# Conserved charge as a generator of symmetry, Peskin & Schroeder

1. Apr 21, 2010

### topper

Hey!

I am stuck at a passage in the QFT book of Peskin & Schroeder and I need your help :)
It is about page 698, last break. The sentence is:

"At long wavelength, the Goldstone bosons become infinitesimal symmetry rotations of the vacuum, Q |0> , where Q is the global charge associated with $$J^{\mu}$$"

I got two questions:

1. In what sense is the action of Q on a general state corresponding to the gauge transformation of this state? What came into my mind is that [Q,$$\phi$$] = $$\delta\phi$$ (the symmetry generator for the field). But how do I see the argument made in the quote above?

2. What do long wavelength have to do with it?

I hope my question is understandable, I'm a bit in a hurry but I will check the post for misunderstandings later,

thank you,

topper

2. Apr 21, 2010

### topper

Does anyone know anything about it or is my question not good?

Thank you :)

3. Apr 22, 2010

### tom.stoer

Do you understand in which sense Q is a generator of a symmetry and how it acts on the vacuum?

As an exercise you can study the n-dim. harmonic oscillator, the SU(n) generators $$T^a$$ and the charges

$$Q^a = \sum_{i,k=1}^n a^\dagger_i (T^a)_{ik}a_k$$

You can check that they are conserved by calculating the commutator with the Hamiltonian $$[H,Q^a] = 0$$. You can calculate their algebra $$[Q^a,Q^b]$$ based on the su(n) algebra $$[T^a,T^b]$$ and you can check how they act on the vacuum $$|0\rangle$$. The last step is to construct a finite SU(n) transformation

$$U(\alpha^a) = e^{i\alpha^aQ^a}$$

and to check that it is a symmetry of the vacuum. In addition you can check how $$a^\dagger_i$$ and $$a_i$$ transform with respect to $$U(\bf{\alpha})$$.

Of course this model does not have Goldstone bosons but from an algebraic perspective it is a rather similar toy model.

Last edited: Apr 22, 2010
4. Apr 22, 2010

### topper

Thank you very much, I will do that tonight, if there are any more questions I will let you know :)

Best regards,
topper

5. Apr 22, 2010

### tom.stoer

Just as a remark: the number operator

$$N = H - \frac{n}{2}$$

corresponds to an additional generator $$T^0 = 1$$ via

$$N = Q^0 = \sum_{i,k=1}^n a^\dagger_i (\delta_{ik})_{ik}a_k$$

for an U(1) symmetry resulting in U(n) = U(1)*SU(n).

The algebra of the generators introducing the su(n) structure constants $$f^{abc}$$ reads

$$[T^a,T^b] = if^{abc}T^c$$

Last edited: Apr 22, 2010
6. Apr 25, 2010

### topper

There are a few things I still don't understand:

Maybe this first question is stupid, but I don't get it:

1. How does a general state (of the Hilbert space) of any QFT transform under a gauge transformation? Where is the connection to the charge? I think there are just too many concepts I don't really understand and they leave a mess :)

In your toy model, it is Q |0> = 0, so that the SU(N) transformation you constructed leaves the vacuum state invariant.

2. Why can you construct a general SU(N) transformation using the charge?

3. Back to the Higgs Mechanism: here, Q |0> is non zero for the charges corresponding to the broken symmetry generators. Why is that so?

Thank you and best regards,
topper

Last edited: Apr 25, 2010
7. Apr 26, 2010

### tom.stoer

First we have to distinguish if we talk about a global or a local (gauge) symmetry. In the el.-weak SU(2)*U(1) case we are talking about the later one, whereas my toy model clearly has a global symmetry, only. The SO(N) rotational symmetry in N dimensions is quite obvious, whereas the (enlarged) SU(N) symmetry is somehow hidden - but that doesn't matter here.

Now any symmetry is related (via the Noether theorem) to a (set of) conserved charge(s); in SU(N) these are the N²-1 charges $$Q^a$$. And - the other way round - once you have constructed these charges, you automatically have the qm. generators of your symmetry:
translational symmery => conserved momentum => momentum operators generate translations
rotational symmery => conserved angular momentum => angular momentum operators generate rotations

With gauge symmetries it's a bit more invloved since they are local symmetries.

Look at the Maxwell- or Yang-Mills equations. If you do a partial gauge fixing $$A_0 = 0$$ one of the equations is the (SU(N) generalization of) the Gauss law

$$G^a = (D_i E_i)^a + gj^a_0 = 0$$

Here we have the SU(N) matter current $$j_\mu$$, the covariant derivative $$D_\mu$$ and the SU(N) electric field $$E^a_\mu = F^a_{0\mu}$$. Since $$A^a_0$$ is not dynamical, i.e. there is no canonical momentum because $$\partial_0 A^a_0$$ does not exist in the Lagrangian, the corresponding equation of motion is just a constraint, namely the Gauss law. It is related to charge conservation. You can see that by performing a volume integral; with vanishing electric field at infinity it just gives you the charge

$$\int d^3x G^a(x) = \int d^3x gj_0(x) = gQ^a$$

In addition the Gauss law is related to the SU(N) symmetry. It is (neglecting surface terms) the generator of (small) gauge transformations. The Gauss law incorporates a local SU(N) symmetry, namely

$$[G^a(x), G^b(y)] = if^{abc}\delta^{(3)}(x-y)$$

So you can generate a local gauge transformation with gauge function $$\theta^a(x)$$ via the global generator

$$G^a[\theta] = \int d^3x \theta^a(x)G^a(x)$$

All this is not relevant in the toy model but it shall explain how symmetries, Gauss law, eqations of motion etc. are related to each other.

Now what does it mean that the Gauss law is a constraint? It must hold for any time, so it is a "local constant of motion" (a constant of motion in any point of space labelled by x) leading to a global constant of motion = a conserved charge. That means that

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

$$\to [H, Q^a] = 0$$
$$\to [H, G^a[\theta] ]= 0$$

These are the qm relations for the symmetry of the Hamiltonian.

Now in SU(N) gauge theories the quantization is rather involved. You cannot simply promote all equations to operator equations because that would violate the canonical commutation relations. So in quantum fiedl theory with SU(N) gauge symmetry the equation

$$G^a = 0$$

must be replaced by

$$G^a |\text{physical state}\rangle = 0$$

where $$G^a$$ is an operator-valued quantity. That means that the algebra of SU(N) commutation relations holds at the operator level, whereas the Gauss constraint is implemented in a subspace, the so-called physical Hilbert space. It is this subspace that contains the physical degrees of freedom. This procedure kills two unphysical polarizations of the gauge bosons, namely one by imposing the temporal gauge in the very beginning and a second one by imposing the gauge law. This rather complicated approach is not necessary in QED because there you can invert the Gauss law and eliminate the longitudinal photon explicitly; it leaves behind the well known Coulomb potential. This is not possible in SU(N) theories directly, as the covariant derivative contains the gauge fields (this is not the case in the U(1) symmetry case!) and is not invertible directly.

Now we come to the big difference between my toy model and the SU(N) gauge symmetry. In my toy model there is no gauge constraint and therefore there is no requirement that the charges must vanish. My toy model allowes for non-zero charge (e.g. non-zero angular momentum). The singulett state is just the state with lowest angular momentum, but in SU(N) gauge theory the singulett states are the only ones that are physically allowed. This is called "color neutrality" in QCD: only "white = color-neutral states" are allowed; colored states are unphysical.

Therefore the answer to your question "How does a general state (of the Hilbert space) of any QFT transform under a gauge transformation" is simple:
- for unphysical states we do not care!
- physical states are annihilated $$Q^a|\text{physical state}\rangle = 0$$

The question "Where is the connection to the charge?" is answered by Noether theorem, gauge symmetry and Gauss law.

Your remark "In your toy model, it is Q |0> = 0, so that the SU(N) transformation you constructed leaves the vacuum state invariant" is correct if you make the rekation to gauge theory. Strictly speaking it is not necessary as there is no requirement of vanishing angular momentum.

You question "Why can you construct a general SU(N) transformation using the charge?" is answered by the commutation relations of the Gauss law. You can use

$$[A_i^a(x), E_k^b(y)] = \delta_{ik} \delta^{ab} \delta^{(3)}(x-y)$$

to derive them. Here $$ik$$ are 3-space indices and $$ab$$ are SU(N) gauge indices.

Strictly speaking it's not the charge of SU(N) theories but the Gauss law that generates these transformations (and even this is not fully correct), but for global transformations where the gauge function is a constant this difference does not matter. So in the toy model it's just charge coming from the Noether theorem that generates the SU(N) trf.

Now comes the interesting point regarding the spontaneous symmetry breaking. I do not know how to impose the Gauss law correctly. I do not even know if all my formulas go through w/o any modification. "Back to the Higgs Mechanism: here, Q |0> is non zero for the charges corresponding to the broken symmetry generators. Why is that so?" It is clear that the Gauss law implementation on physical states has to be modified due to the spontaneous symmetry breaking. But this applies to the constraint on physical states (including the vacuum) only. The operator algebra still holds because otherwise the symmetry would be anomalous and therefore inconsistent (an anomalous symmetry is a symmetry which is broken already at the operator level; in case of gauge symmetries this is forbidden).

In your first post you mentioned Goldstone bosons (instead of the Higgs). That's why I started with this hole story, because for Goldstone bosons it's rather simply: you do not have a (local) gauge symmetry but a global symmetry only. Therefore you do not have a Gauss law and there is no requirement that the charges must annihilate the physical states. The broken symmetry simply means that the vacuum is not annihilated which is something like a non-vanishing angular momentum. The SU(N) generators generate rotations along the vacuum, meaning that

$$e^{i\theta^a Q^a} |\text{vac}\rangle = |\text{vac}^\prime\rangle$$

But this "direction of rotation" is just the direction of the Goldstone mode oscillations.

Last edited: Apr 26, 2010
8. Apr 26, 2010

### tom.stoer

A very good (but rather complicated) paper regarding gauge fixing in SU(N) gauge theories in the canonical formalism is

QCD in the Axial Gauge Representation
Lenz F., Naus H. W. L. and Thies M.
Abstract: Within the canonical Weyl gauge formulation, the axial gauge representation of QCD on a torus is derived. The resolution of the Gauss law constraint is achieved by applying unitary gauge fixing transformations. The result of this formal development is a Hamiltonian explicitly formulated in terms of unconstrained degrees of freedom. Novel features of this Hamiltonian are the non-perturbative dynamics of two-dimensional degrees of freedom appearing in the gauge-fixing procedure, such as Jacobian and centrifugal barrier. These two-dimensional fields appear to be essential for the infrared properties of the theory. The global residual gauge symmetries of QCD are established in this representation. It is shown that SU(N) gauge theories may exhibit at most N-1 massless vector (gauge) bosons. The implications for the phase structure of non-abelian gauge theories (QCD, Georgi-Glashow model) are discussed.
Annals of Physics, Volume 233, Issue 2, 1 August 1994, Pages 317-373

+ references therein; unfortunately not available online

9. Apr 26, 2010

### topper

Thank you very much for your insightful comment.

You wrote :

1. How do you define the physical Hilbert space in general?

Is the equation

$$G^a |\text{physical state}\rangle = 0$$

a proper definition of a physical state or do we neglect something?

10. Apr 26, 2010

### tom.stoer

In the case of unbroken SU(N) gauge symmetry this is exactly the equation for the physical Hilbert space; it just the kernel of the Gauss law.

Why? The Gauss law cannot be an operator identity

$$G^a(x) = 0$$

because in that case the commutation relations would be

$$|G^a(x), G^b(x)] = [0,0] = 0$$

which contradicts the reasoning from above.

But on the other hand it has to be implemented in the quantum theory. The only chance one has a constraint on physical states. This means that only gauge invariant states are physical states. And this is rather natural as the gauge symmetry kills two unphysical degrees of freedom as explained above (the vector potential has four components, only two are the transversal ones, so two have to be eliminated; one by the temporal gauge, one by the Gauss law).

The gauge symmetry means that you have unphysical derees of freedom in your theory which is a big difference to angular momentum. For angular momentum you are not forced to restrict the theory to the subspace with vanishing angular momentum. For gauge theories this is a must because otherwise unphysical degrees of freedom would propagate and violate unitarity and things like that. So the analogy with the global SU(N) or rotational symmetry is not very strict. In the case of a global symmetry you can restrict youself to states with Q|phys> = 0, in gauge theories this is a must.

Btw.: there are different gauge fixing procedures for non-abelian theories, e.g. BRST, but a) i Like the canonical formalism most and b) I think that BRST is fully understood ionlyin the perturbative sector.

To come back to you question: quantizing a theory means that you have to promote the classical equations either to operator equations or to (constraint) equations holding only in a certain subspace. The next question is when to do what. The answer can be found in the Dirac constraint quantization formalism which is in general rather involved and requires a detailed knowledge regarding the canonical formalism in classical mechanics or field theory.
Short summary: in theories w/o second class constraints all constraints are implemented as equations for physical states. A constraint is first class if its commutator with any other constraint is either identically zero or a linear combination of other constraints; therefore the Gauss law is first class as the commutator only involves a linear combination of the Gauss constraints.
In a theory with first class constraints the physical sbspace is defined as the kernel of all constraints. In a theory w/o constraints the physical space is identical to the kinematic space. It is only the gauge symmetry (or some other symmetry which reflects unphysical = redundant degrees of freedom) which forces you to restrict to a certain subspace.

11. Apr 26, 2010

### topper

Ok thank you very much. That was very helpful :)

One last question:

A one particle state in QED is e.g.

$$|k,\mu> = a^\dagger^\mu_k |0>$$

I could now write the a in terms of $$A^\mu$$ and its conjugate momentum and then apply a gauge transformation. Do I get the same result as if I do

$$e^{i\alpha Q}|k,\mu>$$

?

12. Apr 26, 2010

### tom.stoer

It looks as if $$a_k$$ is a one-particle operator for photons with momentum k before gauge fixing. Of course you can express the Fourier modes $$a_k$$ in terms of the original potential $$A(x)$$, but keep in mind that you only have two physical degrees of freedom.

So it depends if you talk about the situation before or after gauge fixing. Before gauge fixing you can do any calculation in x- or in k-space - whatever you like - and you will certainly get the same expressions for the charge and the symmetry transformation. After the gauge fixing plus implementation of the Gauss law there are only physical degrees of freedom and the charge operator annihilates all physical states; so the gauge transformations reduce to the identity.

In addition in QED neither the vacuum nor the one-particle photon state carry charge (the photon is neutral); regradless if you talk about a gauge fixed quantity or, not Q acting on these states gives always zero:

$$Q|0\rangle = 0$$
$$e^{i\alpha Q}|0\rangle = 1|0\rangle$$

13. Apr 26, 2010

### topper

Ok, I think I got it.

I will now read in some books and (hopefully) tighten my knowledge.

14. Apr 26, 2010

### tom.stoer

you're welcome; hope it helps

15. Apr 27, 2010

### DrDu

"At long wavelength, the Goldstone bosons become infinitesimal symmetry rotations of the vacuum, Q |0> , where Q is the global charge associated with LaTeX Code: J^{\\mu} "

I think this statement is too sloppy. In a broken global symmetry there are no operators generating a rotation of the vacuum. Different vacua live in different Hilbert spaces.
However, a Goldstone boson corresponds to a local rotation of the vaccuum ( the resulting state being not the vacuum any more). In the long wavelength limit, and in the absence of long range forces, this rotation does hardly cost energy as the region in which the vacuum is rotated becomes very large and surface effects become negigible.