# Charge as a coefficient of the Lie algebra?

1. Sep 17, 2012

My friend was telling me that charge arises as a coefficient of the Lie Algebra.

Can someone give me a demonstration of this please, I was most fascinated by it because I had never heard of this before.

Regards!

2. Sep 17, 2012

### HallsofIvy

Do you know anything at all about "Lie Algebras"? From your reference to "the" Lie Algebra, as if there were only one, I suspect not. A "Lie Algebra" is a mathematical concept that can be applied to give aq simple structure to some physical laws.

A "Lie Group" is a topological group whose topology is that of a manifold. A "Lie Algebra" is a collection of tangent spaces on a Lie Group. I doubt that that helps!

3. Sep 17, 2012

Hi Halls, I don't want anything too rigorous, may you just give me some equations which show it is a coefficient of the Lie Algebra. It will be my task to understand it..

4. Sep 17, 2012

I should have said, Lie ''algebra's'' by the way :P

5. Sep 18, 2012

I don't mean to sound condescending, but I really don't want buzzwords.

May I ask again for a mathematical demonstration.

6. Sep 28, 2012

Can someone demonstrate for me how charge is the coefficient of the lie algebra?

7. Sep 28, 2012

### vanhees71

This is due to one of Noether's theorems: A conserved quantity generates the the one-parameter subgroup of the corresponding symmetry transformation. That's true in classical physics, where the Lie algebra and group are realized as canonical transformations on phase space as well as in quantum theory, where both algebra and group are realized by the commutator algebra of self-adjoint operators or unitary (ray) representations, respectively.

In the context of quantum-field theory, usually you have a Lagrangian that is symmetric under some (global) gauge group with corresponding conserved currents. E.g. the (free) Dirac field has a conserved vector current,
$$\hat{j}^{\mu}(x)=:\hat{\overline{\psi}}(x) \gamma^{\mu} \psi(x):.$$
It's indeed easy to prove that the current is conserved, i.e., it obeys the continuity equation
$$\partial_{\mu} \hat{j}^{\mu}=0.$$
Then and only then the time-independent (due to the conservation law according to Noether's theorem!) charge operator is a Lorentz invariant (scalar) quantity given by
$$\hat{Q}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \hat{j}(x).$$
It is easy to show that this operator indeed generates phase transformations. For details see

http://fias.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf

8. Sep 28, 2012

### naima

Look at wiki
Charges are considered as operators. In fact they are the Lie algebra generators of your symmetry.

9. Sep 28, 2012

I wondered if it had to do with Noether theory. Good thing I am pretty well read on it Noether charges, the name charge as a coefficient of the Lie Algebra... just seemed very exotic.

10. Sep 28, 2012

Wouldn't it be more accurate to say that charges are described by operators, rather than saying charges are actually operators...?

11. Sep 29, 2012

### vanhees71

I do not understand, why you call it a "coefficient of the Lie algebra". It's an element of a Lie algebra, represented by a self-adjoint linear operator. By exponentiation you get a one-parameter Lie group which represents the symmetry transformation of which the charge is the conserved quantity according to Noether's theorem. In my example this symmetry transformation multiplies the Dirac field operator with a phase factor, and the symmetry is the invariance of the physical dynamics under the multiplication of the field operators by arbitrary phase factors.

12. Sep 29, 2012

### Meselwulf

In complex physics theories involving gauge fields, charge may be described in a number of ways; charge is a generator of the Noether Algebra of a continuous symmetry so a charge can be called a Noether charge. But that won’t mean a lot to many who may be reading this. The reason why a Noether charge is described in such a way, is because as object which flows in a current has a charge and this charge is conserved in Noether’s theorem. A better kind of charge we could attempt to understand, is that the electric charge of a particle is the generator of the U(1) symmetry group which is called electromagnetism.

The author may have simply got his terminology wrong? But there are to my knowledge, several kinds of charges which may in fact be coefficients of some kind of algebra.

13. Sep 29, 2012

### tom.stoer

You can make a simple example in classical mechanics or quantum mechanics (w/o a current density, just with global objects).

The total angular momentum (of a rotationally invariant system) is a conserved quantity.

$$\frac{dL^i}{dt} = 0$$

The commutation relations derived from Poisson brackets or commutators of x and p are

$$[L^i, L^k] = i\,\epsilon^{ikl}\,L^l$$

Now if we look at simple 3*3 so(3) matrices ti they generate the same algebra, i.e. the coefficients εikl in the commutation relation of the matrices are identical

$$[t^i, t^k] = i\,\epsilon^{ikl}\,t^l$$

The ti are generators of rotations acting on 3-vectors, the Li are generators of rotations acting on Hilbert space states. So the vectors spaces on which these objects act are quite different, but their intrinsic property defined by εikl are the same.

You can construct an SO(3) rotation acting on 3-vectors using rotation angles θi

$$D[\theta]=\exp(it^i\theta^i)$$

And you can construct a rotation operator U acting on Hilbert space states

$$U[\theta]=\exp(iL^i\theta^i)$$

In that sense these objects correspond to each other; D[θ] rotates a 3-vector, U[θ] rotates a Hilbert space state (and when extracting the wave function ψ(r) the action of U is shifted from the state |ψ> to the 3-vector r).

According to Noether's theorem the Li are conserved charges; they generate the same symmetry SO(3) from which they are derived. In that sense the Li act as quantum mechanical operators defining a Lie algebra so(3).

The same applies to other symmetries (e.g. the n-dim. harmonic oscillator has an SU(n) symmetry which is much larger than the usual SO(n) rotational symmetry) and even to local (gauge) symmetries like quantum electrodynamics and quantum chromodynamics.

btw.: the terminology "coefficients" is missleading; as you can see the coefficients in the SO(3) rotation are the angles θi, not the charges Li.

Last edited: Sep 29, 2012
14. Sep 29, 2012

### Meselwulf

Oh there is a much more simpler case than you can demonstrate.

Generators of a group are simply infinitesimal rotations. It is in fact an element of the group of which is quite close to unity, unity itself however, would relate to zero rotation.

$$U = 1 + i\epsilon T$$

where $$T$$ is the generator and the $$\epsilon$$ is a small parameter.

$$UU^{\dagger}$$ commutes, there for we get the result

$$(1 + i\epsilon T)(1 - i\epsilon T^{\dagger})$$

This is conserved because

$$i \epsilon (T - T^{\dagger}) = 0$$

This makes it Hermitian. You can even understand it as a small matrix

$$det |M| = 1 + \epsilon \cdot m$$

conserved quantities are reserved for the generators such as angular momentum.

15. Sep 29, 2012

### tom.stoer

The infinitesimal generators follow quite easily from

$$U[\theta]=\exp(iL^i\theta^i) \simeq 1 + iL^i\theta^i$$

but this is not what I wanted to show.

My intention was to explain the relation between the Lie-algebra valued matrices ti and the qm operators Li

You haven't shown that T is conserved; you haven't defined T at all. As a matrix-valued generator it is trivially conserved, as a qm operator in order to show that it is conserved you need a symmetry of the system; w/o rotational invariance the angular momentum still generates rotations but it isn't conserved.

Last edited: Sep 29, 2012