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In QM, the "charge" observable corresponds to representations in which the group transformation advances the phase of the state. Similarly, time evolution via the Schrodinger equation also advances the phase of the state. Is there any significance to this? Or are these two different phases?

As a concrete example, take a representation with 2 charge irreducible sub-representations with charges q and 2q. Assume the state is an energy eigenstate. Time evolution will rotate the phase by $Et/\hbar$. Operating with the charge transformation will rotate phases of the subrepresentations differently: by $q*\theta$ and $2q*\theta$.

I know the charge U(1) is internal, as opposed to spacetime symmetries where the group transforms the wavefunction (in the Schrodinger picture). But since it seems to behave like time-evolution, I'm wondering if there is a deeper significance.

Qmechanic
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HoosierDaddy
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2 Answers2

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How else would a one-parameter transformation look? Time is a one-parameter transformation of the physics generated by a self-adjoint operator - the parameter is time, the generator is the Hamiltonian, and this $\mathrm{U}(1)$ transformation here is also a one-parameter transformation of the physics - the parameter is the abstract angle $\theta$, the generator is the charge operator.

By Stone's theorem, there is just one way in which such transformations can act, which is as $U(t) = \mathrm{e}^{\mathrm{i}Tt}$, where $T$ is the generator and $t$ the parameter. There's actually nothing special about "time evolution" in this sense - the Schrödinger equation $$ \partial_t \psi = \mathrm{i}H\psi$$ is just an instance of the general equation for the infinitesimal form of a one-parameter transformation in the sense of Stone's theorem. See also this answer of mine for a discussion of this in the context of rotations.

ACuriousMind
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Analogously, a wave defined by $\phi(x,t)=e^{\frac{i}{\hbar}(px+Et)}$ is translated in either position or time (or a combination of both) by multiplication by the same phase.

The U(1) charge is similar to (angular) momentum in a circular internal dimension, and the U(1) global symmetry corresponds to translation (rotation) in the internal dimension.

Because the momentum in the internal dimension is fixed for each particle type, the wave can always (even if not an eigenstate of spacetime translation) be written $\phi(x,t,θ)=\phi(x,t)e^{iqθ}$ where $q$ is a constant. The $\phi(x,t)$ form is what you usually see, which makes the similarity to spatial translation less obvious. See this answer for more.

benrg
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