# Gauge invariance requires gauge bosons, why not for neutral fermions?

1. Sep 9, 2012

### bcrowell

Staff Emeritus
My understanding is that for electrons, there is a standard argument that the electromagnetic interaction between them is required, not optional. Since they're identical particles, we should be able to take the wavefunction of two electrons and mix up their identities by any amount we like, and the amount of mixing can vary locally. This is a gauge symmetry. The result of the gauge transformation is that the individual wavefunctions pick up derivatives that don't match up properly with their physical energy and wavelength. To make it work, we have to take the gauge function and make it into a physical field that carries energy and momentum. So if we didn't already know about the electromagnetic field, we'd be forced to invent it.

But what about neutral fermions? Some examples would be neutrons (which I guess you could object are composites of charged fundamental fermions), neutrinos (which you could say do participate in electroweak interactions), and sterile neutrinos (what's their excuse, assuming they exist?).

It seems too good to be true that we could look at a neutral fermion like the neutron and infer that it's composite. And I don't see what fails in the original argument just because the fermion we're talking about is composite.

I dont' speak field theory fluently, so I'd appreciate any attempts to dumb down answers to the level at which I presented the question.

Last edited: Sep 9, 2012
2. Sep 9, 2012

Uhm, no.

3. Sep 9, 2012

### Chopin

To elaborate on that answer a bit more, I think your confusion is coming from the fact that your definition of gauge symmetry is not correct. A gauge symmetry does not involve mixing up the wavefunctions of separate particles, it comes from making a change to the wavefunction of a single particle which has no physically observable consequences.

The gauge symmetry of electrodynamics is present even in classical field theory, so it helps to think about it there. If we have an electric and magnetic field $E$ and $B$, we know from Maxwell's Equations that they can be represented by a potential field $(\phi, A)$, where $E=-\nabla \phi - \frac{\partial A}{\partial t}$ and $B = \nabla \times A$. The gauge invariance of these equations means that we can modify the potentials in such a way that the electric and magnetic fields are unaffected, namely $A' = A + \nabla f$ and $\phi' = \phi - \frac{\partial f}{\partial t}$, for any scalar field $f$. You can check that making this substitution leads to potentials which give the exact same fields $E$ and $B$, so this substitution leads to physically measurable quantities which are the same. Therefore, there is no "right" choice of a potential field--all are equally valid.

This should also make it clear that the gauge invariance does not require the existence of electrons, or vice versa--in this case, we're talking about the invariance only of the photon field, and are making no reference to a coupling between it and any other type of particle. This holds in quantum theory as well--you could create a free field theory of massless photons, which are gauge invariant without any connection to any other species of particle.

However, what you might be thinking about, which is somewhat related, is the U(1) symmetry of electrons. This symmetry essentially says that it doesn't matter what you call an electron and what you call a positron. You can make a linear superposition of an electron and a positron, and call this new combination an "electron", and the new field will still obey the same equations of motion. This is referred to as an "internal" symmetry, and by Noether's Theorem, it leads to a conserved quantity--in this case, the conservation of electron number (also known as electric charge).

It's possible to connect these two things, when you set up the interaction between the electron field and the photon field. Essentially, you take the global U(1) symmetry, which is always true for electrons, and use it to make a local U(1) symmetry, by coupling it to the photon field. At this point, you're correct that the photon field is required in order to preserve the symmetry--when you make a local transformation of the electron field, a corresponding change is necessary to the photon field in order to preserve the physics. In essence, you set up the interaction so that the gauge symmetry of the photon and the U(1) symmetry of the electron are tied together into one big symmetry. This is referred to in the literature as the "minimal coupling prescription".

4. Sep 9, 2012

### bcrowell

Staff Emeritus
I don't think this is right. The action of a gauge transformation is $\Psi\rightarrow\Psi e^{i\phi}$. That's not the same as mixing $e^{i\omega t}$ with $e^{-i\omega t}$.

The picture I was describing in #1 was basically this, expressed in words: http://en.wikipedia.org/wiki/Mathem...eory#An_example:_Scalar_O.28n.29_gauge_theory Actually I'd forgotten since reading that description that it was written for a scalar field. So I guess there's no reason for the word "fermions" to appear in the subject line of this thread.

Essentially it boils down to this. The WP article says: "This term introduces interactions between the n scalar fields just as a consequence of the demand for local gauge invariance." Why does this logic differ from one field to another?

Last edited: Sep 9, 2012
5. Sep 9, 2012

### Chopin

No, but it is the same as mixing $\Psi$ with $\Psi^*$, which is what the U(1) symmetry corresponds to.

Specifically, if you have the Lagrangian $\mathcal{L} = \frac{1}{2}\partial_\mu \Psi^* \partial^\mu \Psi - \frac{1}{2}\mu^2\Psi^*\Psi$, then making the substitution $\Psi \rightarrow \Psi e^{i\phi}$ (which implies $\Psi^* \rightarrow \Psi^* e^{-i\phi}$) leaves the Lagrangian unchanged.

Any spinor field will have a global U(1) symmetry, which you get by mixing together the field with its conjugate. You get that for free, without needing to introduce any other fields--it's just a property of the free-field Lagrangian for the field. But in order to introduce a local symmetry, you need to introduce a gauge field.

Last edited: Sep 9, 2012
6. Sep 9, 2012

### Chopin

Then you're postulating that $\Psi^* = \Psi$, so you don't even have a global U(1) symmetry, much less a local one.

7. Sep 10, 2012

### strangerep

Actually, the basic logic for constructing interactions by the minimal coupling prescription using a local gauge invariance principle is essentially the same in all cases.

But interactions constructed by minimal coupling under a local gauge invariance principle are quite distinct from the so-called "exchange interaction" which arises solely due to a permutation symmetry of identical particles. Cf. Ballentine ch18, in particular pp 497-498.

Last edited: Sep 10, 2012
8. Sep 10, 2012

### Ben Niehoff

I don't think there's any reason you have to have gauge symmetry. It's just something we impose as a convenient way to get interactions. But if you want, you should be able to add an explicitly gauge-symmetry-breaking term, such as a mass,

$$\mathcal{L} = -\frac{1}{4g^2} \operatorname{tr} |F^a|^2 + \frac12 m^2 \operatorname{tr} |A^a|^2.$$
Such a theory doesn't have a local gauge symmetry, but is still perfectly well defined, as far as I know.

In general, just because a global symmetry exists does not mean you have to gauge it. It just means you can gauge it, by adding a gauge field, if you wish.

The difference between Majorana and Dirac fermions is that Majorana fermions are their own antiparticles. There's nothing that forces Dirac fermions to interact with the electromagnetic field merely in order to be consistent.

9. Sep 10, 2012

### vanhees71

Massless vector fields are necessarily realized as gauge fields, if you don't want to have strange realizations of the Poincare group leading to continuous spin-like quantum numbers.

A theory of heavy vector bosons with a mass term is usually not renormalizable, and a naive mass term for sure violates all non-abelian local gauge symmetries. So here, if you want a renormalizable gauge theory with massive gauge bosons you must use the Higgs mechanism (more justly called the Anderson-Higgs-Kibble-Hagen-Guralnik-Englert-Brout mechanism ;-)).

An exception is the case of abelian gauge invariance. There you can find a paricular realization of a massive vector boson, coupling to a conserved current such that local gauge invariance stays intact with a naive vector-meson mass term. You just have to introduce another free auxiliary bosonic scalar field, the socalled Stückelberg ghost. Together with the Faddeev-Popov ghosts, this leads to the right three physical degrees of freedom for a massive vector field and a renormalizable model (the socalled Stückelberg model).

10. Sep 10, 2012

### dextercioby

You needn't have a gauge field to make spinors interact, it's enough to have a scalar field.

11. Sep 10, 2012

### atyy

As strangerep says, identical particles do not require gauge symmetry. The symmetry is the exchange symmetry, or if you consider anyons, the braid symmetry.

http://math.ucr.edu/home/baez/braids/node2.html
"The high-handed manner in which I've thrown out the symmetric group and started working with braid group statistics should disturb you, but again I can cite fancy mathematical physics papers which should allay your fears. A very nice one is Local Quantum Theory and Braid Group Statistics'' by Froehlich and Gabbiani, which gives a proof of the generalized spin-statistics theorem that holds in 2 and 3 dimensions."

12. Sep 10, 2012

### Ben Niehoff

As Vanhees points out, this theory is not renormalizable (something I forgot to consider). So it is a perfectly well-defined effective field theory. :D

Dexter mentions alternative ways to have spinors interact. Another way is via four-fermion terms (which can also run into issues with renormalizability).

At any rate, you can write down any Lorentz-invariant Lagrangian you feel like. The fact that electrons cannot be distinguished is (part of) what gives them Fermi statistics (as opposed to classical statistics). It has nothing to do with gauge symmetry.

13. Sep 10, 2012

### bcrowell

Staff Emeritus
OK, thanks, all, for the answers. Let me see if I can make some sense out of all this. Here's one quote from the thread and then four from the article, which I'll number 1 to 5:

From the article:

They introduce an operator G on identical particles, which basically mixes a state with other states in which the particles have been relabeled.

I think the answer to my original question is that [4] is optional, not mandatory.

Doesn't [1] contradict [3]? Either 1 is false, or 3 is false, or there's something I'm not understanding.

I have noticed in the past that people seem to disagree about what should be called a gauge transformation. E.g., I've seen people get very upset about the claim that diffeomorphisms are the gauge group of GR.

14. Sep 10, 2012

### Chopin

I think I understand your question now. Tell me if this answer is what you were looking for.

When you set up a Lagrangian for a charged field (either scalar or spinor), one observes that, due simply to the structure of the field itself, there is a global U(1) symmetry, that is, you can rotate the entire field around, and it maintains the physics of the Lagrangian. This happens without needing to do anything else--no other fields are necessary.

Now, if you want, you can use this global symmetry to create a local symmetry. To do this, you have to introduce an auxiliary field, which you can couple to the original field in such a way that the whole system has a local U(1) symmetry. The original field simply has its already-existing global symmetry converted into a local one, and the new field has to be of a type which has a gauge symmetry, so that it can absorb the coupling with the other field. The simplest field which has this symmetry is a massless vector field, aka the photon field.

So I guess the answer to your original question is that if you start with a neutral particle, you've already broken the global U(1) symmetry, so there's no way to perform this procedure. So a neutral particle cannot be coupled to a gauge field the way a charged particle can, since there's no global symmetry to "localize".

Last edited: Sep 10, 2012
15. Sep 10, 2012

### atyy

Also, the global U(1) symmetry is not strictly related to identical particles. The global U(1) symmetry enforces charge conservation, but presumably one can have identical particles without charge conservation. So there are 3 distinct things - identical particles (commutation or anti-commutation of the creation and annihilation operators), global symmetry (charge conservation), and gauge symmetry (redundancy of description).

16. Sep 10, 2012

### bcrowell

Staff Emeritus
Thanks, Chopin and atyy. That's basically it. However, what's still bugging me is that the wikipedia article is telling a specific story which specifically involves identical particles (the group G referred to in quote [3] in #13) and refers to the group G as a "gauge group." And yet:

I asked in #13 whether I was correct in seeing this as a contradiction, and, if so, who was right. I still don't think I'm getting a straight answer, just more statements that seem to me to contradict the WP article. This leaves me uncertain about whether the WP article is just plain wrong, or it's just a difference in what's covered by the term "gauge," or whether the apparent contradiction is only because I'm misunderstanding something. There's a physics story in the WP article. It's a story that specifically connects identical particles to gauge.

17. Sep 10, 2012

### The_Duck

bcrowell, maybe you are mixing up different notions of "identical"? What gives rise to Fermi or Bose statistics is the fact that if you switch an electron at x=1 with an electron at x=2, no one can tell the difference, because "all electrons are identical." This is what people are usually referring to when they talk about "identical particles." This is *not* a gauge symmetry.

This symmetry actually isn't immediately apparent in the Lagrangian of a quantum field theory. Quantum field theories are written down in terms of fields and not particles. The exchange symmetry results, not from the structure of the Lagrangian, but from the commutation relations of the fields, when those fields are viewed as operators on the Hilbert space of the theory. When we talk about symmetries of the Lagrangian, we aren't talking about this symmetry. (Of course, it's generally understood which fields are fermions and which are bosons when we write down a Lagrangian).

Gauge symmetries are related to a different sort of undetectable change. In QCD, for instance, I can turn all red quarks into blue quarks and all blue quarks into red quarks, and you wouldn't be able to tell the difference. Note that here we are not switching different particles, but rather changing what we call each particle. So this is *not* what people are talking about when they discuss "identical particles." This symmetry is a result of the way the Lagrangian is constructed from the fields.

18. Sep 10, 2012

### atyy

Is it this WP article http://en.wikipedia.org/wiki/Mathem...eory#An_example:_Scalar_O.28n.29_gauge_theory? I couldn't find identical particles mentioned there. I have heard the story about identical particles and gauge symmetry before, but I believe it is wrong.

So in the statement "Since they're identical particles, we should be able to take the wavefunction of two electrons and mix up their identities by any amount we like, and the amount of mixing can vary locally. This is a gauge symmetry." - the part that is right is that if local phase invariance is required, then you must couple to a gauge field - the part that is wrong is that identical particles require local phase invariance.

Actually, you can see it with just one particle in QM. For one particle, the wavefunction ψ(x) and exp(iθ)ψ(x) represent the same state, because a state is a direction in Hilbert space, and both vectors have the same direction.

If you couple one particle to a gauge field, then ψ(x) and ψ'(x)=exp(iθ(x))ψ(x) both represent the same state - provided you also make the change in the Hamiltonian from A(x) to A'(x)=A(x)-dθ(x)/dx. (Eq 17, 18 of http://www.scholarpedia.org/article/Gauge_invariance).

OTOH, for identical bosons, that means that the wave function ψ(x1,x2) stays the same under interchange of x1 and x2. In the sense that ψ(x1,x2) and ψ(x2,x1) for 2 identical bosons label the same state, that is a redundancy of description and so it is a "gauge redundancy". However, it is not the same gauge redundancy as identifying ψ(x) and ψ'(x)=exp(iθ(x))ψ(x) due to coupling a particle to a gauge field A(x).

Last edited: Sep 10, 2012
19. Sep 10, 2012

### Chopin

I'm not sure I even understand your question here. You can have a U(1) symmetry for both scalar fields (Bose statistics), and spinors (Fermi statistics). So the commutation relations of the field operators have nothing to do with this symmetry.

Additionally, the symmetry is apparent in the field Lagrangian itself, before we've even gotten around to quantizing the theory. So, "particles" don't really even exist yet. Therefore, the gauge transformation doesn't really have anything to do with exchanging individual particles, or anything like that--it's just a formal manipulation on the field itself.

20. Sep 10, 2012

### samalkhaiat

Let us examine this "convenient way"

The local gauge principle makes sense if (as necessary requirements) you have
i) a global internal (compact otherwise arbitrary) group of symmetry G, i.e. a matter field Lagrangian $\mathcal{L}(\phi_{i}, \partial \phi_{i})$ invariant under G.
ii) matter fields $\phi_{i}(x)$ transforming irreducibly under G;
$$\bar{\phi}_{i}(x) = \exp (i \alpha_{A}T^{A})_{ij} \phi_{j}(x) \ \ (1)$$
where $\vec{\alpha}$ are the arbitrary constant parameters, and $\vec{T}$ are the group generators appropriate for the matter fields.
The condition (ii) is the only requirement on the matter fields. So, being electrically charged, neutral bosons or fermions may have no relevance to the gauge group G. Of course, (i) and (ii) are not sufficient conditions. For example, we can put the proton and the neutron in the fundamental representation of the (strong) isospin group SU(2). But, the theory which results from gauging this group does not give a good description of the interaction of hadrons. Ironically, in 1960, J.J. Sakurai predicted the existence of triplet vector mesons, $J^{PC}= 1^{--}$, as the SU(2) gauge field needed to couple to isospin current of hadrons. Few years later, the triplet vector meson was discovered. So, history shows that gauging a global symmetry is GOOD even when it is WRONG.

THE GAUGE PRINCIPLE:
Once you have (i) and (ii), you then try to construct a new Lagrangian, invariant under the local version of eq(1), i.e., we simply replace the constant parameters of G by arbitrary functions of the coordinates, $\alpha_{A}(x)$. Mathematically, this produces a structure on space-time called G-bundle, i.e. to each space-time point, an independent Lie group G is assigned. This simple procedure forces us to introduce a COMPENSATING fields (i.e., predict new particles) called the gauge potentials $A_{J}(x)$.
So, what does the gauge principle buy us? Well, it is amazing how little and simple is the input and how large is the output. By simply letting $\alpha_{A}\rightarrow \alpha_{A}(x)$, a new particle(s), $A_{J}(x)$, shows up and the following questions are answered:
1) How do the gauge potentials transform under Lorentz group and under the gauge group. i.e. what is the meaning of the index $J$?
2) What is the form of (Lorentz-invariant) interaction between the matter fields $\phi_{i}(x)$ and the gauge potentials $A_{J}(x)$?
3) How can locally invariant Lagrangian be deduced from the globally invariant Lagrangian?
4) Massive vector field theories are not renormalizable, why is that?
5) What is the dynamical nature of the gauge field? i.e. what is the form of the invariant Lagrangian of the gauge fields associated with compact and non-compact gauge groups?
6) What is the explicit form of the transformation of the gauge field tensor?
7) What restriction on the gauge group allows the gauge field to couple universally to all matter fields?
8) Which type of gauge groups require more than one coupling constant?

Isn’t that amazing? You get all that by just letting the group parameters to become functions on space-time. I do not know of any other principle which allows us to answer this many questions, do you?

Sam