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Gauge symmetries do not give rise to conservation laws via Noether's theorem, and they represent redundancies in our description of the system. So why do we want to keep them after quantization? For example:

But gauge symmetries are not symmetries at all. They are redundancies in our description of the system. As such, we can't afford to lose them and it is imperative that they don't suffer an anomaly in the quantum theory.

(From David Tong's lecture notes on string theory, Chapter 5, Page 108).

If they are redundancies, then why do we want to formulate a theory with them in it? They give rise to all sorts of inconveniences, e.g. having to define a new functional integral that only integrates over physically distinct configurations.

I have seen how, by assuming a symmetry, e.g. U(1) symmetry, one can derive some beautiful results, such as the Maxwell Lagrangian, as done by Peskin and Schroeder. But why demand that there be gauge symmetries in the first place?

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  • $\begingroup$ gauge symmetries do result in conservation laws, is that a typo? $\endgroup$ – innisfree Nov 7 '13 at 16:08
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    $\begingroup$ @innisfree - A local symmetry does not imply a global symmetry. In particular, local symmetries must preserve the boundary values of the field. For example, in the case with $U(1)$, a local gauge transformation is given by a field $\alpha(x)$ that must satisfy $\lim_{x\to\infty} \alpha(x) = 0$. A global symmetry does not satisfy this condition. Thus, local and global symmetries are different. $\endgroup$ – Prahar Nov 7 '13 at 16:49
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    $\begingroup$ @innisfree - Not quite. Renormalization is somewhat relevant in the sense that it must be done so as to preserve gauge symmetries. However, if a renormalization scheme is chosen to preserve gauge symmetries, there is no guarantee that it preserves other symmetries in the theory. This is the origin of anomalies in the quantum theory. Ward Identities are the analog of Noether's theorem for quantum systems. In classical systems, global symmetries relate to conserved currents and charges. In quantum systems, global symmetries relate to Ward identities. $\endgroup$ – Prahar Nov 7 '13 at 17:01
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    $\begingroup$ @Prahar : " For example, in the case with $U(1)$, a local gauge transformation is given by a field $\alpha(x)$ that must satisfy $\lim_{x \to \infty} \alpha(x)=0$." Why exactly this constraint is needed ? $\endgroup$ – Trimok Nov 7 '13 at 18:35
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    $\begingroup$ @Trimok - This is required so as to not change the boundary conditions of the fields. Recall that a theory is defined with some fixed boundary conditions, such as $\phi(x) \to \phi_0$ as $x \to \infty$. Under a gauge transformation $\phi(x) \to \exp [i \alpha(x)] \phi(x)$. To maintain the boundary condition, we must require $\alpha(x) \to 0$ as $x \to \infty$. $\endgroup$ – Prahar Nov 7 '13 at 19:23
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Gauge symmetries, as the note says, are redundancies in our description of nature. For example, a photon has two physical degrees of freedom (the two polarizations). However, we choose to describe a photon using a 1-form field $A_\mu$ which has 4 degrees of freedom. The two extra degrees of freedom here are related to gauge symmetries. From here on, there are two important questions to be answered:

Why must gauge symmetries be preserved in the quantum theory?

One of the things that should be preserved in going from the classical to a quantum theory is the degrees of freedom of the theory. (The way the degrees of freedom behave can change, but not their number). Taking the example of the photon, the total unphysical degrees of freedom remains unchanged upon quantization (the quantum and classical photon is described by the same $A_\mu$). Thus, in order to maintain the right number of physical degrees of freedom (2 for the photon) gauge invariance must be preserved upon quantization.

If gauge symmetries are redundancies, why introduce them in the first place?

The important point here is the requirement that our theory be Lorentz invariant. Now, there are two ways to be certain that our theory (or more precisely, the $S$-matrix) is Lorentz invariant

  1. Make the action manifestly Lorentz invariant. This implies describing the theory in terms of Lorentz covariant objects such as scalars $\phi$, 1-forms $A_\mu$ or spinors $\psi$ (and other representations of the Lorentz group.)

  2. Don't make the action manifestly Lorentz invariant, i.e. formulate it in terms of fields that are not representations of the Lorentz group, but maintain overall Lorentz invariance by carefully putting together the fields in a Lorentz invariant way.

You may immediately see the advantages of the first technique. Lorentz invariance is manifest, and as long as the indices match up nicely, we never have to worry about getting funny Lorentz non-invariant answers. With the second technique, one has to check for Lorentz invariance at every step of the calculation.

The disadvantage of the first method is the following: Every physical object must be embedded in representations of the Lorentz group. Thus, if we want to describe a spin-1 photon, it must be embedded in the spin-1 representation of the Lorentz group, $A_\mu$. This leads us to a necessary introduction of gauge invariance. (since the 4 dof of $A_\mu$ has to be reduced to the 2 of the photon)

So as a summary, while gauge invariance does create some inconveniences, it allows to circumvent the even more inconvenient formulation of the theory in a Lorentz non-manifest way.

PS - In recent developments, people have tried embedding the dof of the photon into the spinor, which continues to allow for manifest Lorentz invariance, but also circumvents the problem of gauge invariances. This is called the spinor-helicity formalism.

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    $\begingroup$ Thank you for your response, it has helped me tremendously. $\endgroup$ – user32361 Nov 7 '13 at 17:56
  • $\begingroup$ Glad it helped. $\endgroup$ – Prahar Nov 7 '13 at 23:15
  • $\begingroup$ @Prahar - You did not mention the Noether theorem in your answer and from the comments under the question I can assume you would say: Gauge invariance of the Lagrangian has nothing to do with Noether theorem and conserved quantities. Is that right? $\endgroup$ – Statics Sep 23 '16 at 18:15
  • $\begingroup$ @Statics yes, that is right. $\endgroup$ – Prahar Sep 23 '16 at 18:59
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    $\begingroup$ What references are you talking about? I've only seen talk of conserved charges for global symmetries and more recently, conserved charges for local symmetries that are large. There is something called Noethers second theorem which does describe properties of systems with gauge symmetry. But these are properties of off-shell correlators. On-she'll Noethers second theorem does not say anything non-trivial I think for gauge transformations. I will check this later and confirm, so take this last part with caution. $\endgroup$ – Prahar Sep 24 '16 at 2:47
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To answer your last question, because quantizing electromagnetism in terms of $\vec{E}$ and $\vec{B}$ doesn't make Lorentz invariance manifest, and is, well, the pits. Also, it is hard to capture things like the Aharonov-Bohm effect and other topologically interesting things (especially in non-Abelian fields) if you get rid of the geometry (gauge field).

To answer your first question... why keep the gauge symmetry after quantization? Well, if it was in place before quantization, and you made use of the symmetry in any of your calculations, e.g., to derive a result that forbids a certain decay, then if you find that gauge symmetry is not preserved (an anomaly), your calculations aren't consistent. If the symmetry isn't there after quantization, then it wasn't really ever there to begin with (you just started with the offending terms in your Lagrangian having coefficients erroneously equal to zero).

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