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In QFT we upgrade global symmetries to local symmetries and in order to keep the Lagrangian invariant we must add another gauge field. This produces the forces in the standard model. I understand the mathematical structure as described in Peskin and Schroeder and in Schwartz (I'm not up to the point of understanding connections and Fibre bundles) but I can't seem to understand the physical interpretation of an induced local symmetry. How should this be interpreted? An example like $U(1)$ would be greatly appreciated.

Edit:

Most of the answers given are just reexplanations of why we have gauge fields, which I get. I get that they work, I understand why we do them. I just don't get what they physically mean.

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8 Answers 8

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You are not really inducing a local symmetry as you are requiring it.

If you have a Lagrangian $\mathcal{L}$ of the type $(\partial^\mu \phi)^\dagger\partial_\mu \phi + \phi^\dagger \phi$, you trivially have a global $U(1)$ symmetry, meaning you can shift the field $\phi$ by a constant phase $\varphi$ across all of space without changing $\mathcal{L}$: $\phi \rightarrow \underbrace{g}_{\in \,U(1)}\phi = \mathrm{e}^{\mathrm{i}\varphi}\phi.$

The question now is "but what if the phase $\varphi$ were a function of space $\varphi(x)$ and not a constant over all space"?
The reason you want to take this step, essentially, is because it works. I.e. it gives you the strong force and electroweak (for particle physics, + other examples in other fields) forces as we know them and as compatible with what found in experiments. The $\varphi(x)$ will now result in an extra term $\propto \partial_\mu \varphi$ in the transformed Lagrangian which you then "cancel" by adding a gauge field coupling $A_\mu$.

Because "it works", you can come up with convincing stories as to why it should work. You could claim that any experiment always ever samples a finite size of space, so you can never truly confirm a global symmetry.

The key thing, however, is that local symmetries are not real symmetries. They are regarded as redundancies of the theory. Much like the arbitrary phase of a quantum mechanical state is not a real symmetry of the system, but just a redundancy of the state being defined as a ray in a projective Hilbert space.
Global symmetries are real symmetries in that they give you conserved charges via Noether's theorem, and (spontaneously) breaking a global symmetry gives you gapless modes known as Goldstone bosons.

Aside

Usually you write your state/wavefunction/field in space and time $(x,t)$. Space-time is then a four-dimensional manifold. But what if I have internal symmetries that thus do not depend on $x$ and $t$? How can I mathematically account for them?
You extend your base space $B$ (specific spacetime manifold). If you want to introduce spin, then you write every point in space-time as a $2$-vector. If you want to have a global phase $U(1)$, then you attach a constant phase shift $\mathrm{e}^{\mathrm{i}\varphi}$ to all points in space. If you want to have a local phase $U(1)$, then you attach a space-dependent phase shift $\mathrm{e}^{\mathrm{i}\varphi(x)}$ to each point in space.

These "things" that you are extending your base space $B$ by are collectively referred to as a generic fibre $F$. The fibre bundle $E$ is then $B \times F$, i.e. you attach the specific fibre to each point in the base space.

The gauge covariant derivative is defined as $D_\mu = \partial_\mu - \mathrm{i}qA_\mu$, but in differential geometry and general relativity the second term (a "correction" to the flat-space $\partial_\mu$) is referred to as a connection. Because it connects different tangent spaces at different points, so as to compare vectors at different points to perform differentiation.

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  • $\begingroup$ Could you explain further what you mean by internal symmetries that don't depend on x and t? That's what I'm not getting, what does it mean for the field to be redundant, for it to have a "local" symmetry (yes I know its not strictly a symmetry but still). $\endgroup$
    – K Pomykala
    Commented Jul 11, 2020 at 21:32
  • $\begingroup$ How would you explicitly account for different spins when your wavefunction is a scale function of space and time only? $\endgroup$
    – SuperCiocia
    Commented Jul 11, 2020 at 22:21
  • $\begingroup$ *scalar function $\endgroup$
    – SuperCiocia
    Commented Jul 11, 2020 at 23:34
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Gauge symmetries aren't symmetries; they are redundancies in our description.

We don't ever need to work with a theory written in a gauge invariant form; you could always choose a gauge and call this "the Lagrangian". Gauge symmetry amounts to a bookkeeping device to track different equivalent representations of the theory. This is useful because in some representations (Coulomb gauge), the physical degrees of freedom and unitarity are manifest, while in others (Lorenz gauge) symmetries like Lorentz invariance are manifest. There is no one representation that lets us have everything we want -- otherwise perhaps we would just use that and forget about gauge invariance.

As an example, we can consider QED, without coupling to matter. The physical content of the theory is that the photon is massless and has two propagating polarization states. We can quantize the theory in Coulomb gauge ($A_0=0$), and all the relevant physics is present in this quantization. In fact we could start with the theory written in this gauge, never introduce gauge invariance at all, and get all of the correct answers to any question about physical observables. However, Lorentz invariance is not easy to see in Coulomb gauge -- for instance, the photon propagator is not a tensor. Lorentz invariance is so important that it is convenient to be able to move to a different gauge (eg, Lorenz gauge, $\partial_\mu A^\mu=0$) where Lorentz invariance is easy to see, at the cost of having unphysical modes whose effects need to cancel out of observable quantities. Ultimately, gauge symmetry enables us to systematically keep track of equivalent representations of the same physics, and some representations will be more useful for some purposes than others.

Chapter 6 of David Tong's lecture notes on QFT is a freely accessible resource which discusses these points with far more detail: https://www.damtp.cam.ac.uk/user/tong/qft.html

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    $\begingroup$ Agreed! In some sense, you can see how the locality, and then the Gauge redundancy, is somehow a consequence of Lorentz invariance (for $m=0$ vector fields). $A^\mu$ is not a good four-vector since $A^0=0$ holds in all the reference frames. Then under LT $U(\Lambda) A_{\mu}(x)U^{-1}(\Lambda)=\Lambda_\mu^\nu A_\nu(x)+\partial_\mu \alpha(x)$. It is clear that $\alpha$ must be a local function. Then even if $A^\mu$ is ill-defined as a four-vector (only in the massless case) there's no problem the $F^{\mu\nu}$ antisymmetric tensor. $\endgroup$
    – Lox
    Commented Aug 26, 2020 at 7:21
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    $\begingroup$ If I may, Weinberg "Quantum theory of fields. Vol. 1" section 5.9 gives a perfect explanation and is one of the best references, although a little bit technical. $\endgroup$
    – Lox
    Commented Aug 26, 2020 at 7:21
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I think the physical meaning of the local gauge symmetry is the Gauss law. For instance, in electromagnetism, $\nabla \cdot E = \rho$ holds at every point in space for all times. This is the same as gauge invariance because $\nabla \cdot E - \rho$ generates the local gauge symmetry (either in the sense of being a quantum operator or in classical mechanics via the Poisson bracket).

For more detail: in canonical quantization of the gauge field, $E$ and $A$ are conjugate. Thus we find

$[\nabla \cdot E(x),A(0)] = \partial_\mu [E^\mu(x),A_\nu(0)] = \partial_\nu \delta(x),$

thus $\nabla \cdot E(x)$ generates a local gauge transformation of $A$ with parameter $\delta(x)$. To obtain the proper transformation rule on charged matter, we modify $\nabla \cdot E$ to $\nabla \cdot E - \rho$. Gauge invariance of a state $|\psi\rangle$ is thus equivalent to the Gauss law $(\nabla \cdot E - \rho)|\psi\rangle = 0$.

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  • $\begingroup$ Gauss's law does not follow from or even require gauge invariance. $\endgroup$
    – my2cts
    Commented Aug 26, 2020 at 9:50
  • $\begingroup$ @my2cts I added some more details. See what you think. $\endgroup$ Commented Aug 26, 2020 at 17:03
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I think the local character comes from analyzing known phenomena. A widely studied theory like Electromagnetism has a gauge invariance in the classical formulation. We see therefore that changing gauge and having the same theory might persist going quantum, having it in the quantum regime assures us that it will persist classically. Then, we see that in order for that gauge freedom to appear we need another symmetry in the Lagrangian, local U(1). Thus the origin of local symmetries and physics is tied to a gauge freedom.

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  • $\begingroup$ That's just the origin of them, which I get. I want to know what they physically mean however. $\endgroup$
    – K Pomykala
    Commented Jul 11, 2020 at 21:27
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I am trying too to understand how this works, and I think the following argument from Mandl and Shaw "Quantum field theory" could help you, even if it probably won't answer completely your qestion as I find it still an open problem for myself. Anyway at least in the case of electromagnetism it worked well for me as there is a classical theory to wich you can refer unlike with strong or weak interactions.

The following argument is presented in Maggiore's book at page 70, and it starts from the recquirement of local phase simmetry and leads to the interaction. Instead in Mandl and Shaw's book the argument is exactly the opposite, it starts from introducing the interaction with the minimal substition like in the classical theory. (any error that is following is obviously mine and not of the authors of the books)

The Dirac lagrangian is invariant under a global phase trasnformation as we have $$L_0=\bar{\psi}(i\hbar\gamma^{\mu}\partial_{\mu}-mc)\psi$$ and the fields transform in the folllowing way $$\psi \rightarrow e^{i\theta}\psi$$ $$\bar{\psi} \rightarrow e^{-i\theta}\bar{\psi}$$

so you have a global simmetry such as the moltiplication for a global phase, and ask your self what if I want to promote this to a local simmetry? i.e. a simmetry under $$\psi \rightarrow e^{i\theta(x)}\psi$$ $$\bar{\psi} \rightarrow e^{-i\theta(x)}\bar{\psi}$$ This is not anymore a simmetry as the derivative acts in a non trivial way on the local phase, so you can ask yourself, how can I extend the derivative to be covariant under local phase transformations of the fields? In other words, we look for a derivative $D_ {\mu}$ such that transofrms in this way $$D_{\mu}\psi \rightarrow e^{i\theta(x)}D_{\mu}\psi$$ so that the moltiplication by a local phase is indeed a simmetry of the lagrangian. In this way it is enough to substitute the ordinary partial derivative with the covariant one whenever it occurs. In the end in this case the lagrangian will be the following $$L=\bar{\psi}(i\hbar\gamma^{\mu}D_{\mu}-mc)\psi$$ $$D_{\mu}=\partial_{\mu} + \frac{iq}{\hbar c}A_{\mu}$$ you then see that this recquirment leads to the presence of the interaction term $L=L_0+L_I$ $$L_0=\bar{\psi}(i\hbar\gamma^{\mu}\partial_{\mu}-mc)\psi$$ $$L_I=-\frac{q}{c}\bar{\psi}\gamma^{\mu}\psi A_{\mu}$$ that coulps the Dirac field current with the electromagnetic field. So the recquirement of the local phase invariance it seems to turn on the interaction.

In their book Mandl and Shaw follow the oopsite direction. Once you have done the free Dirac theory, you are interested in introducing the interactions and that can be done with a minimal subsitution generalised to a four vector field $A_{\mu}$, with the introduction of the covariant derivative above $$D_{\mu}=\partial_{\mu} + \frac{iq}{\hbar c}A_{\mu}$$ once you do that and get the lagrangian $L=L_0+L_I$ it is possible to show that this is not invariant under gauge transformations $$A_{\mu} \rightarrow A_{\mu} + \partial_{\mu}\theta(x)$$ as you get $$L \rightarrow L' = L + \frac{q}{c}\bar{\psi}\gamma^{\mu}\psi\partial_{\mu}\theta(x)$$ this is a problem as you want that the theory is invariant under gauge transformations, because these represents redundancies of your description of the system that we accept in order to have a manifestely lorentz covariant theory. But the phsyical degrees of freedom must be always the same, we should have the possibility of explicit the system only in terms of these ones (such as in the radiation gauge). So in order to obtain again a gauge invariant lagrangian theory we can assume that the fields transform in the following way under gauge trasformations

$$\psi \rightarrow e^{i\frac{q}{\hbar c}\theta(x)}\psi$$ $$\bar{\psi} \rightarrow e^{-i\frac{q}{\hbar c}\theta(x)}\bar{\psi}$$

and with both these and the $A_{\mu}$ transformation laws you find a gauge invariant lagrangian as the single terms transforms in this way $$L_0 \rightarrow L_0'=L_0 -\frac{q}{c}\bar{\psi}\gamma^{\mu}\psi\partial_{\mu}\theta(x)$$ $$L_I \rightarrow L_I'=L_I +\frac{q}{c}\bar{\psi}\gamma^{\mu}\psi\partial_{\mu}\theta(x)$$ and so the total lagrangian is invariant. I think in this way is easier to see the reason for "promoting" the global phase simmetry to a local one, but this argument works only in this case as you have a classical theory to which you can refer wehre you know how to introduce the interaction, and that is not the case for other theories such as QCD. For the other interactions I can't give a real answer as I am still trying to get it, but I hope this has helped you a little bit at least in the U(1) case.

Edit: I have finally found a question that I read times ago which could be useful to you. Here in the accepted answer it is showed another aspect about the importance of gauge fields and how the recquirement of the invariance under local transformations enable us to couple fields with conserved currents. I hope it can help.

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Global symmetries of the action imply conserved quantities. Examples include translation invariance implying conserved total momentum, time invariance implying conserved total energy, $U(1)$ invariance implying conserved total charge etc..

This is a nice feature to have, but it lacks something very important. We do not know where the momentum, energy, charge etc., is. We only know that whatever happens, the same amount will exist before and after.

Converting a global symmetry into a local symmetry allows us to locate the charge. This is manifest in Gauss' law in electromagnetism or its generalization in non-abelian gauge theories. Knowing the electric field along a closed surface tells us how much charge resides inside said surface.

Gauss' law is a consequence of gauge invariance, because the operator $\nabla \cdot E-\rho$ is what generates local symmetry transformations on quantum states. For physical (gauge invariant) states, this has vanishing expectation value.

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  • $\begingroup$ I think this is incorrect. Global symmetries imply distributions of conserved quantities that describe exactly where the charge, current, energy, momentum etc. is located. $\endgroup$
    – my2cts
    Commented Aug 26, 2020 at 9:47
  • $\begingroup$ Yes you can determine the location of charge for future times in a theory with only a global symmetry, but you need to already know the location of the charge initially. In the local symmetry case, you can find the charge indirectly by measuring a nonlocal quantity, the electric flux. You don't need to know anything about the system, you can just start measuring electric fluxes to narrow down the location of charges. $\endgroup$
    – fewfew4
    Commented Aug 26, 2020 at 14:04
  • $\begingroup$ Are you familiar with the Noether theorem? The Noether charge-current tells you the charge current distribution at any time and place given the wave function. $\endgroup$
    – my2cts
    Commented Aug 26, 2020 at 15:09
  • $\begingroup$ Of course I am familiar with Noethers theorem. The point I was making is that you need initial data to determine this "Noether charge-current". Initial data constitutes measuring the location and velocity of the charge. From this perspective, the theory doesn't allow you locate the charge unless you already know where it was. When there is an electric field, or local symmetry, you can find the charge at any given time without knowledge of the dynamics of the charge. $\endgroup$
    – fewfew4
    Commented Aug 26, 2020 at 15:41
  • $\begingroup$ How should I know if I don't ask? Any way, I agree that you have to solve the equation of motion first. Then the Noether theorem gives you the charge distribution etc. No local symmetry needed. $\endgroup$
    – my2cts
    Commented Aug 26, 2020 at 19:48
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Besides the fact that gauge theory works, the reason we promote global symmetries to local ones is because we value locality. It would be strange if to realize a symmetry, we need to perform some change everywhere in the universe. Relaxing this constraint leads to gauge theory.

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  • $\begingroup$ The global symmetries generate the conservation laws. The local symmetry does not impact the conservation laws in any way, so there is no point in "relaxing" the symmetry. $\endgroup$
    – my2cts
    Commented Aug 26, 2020 at 9:55
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I propose that the local, that is, gauge symmetry has no physical meaning. This is based on the fact that classical electromagnetism can be better formulated without gauge invariance. See my peer reviewed paper at https://arxiv.org/abs/physics/0106078.

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  • $\begingroup$ Have you taken into consideration how this alternative can be implemented quantum mechanically? It is widely known that gauge invariance is necessary to describe a spin $1$ massless particle in a Lorentz invariant and unitary way. See my answer physics.stackexchange.com/questions/554959/… as well as section 5.9 of Weinberg's Quantum Theory of Fields. $\endgroup$
    – fewfew4
    Commented Aug 26, 2020 at 16:43
  • $\begingroup$ For instance, in your paper you postulate that the creation operator of the photon is a scalar operator. But to properly describe a spin $1$ particle it must transform with respect to the spin $1$ representation of the little group. What you've postulated is equivalent to a spin $0$ creation operator. $\endgroup$
    – fewfew4
    Commented Aug 26, 2020 at 16:58
  • $\begingroup$ The last part of the paper, starting with equation 14, deals with this. I don't recognize your statement. I do not postulate that a scalar operator creates a photon. It requires a vector operator. $\endgroup$
    – my2cts
    Commented Aug 26, 2020 at 19:32
  • $\begingroup$ It was equation 18 I was referring to. $\endgroup$
    – fewfew4
    Commented Aug 26, 2020 at 19:41
  • $\begingroup$ @LucashWindowWasher "It is widely known that ..." Rather than that gauge invariance is required, in fact it is formally impossible to write a gauge invariant propagator. It takes a different form for each gauge. The situation is saved because in the end the predictions are the same. So gauge invariance is not making quantisation easier or possible, it is instead an obstacle that has been overcome by quantizing only non-invariant theories and showing that they all give the same result. $\endgroup$
    – my2cts
    Commented Aug 26, 2020 at 19:41

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