Noether theorem, gauge symmetry and conservation of charge

Question

I'm trying to understand Noether's theorem, and it's application to gauge symmetry. Below what I've done so far.

First, the global gauge symmetry. I'm starting with the Lagragian

$$L_{1}=\partial^{\mu}\Psi\partial_{\mu}\Psi^{\ast}-m^{2}\left|\Psi\right|^{2}$$ with classical complex fields. This Lagragian is invariant with respect to the global gauge symmetry $\Psi\rightarrow\tilde{\Psi}=e^{\mathbf{i}\theta}\Psi$, ... such that I end up with $$\delta S=\int dv\left[\dfrac{\delta L_{1}}{\delta\Psi}\delta\Psi+\dfrac{\delta L_{1}}{\delta\Psi^{\ast}}\delta\Psi^{\ast}+\mathbf{i}\left(\Psi\partial^{\mu}\Psi^{\ast}-\Psi^{\ast}\partial^{\mu}\Psi\right)\partial_{\mu}\delta\theta\right]=\int dv\left[\partial_{\mu}j^{\mu}\right]\delta\theta$$ provided the equations of motion ($\delta L / \delta \Psi = 0$, ...) are valid. All along I'm using that $$\dfrac{\delta L}{\delta\phi}=\dfrac{\partial L}{\partial\phi}-\partial_{\mu}\dfrac{\partial L}{\partial\left[\partial_{\mu}\phi\right]}$$ and that $\int dv=\int d^{3}xdt$ for short. The conserved current is of course $$j_{1}^{\mu}=\mathbf{i}\left(\Psi^{\ast}\partial^{\mu}\Psi-\Psi\partial^{\mu}\Psi^{\ast}\right)$$ since $\delta S / \delta \theta =0 \Rightarrow\partial_{\mu}j_{1}^{\mu}=0$.

Here is my first question: Is this really the demonstration for conservation of charge ? Up to now, it seems to me that I only demonstrated that the particle number is conserved, there is no charge for the moment...

Then, I switch to the local gauge symmetry. I'm starting with the following Lagrangian

$$L_{2}=\left(\partial^{\mu}+\mathbf{i}qA^{\mu}\right)\Psi\left(\partial_{\mu}-\mathbf{i}qA_{\mu}\right)\Psi^{\ast} -m^{2}\left|\Psi\right|^{2} -\dfrac{F_{\mu\nu}F^{\mu\nu}}{4}$$ with $F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$. This Lagrangian is invariant with respect to the local gauge transformation $$L_{2}\left[\tilde{\Psi}=e^{\mathbf{i}q\varphi\left(x\right)}\Psi\left(x\right),\tilde{\Psi}^{\ast}=e^{-\mathbf{i}q\varphi\left(x\right)}\Psi^{\ast},\tilde{A}_{\mu}=A_{\mu}-\partial_{\mu}\varphi\right]=L_{2}\left[\Psi,\Psi^{\ast},A_{\mu}\right]$$

Then I have

$$\delta S=\int dv\left[\dfrac{\delta L_{2}}{\delta\Psi}\delta\Psi+\dfrac{\delta L_{2}}{\delta\Psi^{\ast}}\delta\Psi^{\ast}+\dfrac{\delta L_{2}}{\delta A_{\mu}}\delta A_{\mu}\right]$$ with $\delta\Psi=\mathbf{i}q\Psi\delta\varphi$, $\delta A_{\mu}=-\partial_{\mu}\delta\varphi$, ... such that I end up with $$\dfrac{\delta S}{\delta\varphi}=\int dv\left[\mathbf{i}q\Psi\dfrac{\delta L_{2}}{\delta\Psi}+c.c.+\partial_{\mu}\left[j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu}\right]\right]$$ with $j_{2}^{\mu}=\partial L_{2}/\partial A_{\mu}$ and $F^{\nu\mu}=\partial L_{2}/\partial\left[\partial_{\nu}A_{\mu}\right]$

Then, by application of the equations of motion, I have

$$\partial_{\mu}\left[j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu}\right]=0\Rightarrow\partial_{\mu}j_{2}^{\mu}=0$$ since $\partial_{\mu}\partial_{\nu}F^{\nu\mu}=0$ by construction. Of course the new current is $$j_{2}^{\mu}=\mathbf{i}q\left(\Psi^{\ast}\left(\partial^{\mu}+\mathbf{i}qA^{\mu}\right)\Psi-\Psi\left(\partial^{\mu}-\mathbf{i}qA^{\mu}\right)\Psi^{\ast}\right)$$ and is explicitly dependent on the charge. So it seems to me this one is a better candidate for the conservation of charge.

NB: As remarked in http://arxiv.org/abs/hep-th/0009058, Eq.(27) one can also suppose the Maxwell's equations to be valid ($j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu} = 0$, since they are also part of the equation of motion after all, I'll come later to this point, which sounds weird to me), and we end up with the same current, once again conserved.

Nevertheless, I still have some troubles. Indeed, if I abruptly calculate the equations of motions from the Lagrangian, I end up with (for the $A_{\mu}$ equation of motion)

$$j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu}\Rightarrow\partial_{\mu}j_{2}^{\mu}=0$$ by definition of the $F^{\mu \nu}$ tensor.

So, my other questions: Is there a better way to show the conservation of EM charge ? Is there something wrong with what I did so far ? Why the Noether theorem does not seem to give me something which are not in the equations of motions ? said differently: Why should I use the Noether machinery for something which is intrinsically implemented in the Lagrangian, and thus in the equations of motion for the independent fields ? (Is it because my Lagrangian is too simple ? Is it due to the multiple boundary terms I cancel ?)

Thanks in advance.

PS: I've the feeling that part of the answer would be in the difference between what high-energy physicists call "on-shell" and "off-shell" structure. So far, I never understood the difference. That's should be my last question today :-)

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle

Comments

Related: physics.stackexchange.com/q/48305/2451 and physics.stackexchange.com/q/26990/2451 and links therein.

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user Qmechanic

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Indeed, related, but not entirely satisfactory for me :-). Why is it widely said that the global gauge symmetry is the one responsible for charge conservation ? What about the local one ? What about the redundancy of the local symmetry in term of Noether theorem ? I've spend hours on the site today without finding a correct answer. But I would be very glad to see it of course :-)

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle

Answer 1

Comments to the question (v1):

1. Last thing first. On-shell means (in this context) that equations of motion (eom) are satisfied. Equations of motion means Euler-Lagrange equations. Off-shell means strictly speaking not on-shell, but in practice it is always used in the sense not necessarily on-shell. [Let us stress that every infinitesimal transformation is an on-shell symmetry of an action, so an on-shell symmetry is a vacuous notion. Therefore in physics, when we claim that an action has a symmetry, it is always implicitly understood that the symmetry is an off-shell symmetry.]

2. OP wrote: Here is my first question: Is this really the demonstration for conservation of (electric) charge? For that particular action: Yes. More generally for QED: No, because the $4$-gauge-potential $A_{\mu}$, the Maxwell term $F_{\mu\nu}F^{\mu\nu}$, and the minimal coupling are missing in OP's action. It is in principle not enough to only look at the matter sector. On the other hand, global gauge symmetry for the full action $S[A,\Psi]$ leads to electric charge conservation, cf. Noether's first Theorem. [Two comments to drive home the point that it is necessary to also consider the gauge sector: (i) If we were doing scalar QED (rather than ordinary QED), it is known that the Noether current $j^{\mu}$ actually depends on the $4$-gauge-potential $A_{\mu}$, so the gauge sector is important, cf. this Phys.SE post. (ii) Another issue is that if we follow OP's method and are supposed to treat the $4$-gauge potential $A_{\mu}$ as a classical background (which OP puts to zero), then presumably we should also assume Maxwell's equations $d_{\mu}F^{\mu\nu}=-j^{\nu}$. Maxwell's equations imply by themselves the continuity equation $d_{\mu}J^{\mu}=0$ even before we apply Noether's Theorems.]

3. There is no conserved quantity associated with local gauge symmetry per se, cf. Noether's second Theorem. (Its off-shell Noether identity is a triviality. See also this Phys.SE question.)

4. Perhaps a helpful comparison. It is possible to consider an EM model of the form

   $$S[A]~=~\int\! d^4x~ \left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+J^{\mu}A_{\mu}\right),$$ where $J^{\mu}$ are treated as passive non-dynamical classical background matter sources. In other words, only the gauge fields $A_{\mu}$ are dynamical variables in this model. Before we even get started, we have to ensure local (off-shell) gauge symmetry of the action $S[A]$ up to boundary terms. This implies that the classical background sources $J^{\mu}$ must satisfy the continuity equation $d_{\mu}J^{\mu}=0$ off-shell. Thus a conservation law is forced upon us even before we apply Noether's Theorems. Note that global gauge symmetry is an empty statement in this model.

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user Qmechanic

Comments

1. Thanks for this answer 2. I still do not understand why adding new elements in the Lagrangian would change the current, since all terms in the Lagrangian simply add up, as in this post: physics.stackexchange.com/q/48305. Finally, 3. I never impose $j = \partial F$ in the local gauge argument, that's what come out from the eom, contrary to what is claimed in this post : physics.stackexchange.com/q/26990.

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle

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1. You're welcome. 2. Just because your method gives the correct result it doesn't mean that your method is correct. 3. Be aware that Lubos Motl's answer operates with two different definitions of $j^{\mu}$.

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user Qmechanic

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Exactly ! And I would like to understand why :-) Is there a book where these questions are discussed ? Itzyckson and Zuber detailed too much cases for me. I thought the reference by Brading and Brown was fine (as I gave in my question). Apparently, you told me there are a lot of mistakes around there... but I did the calculation in details, so where is (are :-) the mistakes ? I've try reading Aitchison and Hey, but the Noether theorem is not really related to their presentation. Weinberg is too sloppy for me. I've also check Nakahara and Frankel (geometry and physics): only the ...

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle

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... first Noether theorem is treated. So I start scratching my head, calculate, and ask you :-) !

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle

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I just see your last edit (point 4). Thank a lot for this illuminating remark ! I was exactly considering this point when I was considering the local gauge symmetry, but I never thought that the gauge symmetry was "off-shell". Now I understand a bit better your previous remarks. Great thanks again. But, if I understand your point, it means that every symmetry is "off-shell", since it must be plug to the Lagrangian before any calculation can be hope to be correct. Am I right ? Isn't it a bit unfair, since changing the symmetry must change the Lagrangian, then change the eom, ...

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle

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... then change the "on-shell" structure, isn't it ? Or is there again something unclear for me ? Thanks again for this remark by the way. I'm progressing in QFT a bit thanks to you.

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle

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Heu, my bad... I think I may have understood this "off-shell" problem for the specific Lagrangian you gave... it's not local gauge invariant indeed, so the gauge symmetry is necessarily "off-shell" in this case. Am I right ? But would you say that for the Hamiltonian I gave in the question, the local gauge symmetry is "off-shell" ? I don't think so. Am I right ?

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle

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1. Yes, every infinitesimal transformation is an on-shell symmetry of the action, so an on-shell symmetry is a vacuous notion. 2. Well, we are only considering EM actions with local (off-shell) gauge invariance to begin with. 3. I suggest for simplicity to keep the discussion Lagrangian. Hamiltonian formulation belongs to another Phys.SE post.

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user Qmechanic

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Oups, my bad again. I'm too used to discuss only Hamiltonian that I made the mistake. I was talking about the Lagrangian I gave for the local gauge transformation.

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle

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I discussed with a friend more involved in QFT today, and he gave me the following argument about on/off-shell: when it's about the perturbation expansion, one cares about the properties (symmetries) the Lagrangian verifies and that every term in the expansion should verify too (and what you may impose by hand), and the extra requirement of conserved quantities (current) that the system intrinsically verify, since it comes from the equations of motion. The first quantities (coming e.g. from global gauge symmetry) are called off-shell, whereas the second one are called on-shell.

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle

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Would you please confirm (or correct) this statement ?

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle

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The notion of on-shell and off-shell is explained in the beginning of the answer. They are important notions in the formulation of Noether's 1st and 2nd Theorem.

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user Qmechanic

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Ok, thanks. I was wondering why is it so important in QFT to make the difference between on and off shell ?

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle

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I posted a new question about that actually, since I think a definition is not an explanation :-). It's there: physics.stackexchange.com/q/59333

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle

Answer 2

Is this really the demonstration for conservation of charge ?

Yes. The charge is defined as $Q = \int d^3x~j^0$, so $\partial_\mu j^\mu = 0$ shows that it is conserved.

Up to now, it seems to me that I only demonstrated that the probability flow is conserved, there is no charge for the moment...

What you demonstrated is that the current is conserved. I don't think you should call this a "probability flow"; it sounds like you're confusing $\Psi$ with a wavefunction, when in fact it's a quantum field.

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user Matt Reece

Comments

Oups, you're entirely right, I was confusing $\Psi$ and, say $\left| \Psi \right)$ ! I should have say: it sounds I've just prove the conservation of particle number (which I know can be called a (Noether) charge, but I'm trying to understand EM charge of course). I edit the question. Thanks for the remark. I will try to add some edits about Qmechanic remarks as well. Thanks again.

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user FraSchelle

Answer 3

The OP asked me to give an answer to this question. Well, all the questions seem to be about the "necessity" of the Noether's theorem.

So the answer is that Noether's procedure is the way to derive the current from a known symmetry. This is very useful because we usually know very well how a symmetry acts – because we know how field transform under it or how things rotate or shift under spacetime operations etc. On the other hand, the precise form of the conserved current becomes much less obvious, especially once we start to add various interactions. There's "pretty much" just one solution what the current may be to be conserved and Noether's procedure is a way to get this right form. Well, yes, the form of the current is "contained" in the Lagrangian or the equations of motion but it is not obvious how to "extract it" – and that's why we cherish Noether's procedure. If you have a different algorithm how to extract it, tell us, but I know that there can't be any different procedure that wouldn't be equivalent to Noether's one at all.

Now, back to the first example in the question.

For the non-interacting fields, the number of particles – their quanta – is conserved completely. In fact, every free field of species $s$ in every state given by a momentum $k$ and polarization $\lambda$ etc. is conserved, $N_{s,\lambda,\dots}(k,\dots)={\rm const}$. But this is clearly just a special situation when the interactions don't exist and this case isn't physically interesting.

The interesting theories only begin once we have some interactions. They destroy almost all of these "conservation laws". In particular, it is not true that the number of particles is conserved in quantum field theory. We may create electron-positron pairs out of pure energy, and so on. Only some quantities such as charges, energy/momentum, angular momentum are conserved, they are in one-to-one correspondence with the symmetries, and the corresponding currents (which includes the stress-energy tensor) may be derived via the Noether's procedure.

This post imported from StackExchange Physics at 2016-06-21 10:18 (UTC), posted by SE-user Luboš Motl