Can one have $\mathcal{PT}$-symmetry in a QFT theory proportional to an imaginary field? There is a lot of fuss nowadays around $\mathcal{PT}$-symmetry in non-relativistic quantum mechanics. Recently I came across this paper where the authors generalize the non-relativictic Hamiltonian
$$
H = p^2 + x^2(ix)^\epsilon,
$$
with $\epsilon\geq0$ to the Lagrangian
$$
\mathcal{L} = \frac{1}{2}(\nabla \phi)^2 + \frac{1}{2}\phi^2(i\phi)^\epsilon.
$$
As a humble peasant, I have to questions:


*

*As far as I understand, the time reversal operator, in a QFT, is no longer taken as a unitary operator but instead as a Lorentz transformation


$$
g_{\mu \nu} \Lambda^\mu_{\;\rho} \Lambda^\nu_{\;\sigma} = g_{\rho \sigma}
$$,
where, for instance (I use Schwartz book notation on QFT)
$\Lambda = \operatorname{diag}(-1,1,1,1)$. In this sense, there is no point in including the $i$ for time-reversal in a valid QFT theory, right?


*If we ignore that $\mathcal{L}$ has this "$\mathcal{PT}$-symmetry", are there actually physical theories that include a term proportional to $i$ in a Lagrangian density?

 A: Short answer:  ${\cal PT}$-symmetry can be a useful tool for demonstrating the existence of unitary field theories with complex Hamiltonians, but there are deep problems with relating those field theories to any kind of useful phenomenology.
Long version:
You can write down a field theory with a complex Lagrangian and try to solve study it classically as a field theory (ignoring the fact that the derivatives would normally arise from "first quantization"), then imposing "second quantization" by making the fields noncommuting operators.  The obvious problems with this are that the theory is typically going to be nonunitary and unstable—meaning, respectively, that there is no uniform vacuum state around which you can perform perturbations, and that there is no conservation of probability.  Both of these problems are natural consequences of having a Hamiltonian (or, equivalently, Lagrangian) that is not real (for the classical theory) or Hermitian (for the second-quantized version).
However, these problems are not absolutely necessary consequences of having a complex Lagrangian.  In fact, what ${\cal PT}$ symmetry is useful for is ensuring that there is some basis in which the states of the theory form a well-defined Hilbert space, with Hamiltonian that is bounded below and has real eigenvalues.  (Actually, a bit more than ${\cal PT}$ symmetry is required to ensure these properties; you need to restrict attention to only a certain wedge of complex values for the field operators, for example.  However, these are technical issues, which—although they are important in practice—do not affect the overall qualitative answer.)  Of course, having only real eigenvalues ensures that the Hamiltonian is Hermitian in some other basis; however, the transformation that takes you to that basis in which the $H$ is Hermitian is a similarity transformation, but it is not a unitary transformation.
In a single-particle ${\cal PT}$-symmetric quantum theory, it is typically possible to find (although not necessarily in closed form—possibly only as an infinite series expansion) the transformation to the coordinates in which $H$ is Hermitian and everything is well behaved.  For a ${\cal PT}$-symmetric quantum field theory, this is a much more difficult problem, and even if you could find the field theory in which ${\cal L}$ was real/Hermitian, the Lagrangian for the transformed field theory would be nonlocal.  That would make working with the transformed field theory extremely difficult—practically impossible, in fact, unless you got extremely lucky.  So there is definite value in working with the ${\cal PT}$-symmetric version with the complex ${\cal L}$; in the complex version, you might be able to demonstrate renormalizability, for example—which would almost certainly not be feasible in the Hermitian-but-nonlocal basis.
What the ${\cal PT}$-symmetric version of the theory lacks, however, is anything resembling a probabilistic interpretation.  (All quantum field theories that require renormalization have technical issues with the usual probabilistic interpretation of quantum mechanics, but the problems for the ${\cal PT}$-symmetric theories with complex ${\cal L}$ are far worse.)  In particular, there is no straightforward correspondence between correlations functions of the field operators and external incoming and outgoing particle states.  (Technically, this means that there is nothing like the LSZ reduction formula for the complex theory.)  That connection, between the external states and the field operators only exists for a theory when it is written with a manifestly Hermitian Lagrangian.
Therein lies the crux of the problem, and the answer to your question.  There is no tractable way to  develop a theory of particle-like scattering states with an interaction Lagrangian that is not real/Hermitian.  Thus the ${\cal PT}$-symmetry formalism is interesting as a way of demonstrating that a particular field theory has a Hermitian formulation, but it does not provide enough guidance to make such theories phenomenalistically useful.  For particularly simple models, like the Lee model (which is exactly solvable but requires renormalization), the ${\cal PT}$-symmetry provides a useful piece of information, explaining why the theory remains well defined, even though the renormalization appears to carry some of the fields into complex values.  However, even for other exactly solvable models [such as the (1+1)-dimensional chiral Schwinger model, which looks like it should be amenable to ${\cal PT}$-symmetry methods], the ${\cal PT}$-symmetry has proven too complicated to realized in a useful fashion.
(Unfortunately, the most prominent proponents of the ${\cal PT}$-symmetry method have sometimes presented their work rather disingenuously, by not mentioning the fact that their ${\cal PT}$-symmetric theories are not amenable to the usual probabilistic interpretation.  This problem with probability occurs even in the nonrelativistic quantum-mechanical theories.)
