QED: Lagrangian, and Action principle

[JustinLevy]

I'm probably making a mistake, but looking at the free field lagrangian for QED
$$\mathcal{L} \propto (-F^{\mu\nu}F_{\mu\nu}) \propto (\mathbf{E}^2 - \mathbf{B}^2)$$
it appears to me that the action is not bounded from above, nor from below.

Does that mean the equations of motion we obtain by finding the path of extremal action is actually just a "saddle point"?

Regardless of the answer to that, what does it / would it mean for QED if the action is not bounded from below?

 

[JustinLevy]

No one?

Can someone at least confirm that the action is not bounded from below or above? I see some books referring to the action principle as a minimizing principle, while others do comment that it is actually just the extremal/stationary action that gives the classical equations of motion. However I haven't been able to find a textbook yet that explicitly says the action is not bounded from below.

Hopefully that is an easier question, being a yes/no:
Is the action for electrodynamics indeed unbounded from below and above?

 

[tom.stoer]

my speculation wouldbe the following: yes, that seems to be the case, but it could be that this obstacle is related to the gauge symmetry and disappears as soon as one fixes the gauge

 

[JustinLevy]

The gauge can't affect anything because the action is gauge independent.

 

[tom.stoer]

:-) That's not true.

If you vary the action in order to derive its extrema, you must distinguish between physical and unphysical directions (in the variation). Gauge fixing the action means expressing it terms of physical plus unphysical variables. The variation is taken only w.r.t. physical degrees of freedom. So it could be that if you exclude the gauge direction, the action is convex in the other, physical directions.

But I agree that this is somehow strange in QED as both E and B are manifest gauge invariant (so there are no unphysical directions in E and B).

 

[JustinLevy]

Yes interaction terms like j.A are not gauge independent, yet they yield the gauge independent equations of motion for "physical" directions. But the free field term is completely gauge independent, no?

I think you are hitting upon something important here, but I am misunderstanding.

You are saying, if we could separate the terms in the action due to the unphysical pieces and the physical pieces, that it may only be the unphysical terms which allow the action to be unbounded? I don't understand how the action could be in terms of gauge independent quantities, yet the question of whether it is bounded or not is gauge dependent?

 

[tom.stoer]

Yes, I agree, I think I am wrong.

 

[tom.stoer]

What about the following idea:

The Lagrangian looks like E²-B²

But E² is nothing else but v² in classical mechanics; so deriving the Hamiltonian we get

E² + B²

which corresponds to

p² + V(x)

But now everything is fine, isn't it?

Look at the harmonic oscillator; again the action is not bounded from below.

Isn't this one reason why one does a Wick rotation in the path integral?

 

[JustinLevy]

Okay, so the idea is that all that matters is that the Hamiltonian is bounded from below ("stable vacuum")? I guess that makes sense.