# (global) Gauge invariance and field theory

1. Jan 9, 2009

### jensa

Hi everyone,
This is my first post and I hope to get some better understanding of something that has been bugging me.

I understand (global) gauge invariance in the sense that $$|\psi\rangle$$ denotes the same (physical) state as $$e^{i\varphi}|\psi\rangle$$, or more generally, the physical state is a ray representation of the state $$|\psi\rangle$$. The choice of $$\varphi$$ to represent the physical state is arbitrary and therefore any theory should be invariant under changes of this phase - gauge invariance. We call a global gauge-transformation the $$U(1)$$ rotation $$|\psi\rangle\rightarrow e^{i\varphi}|\psi\rangle$$ on all states $$|\psi\rangle$$. By definition gauge-transformations do not transform physical states (the ray representation is invariant)

In many-particle physics a "gauge-transformation" is usually represented by the $$U(1)$$ rotation on the field operators $$\hat{\psi}\rightarrow e^{-i\varphi}\hat{\psi}$$ (I intentionally suppress coordinate dependence because I don't want to go into local invariance, yet). Now for a system of fixed number of particles this transformation is equivalent to the notion of a gauge transformation that I discussed above if we define:
$$|\psi\rangle \equiv \hat{\psi}^\dag|0\rangle$$
However, as soon as the number of particles is not fixed, i.e. suppose we have a state $$\alpha|0\rangle+\beta|\psi\rangle$$, this operation does no longer correspond to a global phase rotation. Indeed, the transformation corresponds to a change in the relative phase so it does change the physical state, right?

It seems to me that what people call gauge-symmetry/gauge-invariance in field theory (referring to $$\hat{\psi}\rightarrow e^{i\varphi}\hat{\psi}$$) is really just the symmetry of particle conservation.

My question then is this, on what basis do we promote this symmetry to a local gauge theory if we do not have conservation of particle number? I was under the impression that it is essential for any gauge theory that the symmetry corresponds to a redundant description of our states, although I must say I am rather ignorant in this field.
Could someone enlighten me?

2. Jan 10, 2009

### olgranpappy

For one thing, what is written directly above is not such a useful or standard definition. Indeed, since the RHS depends on space, the "ket" should depend on space, which is usually something we don't want our kets to do.

Also, one usually likes to differentiate between psi's that label wavefunctions and psi's that are field operators. E.g., with upper and lower case. I think the relation between operators and wavefunctions you are looking for is

$$|\Psi\rangle=\int dx_1\ldots dx_N \Psi_N(x_1,\ldots,x_N)\hat\psi^\dagger(x_N)\ldots\hat\psi^\dagger(x_1)|0\rangle\;,$$
where $\Psi_N(x_1,\ldots,x_N)$ is a usual N-particle wavefunction, and
where $|0\rangle[\itex] is the vacuum. Also, welcome to Physics Forums. Have fun. 3. Jan 10, 2009 ### strangerep Personally, I'm uncomfortable with that way of expressing it (widespread though it is). I prefer to think of it that physical measurements cannot distinguish between $$|\psi\rangle$$ and $$e^{i\varphi}|\psi\rangle$$ -- in the sense that a "measurement" corresponds to a projection operator $$|\psi\rangle \langle\psi|$$ which is obviously unchanged if you multiply on the left by $$e^{i\varphi}$$ and on the right by $$e^{-i\varphi}$$. Again, I'd prefer to say that gauge transformation do not affect physical results, (which is not quite the same thing). If they're field "operators", then the appropriate transformation is $$\hat{\psi}\rightarrow e^{-i\varphi}\hat{\psi} e^{i\varphi}$$ Not quite. Using my corrected transformation rule above, you also have to assume that the vacuum is invariant, i.e., $$e^{-i\varphi}|0\rangle = |0\rangle$$. More generally, the "ray representation" concept is not entirely compatible with the vector space structure of Hilbert space, hence one tries to work with unit-normalized states as much as possible to avoid such messiness. But a similar issue arises in a 1-particle Hilbert space. If $$\Psi,\Phi$$ are two states, then $$\lambda\Psi, \mu\Phi$$ should correspond to the same "physical" states, but $$\lambda\Psi + \mu\Phi$$ is in general not a multiple of $$\Psi + \Phi$$. So I think the problem is more about the incompatibilities of ray representations with vector space structure, and not really about particle conservation (which, in the way you've described it, is just one manifestation of the former). I hope that helps, though I have no idea whether it was "enlightening". 4. Jan 10, 2009 ### Hurkyl Staff Emeritus Speaking just from an algebraic viewpoint (i.e. I don't know the relevant physics).... Just because you constructed the fock space out of tensor products and direct sums, that doesn't mean that when considering them with a U(1) action that you have to take the corresponding actions.... if you want a more natural way of tweaking things, you can include a "twist" to your spaces. If we let T denote the one-dimensional space C with the obvious multiplication action of U(1), and if H is your one-particle Hilbert space, then the U(1) action on [itex]H \otimes H \otimes T^*$ is given by

$$|\psi\rangle \otimes | \psi' \rangle \otimes 1 \mapsto (e^{-i \phi} |\psi\rangle \otimes (e^{-i \phi} | \psi' \rangle ) \otimes ( e^{i \phi} 1 ) = e^{-i \phi} |\psi\rangle \otimes | \psi' \rangle \otimes 1$$

And you can construct the Fock space as

$$\bigoplus_{n = 0}^{+\infty} T \otimes (H \otimes T^*)^{\otimes n}$$

The underlying Hilbert space is naturally isomorphic to the Fock space (because the underlying space of T is just C, and C is the identity object for the tensor product), but its U(1) action is simply multiplication.

5. Jan 10, 2009

### olgranpappy

...Continuing on with my post (sorry for splitting the response into two parts)...

To "gauge" a symmetry means to make it depend on space and time. For example, you said
that
$$\hat\psi(x)\to e^{i\alpha}\hat\psi(x)\;,$$
which alpha independent of space is a symmetry. Certainly, when you let alpha depend on space the transformation
$$\hat\psi(x)\to e^{i\alpha(x)}\hat\psi(x)\;,$$
is no longer a symmetry of your original theory.

But! if you introduce some auxilliary fields (usually called A(x) or something, I.e., change your original theory), and if you then let the A(x) transform too, you can make your theory invarient under the "gauged" transformation. That's why the A(x) fields are usually called gauge fields. E.g., the vector and scalar potentials of EM.

For one thing, many theories (e.g., QED) do NOT conserve particle number. E.g., positron electron pair creation. This is okay. Whether or not your theory conserves particle number can be seen by trying to commute the particle number operator with the hamiltonian. Since you have not told us what hamiltonian you are considering explicitly then I can't say for sure whether your theory conserves particle number.

6. Jan 11, 2009

### jensa

Thanks everyone for your responses. I will try to address them each starting with olgranpappy:

Well I see I probably wasn't very clear in my post. I simply wanted to reduce the problem into its simplest form in order not to cause confusion, apparently it had the opposite effect :)

I really did not want to go into local gauge invariance just yet as my problem with this already appears at the global level. Therefore I am considering a second quantized operator $$\hat{\psi}$$ which produces a particle with the single particle state $$|\psi\rangle$$. Let's even say this state is the only possible state in our system and that the particle is a fermion so that the single particle state can only be occupied or not occupied. In that case our Hilbert space is spanned by the two basis states $$|0\rangle$$ and $$|\psi\rangle$$).

For definiteness and for the sake of following more standard notation, let us say this single only possible state is a state with a plane wave wave-function $$\langle x|k\rangle\sim e^{i k x}$$. The second quantized operator is denoted $$\hat{c}^\dag_k$$ and "creates" the single particle state $$|k\rangle\equiv c^\dag_k|0\rangle$$ out of the vacuum state $$|0\rangle$$.

Now from my original post. I would say that gauge invariance is the arbitrariness in choosing (normalized) basis states, either we take $$|0\rangle,|k\rangle$$ as a basis or we choose $$e^{i\varphi}|0\rangle,e^{i\varphi}|k\rangle$$.

The two different bases correspond to the same physical bases (meaning we could never see any difference between the two in any physical experiment).

I would like to go from this concept of (global) gauge invariance to what in field theory is considered (global) gauge invariance which denotes the invariance under transformations $$\hat{c}^\dag_k\rightarrow e^{i\varphi }\hat{c}^\dag_k$$. My main observation is that this transformation is not the same as the one where both $$|0\rangle$$ and $$|k\rangle$$ is transformed as described above.

Second, and probably most important: It seems like it IS possible to see the effect of a "gauge-transformation" of the form $$\hat{c}^\dag_k\rightarrow e^{i\varphi }\hat{c}^\dag_k$$, in some sort of interference experiment, if you do not have a fixed number of particles, (e.g. a state like $$\alpha|0\rangle+\beta|k\rangle$$ since it corresponds to a change in the relative phase. I do not understand in what sense this transformation corresponds to a gauge transformation if it would be physically observable?

Yes I am aware of this procedure of imposing a local symmetry, although I have always found this quite un-motivated. The explanation for doing this usually is just "because it works". Does someone have a good explanation of why we SHOULD promote a global symmetry to local one?

Ok, again I apologize for not being clearer. Yes it is possible to have terms in your Hamiltonian which do not conserve particle number and still is invariant under global U(1) rotations. In this case it is rather charge that is conserved, right? In these cases you would have to repeat my argument and replace particle number by charge.
In other words, my question would be on what grounds do we promote the global symmetry to a local one if it simply corresponds to symmetry of conserved charge? Is there a fundamental reason? I ask because there exist many examples where the notion of fibre bundles, covariant derivatives, field curvature pop up in standard QM from the arbitrariness of choosing a certain phase for the basis of Hilbert spaces (See for example Berry phase). So it seems to me that the motivation for the promotion of a global symmetry to a local one should be related to this arbitrariness and the fact that the global symmetry is not in fact a physical symmetry but rather a redundancy of our description.

7. Jan 11, 2009

### jensa

Fair enough, I agree that this way of putting it is more appealing in many ways.

Well, Of course you are right. This is kind of my point! Any operator, $$\hat{A}=\sum_{nm}A_{nm}|\psi_n\rangle\langle\psi_m|$$ expressed through the basis elements $$\{|\psi_n|\rangle\}$$. is invariant under U(1) rotations $$e^{i\varphi}|\psi_n\rangle$$ since they transform as $$\hat{A}\rightarrow e^{-i\varphi}\hat{A}e^{i\varphi}=\hat{A}$$. In other words, if "gauge-transformations" in field theory were defined in this way then everything would be consistent. The problem is that they aren't! The standard way to denote a gauge transformation in field theory is to transform the operators as $$\hat{\psi}\rightarrow e^{i\varphi}\hat{\psi}$$, and it is not consistent with the phase rotation of the states (which I think is what you are saying).
Consider the problem from my last post. In terms of the basis elements $$|0\rangle,|k\rangle$$ we have $$c^\dag_k=|k\rangle\langle 0|$$ the transformation rule $$\hat{c}^\dag_k\mapsto e^{i\varphi}\hat{c}^\dag_k$$ corresponds to transforming the state $$|0\rangle$$ differently from the state $$|k\rangle$$ by a relative phase $$\varphi$$.

This will not be apparent if we restrict ourselves to operators which do not change the number of particles (i.e. of the type $$c_k^\dag c_k$$, which are the only ones with non-zero expectation values in systems with a definite number of particles.

So I am more or less stating that the two notions of gauge invariance are not quite compatible, which becomes clear if you consider systems where the particle number is not fixed.

Yes, but I think it is essential that we are talking about GLOBAL phase rotations (or just multiplication by complex number $$\lambda$$) so that two states $$\Psi,\Phi$$ are transformed in the same way $$\Psi + \Phi\mapsto \lambda\Psi + \lambda\Phi=\lambda(\Psi+\Phi), \ \lambda\in \mathbb{C}$$:

8. Jan 11, 2009

### jensa

This is a very interesting answer! Let me see if I understand you correctly:

First of all the transformation $$\hat{c}_k^\dag\mapsto e^{i\varphi}\hat{c}_k^\dag$$ can be identified with the action of U(1) on the single-particle Hilbert space $$H$$. The problem then is that different subspaces $$H^{\otimes n}$$ and $$H^{\otimes m}$$ transform differently (states in these subspaces obtain different phase factors $$e^{in\varphi}$$ vs $$e^{im\varphi}$$). The action of the U(1) rotation on the many-particle Hilbert space is therefore not trivial (i.e. not just overall phase factor) and in particular it is not at all obvious that the transformed state is the same physical state as the original one (or more appealingly we can say that it is not obvious that we could not physically distinguish between the original state and the transformed state), so in what sense is this a gauge-transformation?

Your answer to this, as far as I understand, is that we can make a Fock-space construction for which the U(1) action does indeed only correspond to a global phase rotation. You say that this Fock space is isomorphic to the underlying Hilbert space so that really we can identify the transformation $$\hat{c}^\dag_k\rightarrow e^{i\varphi}\hat{c}^\dag_k$$ (which corresponds to phase rotations on single particle Hilbert space H) with the global phase rotations $$\in U(1)$$ of states in the many-particle Hilbert space.

This would be the answer to my question, although I feel I need to let this sink in to really understand it. Do you happen to have any reference to where this sort of thing is discussed?

Thank you very much Hurkyl!

9. Jan 11, 2009

### olgranpappy

yup.

Have you ever heard of such a thing as "super selection rules"? These rules forbids forming certain linear combinations of states. E.g., the sum of an angular momentum 1/2 state and an angular momentum zero state. This is forbidden because then an overall rotation by 2\pi would not just be an overall phase factor, but rather actually change the state, which is not allowed. These rules are in addition to the usual rules of quantum mechanics (that's why they are called super), but apparently quite necessary. I think Weinberg talk about this in his field theory books... probably other better references too, but I can't think of any off the top of my head. Maybe google.

On the other hand, it is not unheard of to form useful states that differ in the number of particles, for example, the BCS ground state.

10. Jan 12, 2009

### jensa

Yes this seems to be the correct explanation. I googled a bit and found several interesting sources. Indeed there is assumed to exist a superselection rule for charge which gives rise to exactly the U(1) gauge group. I think a short summary is that within the full algebra that can be constructed from the annihilation and creation operators (called field algebra apparently) there exist a sub-algebra of all the elements that you can form that are hermitian (and thus correspond to observables). This sub-algebra is called the observable algebra. The requirement of Hermiticity actually implies only operators which conserve the number of particles. The superselection rule, as I understand it, is then put on top stating that there cannot exist superpositions of states with different number of particles (or charge) because they are not eigenstates of any observables. Weinberg also says something about the fact that the issue of superselection rules can not be determined by symmetry arguments alone... so quite possibly it can only be determined by testing.

Yes. To be honest this is one of the reasons why I have taken an interest in all of this. The BCS ground-state is exactly one of those states which are superpositions of different number of particles and therefore "gauge-transformations" seem to transform the ground state to a different ground state. Sometimes in literature this is even used to identify the space of degenerate ground states like one usually does within the ordinary theory of spontaneously broken symmetry, where this space would be associated with the goldstone modes. However, we are told that because the symmetry is a gauge symmetry the would-be-goldstone-boson can be absorbed into the gauge-field through the Higgs mechanism and the gauge bosons then acquire a mass (yielding meissner effect etc)

So does the BCS ground-state violate charge superselection. Is "gauge-invariance" really broken in superconductors? These are the kind of questions I want to understand the answer to. Perhaps the violation of the superselection rule is a consequence of us working in the grand canonical ensemble so that the number of particles is conserved only on the average?

Anyway thank you very much olgranpappy for pointing me in the right direction. I will try to read up on this stuff and hopefully I will get to a point where I at least think I understand it:)

Last edited: Jan 12, 2009
11. Jan 12, 2009

### atyy

12. Jan 12, 2009

### jensa

Thank you atyy for the link. I actually read that paper some months ago at a time when I also had a spike of interest in this. Unfortunately I did not feel completely satisfied by it. I remember feeling like the author spent more time pointing out misconceptions about the issue rather than presenting a coherent exposition of how it should be treated. But I really should read the paper again, maybe I was just too lazy/stupid to understand it. I am the kind of person who really needs people to spell it out for me:)

Anyway, even though I don't understand the details yet, I wanted to share what I found so far regarding the charge super-selection rule and its relation to the BCS ground state in case someone else is interested and reads this thread.

In a paper from 1952 by Wick, Wightman and Wigner (Phys. Rev. 88, 101) the charge selection rule was presented at the level of a conjecture. I don't know if its status has been elevated since but I have seen papers which propose set-ups to experimentally test this rule so I don't think so.

Starting with a paper by Aharonov and Susskind from 1967 (Phys. Rev. 155, 1428) it became apparent that it is possible to have coherent superposition of different charge states for subsystems without violating the charge superselection rule if the total system still preserves total charge conservation.

The BCS ground state thus does not violate the super-selection rule as long as we consider the superconductor to be part of a larger system with which it can exchange particles (which we implicitly do when we are working in the grand canonical ensemble). Sine the E_M gauge-invariance is a consequence of the charge super-selection rule the BCS ground state does not violate gauge-invariance either. One article discussing this can be found here: http://arxiv.org/abs/cond-mat/0102429

I would like to thank all of the nice people at the forum who have helped me with this. In particular olgranpappy for pointing me to super-selection rules.

Last edited: Jan 12, 2009
13. Jan 12, 2009

### Hurkyl

Staff Emeritus
A question for my own edification.... I understand from C*-algebra theory that if I have a situation such as:

1. A 2-D Hilbert space spanned by |-1> and |1>
2. An operator Z satisfying Z|n> = n|n>
3. Every observable is a function of Z

Then it turns out that the state space is the interval [-1, 1]. Equivalently, every state is a mixture of the pure states described by |-1> and |1>. (So the ket |-1> + |1> actually refers to the statistical mixture of 50% |-1> and 50% |1>)

This phenomenon is exactly what 'superselection' refers to, correct?

14. Jan 12, 2009

### jensa

Like I said, I don't quite understand the details of this "super-selection", but from what I understand it refers exactly to what you describe.

15. Jan 12, 2009

### olgranpappy

You're welcome. Thank you for putting together a list of interesting papers. Cheers.

16. Jul 17, 2010

### slbnju

I would like to recommend the book of Wen's for this problem. In
the chapter of interacting boson system of "Quantum field theory
of many body system" by X. G. Wen, he gave a viewpoint from the
fluctuation of order parameter. In fact, the nonzero order
parameter leads to some degenerate Goldstone modes. To the
classical solutions of GL equation, we restrict the fluctuation
order parameter. The fluctuation of Cooper pairs cost no energy,
that is, it costs no energy to add or remove a pair. However, the
quantum fluctuation lifts the degeneracy and the system will
possess U(1) symmetry again. Any way, if excitation costs less
energy of no interest under consideration compared to the gap, we
can still say this system breaks U(1) symmetry. If the system has
a definite phase, it cannot fluctuate to another values during a
finite time period so the system is U(1) breaking.