Why is gauge symmetry not a true symmetry?

[TimeRip496]

A symmetry of a physical system is a physical or mathematical feature of the system that is preserved or remains unchanged under some transformation. For example, the speed of light is an example of symmetry and its value will always will always remain the same no matter where and what coordinate system(e.g. cartesian, polar, etc) one use.

But as for gauge symmetry, we can only only use one coordinate though we get to choose what coordinate system we use(gauge fixing?) and we can made the transition between different coordinate system. Is this what gauge symmetry? If it is, is it why gauge symmetry is not a symmetry cause we can't preserve invariance when we move between different coordinate system?

Thanks for your help.

 

[jfizzix]

A symmetry is a transformation that leaves something invariant.

For example, when Newton's equations of motion are invariant by a translation of spatial coordinates (i.e., by changing what you call your zero coordinate point), time coordinate, or a rotation in spatial coordinates, we would say those equations have translational symmetry, time translational symmetry, and rotational symmetry.

Maxwell's equations of electromagnetism have gauge symmetry in that they are invariant under a gauge transformation.

You may be interested to know that there is a profound connection between these continuous symmetries, and conservation laws. The idea, explained by Emmy Noether's theorem is that for each continuous symmetry of the equations of motion, there is a corresponding conserved quantity.

In the ones listed here:

• space translation symmetry leads to conservation of momentum
• time translation symmetry leads to conservation of energy
• rotational symmetry leads to conservation of angular momentum
• gauge symmetry leads to conservation of electric charge (this one's harder to show)

The constancy of the speed of light is not a symmetry, but it is evidence of Lorentz symmetry (that the equations of motion in relativity are invariant under a Lorentz transformation)

 

[bsmile]

The symmetry does not only apply to coordinates. For the gauge symmetry, it applies to a generic physical quantity, which after the transformation, the action is left unchanged.

 

[TimeRip496]

A symmetry is a transformation that leaves something invariant.

For example, when Newton's equations of motion are invariant by a translation of spatial coordinates (i.e., by changing what you call your zero coordinate point), time coordinate, or a rotation in spatial coordinates, we would say those equations have translational symmetry, time translational symmetry, and rotational symmetry.

Maxwell's equations of electromagnetism have gauge symmetry in that they are invariant under a gauge transformation.

You may be interested to know that there is a profound connection between these continuous symmetries, and conservation laws. The idea, explained by Emmy Noether's theorem is that for each continuous symmetry of the equations of motion, there is a corresponding conserved quantity.

In the ones listed here:

• space translation symmetry leads to conservation of momentum
• time translation symmetry leads to conservation of energy
• rotational symmetry leads to conservation of angular momentum
• gauge symmetry leads to conservation of electric charge (this one's harder to show)

The constancy of the speed of light is not a symmetry, but it is evidence of Lorentz symmetry (that the equations of motion in relativity are invariant under a Lorentz transformation)

But is there a difference between gauge symmetry and symmetry? Or is gauge symmetry a symmetry, just like translational/rotational symmetry? Because I search online and I see people saying that gauge symmetry is not a true symmetry. This is the part which I don't understand.

 

[Mentz114]

But is there a difference between gauge symmetry and symmetry? Or is gauge symmetry a symmetry, just like translational/rotational symmetry? Because I search online and I see people saying that gauge symmetry is not a true symmetry. This is the part which I don't understand.

I assume this applies to a Lagrangian or action of some kind. The important symmetries are those that generate conserved quantities and represent physical degrees of freedom. Symmetries that do not generate conserved quantities are 'mopped-up' by adding constraints to the action - i.e. gauge fixing. I think those are the ones referred to as 'not true' symmetries. Dirac wrote an important book on this subject but I can't even recall the title right now.

 

[atyy]

One way to see that gauge symmetry is not a symmetry is that it cannot be "spontaneously" broken.

The laws of physics are invariant under translation by an arbitrary distance. Yet a crystal lattice is not invariant under translation by an arbitrary distance - it is only invariant if one translates it by the lattice spacing. Thus translational symmetry of the underlying laws of physics is "spontaneously" broken by a crystal.

A gauge symmetry is not a symmetry because it is just a way of calling the same thing by more than one name.

 

[bsmile]

Is this the way you define a "spontaneous symmetry broken"? This is quite confusing to me, can anybody explain to me further on it, or you might try to use a metaphor to give one an idea what it is?

One way to see that gauge symmetry is not a symmetry is that it cannot be "spontaneously" broken.

The laws of physics are invariant under translation by an arbitrary distance. Yet a crystal lattice is not invariant under translation by an arbitrary distance - it is only invariant if one translates it by the lattice spacing. Thus translational symmetry of the underlying laws of physics is "spontaneously" broken by a crystal.

A gauge symmetry is not a symmetry because it is just a way of calling the same thing by more than one name.

 

[stevendaryl]

A gauge symmetry is not a symmetry because it is just a way of calling the same thing by more than one name.

Hmm. Is there a mathematical or physical criterion for knowing when two descriptions of the universe are just different ways of describing the same situation?

Let's take the paradigm example of a gauge symmetry, which is the electromagnetic field. If you start with Maxwell's equations for the electric field $\vec{E}$ and the magnetic field $\vec{B}$, then you can introduce a pair of nonphysical fields $\Phi$ and $\vec{A}$, and define $\vec{E}$ and $\vec{B}$ in terms of them:

$\vec{E} = - \frac{d}{dt} \vec{A} - \nabla \phi$
$\vec{B} = \nabla \times \vec{A}$

Since $\vec{E}$ and $\vec{B}$ were the physical fields, then you can see that $\vec{A}$ and $\phi$ are not uniquely defined; the physical fields are unchanged by a replacement:

$\vec{A} \rightarrow \vec{A} + \nabla \chi$
$\phi \rightarrow \phi - \frac{d}{dt} \chi$

So the above transformation is nonphysical, since it makes no change to the physical fields.

However, you could imagine an alternate history of science in which Maxwell first discovered the laws of electromagnetism in terms of $\vec{A}$ and $\phi$, and then later discovered that his laws were invariant under the above transformations. What reason would physicists have for concluding that this was a gauge symmetry, and not a true physical symmetry?

 

[atyy]

What reason would physicists have for concluding that this was a gauge symmetry, and not a true physical symmetry?

The physical reason is that gauge symmetries that are not recognized by physicists cause mathematicians to buzz annoyingly around the physicists until the physicists cannot stand it anymore :) Then they start to use words like principal bundle :P

If the physicists had started with QED in the path integral formulation, they would need a gauge fixing condition to get the correct results, so they would be able to recognize that it is a redundancy of the description.

How about classically? Let's take the simple case of electrostatics. Maybe take neurobiology. Maybe there would be huge debates about whether the membrane potential of a neuron is really 0 mV (the Hodgkin and Huxley convention) or -70 mV (the modern convention). Maybe it would be never resolved.

 

[atyy]

Is this the way you define a "spontaneous symmetry broken"? This is quite confusing to me, can anybody explain to me further on it, or you might try to use a metaphor to give one an idea what it is?

You are at a Chinese dinner, so the table is round. Is your pair of chopsticks on your right or left? You will eat fine either way, but in any particular dinner, someone will choose one, forcing everyone else to make the same choice.

[Apparently this metaphor goes back in some version to Abdus Salam http://philsci-archive.pitt.edu/563/1/SSB.pittarchive.mss.pdf and http://www.nytimes.com/1996/11/23/world/abdus-salam-is-dead-at-70-physicist-shared-nobel-prize.html?_r=0]

 

[stevendaryl]

If the physicists had started with QED in the path integral formulation, they would need a gauge fixing condition to get the correct results, so they would be able to recognize that it is a redundancy of the description.

This is worth understanding. Without gauge-fixing, what goes wrong in QED? Presumably, there are too many degrees of freedom for the Lagrange equations of motion to give unique equations of motion for the fields?

 

[atyy]

This is worth understanding. Without gauge-fixing, what goes wrong in QED? Presumably, there are too many degrees of freedom for the Lagrange equations of motion to give unique equations of motion for the fields?

The way I have always heard it explained is that the path integral sums over all paths, so it is essentially a sort of counting. Without the gauge fixing, one overcounts.

A quick google suggests the discussion on p277 of http://eduardo.physics.illinois.edu/phys582/582-chapter9.pdf.

 

[atyy]

gauge symmetry leads to conservation of electric charge (this one's harder to show)

The rest of your post is right, but this point isn't. It is the global U(1) symmetry that leads to conservation of electric charge, not the gauge "symmetry".

*Sometimes people call the global symmetry a gauge symmetry. However in the context of this thread, the OP is distinguishing the gauge redundancy from the "physical" symmetries, so in the language of the OP, it is not the gauge symmetry that leads to charge conservation.

 

[radium]

Symmetries transform one state into another state related by some operation. A continuous symmetry comes with a conserved charge/current. You can also have discrete symmetries with no such conserved charge like parity, time reversal, etc.

A gauge transformation is a local transformation that depends on the space time position. In the standard model, it is an element of some Lie algebra depending on the theory (QED is U(1) QCD SU(3), etc). You can also have "discrete" lattice gauge theories, the simplest being a Z2 "Ising" gauge theory. This implies you have some nontrivial topological order and you get corresponding nontrivial topological excitations.

Gauge symmetry is not a real symmetry since a gauge transformation does not relate different states. It shows two states are actually the same. So if you have two states and you can access one from the other, they are actually the same state. To include both would make the Hilbert space over complete. So fixing a gauge projects onto the physical degrees of freedom. In the path integral, this chooses a gauge orbit so you are not over counting states.

Another way to see this is that initially, when you try to solve for the photon propagator, you need to add some xi term, serving as a Lagrange multiplier (some constraint) for the propagator to be invertible. That is what gauge fixing is.

In the abelian U(1) case, gauge fixing is quite easy. However, for non abelian theories if you want to pick a gauge which makes Lorentz invariance manifest, you run into problems since the path integral will give you the inverse of a determinant depending on the gauge field. That's where ghosts come in, etc.

 

[ShayanJ]

One way to see that gauge symmetry is not a symmetry is that it cannot be "spontaneously" broken.

The laws of physics are invariant under translation by an arbitrary distance. Yet a crystal lattice is not invariant under translation by an arbitrary distance - it is only invariant if one translates it by the lattice spacing. Thus translational symmetry of the underlying laws of physics is "spontaneously" broken by a crystal.

A gauge symmetry is not a symmetry because it is just a way of calling the same thing by more than one name.

I'm OK with your argument. It really makes sense to think about gauge symmetries that way, only as redundancies in our descriptions. But the strange thing here is that these non-physical symmetries are actually determining how the physics should be. The gauge fields that we add to our classical Lagrangians are the result of our demand that those Lagrangians should be gauge-invariant. This is really strange that non-physical properties are determining how physics should be!

 

[atyy]

I'm OK with your argument. It really makes sense to think about gauge symmetries that way, only as redundancies in our descriptions. But the strange thing here is that these non-physical symmetries are actually determining how the physics should be. The gauge fields that we add to our classical Lagrangians are the result of our demand that those Lagrangians should be gauge-invariant. This is really strange that non-physical properties are determining how physics should be!

The traditional "gauge principle" used to determine "derive" coupling between electric charge and electric potential is also misleading terminology, although it is what everyone uses. In fact, the gauge principle taken at face value cannot determine the coupling uniquely. Rather the "gauge principle" is more appropriately called a "minimal coupling" principle, similar to the equivalence principle of general relativity.

 

[ShayanJ]

The traditional "gauge principle" used to determine "derive" coupling between electric charge and electric potential is also misleading terminology, although it is what everyone uses. In fact, the gauge principle taken at face value cannot determine the coupling uniquely. Rather the "gauge principle" is more appropriately called a "minimal coupling" principle, similar to the equivalence principle of general relativity.

So we're just pretending that the demand of gauge invariance is determining the interaction term?

 

[stevendaryl]

So we're just pretending that the demand of gauge invariance is determining the interaction term?

The way I've heard it argued, local gauge invariance implies the existence of gauge fields of some kind, whether or not it determines the interaction term. So rather than starting with an electromagnetic field $A^\mu$ and trying to deduce its interaction with some matter field $\phi$, you start with the free-field $\phi$. Note that the equations of motion are invariant under a global phase change: $\phi \rightarrow e^{i \chi}\phi$. Now, you insist (and I'm not sure why you feel that you have the right to insist this) that the equations should also be invariant under a local phase change, where $\chi$ varies from point to point. But if $\chi$ varies, then you end up with extra terms in your Lagrangian of the form $(\phi^* \partial_\mu \phi) \partial^\mu \chi$ (and its complex-conjugate). So the invariance is spoiled unless you have another field that also changes with $\chi$. Adding a gauge field $A^\mu$ that changes with $\chi$ according to $\delta A^\mu \propto \partial^\mu \chi$ is the simplest choice, but the local gauge symmetry forces you to have some kind of interaction term.

 

[atyy]

So we're just pretending that the demand of gauge invariance is determining the interaction term?

It is the name for a programme for guessing the correct interaction https://www.physicsforums.com/threads/photons-and-qed.764368/page-2#post-4813808.

 

[ShayanJ]

Anyway, my original comment is still applicable. As gauge invariance is not a physical symmetry and is only a redundancy in our description, this seems really strange that a non-physical thing is determining the physics. For me, either gauge invariance is somehow physical or there is something really physical that can be used instead of gauge invariance.

 

[radium]

Yes, you can also think of the gauge field as arising from the condition that you want derivatives to transform the same way as the fields. As you may know from differential geometry, derivatives are locally defined objects, it does not make sense to compare two objects on different points of a manifold. however, if you have a connection, you can transport geometric objects along smooth curves on the manifold parallel to the connection. The gauge field is this connection, giving you the covariant derivative which transforms the same way as the fields.

This is a nice geometric way of thinking about it.

Gauge invariance does produce physical phenomena. They have been observed in experiments. It is why the hall coefficient is quantized (or in fractions for the FQHE but that has a lot of other stuff going on) and is the second most accurate way to measure the fine structure constant. However, the thing that all of these phenomena have in common is that while you will never measure the actual phase, you can detect differences interference from the phase (Aharonov Bohm effect), and quantization conditions from gauge invariance, which can be seen by considering the Wilson loop.

In the Aharanov Bohm effect, you do not have a field where the electrons are, but you do have a closed solenoid they travel around. The interference is there even though the electrons never see the field, the interference comes from the fact that you cannot globally set the vector potential to zero in the area around the solenoid because of stokes' theorem involving loops encircling the solenoid. So you can define two different potentials(they need to be pure gauge so the field outside the solenoid is locally zero) to be valid in different regions which must then match at the boundary of those regions. This will impose quantization condition on the difference of the two gauge potentials, however their actual values can never be measured.

 

[stevendaryl]

Anyway, my original comment is still applicable. As gauge invariance is not a physical symmetry and is only a redundancy in our description, this seems really strange that a non-physical thing is determining the physics. For me, either gauge invariance is somehow physical or there is something really physical that can be used instead of gauge invariance.

The point is that the free-field theory is not gauge-invariant, so you have to add interaction terms in order to get something gauge-invariant. You can only view a gauge symmetry as non-physical if the theory is actually gauge-invariant, after all.

It may or may not be helpful to look at the Kaluza-Klein model. That's basically just General Relativity applied to a spacetime that is 5-dimensional, rather than 4, with the extra dimension assumed to be topologically a circle. The KK model can then be re-interpreted as a 4-D theory with some additional fields. I'm not sure of the details, but it's something along the lines of interpreting $A^\mu \equiv g^{\mu 5}$ where $g$ is the metric tensor, and $\mu$ ranges over the first 4 dimensions, and the index 5 is the circle dimension. The reinterpreted theory is almost equivalent to electrodynamics in 4-D spacetime (although there is an additional scalar field, $\phi \equiv g^{55}$). In this reinterpretation of electrodynamics as GR in 5 dimensions, the electromagnetic field is interpreted as connection coefficients, and gauge transformations are interpreted as coordinate transformations (I think).

 

[vanhees71]

Of course, local gauge invariance is a symmetry principle, and it's a pretty strong one. Of course, it's true that for a theory with a local gauge theory you have non-physical field-degrees of freedom. The most simple example is electromagnetism, which is a gauge theory based on the abelian compact gauge group U(1). For the case of free abelian gauge fields you introduce 4 field-degrees of freedom (in the standard realization), a four-vector field $A_{\mu}$, but as already the analysis in terms of classical field theory (i.e., classical electrodynamics) shows, there are only two physical field-degrees of freedom.

Mathematically it's not the field $A_{\mu}$ that represents a physical situation but a factor space, i.e., these fields modulo a gauge transformation. This implies that the field equations of motion have no unique solutions but the non-uniqueness is irrelevant for the physical meaning of these fields. In the classical theory what's measurable are only the field strengths, i.e., $F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$, and you can express everything in terms of this gauge-invariant field-strength tensor. Only for practical purposes it is very often simpler to introduce the four-vector potential, which then leads to simpler equations of motion for a given physical situation together with the freedom to "fix the gauge", i.e., to invoke a constraint to fix the non-uniqueness of the solution without affecting the physical meaning of the solution, which finally is given in terms of the field-strength tensor, which is gauge independent.

For quantum field theory it's most simple to think in terms of the path integral. There you integrate over all field configurations first. For the free electromagnetic field, you end up with a vanishing functional determinant in the denominator. To fix this you have to fix the gauge and make sure that everything physical (S-matrix elements) are gauge invariant to make sense.

In the operator formalism ("canonical quantization") you immediately see that the non-fixed theory doesn't make sense, because you either break manifest Poincare invariance by fixing the gauge completely (e.g., radiation gauge for the free em. field) and quantize only the non-redundandent two transverse physical degrees of freedom. This is inconvenient when it comes to perturbation theory for the interacting case. The other way is to start with a pseudo-Hilbert space with an indefinite scalar product and fix the gauge in terms of defining physical and non-physical states. Then it turns out that you get a good Hilbert space with positive definite scalar product ("Gupta Bleuler formalism") with a unitary S-matrix.

Breaking a local gauge theory in any way explicitly (be it simply by introducing gauge-symmetry violating terms in your Lagrangian or by anomalies) leads to the destruction of the theory, because you populate unphysical degrees of freedom, violating unitarity of the S-matrix. So you must preserve gauge invariance, because otherwise the theory looses any physical sense.

Last but not least, it's clear that a local gauge symmetry cannot be spontaneously broken. The reason is that a gauge transformation on any physical state, projected again to the physical part of the Hilbert space leads back to the same physical state you started with. So, if you have a ground state and apply a gauge transformation you get still the same ground state, i.e., the ground-state energy is not degenerate and thus there is no spontaneous symmetry breaking and consequently also no massless Goldstone-Nambu modes. Still, the would-be Goldstone modes of course do not get lost, but provide the additional field-degree of freedom necessary to have massive vector rather than massless vector fields. You can fix the gauge in a way that all the would-be Goldstone bosons are absorbed into the gauge field, the socalled "Unitary gauge", because then the Lagrangian contains only physical degrees of freedom (where you have to keep in mind that the gauge fields together with the Faddeev-Popov ghosts of the gauge-fixed Lagrangian then build the physical particles, because the FP ghost fields just compensate unphysical gauge-field degrees of freedom).

The disadvantage of the unitary gauge is that the proper vertex functions are not manifestly renormalizable (provided you have a naively renormalizable theory with coupling constant only with mass dimensions $\geq 0$). Choosing another gauge (e.g., 't Hooft's class of $R_{\xi}$ gauges) lead to a manifestly renormalizable theory at the prize that you have would-be Goldstone modes left in your Feynman rules. However, they conspire together with the gauge fields and the FP ghosts that you are only left with the physical degrees of freedom, i.e., massive vector fields.

Rather than talking about "spontaneous symmetry breaking" of a local gauge theory (which is, however, common in the usual physicist's slang) one should simply call it a "Higgsed" local gauge symmetry.

 

[samalkhaiat]

Anyway, my original comment is still applicable. As gauge invariance is not a physical symmetry and is only a redundancy in our description, this seems really strange that a non-physical thing is determining the physics. For me, either gauge invariance is somehow physical or there is something really physical that can be used instead of gauge invariance.

What is more “physical” than a principle that gives birth to all known force carriers (the vector bosons)? With no extra input other than local gauge invariance, we are able to determine the form of all known interactions, i.e., demanding local gauge invariance leads to $A_{\mu}^{a}J^{\mu}_{a}$ as the only possible renormalizable and Lorentz invariant (matter-vector fields) interaction Lagrangians. Furthermore, the explicit form of the matter fields physical currents $J^{\mu}_{a}$ is determined by the very same local gauge invariance through the second Noether theorem. If this is not a “physical symmetry” (whatever that means) then we should be able to replace it. The question then is: Can we do physics without it?
My advice to you (if you like) is to ignore this “redundancy” thing and think of gauge transformations as the symmetries of constraint systems.

 

[samalkhaiat]

Symmetries transform one state into another state related by some operation.

This is neither necessary nor sufficient condition for symmetries. I really don’t understand the meaning of the term “physical symmetry”. Symmetries are mathematical transformations, on the independent and/or dependent variables of the system, which leave the action integral unchanged. This, in turn, leads (for continuous transformations) to (on-shell) conserved dynamical quantities.
Nothing in the above definition of symmetry says that the equations of motion of the system must all be independent of each other. Local gauge transformations do satisfy the above definition of symmetries and they, therefore, are the symmetries of constraint physical systems.

 

[samalkhaiat]

In the classical theory what's measurable are only the field strengths, i.e., $F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$, and you can express everything in terms of this gauge-invariant field-strength tensor.

It would be so nice if this is true. Unfortunately, the interaction Lagrangian $A_{\mu}J^{\mu}$ cannot be expressed in terms of the $F_{\mu\nu}$.

 

[ShayanJ]

What is more “physical” than a principle that gives birth to all known force carriers (the vector bosons)? With no extra input other than local gauge invariance, we are able to determine the form of all known interactions, i.e., demanding local gauge invariance leads to $A_{\mu}^{a}J^{\mu}_{a}$ as the only possible renormalizable and Lorentz invariant (matter-vector fields) interaction Lagrangians. Furthermore, the explicit form of the matter fields physical currents $J^{\mu}_{a}$ is determined by the very same local gauge invariance through the second Noether theorem. If this is not a “physical symmetry” (whatever that means) then we should be able to replace it. The question then is: Can we do physics without it?
My advice to you (if you like) is to ignore this “redundancy” thing and think of gauge transformations as the symmetries of constraint systems.

This seems better.
Is this paper a good explanation of what you say?

 

[samalkhaiat]

This seems better.
Is this paper a good explanation of what you say?

I don't really know the paper. But, my answer is yes, provided that they give proper answer to the questions they posed:

How many gauge transformations (with independent gauge parameters) are there for a given action?
What is the structure of the gauge generators (how many time derivatives they contain) for a
given action? What is the structure of an arbitrary symmetry of the action of a singular theory?
Is there a constructive procedure to find all the gauge transformations for a given action?
How can one relate the constraint structure in the Hamiltonian formulation with the symmetry
structure of the Lagrangian action?

I must say though, that understanding the second Noether theorem is the key point in here.

 

[vanhees71]

It would be so nice if this is true. Unfortunately, the interaction Lagrangian $A_{\mu}J^{\mu}$ cannot be expressed in terms of the $F_{\mu\nu}$.

That's true, but in the classical theory all you need are the equations of motion, and these depend only on the field strengths $F_{\mu \nu}$ or $(\vec{E},\vec{B})$ in the 1+3-formalism.

 

[stevendaryl]

That's true, but in the classical theory all you need are the equations of motion, and these depend only on the field strengths $F_{\mu \nu}$ or $(\vec{E},\vec{B})$ in the 1+3-formalism.

The equations of motion don't involve potentials, but the Lagrangian does. I guess there is a sense in which the Lagrangian is less physical than the equations of motion, because the lagrangian isn't unique.

 

[vanhees71]

That's true, but in quantum theory the field strength tensor is also not enough (at least not within local QFT). An example is the Aharonov-Bohm effect, where a non-integrable phase $\propto \oint_{\partial F} \mathrm{d} \vec{r} \cdot \vec{A} = \int_{F} \mathrm{d}^2 \vec{f} \cdot \vec{B}$ leads to observable effects. Of course, although anything observable is gauge invariant, and that's the case here too, because the phase is gauge invariant when written in the second form as the magnetic flux, but in this form it looks non-local, because it integrates over a surface, where the particles don't feel any force (in a classical picture), while the first way shows that the interaction is local in the sense of QFT, because it's along an arbitrary trajectory around the "classically forbidden region".

Last but not least the analysis of the representation theory of the Poincare group a la Wigner (1939) leads to the conclusion that massless vector fields must be represented as gauge fields, and the photon is to very high accuracy known to be massless. In this sense it is quite natural to describe electromagnetism as a gauge theory.

 

[ShayanJ]

Last but not least the analysis of the representation theory of the Poincare group a la Wigner (1939) leads to the conclusion that massless vector fields must be represented as gauge fields, and the photon is to very high accuracy known to be massless. In this sense it is quite natural to describe electromagnetism as a gauge theory.

As I understand it, we know that a massless spin 1 particle has two degrees of freedom but a massless vector field has three, so we need gauge invariance to eliminate one degree of freedom. It seems to me its a natural thing to suggest that we should look for a mathematical structure that by itself accommodates only two degrees of freedom and also behaves like a vector field in the ways we want, but now that we don't have such a thing, we use vector fields. Maybe gauge invariance is needed because of this "lack of technology". But if there is no such structure in mathematics and it doesn't make sense to look for such a thing, then maybe you should think that this "lack of technology" and the need for gauge invariance have some physical meaning, as Sam suggested.

 

[vanhees71]

No, it's really the math of the Poincare group. In short the argument goes as follows: You start with the representation of space-time translations, which leads to eigenvectors of energy and momentum. Now you want irreducible representations (defining what's an elementary particle). This leads you to look for the Casimir operators of the Poincare group. With energy and momentum you can build $p_{\mu} p^{\mu}=m^2$, where $m$ is the (invariant) mass of the particle.

Now since the representation is irreducible, you can use the energy-momentum states at fixed mass, and further you can define a standard momentum $K$ with $K^2=m^2$ and all other momentum eigenstates posssible must be reachable by application of the unitary operations on the eigenvectors of energy-momementum eigenvalues $K$. All the basis vectors at fixed standard momentum $K$ must be reachable with POincare transformations either, because we have an irreducible representation. In other words these vectors build a representation of the subgroup which leave the standard momentum unchanged.

Now take the two physically relevant cases $m^2>0$ and $m=0$ ($m^2<0$, socalled tachyons have trouble, and as far as I know there are no sensible models with interacting tachyons known, let alone that no tachyon has been ever seen in experiment). For $m>0$, you can choose the standard momentum $K=(m,0,0,0)$. Then the subgroup of the Lorentz group which leaves this momentum invariant is the SO(3) acting on the spatial part of the momentum, and since $\vec{K}=0$, these rotations all leave $K$ unchanged. Now you only need to find the irreps. of SO(3) (or its covering group SU(2)), which leads to particles with spin quantum number $s \in \{0,1/2,1,3/2,\ldots \}$. Then the spin-$z$ component (i.e., the angular momentum of a particle with a standard momentum, which here means it's the particle at rest) takes the $2s+1$ values $\sigma_z \in \{-s,-s+1,\ldots,s-1,s \}$. E.g., a massive vector particle belongs to the representation with $s=1$ (by definition) and thus has three spin-degrees of freedom. Particle states with non-zero momentum can be defined by applying rotation free Lorentz boosts on the states with standard momentum (defining the socalled Wigner basis and with it the full unitary irrep of the Poincare group).

What's different for $m=0$ is that the standard momentum must be light-like. The usual choice is to use a particle running in $z$ direction, i.e., the standard momentum is taken as $K=(k,0,0,k)$. Now the subgroup of the Lorentz group leaving this standard vector unchanged (the socalled "null rotations") turns out to be non-compact and equivalent to the group ISO(2) of the Euclidean plane. This group consists of translations (two degrees of freedom) and a rotation in the plane O(2). A generall irrep of this little group to the standard momentum $K$ will define an infinite dimensional Hilbert space (as in quantum theory of a particle in a plane the generators of the translations have the entire $\mathbb{R}^2$ as a spectrum), but here this implies that you'd have something like a particle with an infinite number of intrinsic quantum states like the spin of massive particles. Such a thing is also not known to exist in nature, and thus we restrict ourselves to representations, where the translations are represented trivially. Then you are only left with the O(2) rotations, i.e., in the physical space the rotations around the 3 axis. Since in the end you want to have representations of the entire Poincare group, including all rotations and boosts of the Lorentz group, the rotations around the 3-axis must also be represented by $\exp(\pm \mathrm{i} \lambda \varphi)$ with $\lambda \in \{0,\pm1/2,\pm 1,\ldots \}$. Thus all massless particles have only two polarization degrees of freedom (or one if you have a scalar particle) (here we used the helicity $\lambda$, i.e., the projection of the particles angular momentum to the direction of its momenum).

Now we want local quantum field theories, i.e., we like to work with field operators that transform like classical fields, e.g., for a vector field we want to have $$A'^{\mu}(x')={\Lambda^{\mu}}_{\nu} A^{\nu}(x),$$
where $\Lambda$ is an arbitrary proper orthochronous Lorentz transformation. This shows that for the massless case, in order to have the null rotations represented trivially, the $A^{\mu}$ must be defined modulo to gauge transformations, i.e., with $A_{\mu}$ also $A_{\mu}+\partial_{\mu} \chi$ with an arbitrary scalar field $\chi$ represents the very same field.

For details, see Appendix B of my lecture notes on QFT:

http://fias.uni-frankfurt.de/~hees/publ/kolkata.pdf

 

[samalkhaiat]

No, it's really the math of the Poincare group. In short the argument goes as follows: You start with the representation of space-time translations, which leads to eigenvectors of energy and momentum. Now you want irreducible representations (defining what's an elementary particle). This leads you to look for the Casimir operators of the Poincare group. With energy and momentum you can build $p_{\mu} p^{\mu}=m^2$, where $m$ is the (invariant) mass of the particle.

Now since the representation is irreducible, you can use the energy-momentum states at fixed mass, and further you can define a standard momentum $K$ with $K^2=m^2$ and all other momentum eigenstates posssible must be reachable by application of the unitary operations on the eigenvectors of energy-momementum eigenvalues $K$. All the basis vectors at fixed standard momentum $K$ must be reachable with POincare transformations either, because we have an irreducible representation. In other words these vectors build a representation of the subgroup which leave the standard momentum unchanged.

Now take the two physically relevant cases $m^2>0$ and $m=0$ ($m^2<0$, socalled tachyons have trouble, and as far as I know there are no sensible models with interacting tachyons known, let alone that no tachyon has been ever seen in experiment). For $m>0$, you can choose the standard momentum $K=(m,0,0,0)$. Then the subgroup of the Lorentz group which leaves this momentum invariant is the SO(3) acting on the spatial part of the momentum, and since $\vec{K}=0$, these rotations all leave $K$ unchanged. Now you only need to find the irreps. of SO(3) (or its covering group SU(2)), which leads to particles with spin quantum number $s \in \{0,1/2,1,3/2,\ldots \}$. Then the spin-$z$ component (i.e., the angular momentum of a particle with a standard momentum, which here means it's the particle at rest) takes the $2s+1$ values $\sigma_z \in \{-s,-s+1,\ldots,s-1,s \}$. E.g., a massive vector particle belongs to the representation with $s=1$ (by definition) and thus has three spin-degrees of freedom. Particle states with non-zero momentum can be defined by applying rotation free Lorentz boosts on the states with standard momentum (defining the socalled Wigner basis and with it the full unitary irrep of the Poincare group).

What's different for $m=0$ is that the standard momentum must be light-like. The usual choice is to use a particle running in $z$ direction, i.e., the standard momentum is taken as $K=(k,0,0,k)$. Now the subgroup of the Lorentz group leaving this standard vector unchanged (the socalled "null rotations") turns out to be non-compact and equivalent to the group ISO(2) of the Euclidean plane. This group consists of translations (two degrees of freedom) and a rotation in the plane O(2). A generall irrep of this little group to the standard momentum $K$ will define an infinite dimensional Hilbert space (as in quantum theory of a particle in a plane the generators of the translations have the entire $\mathbb{R}^2$ as a spectrum), but here this implies that you'd have something like a particle with an infinite number of intrinsic quantum states like the spin of massive particles. Such a thing is also not known to exist in nature, and thus we restrict ourselves to representations, where the translations are represented trivially. Then you are only left with the O(2) rotations, i.e., in the physical space the rotations around the 3 axis. Since in the end you want to have representations of the entire Poincare group, including all rotations and boosts of the Lorentz group, the rotations around the 3-axis must also be represented by $\exp(\pm \mathrm{i} \lambda \varphi)$ with $\lambda \in \{0,\pm1/2,\pm 1,\ldots \}$. Thus all massless particles have only two polarization degrees of freedom (or one if you have a scalar particle) (here we used the helicity $\lambda$, i.e., the projection of the particles angular momentum to the direction of its momenum).

Now we want local quantum field theories, i.e., we like to work with field operators that transform like classical fields, e.g., for a vector field we want to have $$A'^{\mu}(x')={\Lambda^{\mu}}_{\nu} A^{\nu}(x),$$
where $\Lambda$ is an arbitrary proper orthochronous Lorentz transformation. This shows that for the massless case, in order to have the null rotations represented trivially, the $A^{\mu}$ must be defined modulo to gauge transformations, i.e., with $A_{\mu}$ also $A_{\mu}+\partial_{\mu} \chi$ with an arbitrary scalar field $\chi$ represents the very same field.

For details, see Appendix B of my lecture notes on QFT:

http://fias.uni-frankfurt.de/~hees/publ/kolkata.pdf

Okay, the thread below translates Vanhees71's words into math. Have a look at it.

www.physicsforums.com/showthread.php?p=2223048#post2223048

 

[king vitamin]

Symmetries are mathematical transformations, on the independent and/or dependent variables of the system, which leave the action integral unchanged.

If you use this as your definition of symmetry, then gauge transformations count, but so do anomalous symmetries. So it's not a good definition for quantum systems. E.g. pure Yang Mills is classically scale invariant but it isn't a symmetry for the quantized theory.

I believe saying that gauge transformations aren't symmetries or are "redundancies" makes sense because they do not transform distinct states into each other. If you take an arbitrary state and apply a gauge transformation to it, you end up with the same physical state. Whereas global symmetries do transform between states, for example it transforms between states in an irreducible multiplet. Or as another example, maybe every energy eigenstate is in a 1D irrep of a symmetry, but a linear combination of two states in different irreps will transform to a different linear combination.