# Is the $U(1)$ in the standard model identified with quantum-mechanical phase?

I think there's a tension between two claims I've read:

1. The standard model is Yang-Mills theory with gauge group $$SU(3) \times SU(2) \times U(1)$$. Here the $$U(1)$$ factor is data on the same level as the other two factors, and has nothing at all to do with quantum mechanics. For instance, it's present even in the classical theory.

2. I've heard it claimed that charge conservation corresponds via Noether's theorem to the gauge symmetry of quantum mechanics given by the fact that the overall phase of a quantum mechanical wavefunction can never be measured; only relative phases can be measured. For instance, this is how I interpret the discussion in Connection to gauge invariance section of the Wikipedia article on charge conservation.

From the second claim, it seems like electromagnetism has a "privileged" relationship to quantum mechanics -- its gauge group comes directly from the postulates of quantum mechanics in a way that the gauge groups of the weak and strong forces do not. This makes it puzzling to me that all three gauge groups appear to be treated on "equal footing" in the standard model. It also seems strange (or at least very ironic) that electromagnetism should have such a rich classical limit if it arises directly from the postulates of quantum mechanics.

Let me try to summarize my confusion in a few more pointed

Questions:

1. Is the $$U(1)$$ factor in the Standard Model gauge group really "identified" with the gauge symmetry of quantum mechanical phase?

2. If not, is it still somehow correct to claim that charge conservation is related to the gauge symmetry of the phase of quantum mechanical wavefunctions?

3. If so, does this somehow give electromagnetism a "privileged" role in quantum theory? For instance, is it impossible to formulate a quantum field theory which doesn't include any form of electromagnetism? And how does this work out mathematically?

• The wikipedia article is confused. The electromagnetic vector potential comes from gauging the local U(1) symmetry in whatever hamiltonian you're working with. The statement about QM phase is about a global property, ie global symmetry (no spatial dependence). Moreover, charge conservation is not even necessarily present in all QM systems or QFT, which refers to your third question. Jun 26, 2019 at 15:37
• U(1) gauge invariance involves rephasings of complex quantities and applies equally to classical and quantum objects. It has no privileged position in QM, and vice versa. I cannot follow how you ever imagined the contrary. Jun 26, 2019 at 15:50
• @Aaron Thanks, I think that answers my question. If you care to write your comment up as an answer, I'd happily accept. Jun 26, 2019 at 15:57

Is the U(1) factor in the Standard Model gauge group really "identified" with the gauge symmetry of quantum mechanical phase?

No.

The $$U(1)$$ phase that everyone talks about is the one of the action and of the Lagrangian, that can be spontaneously broken so as to reveal the massless Goldstone bosons. It is hence related to field theories. The generator of this $$U(1)$$ symmetry can be used in Noether's theorem to find the conserved charge and conserved current associated with this symmetry.

For example, a $$U(1)$$ invariant theory has a Lagrangian (and an action, by extension) of the form $$\mathcal{L} \propto (\partial_\mu \psi)^\dagger(\partial^\mu \psi)$$. You can see that under a $$U(1)$$ translation of the field operator $$\psi \rightarrow e^{\mathrm{i}\phi}\psi$$, where $$e^{\mathrm{i}\phi} \in U(1)$$, $$\mathcal{L}$$ remains invariant.
The Nother charge associated with this symmetry is particle-number conservation.

Quantum mechanics is built around the wavefunction $$\Psi$$, as opposed to (quantum) field theory which is defined from field operators $$\psi$$ that define lagrangians, actions etc.
Quantum mechanics is defined over a projective Hilbert space, in which the protagonist is a state ray. This essentially means that if $$|\psi\rangle$$ is a state that adequately describes our system, then any $$c|\psi\rangle$$ (with $$c$$ a pure phase factor) will also do just as well. This is because the key thing in quantum mechanics is to preserve the probability (the "norm") $$\langle\psi|\psi\rangle$$, because it has the physical meaning of probability.

So yeah even in quantum mechanics is just a phase factor. You could call it a $$U(1)$$ symmetry. But it cannot be spontaneously broken. There are no conservation laws associated with it.

Example: a Bose-Einstein condensate.
The phase transition from a thermal state (Boltzmann distributed) to the macroscopically occupied ground state breaks $$U(1)$$ symmetry.

In the context of quantum field theory, it's all good. The Lagrangian has a $$U(1)$$ invariance for $$T>T_c$$, while it does not have it anymore for $$T.

Within the quantum mechanical picture, the wavefunction for $$T is described by a coherent state $$|\theta\rangle$$ whereas that for $$T>T_c$$ is a number-specific eigenstate $$|n\rangle$$. The two states have an uncertainty relationship $$\Delta n \Delta \theta \geq 1$$ where $$n$$ is the number of particles and $$\theta$$ the phase of the wavefunction.
A coherent state has a specific/unique phase and a completely undetermined number of particles, i.e. it breaks the particle-number conservation entails from U(1) symmetry breaking. But even the coherent state wavefunction $$|\theta\rangle$$ is still "invariant" under the usual quantum-mechanical phase transformation $$|\theta\rangle \rightarrow |\theta\rangle e^{\mathrm{i}\phi}$$!

So the two $$U(1)$$ symmetries are not related.

If not, is it still somehow correct to claim that charge conservation is related to the gauge symmetry of the phase of quantum mechanical wavefunctions?

No.

Charge conservation is related to the breaking of a symmetry of the Lagrangian, where you derive the Noether current/charge relation.

You can never break the quantum-mechanical "gauge" symmetry, you can always multiply the wavefunction by a phase factor.

If so, does this somehow give electromagnetism a "privileged" role in quantum theory? For instance, is it impossible to formulate a quantum field theory which doesn't include any form of electromagnetism? And how does this work out mathematically?

No.

QED (Quantum ElectroDynamics) is a quantum field theory defined from a locally invariant $$U(1)$$ symmetry of the Lagrangian (/Action). Local invariance entails a coupling of the field operator $$\psi$$ to gauge field(s) $$A_i^\mu$$.

Other symmetries entail other theories. You can make up any symmetry you want (say $$SU(5)$$) and develop a theory. They might not always correspond to a physical theory found in nature.

On the other hand, $$SU(3)$$ gives eight fields that we associate with gluons and hence the strong force. $$SU(2)_L \times U(1)_Y$$ gives four particles ($$W^+, W^-, Z^0$$ and the photon) which corresponds to electro-weak theory.

How can you get a theory that does not include electromagetism?
A theory that is globally $$U(1)$$ invariant, so that it does not couple to a gauge field $$A^\mu$$.