Physical interpretation of Lorentz group non-compactness in the case of Weyl spinors

Question

If we consider the generators of Lorentz group $J$ and $K$, it is possible to indroduce the operators $J^{\pm}=\frac{J\pm iK}{2}$ which shows the $SU(2)\times SU(2)$ structure of the Lorentz group. Simply inverting the last relation we obtain $$J=J^++J^-$$ $$K=-i(J^+-J^-)$$ If we consider the representation of the Lorentz group $(\frac{1}{2},0)$ on Weyl spinors then $J^+=\frac{\sigma}{2}$ and $J^-=0$ then we have $$J=\frac{\sigma}{2}$$ $$K=i\frac{\sigma}{2}$$ with $J$ hermitian and $K$ anti-hermitian. In fact as Lorentz group is non-compact there are no non-trivial finite dimensional unitary representation and $K$ not being hermitian confirms this. Is it a problem in QFT to have transformations that are not represented by unitary operators? In non-relativistic QM observables are hermitian and so symmetry transformations are unitary as we expect probability to be conserved. Is there a physical interpretation to this related to probability?

Afterthought: If this is a problem, It could be avoided just simply thinking to Weyl field representations of the group which are infinite dimensional and so could have a unitary representation?

Answer 1

Let me suggest a different point of view. It is crucial that symmetry operators acting on a Hilbert space be unitary by the very definition of what a symmetry is. Thus, we expect that the Hilbert space is infinite dimensional. This is already the case in the Klein Gordon field, whose corresponding Hilbert space is a Fock space built from a $1$-particle Hilbert space spanned by states of definite momentum $|p\rangle$.

The fields on the other hand, are built so that they transform as $\tilde{\phi}(x)=D(\Lambda)(\phi(\Lambda^{-1}x))$, with $D$ a finite dimensional representation of the Lorentz group. For example, in the case of Klein-Gordon, the field takes values in $\mathbb{R}$ and the representation is given by $D(\Lambda)=1\in\mathbb{R}$. As you mentioned, this representation cannot be unitary. However, what does unitarity mean hear? For this concept to make sense we need to equip the vector space where our field takes values with an inner product. Thus, the fact is that there is no inner product on this vector space which makes $D$ a unitary representation. That is ok though! In quantum field theory our fields describe operators, not states. There is no meaning to an inner product structure to the operators in a quantum system (well, there is but it is more subtle, see GNS representation).

What is the relationship then between $D$ and the unitary representation $U$ on our Hilbert space? Well, precisely that they are compatible $$U(\Lambda)^\dagger\phi(x)U(\Lambda)=\tilde{\phi}(x)=D(\Lambda)(\phi(\Lambda^{-1}x)).$$

I hope this was useful :)

Answer 2

The situation is the following. You have a $2$-component(or $4$, doesn't really matter) object $\psi_i$. Linear transformations on it will be implemented by $2\times 2$(or $4\times 4$) matrices $A_{ij}$-$$\psi_i\to A_{ij}\psi_j$$

This matrix $A$ is obviously finite dimensional-since the multiplet $\psi_i$ is finite dimensional. The crucial point is, each of these $\psi_i$ is actually a field-$\psi_i(x)$. And since Lorentz transforms also act on $x\to x'$, we must also account for them. Being a field, thus a function, it now necessarily lies in an infinite dimensional space, and therefore must be acted on by infinite dimensional representations, and thus unitarity is preserved.

We thus have $$x\to x'=\Lambda x\hspace{5mm}\psi_i(x)\to \psi_i'(x')=A_{ij}\psi_j(x)$$But we wish to compare fields at the same point $x$ in spacetime(it is to be viewed as an active transformation)-so we invert the argument int he second equation, using $x=\Lambda^{-1}x'$-$$\psi'_i(x)=A_{ij}\psi_j(\Lambda^{-1}x)$$and now the last term is expanded in terms of $\psi(x)$ using a differential operator that goes roughly as $x_\mu\partial_\nu - x_\nu \partial_\mu$. Being a differential operator, it is infinite dimensional; as it must be-since $\psi_i(x)$ is a multiplet of functions.

Another way to derive all this is to note that the generator of the spin-part(the part that acts on the internal indices $i$ in $\psi_i$, if represented as $S_{\mu\nu}$, acts as $$L_{\mu\nu}\psi(0)=S_{\mu\nu}\psi(0)$$ where $L$ is the full generator. Then, we can translate the fields using unitary translation operator, and using the Poincare algebra derive the full generator, which is ofcourse infinite dimensional because translations are involved.

At any rate, unitarity is preserved, the key point being that we use fields.