Recognitions:

## Spontaneously broken gauge symmetry

 Quote by DrDu Haag exclusively talks about global gauge transformations, so the EOM are deterministic.
 Quote by DrDu So Struyve also calls global U(1) transformations gauge transformations. "Operational indistinguishability" is also what Haag had in mind. He talks explicitly of gauge transformations of the 1st kind (cf. eq. 21) which is synonymous to global gauge transformations.
Struyve discusses that a global symmetry can be considered operationally distinguishable or not depending on whether the system is viewed as being the whole universe or a subsystem (p15):

"The other partial transformations do not correspond to a local transformation and can hence be viewed as changing the physical state of the subsystem. However, the observational difference only arises when comparing the fields of the subsystem with those of the environment and not from within the subsystem itself. This is analogous to the familiar view one can take on, for example, translation and rotation symmetry in classical mechanics. While translations or rotations of the whole universe are unobservable, and hence may be regarded as gauge transformations, translations or rotations of (approximately isolated) subsystems relative to their environment will yield (in principle) observable differences."

Presumably Haag must be taking the superconductor to be the whole universe so that the ground states are operationally indistinguishable? OTOH, since the global symmetry is not a gauge symmetry in the sense of producing nondeterministic EOM, if the superconductor is considered a subsytem, then the ground states are operationally distinguishable, say with respect to a second superconductor via the Josephson effect?

 Recognitions: Science Advisor I believe Greiter's Eq 11-13 are correct. The are the same definition of gauge invariance as Scholarpedia's Eq 17 & 18. There isn't any conflict with Haag's analysis, because Greiter's gauge invariance is not the same as Haag's gauge invariance. Rather Haag's gauge invariance is Greiter's global invariance (Eq 98), which is not the same as Greiter's gauge invariance (Eq 11-13).

Recognitions:
 Quote by atyy I believe Greiter's Eq 11-13 are correct. The are the same definition of gauge invariance as Scholarpedia's Eq 17 & 18. There isn't any conflict with Haag's analysis, because Greiter's gauge invariance is not the same as Haag's gauge invariance. Rather Haag's gauge invariance is Greiter's global invariance (Eq 98), which is not the same as Greiter's gauge invariance (Eq 11-13).
You (and Greiter) are mixing up state space and position representation. A state $|\phi>$ transforms into $U|\phi>$. To get the transformation in the position representation, you have to project onto $<x|$, ie $\phi(x)=<x|\phi>$ gets transformed into $\phi'(x)=<x|U|\phi>$. Greiter claims that $U|\phi>=|\phi>$, so that $\phi(x)=\phi'(x)$ (note that the function $\phi$ in eq. 13 is, in contrast to the claim by Greiter, not the wavefunction in position representation) which is at variance with Scholarpedia.

Recognitions:
 Quote by DrDu You (and Greiter) are mixing up state space and position representation. A state $|\phi>$ transforms into $U|\phi>$. To get the transformation in the position representation, you have to project onto  gets transformed into $\phi'(x)=$. Greiter claims that $U|\phi>=|\phi>$, so that $\phi(x)=\phi'(x)$ (note that the function $\phi$ in eq. 13 is, in contrast to the claim by Greiter, not the wavefunction in position representation) which is at variance with Scholarpedia.
If I understand Greiter correctly, since a change of gauge is a different description of the same physical situation, the two wave functions correspond to the same state in different gauges. This seems to match Wen's description of a gauge transformation as being by definition a "do nothing" transformation. Greiter's and Wen's definitions include "local" gauge transformations, so they're different from Haag's "global gauge" transformation.

So

$$U^{\Lambda}|\phi> = U^{\Lambda} \prod (u_{k}+v_{k}e^{i\phi}c^{\dagger}_{k \uparrow}c^{\dagger}_{-k \downarrow})|0> = \prod (u_{k}+v_{k}e^{i\phi^{\Lambda}}c^{\dagger\Lambda}_{k \uparrow}c^{\dagger\Lambda}_{-k \downarrow})|0> = |\phi^{\Lambda}>= |\phi> \text{ ,}$$

where the equality between the gauge-transformed $|\phi^{\Lambda}> \text{ and } |\phi>$ is due to
$$c^{\dagger\Lambda}_{k \uparrow} = e^{i\frac{e}{\hbar c}} c^{\dagger}_{k \uparrow} \text{ ,}$$
$$\phi^{\Lambda}=\phi-\frac{2e}{\hbar c} \Lambda \text{ .}$$

 Recognitions: Science Advisor Ok, so U^Lambda would be strictly the identity transformation in that scheme. I still don't see where this leads to. Who is Wen?

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