6
$\begingroup$

$\bullet$ Is it fair to assume that the classical Hamiltonian or Lagrangian of a system (a particle or a field) is always a real-valued function?

$\bullet$ If not, can you provide counter-examples?

$\bullet$ The cases in which the Hamiltonian represents the total energy of a system, it must be real on physical grounds. Apart from that is there any other criterion which demands that a Lagrangian or Hamiltonian must be real?

$\endgroup$
3
  • $\begingroup$ Using your comment about the Hamiltonian and Energy as a starting point, the Lagrangian is essentially the Laplace transform of H (or -H). So, under normal circumstances L would be real too. Unless you added some imaginary degrees of freedom to the configuration space. Another question might be whether you could add non-real dof to the system while maintaining a real valued H and L. $\endgroup$
    – user196418
    Nov 4, 2018 at 22:04
  • 2
    $\begingroup$ possible duplicate: Hermiticity of the Lagrangian in QFT. $\endgroup$ Nov 4, 2018 at 22:24
  • 2
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/127797/2451 , physics.stackexchange.com/q/46528/2451 and links therein. $\endgroup$
    – Qmechanic
    Nov 4, 2018 at 22:24

2 Answers 2

5
$\begingroup$

Any physical lagrangian or hamiltonian is real. There may be models out there in which $\cal{L}$ and $\cal{H}$ are complex, but in those cases the "complexity" must be for mathematical convenience, and the quantities calculated in such a model will need to be converted to real quantities or simply not correspond to physical observables. One strong reason for this is that the integral of a complex function is, in general , complex as well. Therefore a complex lagrangian would give rise to a complex action, which doesn't make any sense. Even if the hamiltonian doesn't correspond to the energy, a complex-value $\cal{H}$ would result in a complex lagrangian, so the problem remains.

$\endgroup$
2
  • $\begingroup$ Why doesn't a complex action make sense? @Othin $\endgroup$
    – SRS
    Nov 5, 2018 at 16:58
  • 1
    $\begingroup$ You need a real action to get measurable (real) quantities in classical physics. Of course you may still use complex variables, but in the end this will either be some kind of mathematical trick (like making a transformation and, later on, making sure to extract a real quantity out of it) or some complex field not corresponding to any measurable quantity, like a wave function. You would then need to postulate some device that allows physical interpretation like we do, for example, when we interpret the square of the w.f as a probability density. $\endgroup$
    – Othin
    Nov 6, 2018 at 17:53
4
$\begingroup$

I am not sure a classical Lagrangian must be real. The reason is as follows. If we start with some real Lagrangian and add a complex total divergence, we obtain a complex Lagrangian, but this final Lagrangian will have the same equations of motion as the initial Lagrangian, so one can use this final Lagrangian, if it is more convenient for some reasons, and obtain correct results.

Example: let us note that the standard Dirac Lagrangian $\bar{\psi}(i\gamma^\mu\partial_\mu-m)\psi$ is generally complex (and you can consider this theory classical before second quantization), but you can add a complex total divergence and make this Lagrangian real (symmetric Lagrangian $\frac{i}{2} (\bar\psi \gamma^\mu \partial_\mu \psi - (\partial_\mu \bar\psi) \gamma^\mu \psi ) - m \bar\psi \psi$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.