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I'm looking for generalizations of the Euler-Lagrange equations that would be derived from a complex-valued Lagrangian density. I realize that “minimum” and “maximum” don't have obvious meaning for a complex-valued action, so am looking for E-L equations that correspond to (a) constant amplitude of the action, (b) constant phase of the action, or (c) both.

The papers I've found mostly avoid the issue by allowing complex valued field variables within the Lagrangian but ensuring that the Lagrangian itself is real-valued.

This paper might be relevant: Non-standard complex Lagrangian dynamics

Any advice will be welcome.

Qmechanic
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S. McGrew
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    There are instances when, even though the action is real, the complex saddle points (or trajectories) have to be taken into account. See e.g. this. – mavzolej Mar 16 '19 at 22:41
  • @S. McGrew: Do you have any physical system in mind? Which? – Qmechanic Mar 17 '19 at 03:17
  • For what it's worth, see my answer at https://physics.stackexchange.com/q/438956/ – akhmeteli Mar 17 '19 at 03:29
  • @akhmeteli, it's a useful comment. I'm not concerned at this point whether the result is physical or not; just what the result looks like. – S. McGrew Mar 17 '19 at 03:41
  • @Qmechanic, yes, I have a physical system in mind but don't want to fog the question by mentioning at this point what the system is. However, variation of the Feynman path integral would be (I think) vaguely analogous to variation of a complex Lagrangian. – S. McGrew Mar 17 '19 at 03:46
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    The action in FPI is real. – Qmechanic Mar 17 '19 at 03:57
  • Understood, but the integrand is complex. Variation of that integral is what would be vaguely analogous, I think. – S. McGrew Mar 17 '19 at 04:12
  • @Qmechanic, I added a link to a paper that appears relevant. – S. McGrew Mar 17 '19 at 13:33

2 Answers2

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If you have a complex action, you need to decide what you require to be stationary. It can be a) action provided its amplitude is constant; b) action provided its phase is constant; c) amplitude of the action; d) phase of the action; e) real part of the action, etc. In each of these cases this is equivalent to requiring that some real action is stationary, for example, in case c) you can choose action equal to the amplitude of the "old" complex action.

So what happens if you require, say, that both the amplitude and the phase of the complex action are stationary? As each of these requirements is typically sufficient to get equations of motion, both of these requirements together typically yield an overdetermined system of equations. It is possible, however, that this overdetermined system is consistent and makes sense, but I cannot offer an example at the moment.

akhmeteli
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  1. A stationary action principle for a complex action $S_c=S_1+iS_2\in \mathbb{C}$ is equivalent to 2 real stationary action principles for the real and imaginary part, $S_1,S_2\in\mathbb{R} $. In other words, the EL equations for $S_c$ are precisely the EL equations for $S_1$ and the EL equations for $S_2$. It may be possible to organize the EL equations for $S_c$ as complex equations, especially if the Lagrangian is holomorphic.

  2. In the Feynman path integral $Z$, the action $S$ is real, at least in the Minkowskian formulation. However, when evaluating the semiclassical approximation via the method of steepest descent, one typically deforms the contour of integration into the complex plane, which may lead to complex contributions to $Z$. Stationary points in the complex plane may or may not have a straight-forward physical interpretation as solutions to the (analytic continued) EL equations.

Qmechanic
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  • In other words, a complex Lagrangian yields two sets of standard-form EL equations, one for the real part of the Lagrangian and one for the imaginary part of the Lagrangian? In that case, unless the field variables themselves are complex it seems that, as @akhmeteli pointed out, the equations would be overdetermined. – S. McGrew Mar 17 '19 at 18:59
  • Not necessarily. – Qmechanic Mar 17 '19 at 19:04
  • Can you provide a simple example? – S. McGrew Mar 17 '19 at 19:15
  • OK, I can provide an example, but it's too simple: if the real and imaginary parts of the Lagrangian lead to the same equations; and they don't need to be identical to do that. – S. McGrew Mar 18 '19 at 14:01
  • I can see that if the two sets of equations are the same, they won't be overdetermined. Are there other cases you can think of where the sets won't be overdetermined? – S. McGrew Mar 19 '21 at 01:25