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

Regards!

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- Thread starter help1please
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- #1

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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

HallsofIvy

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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

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Thank you for your patience.

- #4

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

- #5

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May I ask again for a mathematical demonstration.

- #6

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

- #7

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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,

[tex]\hat{j}^{\mu}(x)=:\hat{\overline{\psi}}(x) \gamma^{\mu} \psi(x):.[/tex]

It's indeed easy to prove that the current is conserved, i.e., it obeys the continuity equation

[tex]\partial_{\mu} \hat{j}^{\mu}=0.[/tex]

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

[tex]\hat{Q}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \hat{j}(x).[/tex]

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

- #9

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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,

[tex]\hat{j}^{\mu}(x)=:\hat{\overline{\psi}}(x) \gamma^{\mu} \psi(x):.[/tex]

It's indeed easy to prove that the current is conserved, i.e., it obeys the continuity equation

[tex]\partial_{\mu} \hat{j}^{\mu}=0.[/tex]

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

[tex]\hat{Q}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \hat{j}(x).[/tex]

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

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

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Look at wiki

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

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

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- #12

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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

tom.stoer

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

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.

[tex]\frac{dL^i}{dt} = 0[/tex]

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

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

Now if we look at simple 3*3 so(3) matrices t

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

The t

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

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

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

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

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 L

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 θ

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- #14

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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.

[tex]\frac{dL^i}{dt} = 0[/tex]

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

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

Now if we look at simple 3*3 so(3) matrices t^{i}they generate the same algebra, i.e. the coefficients ε^{ikl}in the commutation relation of the matrices are identical

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

The t^{i}are generators of rotations acting on 3-vectors, the L^{i}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 ε^{ikl}are the same.

You can construct an SO(3) rotation acting on 3-vectors using rotation angles θ^{i}

[tex]D[\theta]=\exp(it^i\theta^i[/tex]

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

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

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 L^{i}are conserved charges; they generate the same symmetry SO(3) from which they are derived. In that sense the L^{i}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.

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.

[tex]U = 1 + i\epsilon T[/tex]

where [tex]T[/tex] is the generator and the [tex]\epsilon[/tex] is a small parameter.

[tex]UU^{\dagger}[/tex] commutes, there for we get the result

[tex](1 + i\epsilon T)(1 - i\epsilon T^{\dagger})[/tex]

This is conserved because

[tex]i \epsilon (T - T^{\dagger}) = 0[/tex]

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

[tex]det |M| = 1 + \epsilon \cdot m[/tex]

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

- #15

tom.stoer

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

The infinitesimal generators follow quite easily from

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

but this is not what I wanted to show.

My intention was to explain the relation between the Lie-algebra valued matrices t

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.This is conserved because ...

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