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?


2 Answers 2


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 :)

  • $\begingroup$ I just realized that this answer is essentially the same as the one found in physics.stackexchange.com/q/215481 of ACuriousMind that he linked to in a comment to the other answer. $\endgroup$ Commented Apr 27, 2020 at 18:59
  • $\begingroup$ Thanks to your answer and the linked questions I think I got where I am wrong. If it isn't a problem there is just one part that isn't clear to me, why we must equipe the space where field takes values with the inner product and not a possible space of fields. Sorry if it doesn't make sense $\endgroup$
    – Ratman
    Commented Apr 30, 2020 at 10:38
  • $\begingroup$ Don't worry! Please ask as many questions as you have and if when I have time (and if I know the answer) I'll do my best to help. First, I think that there may have been a confusion due to a mathematical thing. A lot of fields that appear in field theory take the mathematical structure of something known as sections of a vector bundle. Locally, this means that fields are maps $\phi: M\rightarrow V$ with $M$ the space where the field is defined (spacetime, worldline, worldsheet, etc...) and $V$ is a vector space where the field takes values. The latter is the finite dimensional space where $\endgroup$ Commented Apr 30, 2020 at 13:50
  • $\begingroup$ the Lorentz group is represented. It is this space the one that cannot have an inner product in which this representation is unitary. The key point is that this is NOT the vector (Hilbert) space $\mathcal{H}$ of states of the quantum theory. The latter space does have an inner product, which is essential to compute probability amplitudes. Moreover, since we want symmetries to preserve such amplitudes, the representation on $\mathcal{H}$ has to be unitary. Therefore, the space of states is infinite-dimensional. $\endgroup$ Commented Apr 30, 2020 at 13:53
  • $\begingroup$ As for your second question, the key idea is that although the space of fields is certainly a vector space, it is infinite dimensional and demands analytic tools to deal with. In particular, I don't know of any inner product on this space. On the other hand, it so happens that this space is also a $C^\infty(M)$ module. It is well behaved in that it is locally finitely generated. Thus, one might want to try to equip it with a $C^\infty(M)$-linear inner product. This turns out to be equivalent to just setting a normal $\mathbb{R}$-linear inner product on $V$. $\endgroup$ Commented Apr 30, 2020 at 14:01

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.

  • 1
    $\begingroup$ This isn't quite right - you're correct that the "space of field values" is infinite-dimensional but this is a red herring - there is no natural inner product on this space and hence no notion of "unitarity" at all. The crucial point is that the space of states - the infinite-dim. Hilbert space, the Fock space in the free case - is not the same as the space of fields at all, and hence there is no requirement at all for the representation on the finite-dim. field vector to be unitary at all. See also this answer of mine $\endgroup$
    – ACuriousMind
    Commented Apr 27, 2020 at 18:44

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