According to Wen's description if two states $|a\rangle$ and $|b\rangle$ with $\langle a|b\rangle=0$ have same physical properties, they are symmetric.

On the the other hand if we label same physical state with different labels like $|a\rangle=|b\rangle$ now this is a gauge structure.

for example if we have system with a hamiltonian, $$H=\frac{p^2}{2m}+ V\cos(2\pi x /a)$$ and with Hilbert space $$\mathcal{H}=\{|x\rangle\}$$

this system has a translational symmetry $T_a$ with $T_a|x\rangle=|x+ a\rangle$with $[H,T_a]=0$.

Now, $|x\rangle$ and $|x+a\rangle$ are two physically different states which are symmetric.

On the other hand if I study another system with $$H=\frac{p^2}{2m}+ V\cos(2\pi x /a)$$


now this system has a gauge structure, $|x\rangle$ and $|x+a\rangle$ now are same physical states. In this sense this is not asymmetry this is using different labels of same physical state.

Now my confusion is the following,

in the theory of weakly interacting non relativstic bosons,


we have $U(1)$ symmetry. This as far, as i understood, is not a gauge structure as in previous example this is a symmetry $\psi$ $\psi e^{i\phi}$ corresponds to physically different states with same properties, it is not re labeling same physical state because for certain $\mu$ there is a spontenous symmetry breaking such that, ground state will have a definite phase. Thus if this were a gauge structure there would be no breaking of it by definition. Now if we, make a integral transformation with $$\psi=\sqrt{\rho}e^{i\phi}$$ and expand the action around $\rho_0=\mu/g$ superfluid solution up to second order in $\delta\rho=\rho-\rho_0$and integrate out $\delta\rho$, we would have a $XY$ model for nambu goldstone modes,


Now we also have a $U(1)$ gauge theory which is, $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ with $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ where a transforms as

$$A_\mu\rightarrow A_\mu +\partial_\mu\phi$$ under gauge transformation. now, this theory is indeed a gauge theory $A_\mu$ and $A_\mu +\partial_\mu\phi$ are two different labels of same physical state. This is not a symmetry, this is a gauge structure. The above situation was a symmetry.

Now what I do not get is this, if I couple This $U(1)$ gauge theory to $U(1)$ symmetric bosonic system, I get

$$\mathcal{L}=\mathcal{L}(\phi,A_\mu)+ 1/2(E^2-B^2) $$

with partition function is,

$$Z=\int DAD\psi D\bar{\psi}e^{i\int d^3x\mathcal{L}}$$

In this case we say that for symmetry broken case we can choose $\psi=|\psi|$ thus we removed the gappless goldstone mode. and system became gapped thats the anderson higgs mechanism. but I just do not get it.

Because, the theory of $\phi$ does not have a gauge sturtuce, different phases are different physical states, but when we couple it to fluctuating gauge field, we treat this physical $U(1)$ symmetry as if it is a gauge structure and we gauge away the gapless nambu degrees of freedom.

In other words,

when I solve equation of motion for $\psi$ for the $$\mathcal{L}(\psi,A=0)$$ model, I find infinite number of solutions in form of $\sqrt{rho_0}e^{i\phi}$ than I say system chooses a $\phi$ spontaneously and this is spontenous symmetry breaking.

how ever if I do the same for $$\mathcal{L}(\psi,A)$$ with fluctuating $A$, I say that equation of motion has many solutions all of which describe the same physical motion thus it is just relabeling of same physical situation. Than I fix the gauge and remove the $\phi$. I don't get it, in first case, SSB, you treat $\phi$ as physical quantity*(different $\phi$ are different physical states with same physical properties), but in second case you treat it as a non physical label which can be gauged away.

I have a trouble understanding the higgs mechanism, and the difference between $U(1)$ gauge theory and $U(1)$ symmetric theory. and in higgs mechanism it looks like $U(1)$ symmetric theory becomes $U(1)$ gauge theory that's why it becomes gapped.

I probably have some kind of misconception somewhere.

  • $\begingroup$ It might be useful to consider the two topics [(i) difference between $U(1)$ symmetry and $U(1)$ gauge theory; (ii) Higgs mechanism] separately. For (i) you may use the non-relativistic, complex boson model you wrote (with repulsive interaction). You can gauge this by promoting the $U(1)$ symmetry you noted to a local symmetry: demand invariance under $\Psi (r) \mapsto e^{i \phi(r)} \Psi(r)$, which will necessitate introduction of additional terms in the Lagrangian. For (ii) it might be useful to consider the case of relativistic complex $\phi^4$ theory first in the symmetry broken state. $\endgroup$
    – vik
    Aug 14, 2019 at 3:19
  • $\begingroup$ for (i) if I enforce local u(1) invariance i would get coupling to electro magnetic field. now when i do that did the u(1) symmetry become a gauge structure. $\endgroup$
    – physshyp
    Aug 24, 2019 at 16:36

1 Answer 1


I finally figured out the answer, now the $U(1)$ gauge transformation is defined as $$U(1):A_\mu\to A_\mu+\partial_\mu f$$ where $$f(x\to\infty)\to0$$ and the gauge transformation on fields acts as

$$U(1):\psi\to e^{if}\psi$$

thus the states related to each other with the above transformations are completely the same states, they are not symmetry they are just relabeling of the same thing.

on the other hand, $U(1)$ symmetry transformation transforms the system with constant $f$. Thus, it does not full fill the $$f(x\to\infty)\to0$$ condition. And the states related to each other with constant f, are different physical states with the same properties.

That's the difference between $U(1)$ gauge transformation and symmetry transformation. Finally in Higgs mechanism, we break the invariance with constant $f$ but we do not break the local $U(1)$ gauge invariance. And we can remove the gold stone mode by choosing proper $f$ since the gold stone mode is not constant, we can find an $f$ to remove it.


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