Given a quantum theory specified with a Lagrangian and the degrees of freedom to be varied, what is the procedure to determine if the theory is unitary or not?

Concrete example to aid discussion:
(Taken from discussion of some simple models in this Phys.SE post, using path #2 without imposing condition E to obtain a non-unitary theory.)

Start with a Lagrangian for some complex scalar field. $$\mathcal{L}=\partial^\mu \phi^* \partial_\mu \phi -m^2 \phi^* \phi -\lambda (\phi^* \phi)^2$$

Is this unitary? How can this be checked and verified?

Now, write the complex field with two real components $\phi = \phi_1 + i \phi_2$. The Lagrangian is then $$\mathcal{L}= \left(\partial^\mu \phi_1 \partial_\mu \phi_1 -m^2 (\phi_1)^2 -\lambda (\phi_1)^4 \right) -2\lambda (\phi_1)^2(\phi_2)^2 +\left( \partial^\mu \phi_2 \partial_\mu \phi_2 - m^2 (\phi_2)^2 -\lambda (\phi_2)^4 \right)$$

Now complexify the fields (let $\phi_1$ and $\phi_2$ now be complex valued), and do not impose $${\rm Im}(\phi_1)~=~0~=~{\rm Im}(\phi_2).$$ From earlier discussion, this new theory will not be unitary.

What procedure can I go through starting from this Lagrangian to show that this is no longer unitary?

  • $\begingroup$ It is not clear to me when QMechanic says to complexify the fields if the terms $m^2(\phi_1)^2 \rightarrow m^2 \phi_1^* \phi_1$ as usual, or if left in the weird form $m^2(\phi_1)^2$. I assume he meant the later, as it is this weird form that allows the evolution to become non-unitary? $\endgroup$
    – John
    Sep 16, 2014 at 0:39
  • $\begingroup$ My Phys.SE answer was referring to the latter weird form (as you call it), i.e. without complex conjugation. My comment about non-unitarity in that answer was merely referring to that e.g. the kinetic term of such complex theories would not be positive definite. $\endgroup$
    – Qmechanic
    Sep 16, 2014 at 13:54
  • $\begingroup$ @Qmechanic even if the general case is hard, I'd love an answer which shows how some specific cases (the choice in sign of some terms in the Lagrangian, etc.) leads to evolution not being unitary somehow. The only answer currently seems to require all the consequences of the theory to be determined first, which isn't very satisfying. Just like we could point to a term in a Lagrangian and say "that would break Lorentz invariance" isn't there some way in simple cases to determine if something would make the theory non unitary? $\endgroup$
    – John
    Sep 17, 2014 at 4:56
  • $\begingroup$ @PhysStudent your example breaks unitarity because you are taking a lagrangian which isn't hermitian. But the point is that even if you have a lagrangian that is naively hermitian, still the theory may be secretely non-unitary. For example, speaking of $\phi^4$ theory, there is a beautiful new paper here arxiv.org/abs/1409.1581 where they claim that $\phi^4$ in non-integer dimension is not unitary (even though the lagrangian is hermitian). $\endgroup$
    – TwoBs
    Sep 18, 2014 at 1:37

2 Answers 2


I would check if the corresponding Hamiltonian is self-adjoint. The time evolution operator is

$$ U(t,t') = \text{e}^{i(t-t')H} \, .$$

Unitarity is equivalent to requiring that probability is conserved along the time evolution,

$$ \frac{\text{d}}{\text{d}t} \langle \psi |\psi\rangle = i \langle \psi | H |\psi \rangle -i \langle \psi | H^\dagger |\psi \rangle = 0 \leftrightarrow H = H^\dagger \, .$$

Equivalently we have

$$ U^\dagger U = 1 \leftrightarrow H = H^\dagger \, .$$

If you take the legendre transform of your Lagrangian,

$$ H = \partial_\mu \phi \pi_mu + \partial_\mu^* \phi \pi_mu^* - L \, ,$$

you can write the Hamiltonian of your system. Then you can check weather we have

$$ H^\dagger = H \, .$$

If this is not the case, your get imaginary energy levels and decay of probability. If it is, your quantum time evolution is well defined. I would check that the potential is bounded by below so that the system has a ground state as well, but I don't think that this has anything to do with unitarity.

Note that you may be able to get out of all this if your Hamiltonian is only PT symmetric (instead of self-adjoint). See http://arxiv.org/abs/quant-ph/0501052. This is however a very recent proposition to upgrade quantum mechanics (typically to out-of-equilibrium set-ups) which is not mainstream yet and under investigation.

  • $\begingroup$ Maybe I'm misunderstanding, but consider QED, $\mathcal{L}=\bar\psi(i\gamma^\mu D_\mu-m)\psi -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$. Now, I suspect if I flipped the sign in the last term, the theory would be unstable since the photon modes would have negative energy, and thus scattering or any evolution would be non-unitary due to boundedness from below issues. So would your answer be equivalent to some phrase like Unitrary at the "tree level" of Feynman diagrams, but not at higher levels or not non-perturbatively Unitary? How would I calculate this to check? $\endgroup$
    – John
    Sep 18, 2014 at 15:00
  • 1
    $\begingroup$ Self-adjointness of the Hamiltonian is a way to make sure that its eigenvalues are real. Eigenstates evolve as $|E(t)\rangle = \text{e}^{t i\hbar E}|E\rangle$. With complex eigen-values you get exponential growth or decay of the norm. With real eigen-values oscillations. This has nothing to do with the existence (or not) of a state with a minimum energy, i.e. the stability of the Hamiltonian. $\endgroup$ Sep 18, 2014 at 18:54
  • $\begingroup$ Self-adjointness is clearly necessary, but does not appear to be sufficient. While classically all we care about are the "on-shell" equations, for QFT we can't ignore the catastrophic results if the off-shell solutions are very poorly behaved. For instance by flipping that sign, the Hamiltonian is still self-adjoint, but now trying to calculate any scattering will result in nonsense since we can take infinite energy from the vacuum. So to be unitary it appears the evolution must be self-adjoint and bounded in some sense (probably in the sense given in the other answer?). $\endgroup$
    – John
    Sep 22, 2014 at 23:33
  • $\begingroup$ this related question discusses scattering bounds for the theory to remain unitary physics.stackexchange.com/questions/90736/… $\endgroup$
    – John
    Sep 24, 2014 at 3:19

There is so called optical theorem, which states that for the unitary theory must be $$ Im (M_{k_{1}, k_{2} \to k_{1}, k_{2}}) = 2E_{cm}|\mathbf p_{cm}|\sigma_{total}(k_{1}, k_{2} \to all), $$ where $cm$ denotes center of mass frame, $\mathbf {p}_{cm}$ - momentum of one particle at CM frame, $M$ is amplitude of scattering and $\sigma_{total}$ is total cross section. So for basic validation you must use this theorem.

Also there is simple (but not exact) method of checking of unitarity by checking of lagrangian on the dimensional coupling constant (they may also be hidden, like in gauge theories, in polarization vectors). By the naive thinking, the presence of dimensional constant with dimension $E^{-n}, n < 0$ leads to the appearance of energy with positive dimension in matrix element which will lead to infinite amplitude and cross section, so will break the unitarity. But sometimes (like in gauge theories) corresponding constant does not contribute to divergence.

  • $\begingroup$ I think your answer is misleading. Infinite amplitudes and cross-sections are allowed by unitarity. There is a famous bound by Froissart that basically says that for a gaped theory the amplitudes can't grow faster than $s\log^2 s$, where $s=E_{CM}^2$. In fact, QCD saturates it, while being a perfectly healthy theory. Moreover, if the theory isn't gaped, like QED, the total cross-section may be even infinite due to the Coulomb singularity. Finally, about the optical theorem, is practically impossible to check its violation if one is willing to add, order by order, the needed counterterm. $\endgroup$
    – TwoBs
    Sep 15, 2014 at 7:24
  • $\begingroup$ but led me add a positive example where the optical theorem can be used as a check. Take a theory for one Goldstone boson $\pi$ with a shift-symmetry $\pi\rightarrow \pi+c$. The effective lagrangian in the IR is something of the form $L=\frac{1}{2}(\partial_\mu\pi)^2+c(\partial_\mu\pi)^4+\ldots $. Now, the underlying UV theory that gave rise to this IR theory can't be unitary+analytic+crossing symmetric and deliver also $c<0$, because if you calculate the two body elastic forward scattering and look at $c\propto \partial^2_s A(s=0,t=0)\propto\int ds/s^2 \sigma_{tot}(s)>0$. Only $c>0$ is allowd $\endgroup$
    – TwoBs
    Sep 15, 2014 at 7:36
  • $\begingroup$ @TwoBs: Did you write about QCD as free nonabelian gauge theory without fermion interaction? If yes, I can already claim that it satisfies optical theorem. $\endgroup$ Sep 15, 2014 at 13:03
  • $\begingroup$ I am not sure what you mean with by ''free'' non-abelian gauge theory. I was actually referring to actual QCD and, as I said, some of the amplitudes do grow as $E^2 Log^2 E$ (and cross-sections grow logaritmically as $\log^2 s$) contrary to what you said about the need of non-infinite amplitudes and cross-sections. BTW, such maximally fast growth happens in any theory that respect Regge, like QCD where is reached by the exchange of what is called a Pomeron. The point is that QFT doesn't forbid amplitudes that grows as fast as $s\log^2s$, as QCD shows. $\endgroup$
    – TwoBs
    Sep 15, 2014 at 14:00
  • $\begingroup$ I am anyway interested in seeing your example to check the optical theorem in pure Yang-Mills (if that is what you refferred to with ''free'' non-abelian gauge theory without fermions) that confines at some scale $\Lambda$, say the glueball mass. In the forward limit that enters in the optical theorem you have $t\rightarrow 0$ which is lower than the confinement scale and non-perturbative tools are needed to calculate the left-hand side of the optical theorem. $\endgroup$
    – TwoBs
    Sep 15, 2014 at 14:06

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