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Question:
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?

Community
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John
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2 Answers2

5

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.

Steven Mathey
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1

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.

Andrew McAddams
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