially on the phase space of initial data of Maxwell’s equations, preserving the temporal gauge condition $A_0=0$ (see equation 46.6). We will see that the condition of invariance under this group action can be used to impose Gauss’s law.

It is a standard fact from the theory of electromagnetism that

$$\nabla \cdot \mathbf{E}=\rho (\mathbf{x})$$

is the generalization of Gauss’s law to the case of a background electric charge density $\rho (\mathbf{x})$ . A failure of Gauss’s law can thus be interpreted physically as due to the inclusion of states with background electric charge, rather than just electromagnetic fields in the vacuum.

There are two diferent ways to deal with this kind of problem:

• Before quantization, impose Gauss’s law as a condition on the phase space.

• After quantization, impose Gauss’s law as a condition on the states, defining the physical state space ${\mathcal{H}}_{phys}\subset \mathcal{H}$ as the subspace of states satisfying

$$\nabla \cdot \hat{\mathbf{E}}|\psi ⟩=0$$

where $\hat{\mathbf{E}}$ is the quantized electric field.

To understand what happens if one tries to implement one of these choices, consider first a much simpler example, that of a non-relativistic particle in 3 dimensions, with a potential that does not depend on one configuration variable, say $q_3$ . The system has a symmetry under translations in the 3 direction, and the condition

$$P_3|\psi ⟩=0\text{\hspace{1em}(46.12)}$$

on states will commute with time evolution since $[P_3,H]=0$ . In the Schr¨odinger representation, since

$$P_3=-i\frac{\partial }{\partial q_3}$$

if we define ${\mathcal{H}}_{phys}$ as the subset of states satisfying 46.12, this can be identified with the space of wavefunctions of two position variables $q_1,q_2$ . One technical problem that appears at this point is that the original inner product includes an integral over the $q_3$ coordinate, which will diverge since the wavefunction will be independent of this coordinate.

If the condition $p_3=0$ is instead imposed before quantization (i.e. on the phase space coordinates) the phase space will now be five dimensional, with coordinates $q_1,q_2,q_3,p_1,p_2$ , and it will no longer have a non-degenerate symplectic form. It is clear that what we need to do to get a phase space whose quantization will have state space ${\mathcal{H}}_{phys}$ is remove the dependence on the coordinate $q_3$

In general, if we have a group $G$ acting on a phase space $M,$ , we can define:

Definition (Symplectic reduction)

Given a group $G$ acting on phase space $M_i$ preserving the Poisson bracket, with moment map

$$M\to {\mathfrak{g}}^{\ast }$$

the symplectic reduction $M/G$ is the quotient space ${\mu }^{-1}(0)/G$

We will not show this here, but under appropriate conditions the space $M/G$ will have a non-degenerate symplectic form. It can be thought of as the phase space describing the G-invariant degrees of freedom of the phase space $M$ What one would like to be true is that “quantization commutes with reduction”: quantization of $M/G$ gives a quantum system with state space ${\mathcal{H}}_{phys}$ identical to the G-invariant subspace of the state space $\mathcal{H}$ of the quantization of $M.$ Rarely are both M and $M/G$ the sort of linear phase spaces that we know how to quantize, so this should be thought of as a desirable property for schemes that allow quantization of more general symplectic manifolds.

For the case of a system invariant under translations in the 3-direction, $M={\mathbf{R}}^6,G=\mathbf{R},\mathfrak{g}=\mathbf{R})$ and the moment map takes as value (see equation 15.12) the element of ${\mathfrak{g}}^{\ast }$ given by $\mu (\mathbf{q},\mathbf{p})$ where

$$\mu (\mathbf{q},\mathbf{p})(a)=a{p}_3$$

${\mu }^{-1}(0)$ will be the subspace of phase space with $p_3=0$ . On this space the translation group acts by translating the coordinate $q_3,$ , so we can identify

$$M//\mathbf{R}={\mu }^{-1}(0)/\mathbf{R}={\mathbf{R}}^4$$

with the phase space with coordinates $q_1,q_2,p_1,p_2$ . In this case quantization will commute with reduction since imposing $P_3|\psi ⟩=0$ or quantizing $M//\mathbf{R}$ give the same space of states (in the Schr¨odinger representation, the wavefunctions of position variables $q_1,q_2)$ .

This same principle can be applied in the infinite dimensional example of the temporal gauge phase space with coordinates

$$(\mathbf{A}(\mathbf{x}),-\mathbf{E}(\mathbf{x}))$$

and an action of the group ${\mathcal{G}}_0$ of time-independent gauge transformations, with Lie algebra the functions $\phi (\mathbf{x})$ . In this case the condition $\mu =0$ will just be Gauss’s law. To see this, note that the moment map $\mu \in (Lie{\mathcal{G}}_0{)}^{\ast }$ will be given by

$$\mu (\mathbf{A}(\mathbf{x}),-\mathbf{E}(\mathbf{x}))(\phi (\mathbf{x}))={\int }_{{\mathbf{R}}^3}\phi ({\mathbf{x}}^{\prime })\nabla \cdot \mathbf{E}{d}^3{\mathbf{x}}^{\prime }$$

since

$$\begin{array}{rl}\{\mu ,\mathbf{A}(\mathbf{x})\}& =\{{\int }_{{\mathbf{R}}^3}\phi ({\mathbf{x}}^{\prime })\nabla \cdot \mathbf{E}{d}^3{\mathbf{x}}^{\prime },\mathbf{A}(\mathbf{x})\}\\ & =\{{\int }_{{\mathbf{R}}^3}(-\nabla \phi ({\mathbf{x}}^{\prime }))\mathbf{E}({\mathbf{x}}^{\prime })d^3{\mathbf{x}}^{\prime },\mathbf{A}(\mathbf{x})\}\\ & =-\nabla \phi (\mathbf{x})\end{array}$$

(using integration by parts in the first step, the Poisson bracket relations in the second). This agrees with the definition in section 15.3 of the moment map, since $\nabla \phi (\mathbf{x})$ is the infinitesimal change in $\mathbf{A}(\mathbf{x})$ for an infinitesimal gauge transformation $\phi (\mathbf{x})$ . One can similarly show that, as required since $E$ is gauge invariant, $\mu$ satisfies

$$\{\mu ,\mathbf{E}(\mathbf{x})\}=0$$

Previous section: 46.2 The Hamiltonian formalism for electromagnetic fields · Next section: 46.4 Quantization in Coulomb gauge · 中文版本

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