# Observables labelling one-particle states in Quantum Field Theory

I'm studying introductory QFT using the first volume of Weinberg's series, and i'm having problems in understanding how single particle states of the free theory are labelled, i.e. what observables can be used to uniquely identify the state. My doubts come in the first place from the commutation relations of the Poincarè Algebra: since the linear momentum is a vector operator it does not commute with the rotation generator` $$[J_{i}, P_{j}] = i\epsilon_{ijk}P_{k}$$ and it is not possible to diagonalize simultaneously both the linear and angular momenta. The single particle states are chosen as the eigenstates of the 4-momentum, but this is not enough to uniquely identify them and Weinberg adds another label $$\sigma$$ which is then used to write the general action of a quantum Lorentz Transformation on a single-particle state as $$U(\Lambda)\Psi_{p,\sigma} = \sum_{\sigma'}C_{\sigma'\sigma}(\Lambda, p) \Psi_{\Lambda p, \sigma'}$$ First question: what is the physical meaning of $$\sigma$$ ? Being a label to physical states shouldn't it be the eigenvalue of some observable ? My real problem however comes later: with the method of induced representations it is found that the representations of Lorentz transformations are given by the ones of $$SU(2)$$ which are labelled by the total spin of the particles. I can understand the connection between the spin of the particle and the one of the representation in light of the square of the Pauli-Lubanski operator being a Casimir operator for the Lorentz group, what messes me up is that now $$\sigma$$ is explicilty written as the eigenvalue of the third component of the angular momentum, which does not commute with $$\textbf{P}$$. How can i match this identification with the general commutation rules of the Poincarè Algebra ?

## 1 Answer

The discrete index in Weinberg's derivation is a generalization of the spinor index that we see in QM. The generalization is required from theoretical principles, because we don't know under which group the spinors will transform.

It is a mathematical fact (Wigner's paper) that Lorentz group has no finite-dimensional unitary representations. However there is an infinite-dimensional unitary representation of Poincare group (Lorentz group plus boosts). Because of this we are obliged to select some of the indices of $$\Psi$$ that would be not discrete. We take 4-momentum ($$p^\mu$$) out of full Poincare group and add them as a label to our states $$\Psi_p$$. Any coordinate transformation that would change classical 4-momentum $$p'^\mu = \Lambda^\mu_{\; \nu} p^\nu$$ should give us $$\Psi_{p'} = \Psi_{\Lambda p}$$ up to a phase.

Such $$\Psi_p$$ would describe scalar particles that do not have any additional transformation properties under rotations, but from QM we know that particles could also have spin that transforms under $$SU(2)$$ (double cover of rotation group $$SO(3)$$). Relativistic particles should have a property similar to this, so that we could reproduce original non-relativistic results in an approximation scheme. So we are forced to make our $$\Psi_p$$ a spinor by adding a discrete index $$\sigma$$ to it: $$\Psi_{p, \sigma}$$.

To complete the description of the $$\Psi$$ we have to understand what happens to $$\sigma$$ under rotations and boosts. Both $$p$$ and $$\sigma$$ change under those transformations. We have to isolate $$\sigma$$ from $$p$$ in order to gain more insight. Thankfully there are combinations of both boosts and rotations that leave $$p$$ unchanged (rotate $$p$$ to $$p'$$, and then boost to make it $$p$$ again). Those transformations form a group (there is an identity, inverse, and you can combine two of them). It can be checked (see Wigner's classification) that the group has to be different for massive and massless particles: $$SO(3)$$ and $$ISO(2)$$. Both of them have known unitary representations, and we can apply them to transform the $$\sigma$$ indices.

Now with transformations of $$p$$ and $$\sigma$$ we are seemingly in conundrum. We know how $$p$$ changes under general Lorentz transformations, but only know how $$\sigma$$ changes under very special subset of Lorentz transformations (Little group). How do we combine these two together to get full transformation properties? The trick is to relate $$\Psi_{p,\sigma}$$ to $$\Psi_{\Lambda p,\sigma'}$$ by a defining a special Lorentz transformation $$\Lambda'(\Lambda p, p)$$ that depends only on starting and final momenta, and not on the $$\Lambda$$ itself. We then define $$\sigma$$s to be equivalent for states transformed with those $$\Lambda'$$:

$$\Psi_{p, \sigma} \rightarrow \Psi_{\Lambda' p, \sigma}.$$

These notes are far from comprehensive and Weinberg does much better job than me at explaining it, but perhaps more concise form would help to see this procedure differently.