Moving from Dirac equation to Lagrangian density

1. Sep 1, 2009

Hi all,

As a blind follower of QFT from the sidelines (the joys of the woefully inadequate teaching of theory to exp. particle physics students...), I have just realised that I've never actually gone further than deriving the Dirac equation, and then just used the Dirac Lagrangian density as the basis for learning about gauge theories.

What is not clear to me is how to move from the Dirac equation to the Dirac Lagrangian density ($$\bar\psi\left(i\gamma^\mu\delta_\mu -m\right)\psi$$). I've been playing around with a few ideas but am getting absolutely nowhere. There is a glib explanation in P&S, but it struck me as somewhat circular and I wasn't convinced at all. Srednicki doesn't cover it either, as far as I can see (I have the former in front of me, the latter on my desk at work...).

Any help welcome!

Cheers.

2. Sep 1, 2009

javierR

I'm not sure what you're looking for here in order to be satisfied. The Dirac equation is the "equation of motion" for the Lagrangian (upon variation with respect to Psi-bar). This is just variational calculus. Going the other way, you could just look for the Lorentz-invariant Lagrangian that would have the Dirac equation as the eqn of motion. The modern point of view is to start with a Lagrangian/action as the fundamental entity from which field equations follow. In that case, you could ask how could we know to write down the Dirac action. Well if you want massive spin-1/2 fields in a Lorentz-invariant action, the simplest one you can write down is the Dirac action. The building block objects are: $$\bar{\Psi}\Psi$$ scalar, $$\bar{\Psi}\gamma^{\mu}\Psi$$ vector, and so on for higher rank tensors.

3. Sep 1, 2009

olgranpappy

To go from the "first quantized" form to the "second quantized" form of any single-particle Hamiltonian "h". E.g.,
$$h=-\nabla^2/2m\;,$$
we introduce position-space particle creation- and annihilation-operators, $\psi^\dagger({\bf x})$ and $\psi({\bf x})$, respectively. These are not wavefunctions.

The second quantized form is then given by
$$\int d^3 x \left(\psi^\dagger({\bf x})(h\psi({\bf x}))\right)\;.$$

This is so because the action of the above hamiltonian on any N-particle ground state returns the correct first-quantized hamiltonian--in this case
$$\sum_{i=1}^N-\nabla_i^2/2m\;,$$
and thus the second quantized form is the appropriate generalization to the case where there is a changing number of particles (Fock space).

The above generalizes pretty straightforwardly to any single body hamiltonian (e.g., the dirac equation).

To include interactions is also not too hard. I recommend the book by Lowell Brown on QFT.

4. Sep 1, 2009

olgranpappy

Oh... I think I just answered the wrong question. OP is interested in action and lagrangian and I was talking about hamiltonians... ugh. sorry. Again, the answer is in L. Brown's book, thought. check it out. Cheers.

5. Sep 1, 2009

Thanks for the response. Sure, it is indeed being able to 'go the other way' that is not immediately, uniquely apparent.

Right, quite happy with this, just wanted to make sure I wasn't missing anything. Thanks!

6. Sep 1, 2009

RedX

If $$(i\gamma^\mu\delta_\mu -m)\psi=0$$ is the Dirac eqn. then as a first guess the Lagrangian would be $$\psi^{\dagger}(i\gamma^\mu\delta_\mu -m)\psi$$ so that varying with respect to $$\psi^{\dagger}$$ would give the Dirac eqn. However, that would be wrong and the answer is $$\psi^{\dagger}\gamma^{0}(i\gamma^\mu\delta_\mu -m)\psi$$. So an extra gamma^0 got snuck in there for $$\psi^{\dagger}$$ to be related to $$\psi$$ by the adjoint operation.

But if you ask yourself why $$\psi^{\dagger}$$ must be related to $$\psi$$ by the adjoint, why can't it truly be an independent field instead, then the answer is because you want only one field. Or Hermiticity or something.

I think also the other answer of finding the Hamiltonian first is good, and then doing a Legendre transformation. Finding the conserved energy to find the Lagrangian or something.