# Noether currents for local gauge symmetry

hi everyone,
I have been trying to understand gauge theory. I am familiar with the Noether's theorem applied in the context of simpler text book cases like poincare invariant Lagrangians.

This is my question: Are there Noether currents corresponding to the local gauge symmetries too and would there be an infinity of them since there could possibly be an infinite set of linearly independent basis functions that generate the gauge symmetry?

thanks!

Bill_K
There are, in fact, TWO Noether Theorems. The first more familiar theorem applies to global symmetries, the second one to local symmetries.

Noether defines Lagrange Expressions Ei = δL/δxi, that is, the quantities which when set to zero give the Euler-Lagrange equations. The first theorem says, if the action is invariant under a symmetry group of transformations which depend on constant parameters (global symmetries), then if the equations of motion are satisfied there is a Noether current jμ obeying a continuity equation ∂μjμ = 0, one for each group parameter.

The second theorem is quite different. It says, if the action is invariant under a group of transformations which depend on arbitrary functions and their first derivatives (local symmetries), then even if the equations of motion are not satisfied, there exist identities ("Bianchi identities") connecting the Ei to each other. That is, the Ei are not independent.

For further details, see this paper.

dextercioby
Homework Helper
In gauge theory they are called <Noether identities> which for GR bear the name after Bianchi. They state that (using de Witt condensed notation) if the Lagrangian action is invariant under the (for simplicity irreducible) gauge transformations

$$\phi'^{a} = Z^{a}_{~b} \phi^{b}$$

then the following "b" identities hold along the gauge orbits:

$$\frac{\delta S_{L}}{\delta \phi^{a}}Z^{a}_{~b} = 0$$

samalkhaiat
THE SECOND NOETHER THEOREM

Let $\phi_{n}, \ ( n = 1, 2, …, N)$ be the set of all (matter + gauge) fields in the theory $\mathcal{L}( \phi_{n}, \partial_{\mu}\phi_{n})$.
Let us consider the following infinitesimal “local gauge” transformation (repeated indices are summed over)
$$\delta \phi_{n} = R_{n a}(x) \epsilon^{a}(x) + T^{\mu}_{n a}(x) \partial_{\mu}\epsilon^{a}(x), \ \ \ (1)$$
where $\epsilon^{a}(x), \ ( a = 1, 2, … , r)$ are arbitrary, twice differentiable, infinitesimal functions of the coordinates $x^{\mu}$; $R_{na}$ and $T^{\mu}_{na}$ are some Lie algebra valued functions of $\phi$ and $\partial \phi$.

THE STATEMENT OF THE PROBLEM:

What is the necessary and sufficient condition for the theory $\mathcal{L}(\phi , \partial \phi )$ to be invariant under the local gauge transformation eq(1)?

SOLUTION:
$$\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial \phi_{n}}\delta \phi_{n} + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_{n})} \partial_{\mu}(\delta \phi_{n}).$$
Inserting eq(1), we find
$$\delta \mathcal{L} = A_{a}\epsilon^{a} + B^{\mu}_{a}\partial_{\mu}\epsilon^{a} + \frac{1}{2}( F^{\mu \nu}_{a} + F^{\nu \mu}_{a}) \partial_{\mu}\partial_{\nu}\epsilon^{a}, \ \ (2)$$
where
$$A_{a} \equiv \frac{\partial \mathcal{L}}{\partial \phi_{n}} R_{na} + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_{n})} \partial_{\mu}R_{na} ,$$
$$B^{\mu}_{a} \equiv \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_{n})} R_{na} + \frac{\partial \mathcal{L}}{\partial \phi_{n}} T^{\mu}_{na} + \frac{\partial \mathcal{L}}{\partial (\partial_{\nu}\phi_{n})} \partial_{\nu}T^{\mu}_{na},$$
$$F^{\mu \nu}_{a} \equiv \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu}\phi_{n})} T^{\nu}_{na}.$$
Owing to the arbitrariness of $\epsilon^{a}$, the left-hand-side of eq(2) vanishes if and only if each coefficient of $\epsilon^{a}$, $\partial_{\mu}\epsilon^{a}$ and $\partial_{\mu} \partial_{\nu} \epsilon^{a}$ vanishes.
Thus, $\delta \mathcal{L} = 0$ if and only if the following relations hold identically (i.e. irrespective of whether or not $\phi_{n}$’s are solutions of the field equations),
$$A_{a} = 0, \ \ \ \ \ (3)$$
$$B^{\mu}_{a} = 0, \ \ \ \ (4)$$
$$F^{\mu \nu}_{a} = - F^{\nu \mu}_{a}. \ \ (5)$$
Now, let us define the following object
$$\mathcal{J}^{\mu}_{a} \equiv \frac{\delta \mathcal{L}}{\delta \phi_{n}} T^{\mu}_{na} + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_{n})} R_{na}, \ \ \ (6)$$
where
$$\frac{\delta \mathcal{L}}{\delta \phi_{n}} = \frac{\partial \mathcal{L}}{\partial \phi_{n}} - \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu}\phi_{n})}\right),$$
are the Euler derivatives. In terms of $\mathcal{J}^{\mu}_{a}$, we can rewrite the equations (3) and (4) as
$$\frac{\delta \mathcal{L}}{\delta \phi_{n}} R_{na} + \partial_{\mu}\mathcal{J}^{\mu}_{a} = \partial_{\mu}\left( \frac{\delta \mathcal{L}}{\delta \phi_{n}}T^{\mu}_{na}\right), \ \ (3a)$$
$$\mathcal{J}^{\mu}_{a} + \partial_{\nu}F^{\nu \mu}_{a} = 0. \ \ \ \ (4a)$$
The identity eq(3a) is called the Noether Identity. From eq(5) and eq(4a), it follows that the “current” $\mathcal{J}^{\mu}_{a}$ is conserved identically:
$$\partial_{\mu}\mathcal{J}^{\mu}_{a} \equiv 0. \ \ \ \ (7)$$

Thus (Noether’s second theorem) the necessary and sufficient condition for local gauge invariance is given by the following set of identities:
$$\frac{\delta \mathcal{L}}{\delta \phi_{n}} R_{na} = \partial_{\mu}\left( \frac{\delta \mathcal{L}}{\delta \phi_{n}}T^{\mu}_{na}\right), \ \ (3b)$$
$$\partial_{\nu}F^{\nu \mu}_{a} = - \mathcal{J}^{\mu}_{a}, \ \ \ \ (4b)$$
$$F^{\mu \nu}_{a} = - F^{\nu \mu}_{a}. \ \ \ \ \ \ (5b)$$

IMPORTANT REMARKS

1) The presence of $r$ (Bianchi) identities, eq(3b), means that the field equations
$$\frac{\delta \mathcal{L}}{\delta \phi_{n}} = 0,$$
for $N$ unknown $\phi_{n}, \ (n = 1,2, ... ,N)$ are not independent of each other, but that only $(N - r)$ equations are functionally independent. Thus, in order to determine $\phi_{n}$ uniquely, we need to impose $r$ “gauge fixing” conditions.

2) Eq(5b) suggests that $F^{\mu \nu}_{a}$ is the (Non-Abelian) gauge field tensor.

3) If $F^{\mu \nu}_{a}$ is the field strength, then eq(4b) must be the field equation for the gauge field $A^{\mu}_{a}$! But the apparent form of that equation does not look gauge covariant. So, where is the covariant derivative?

4) Eq(6) shows that the current $\mathcal{J}^{\mu}_{a}$ consists of two contributions: one comes from the matter fields Lagrangian $\mathcal{L}( \psi , \partial_{\mu} \psi , A_{\mu})$ and the other comes from the pure gauge field Lagrangian $\mathcal{L}_{YM}( F^{2})$. For example, in QCD
$$\mathcal{L}_{QCD} = i \bar{q}\gamma^{\mu}D_{\mu}q - \frac{1}{4}F^{\mu \nu}_{a}F^{a}_{\mu \nu},$$
$$D_{\mu} = \partial_{\mu} - i A^{a}_{\mu}\frac{\lambda^{a}}{2},$$
the current $\mathcal{J}^{\nu}_{a}$ is the sum of the quarks “current” $J_{a}^{\nu} = \bar{q}\gamma^{\nu}( \lambda_{a}/2) q$ and the gluons “current” $f_{abc}A_{\mu}^{b}F^{c \mu \nu}$:
$$\mathcal{J}^{\nu}_{a} = J^{\nu}_{a} + f_{abc}A^{b}_{\mu}F^{c \mu \nu}. \ \ (8)$$
The presence of the gauge field $A^{a}_{\mu}$ in eq(8) spoils the gauge covariance of the current $\mathcal{J}^{\mu}_{a}$. However, if we insert eq(8) in eq(4b), we recover the gauge covariant field equation
$$D_{\mu}F^{\mu \nu}_{a} = - J^{\nu}_{a}.$$

5) From this equation, we see that the matter field current satisfies the covariant conservation law
$$D_{\mu}J^{\mu}_{a} = 0.$$
Compare this with, eq(7), the ordinary conservation of $\mathcal{J}^{\mu}_{a}$.
This is indeed the universal property of all gauge theories including gravity:
$\mathcal{J}$ is conserved but not covariant, while the matter field current (energy-momentum tensor) is covariant but not (genuinely) conserved.

6) If we put $T^{\mu}_{na} = 0$ in eq(6), we obtain the Noether current of the global symmetry,
$$\mathcal{J}^{\mu}_{a} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_{n})} R_{na}.$$

7) Taking $T^{\mu}_{na} = 0$, the identities eq(3a) and eq(4a) collapse to the following (Noether) identity of the first Noether theorem
$$\frac{\delta \mathcal{L}}{\delta \phi_{n}} R_{na} + \partial_{\mu}\mathcal{J}^{\mu}_{a} = 0.$$
This shows that the Noether current of the global symmetries is conserved if and only if $\phi_{n}$’s are solutions to the field equations (i.e. on shell conservation),
$$\frac{\delta \mathcal{L}}{\delta \phi_{n}} = 0.$$
Compare this to the fact we established earlier that the current (of the local symmetries) in eq(6) is conserved identically (i.e. off shell conservation).

Sam

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For your equation under "SOLUTION:", that should be equal to zero right?

Also FYI you should collect all your posts and make them into a set of notes, I follow them like a hawk.

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samalkhaiat
For your equation under "SOLUTION:", that should be equal to zero right?

We want to find all the necessary and sufficient conditions for $\delta \mathcal{L}$ to be equal to zero. These conditions are eq(3b),(4b) and (5b).

Oh ok wait then, I was having trouble understanding it. You mean given $\mathcal{L}\rightarrow \mathcal{L}'=\mathcal{L}+\delta\mathcal{L}$ what must $\delta\mathcal{L}$ be so that $\delta\mathcal{L}=0$?

samalkhaiat
Oh ok wait then, I was having trouble understanding it. You mean given $\mathcal{L}\rightarrow \mathcal{L}'=\mathcal{L}+\delta\mathcal{L}$ what must $\delta\mathcal{L}$ be so that $\delta\mathcal{L}=0$?
$$\delta \mathcal{L} = \mathcal{L}(\phi + \delta \phi ) - \mathcal{L}(\phi)$$
Given the transformation eq(1), we want to see
1) what conditions make the left hand side vanishes.
2) if the left hand side IS zero, does that lead to same conditions in (1).
this what we mean by necessary and sufficient conditions (or if and only if).

tom.stoer
Are there Noether currents corresponding to the local gauge symmetries too and would there be an infinity of them since there could possibly be an infinite set of linearly independent basis functions that generate the gauge symmetry?
In some sense there is an infinity of Noether charges. Usually a charge is constructed from a conserved current, i.e.

$$\partial_\mu j^\mu = 0\quad\Rightarrow\quad Q = \int dV j^0\quad\text{with}\quad\partial_0 Q=0$$

The last equation can be written as Hamiltonian equation of motion, i.e.

$$[H,Q] = 0$$

This equation can be interpreted in two ways:
- the charge Q is conserved
- the Hamiltonian has a symmetry generated by Q

For local gauge transformations you need a local generator. In typical gauge theories like QED you may chose the A°=0 gauge and keep the correspondiong Euler-Lagrange equation which is a constraint, namely the Gauss law, i.e.

$$G(x) \sim 0$$

This constraint is time independet, and therefore

$$[H,G(x)] = 0$$

Again this equation can be interpreted in two ways:
- the 'charge' G(x) is conserved
- the Hamiltonian has a symmetry generated by G(x)

G(x) generates gauge symmetries via the unitary operator

$$U[\theta] = e^{-i\int dV\,G(x)\,\theta(x)}$$

Note that for non-abelian gauge symmetries this can be generalized to the non-anbelian Gauss law

$$G^a(x) \sim 0$$
$$[H,G^a(x)] = 0$$
$$U[\theta] = e^{-i\int dV\,G^a(x)\,\theta^a(x)}$$

for which the local Gauss law operators satisfy a local version of the gauge symmetry

$$[G^a(x), G^b(x)] = if^{abc}\,G^c(x)\,\delta(x-y)$$

So in some sense the local Gauss law operators play the role of local charges. If you like you can derive an infinity of global charges and rewrite the unitary operators, namely

$$Q[\theta] = \int dV\,G^a(x)\,\theta^a(x)$$
$$U[\theta] = e^{-iQ[\theta]}$$

for (nearly) arbitary gauge functions θa(x).

In this (Hamiltonian) formalism quantum gauge theary is like doing quantum mechanics with an infinity of d.o.f. and an infinity of symmetries.

samalkhaiat
The above post by tom.stoer seems to suggest that there are infinite class of Noether charges associated with gauge symmetries! In this post, I will prove, once and for all, that such claim is (quoting S. Weinberg) an “illusion”. To make the treatment simple and transparent, I will use QED and throw away any field equation when I encounter one (i.e. I will work on-shell only). The treatment though is general in the sense that it does not depend on any specific form of $\mathcal{L}( \psi , \partial \psi , A, \partial A)$.
Ok, under U(1) gauge transformation,
$$\delta \psi = i \epsilon (x) \psi , \ \ \delta A_{\mu}= \partial_{\mu}\epsilon ,$$
the Lagrangian changes according to
$$\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial \psi}(i \epsilon \psi ) + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\psi)}\partial_{\mu}(i \epsilon \psi ) + \frac{\partial \mathcal{L}}{\partial A_{\mu}}\partial_{\mu}\epsilon + \frac{\partial \mathcal{L}}{\partial ( \partial_{\nu}A_{\mu})}\partial_{\nu}\partial_{\mu}\epsilon .$$
Now, if we use the equation of motion for the matter field $\psi$, we can rewrite the above as
$$\delta \mathcal{L} = - J^{\mu}\partial_{\mu}\epsilon + \frac{\partial \mathcal{L}}{\partial A_{\mu}}\partial_{\mu}\epsilon + \frac{\partial \mathcal{L}}{\partial ( \partial_{\nu}A_{\mu})}\partial_{\nu}\partial_{\mu}\epsilon , \ \ (1)$$
where
$$J^{\mu} = - i \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu}\psi)}\psi , \ \ (2)$$
is the conserved Noether current associated with the global U(1) symmetry. Now Eq(1) shows that $\delta \mathcal{L} = 0$ if, and only if, the following relations hold
$$\frac{\partial \mathcal{L}}{\partial A_{\mu}} = J^{\mu}, \ \ \ \ (3)$$
$$\frac{\partial \mathcal{L}}{\partial ( \partial_{\mu}A_{\nu})} = - \frac{\partial \mathcal{L}}{\partial ( \partial_{\nu}A_{\mu})}. \ \ (4)$$
For later use, let us write the time-independent (i.e. conserved) physical charge associated with the global symmetry current $J^{\mu}$ as
$$Q = \int d^{3}x \ J^{0}(x) = \int d^{3}x \frac{\partial \mathcal{L}}{\partial A_{0}}. \ \ (5)$$
This charge is physical in sense that it is non-zero everywhere and imposes an algebraic condition on the scattering amplitude.
Ok, let us go back to our problem. With any gauge function $\epsilon (x)$, we certainly can define a current by
$$J^{\mu}[ \epsilon ] = J^{\mu}(x) \epsilon - \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu}A_{\nu})}\partial_{\nu}\epsilon . \ \ (6)$$
Using the local gauge invariance conditions eq(3) and eq(4), we can rewrite eq(6) as a total divergence
$$J^{\mu}[ \epsilon ] = \partial_{\nu} \left( \frac{\partial \mathcal{L}}{\partial ( \partial_{\nu}A_{\mu})} \epsilon (x) \right). \ \ \ \ (7)$$
This current is identically conserved because it is a total divergence of anti-symmetric tensor. So, it “look-like” that we can construct an infinite class of charges
$$Q[ \epsilon ] = \int_{V}d^{3}x J^{0}[ \epsilon (x) ], \ \ \ (8)$$
which are conserved if $\epsilon (x)$ is bounded at large distances:
$$\lim_{|x| \rightarrow \infty} \epsilon (x) < \infty.$$
We can also show that these charges generate the correct infinitesimal local gauge transformations on the fields:
$$[iQ_{\epsilon}, \psi ] = i \epsilon \psi ,$$
$$[iQ_{\epsilon} , A_{\mu}] = \partial_{\mu}\epsilon .$$

So, what is wrong with these $Q_{\epsilon}$? And do they define a class of infinite charges?
Using eq(7) in eq(8), we find
$$Q[ \epsilon ] = \int_{V}d^{3}x \ \partial_{j}\left( E^{j}(x)\ \epsilon (x) \right),$$
where
$$E^{j}(x) = \frac{\partial \mathcal{L}}{\partial ( \partial_{j}A_{0})},$$
is the electric field 3-vector. Using the divergence theorem, we find
$$Q[ \epsilon ] = \int_{\partial V}dS_{j}\ E^{j}(x) \ \epsilon (x).$$
Now, if $\epsilon \rightarrow 0$ at large distances, then $Q[ \epsilon ]$ itself vanishes! Therefore it is not a physical charge. If, on the other hand, $\epsilon$ approaches some finite constant $g$ at infinity, then
$$Q[ \epsilon ] = g \int_{V}d^{3}x \ \partial_{j}E^{j} = g \int_{V}d^{3}x \frac{\partial \mathcal{L}}{\partial A_{0}},$$
and from eq(5) we find
$$Q[ \epsilon ] = g \int_{V}d^{3}x \ J^{0}(x) = g \ Q,$$
which is not a new charge, it is just our good old electric charge of the global symmetry. So, there is not even one new charge, let alone infinite number of them, associated with the local gauge symmetry.

Sam

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tom.stoer
The above post by tom.stoer seems to suggest that there are infinite class of Noether charges associated with gauge symmetries! ... I will prove, once and for all, that such claim is ... an “illusion”
You misinterpret what I am saying.

My construction is rigorous in the canonical framework. The (infinitly many) charges

$$Q[\theta] = \int dV\,G^a(x)\,\theta^a(x)$$

can be constructued on the full Hilbert space. One can prove that they generate the gauge algebra

$$[Q^a[\theta], Q^b[\theta]] = i\,f^{abc}\,Q^c[\theta]$$

which is a global version of the local algebra generated by Ga(x); so the charges are independent quantum mechanical operators.

In your post you do not prove that these charges do not exist, you simply prove that they must vanish in the physical Hilbert space, which is a different statement.

Therefore it is not a physical charge.

Yes, the charges are "unphysical". But this is obvious, following from the fact that they have been constructed from Gauß law which of course annihilates physical states. And therefore the gauge symmetry itself is "unphysical" and reduces to the identity on physical states

$$G^a(x)|\text{phys.}\rangle = 0\;\to\;Q^a[\theta]|\text{phys.}\rangle = 0$$
$$U[\theta] = 1_\text{on phys. states}$$

The vanishing of these charges simply means that all physical states are gauge singulets.

So, there is not even one new charge, let alone infinite number of them, associated with the local gauge symmetry.

Please note that I have indeed shown that there is an infinite number of independent charges; I have constructed them ;-) They exist in a very precise way as quantum mechanical operators, and they are indeed related to the local gauge symmetry (Gauß law). Your proof that the charges must vanish (i.e. that physical states are in the kernel of these charges) does not mean that they do not exist as operators; the eigenvalues of the charges indeed vanish, but not the charges as operators (which would be inconsistent with the gauge algebra).

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samalkhaiat
You misinterpret what I am saying.

Yes, I think I did! By CHARGE, Weinberg and I mean Lorentz and gauge invariant OBSERVABLE, i.e. things like internal charges (electric, iso-spin, colour,..) and space-time charges ( momentum, angular momentum,…). These charges are the generators of GLOBAL Lie group of symmetries. They are the construct of the first Noether theorem. The second Noether theorem provides us with similar objects (call them charges if you wish) except that they are gauge-dependent and therefore non-observables. They are useful for understanding the relationship between local gauge symmetries and the algebra of FIRST CLASS constraints which I can write a book about.

My construction is rigorous in the canonical framework. The (infinitly many) charges

$$Q[\theta] = \int dV\,G^a(x)\,\theta^a(x)$$

1) The question was about Noether first and second theorems. In the Hamiltonian formalism, YOU don’t have Noether theorem.
2) In your “rigorous construction”, you threw at us this integral without saying anything about the asymptotic behaviour of the integrand! Even for formal integral you need to tell us what the integrand does at the boundary, otherwise it is non-sense. If you do not specify what $\theta ( \infty )$ is, you can’t even show that $Q[\theta ]$ is conserved. Ok, so before you insist on calling them (INFINITE) CHARGES, you should prove they are time independent first. But, if you try to prove that they are conserved, you find out that you DONOT have any new charge. This is exactly what I have shown at the end of my previous post.

can be constructued on the full Hilbert space. One can prove that they generate the gauge algebra

3) No Hilbert space or any QM is needed to answer the original question. Who needs QM, operators to answer a question about classical field theory?

$$[Q^a[\theta], Q^b[\theta]] = i\,f^{abc}\,Q^c[\theta]$$

which is a global version of the local algebra ..

Ok, I think I did mention that in my post.

Sam

tom.stoer
By CHARGE, Weinberg and I mean Lorentz and gauge invariant OBSERVABLE, ... The second Noether theorem provides us with similar objects ... except that they are gauge-dependent and therefore non-observables. They are useful for understanding the relationship between local gauge symmetries and the algebra of FIRST CLASS constraints
I absolutely agree.

In your “rigorous construction”, you threw at us this integral without saying anything about the asymptotic behaviour of the integrand!
Yes, I left that out. The construction depends on the topology and it is often safer to study e.g T³ with an approrpiate limit instead of R³; but I agree that the boundary conditions for charges and field operators are crucial.

Ok, so before you insist on calling them ... CHARGES, you should prove they are time independent first.
The local constraints Ga follow from the vanishing of the momentum ∏° conjugated to A° (A° acts a s aLagrange multiplier in gauge theories). The consistency of ∏° ~ 0 with the time evolution requires Ga ~ 0 which can be proven by calculating [H, Ga] = 0 (I thought this is standard for QED, QCD and other gauge theories).

No Hilbert space or any QM is needed to answer the original question. Who needs QM, operators to answer a question about classical field theory?
You are right. But my arguments are valid in classical field theory as well.