QFT and unitary Lorent representation

[simic4]

Hey guys! A question:

My QFT Lagrangian, fournishes through Noether's thm plus relativistic invariance a supposedly unitary and a supposed representation of the Lorentz group. These are the operators meant to act on my Hilbert space of possible states.

What guarantees that this actually happens, ie: 1. The commutation relations of the lorentz algebra are actually satisfied, 2. The operators are hermitian (guarantiing a unitary representation). 3. They actually do what they are advertized to do ie: for example applying the z-direction angular momentum operator to a single particle state created by your creation operator does give what its suposed to. ie: the dirac field creates s=1/2, the scalar field s=0, etc..

i find it amazing and non trivial if conditions 1-3 are generally satisfied just because of lorentz invariance plus the right commutation relations for the fields! (if true for those two reasons alone then it might motivate the supposedly put in by hand commutations relations!)

maybe there's a good reference i could look at?

thanks!

simic

 

[samalkhaiat]

Hey guys! A question:

What guarantees that this actually happens, ie: 1. The commutation relations of the lorentz algebra are actually satisfied, 2. The operators are hermitian (guarantiing a unitary representation). 3. They actually do what they are advertized to do ie: for example applying the z-direction angular momentum operator to a single particle state created by your creation operator does give what its suposed to. ie: the dirac field creates s=1/2, the scalar field s=0, etc..

Absolutely, one can prove that Noether charges of any (internal & spacetime) Lie group G, do satisfy the Lie algebra of G;

$$[Q_a,Q_b]=C_{ab}{}^{c}Q_{c}$$

and generate the right transformations on the irreducible representations of G (the fields);

$$\delta \phi =[i\epsilon^{a}Q_{a},\phi]$$

This is true even when the "internal" G group is not exact symmetry. That is, when the Lagrangian contains a symmetry-breaking terms. i.e Noether charges change with time.

The remarkable fact that we can prove the above results in the contex of QFT, where Lie braket is commutator, as well as in classical field theory, where Lie braket is Poisson's, is a very good justification for the quantization "law"

$$[A,B]_{PB} \rightarrow i[\hat{A},\hat{B}]_{c}$$

easy to read book is
Field Quantization, Greiner Reinhardt, Springger


regards

sam

 

[simic4]

thanks! And i,ll take a look at the book.

 

[simic4]

A very beautiful result..

simic

 

[samalkhaiat]

A very beautiful result..

Not without troubles though!
In QM, when we write

$$\overline{|\Psi>}=U|\Psi>$$,
$$\bar{A}=UAU^{\dagger}$$

we mean[all books say] that $$U$$ connects states that belong to an irreducible representation of some symmetry group G. And [only 1 or 2 book say] that implicit in the above two equations is the invariance of the vacuum under G.
Indeed, $$\overline{|\Psi>}$$ and $$|\Psi>$$ must be connected to the vacuum through some appropriate creation operators;
$$\overline{|\Psi>} = \bar{A^{\dagger}}|0>$$
$$|\Psi> = A^{\dagger}|0>$$

so it is easy to see that the top two equations are consistent, only if
$$U|0> = |0>$$
or, by writing, $$U=1 + i\epsilon_{a}Q_{a}$$ ;
$$Q^{a}|0> = 0, \forall{a}$$

So, what does this mean in english?
We say that the symmetry is manifest and unitarily implimented on the Hilbert space of states, only when Noether charges annihilate the vacuum.
I.e. only when Q|0>=0, the invariance of the Hamiltonian is manifest in the degeneracies of the energy eigenstates(particles) corresponding to the irreducible representation of G;
$$E(\bar{\Psi}) = <\bar{\Psi}|UHU^{\gagger}|\bar{\Psi}> = <\Psi|H|\Psi> = E(\Psi)$$

When $$Q_{a}|0> \neq 0$$ the particles are nolonger degenerate for, in this case we have
$$E(\bar{\Psi}) \neq E(\Psi)$$
and say that the G-symmetry spontaneously broken (hidden) eventhough H is still invariant under G:

"Spontaneous breakdown of symmetry is the lack of degeneracy in the particle spectra in a symmetric theory"

We said that Q|0> = 0 means that the symmetry is manifest!
So, is it possible to show that

$$Q_{a}|0> = 0 \Rightarrow \partial_{\mu}J_{a}^{\mu} = 0$$

Yes, it is and this is exactly what Coleman proved:

If, for any 4-vector $$J_{\mu}(x)$$

$$Q(t)= \int d^3x J^{0}(x)$$

is well defined operator on the H-space, and $$Q(t)|0> = 0$$, then

$$\partial_{\mu}J^{\mu}= 0$$

and $$Q(t)$$ is time independent.

However, the converse need not be true! (Q|0>=0, is sufficient but not necessary condition for exact symmetry), To see this, consider a Poincare'-invariant theory with internal symmetry group G ([G,Poincare]=0),
Noether theorem then leads to a conserved G-currents $$\partial_{\mu}J^{\mu}_{a} = 0$$ and time-independent charges;

$$Q_{a}= \int d^3x J_{a}^{0}(x)$$

Now, let us calculate the norm of $$Q_{a}|0>$$

$$<0|Q_{a}Q_{a}|0> = \int d^3x <0|J_{a}^{0}(x) Q_{a}|0>$$

From the invariance under translations:

$$J^{0}_{a}(x)= e^{-i\hat{P_{\mu}}x^{\mu}}J_{a}(0) e^{i\hat{P_{\mu}}x^{\mu}}$$
$$[P_{\mu},Q_{a}]=0$$

it follows that
$$<0|Q_{a}^{2}|0> = \int d^3x <0|J_{a}^{0}(0)Q_{a}|0> \rightarrow \infty$$,

unless $$Q_{a}|0> = 0$$.

I.e, either the symmetry is exact and unitarly implemented on the H-Space (Q|0>=0), or it is spontaneously broken (Q|0> does not exist in the H-Space because its norm is infinite).

So, the behavior of the vacuum under a symmetry transformations is crucial in categorising how that symmetry is implemented in the H-Space of the quantum theory.

Coleman summarizes this by saying:
"symmetries of the vacuum are symmetries of the world"

OK, this seems to paint a nice picture, so where are the troubles that I mentioned in the beginning of this post?
The troubles show up when we consider a gauge-constraint systems (I am not talking about the well-known QFT problems)
Take, for example, the EM gauge transformation;

$$A'_{\mu} = A_{\mu} + \partial_{\mu}\Lambda$$

Everything in nature indicates that this is an exact symmetry. So we expect to find a unitary operator $$U$$, such that
$$U|0>=|0>$$
and,
$$A'_{\mu} = UA_{\mu}U^{\dagger} = A_{\mu} + \partial_{\mu}\Lambda$$
are satisfied in the quantum theory of free EM field. But this leads to the contradictory statement;
$$<0|A'_{\mu}|0> = <0|A_{\mu}|0> = <0|A_{\mu}|0> + \partial_{\mu}\Lambda$$
or,
$$\partial_{\mu}\Lambda = 0$$
(this can not be right because $$\Lambda$$ is an arbitrary function)

So, one is led to believe that the EM vacuum is not invariant under the gauge transformation. i.e
$$U|0> \neq |0>$$
or in terms of Q
$$Q_{\Lambda}|0> \neq 0$$
This means that the gauge symmetry is spontaneously broken!
i.e
$$<0|[iQ_{\Lambda},A_{\mu}]|0> = \partial_{\mu}\Lambda \neq 0$$

but this is equivalent to the statement that the field operator has non-vanishing vacuum expectation value;
$$<0|A_{\mu}|0> \neq 0$$
which is wrong because of Poincare' invariance.
So, I am baffled!
As far as I know, this problem remains unsolved even when we choose a small(PHYSICAL) subspace of the H-Space by postulating appropriate subsidiary condition.

If you people want a PhD, just solve the above problem

regards

sam

 

[Perturbation]

Hey, I found a proof that the Noether charges do generate the symmetries of the theory whilst I was flicking through a book of mine (it's in a footnote Chp. 26 pg. 92 of Weinberg "Quantum Theory of Fields" Vol. III).

Consider a Lagrangian (not density) $L=L(q_n, \dot{q}_n)$. If the action has some symmetry $q\rightarrow q+\delta q$ then the Lagrangian will remain unchanged up to some time derivative of some suitable functional F (as when you integrate the perturbed Lagrangian over time you'll end up with $F(t_1)-F(t_0)$, and one can of course choose an F so that this vanishes, i.e. the action remains invariant, just as when one would usually choose the Lagrangian density to change by a four-divergence in relativistic field theory). Thus

$$\delta L=\sum_n\frac{\partial L}{\partial\dot{q}^n}\delta\dot{q}^n+\sum_n\frac{\partial L}{\partial q^n}\delta q^n=\frac{d}{dt}F$$ (1)

Then the charge Q ($=\int d^3x j^0$)associated with the conserved Noether current is

$$Q=-\sum_n\frac{\partial L}{\partial\dot{q}^n}\delta q^n+F$$

Using the usual commutation relations

$$\left[\frac{\partial L}{\partial\dot{q}^n}, q^m\right] =-i\delta^m_n$$

$$\left[q^n, q^m\right] =0$$

You get the commutator

$$\left[Q, q^m\right]=-i\delta q^m-\sum_{nl}\frac{\partial L}{\partial\dot{q}^l}\frac{\partial\delta q^l}{\partial\dot{q}^n}\left[\dot{q}^n, q^m\right]+\sum_n\frac{\partial F}{\partial\dot{q}^n}\left[\dot{q}^n, q^m\right]$$

From (1) we get (expand dF/dt and $\delta\dot{q}$ and equate coefficients of the second time derivatives of the qs)

$$\sum_l\frac{\partial L}{\partial\dot{q}^l}\frac{\partial\delta q^l}{\partial\dot{q}^n}=\frac{\partial F}{\partial\dot{q}^n}$$

And thus

$$\left[Q, q^m\right]=-i\delta q^m$$

And

$$\left[Q, \dot{q}^m\right]=-i\delta \dot{q}^m$$

Thus the Qs generate the symmetries of the theory. The above obviously generalises to field theory where index n of each configuration variable would refer to spin, species, spatial coordinates etc.