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I'm wondering how can one formally justify the electromagnetic response of a system which does not verify local U(1) gauge invariance.

A good example of what I would like to consider is given by the two-body interaction term discussed in relation with superconductivity as I'll elaborate below, but many examples can be found and the question is rather general.

Most of the people starts with a BCS Hamiltonian having generically the following form
$$H_{\text{BCS}}=\sum_{k,k'}\hat{c}_{k\alpha}^{\dagger}\left(\mathbf{i}\sigma_{y}\right)_{\alpha\beta}^{\dagger}\hat{c}_{-k\beta}^{\dagger}U\left(k,k'\right)\hat{c}_{k'\alpha}\left(\mathbf{i}\sigma_{y}\right)_{\alpha\beta}\hat{c}_{-k'\beta}$$

i.e. describing singlet Cooper pairing of electron with fermionic operators $\hat{c}_{k}$ in mode $k$, the greek indices being the spin ones. I think this Hamiltonian is manifestly not U$(1)$ local gauge invariant, due to the $k$ and $k'$ on different operators.

I'm wondering whether it makes sense to talk about the electrodynamic response of a superconductor when one starts with a non gauge invariant Hamiltonian. More generally, does it makes sense to discuss non-gauge invariant Hamiltonian in the context of condensed matter ? How should we understand such non-gauge-invariant Hamiltonians, $H_{\text{BCS}}$ being a simple example ?

More details:

  • The original BCS Hamiltonian has $k=k'$ and $U\left(k,k'\right) \rightarrow -g$ a constant, and so the $s$-wave interaction is both local and global U(1) invariant. $H_{\text{BCS}}$ is only U(1) symmetry global invariant, as far as I can see.
  • The Hamiltonian $H_{\text{BCS}}$ given above is particularly useful to describe some non-conventional effects ($d$-wave pairing for instance) and can be further generalise. I have not doubt about the validity of the results obtained using this Hamiltonian (some of them are even justified experimentally).
  • I'm wondering about the possibility to formally define an electromagnetic response in a non-U(1) gauge invariant theory. It is for instance clear that one can add some gauge invariant part of the above Hamiltonian, such that the constitutive Maxwell equations are preserved (no magnetic monopole and Faraday's law). But it seems also clear for me that one intrinsically imposes from the beginning some different matter-field interaction, isn't it ? Or at least that the covariant substitution is no more a correct prescription...
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I found this to be interesting, because at least pedagogically people will write down long range electronic interactions which totally break gauge invariance. I mean we've all seen someone write down a "general four point interaction": $$\int\psi(1)\psi(2)\bar{\psi}(3)\bar{\psi}(4)V(1,2,3,4)$$. This breaks gauge invariance, which is usually horrible, but frequently this doesn't seem to lead any obvious issue. Why is that? It's bothered me before.

To return specifically to your question, first let's rewrite in real space: $$\int\Delta(r)U(r,r')\bar{\Delta}(r')$$ where $\Delta = c_{\uparrow}c_{\downarrow}$ has $U(1)$ charge $2$. This is not gauge invariant unless $U$ transforms properly. Now we started with a gauge invariant system at some level, so we must have gotten to this point by "integrating out" out some charged degrees of freedom. In this simplest scenario we did second order perturbation theory and $U$ is just some correlator: $$ U(r,r';A) = \langle O(r)\bar{O}(r') \rangle_{A}$$

where $O$ has the right charge. Or it could be something more complicated, the details don't matter. I've explicitly noted this correlator must depend on the gauge field $A$. This is not surprising since $U$ measures a charge being released at $r$ and destroyed at $r'$ - regardless of the details there must be Ahranov-Bohm phases. It is this dependence of $U$ on $A$ that maintains gauge invariance, clearer below.

There is a sort of minimal coupling prescription for $U$:

$$U(r_1,r_2;A) = \int \mathcal{DP}\exp(i\!\int_{r_1}^{r_2}\!\!\!\!A(r')\cdot dr')$$

where the $\mathcal{DP}$ is some measure on the space of paths from $r_1$ to $r_2$, and this measure does not depend $A$. This is minimal in the sense of being, well, minimal and in the sense that if you expand $U$ as a polynomial in derivatives you recover the usual minimal coupling prescription. You can see that this has right gauge transformations properties. In the special case where the motion is essentially semiclassical you just a finite sum over Wilson lines.

Okay so that's the way this interaction should be written, now the real question is when can we ignore all this, since we surely don't want to estimate some path integral measure on the atomic scale when we can barely estimate single interaction energies on the atomic scale. Now violating gauge invariance leads to horrible things, I needn't remind. But clearly there is a sense where if $U(r,r')$ has a small radius of effect and if we probe with a very long wavelength then we shouldn't know that it's not a delta function, and hence we shouldn't know that its violating gauge invariance.

This is not really correct: If we follow our "minimal coupling" prescription we realize what we need is not that the distance between $r$ and $r'$ is small, but that the region explored by the bulk of paths in $\mathcal{DP}$ is small. This makes sense, since the issue comes from Ahranov Bohm phases. It doesn't help if $r$ and $r'$ are close if we get between them by travelling on circuitous paths that are enormously sensitive to the gauge fields. So the correct criterion is if the magnetic flux threading the region with paths is much smaller than a flux quantum, then we can "straighten out" all the paths in our path integral and just write it as:

$$ U(r_1,r_2; A) \approx \tilde{U}(r_1,r_2)\exp(i\int_{r_1}^{r_2}\!\!\!A(r')dr')$$

where the integral is taken over whatever path in the region we choose. Note this approximation is gauge invariant, since the error depends on the magnetic field, and its as simple a gauge invariant as thing as we get. Now in terms of linear response, at this point one could simply plug in an external field and decide whether or not it was sufficiently small to be ignored. Or one could simply make peace with the Wilson line and proceed.

If you wanted to operate more generally but still wanted to ignore that Wilson line you can do it as long as you restrict yourself to a sufficiently smooth gauge. If you are in a gauge where $A$ does not vary appreciably over the range of $U$ then $\int_{r_1}^{r_2}\!\!\!A(r')\cdot dr' \approx A\cdot(r_2 - r_1)$. And the point this is approximately pure gauge: it corresponds to the gauge transformation with field $\chi = A(r)\cdot r$. So we may essentially ignore it.

To have a smooth gauge one must have

  • Slowly varying fields
  • Small magnetic field.

which are fairly intuitive physical requirements. And then additionally we must not introduce gauge transformations which vary quickly. So one way to think of all this is that we can manipulate such apparently non-gauge invariant expressions because we've actually gauge fixed the high frequency modes of the gauge field. Probably could have thought of that without all this work, but such is life.

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Excellent answer once again, thanks a lot. Nevertheless, I'm still a bit annoyed by this answer. I was indeed thinking about the $U$ which is the only one able to correct the absence of gauge invariance. But it means you can do whatever you want in principle. It then weakens the gauge principle to my mind. Also, at the end it seems to me you essentially choose a gauge, which once again weakens (to say the least) the gauge principle. Is there absolutely no alternative ? (Except the long calculation by Nambu, keeping both the electron and phonon fields...) Thanks again for all. –  FraSchelle Jun 28 '13 at 10:19
    
Also, did I mention I really like your argument ? So I do that once more :-) Is there any discussion of this point on literature, especially about the possible breaking of the Wilson path approach you provided ? Thanks in advance. –  FraSchelle Jun 28 '13 at 13:43

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