Brian Bi

## How do I construct a Lagrangian from a given field equation (besides solving the E-L differential equation)?

If anyone wants to show this mathematically, can they use the example of the Schrodinger or Dirac equations? Thank You!
I'm a bit confused by your question, where you say "besides solving the E-L differential equation". I'll just ignore that part. So you want to know how to go from the equations of motion to the Lagrangian, whereas we usually do things the other way around.

I don't think there's a systematic procedure for doing this, although I would love to be proven wrong. One difficulty is that to any physically correct Lagrangian density we can add a divergence and in doing so end up with another Lagrangian density that generates the same equations of motion. Of course, this doesn't mean finding a systematic procedure is impossible; but even if you could find one, it might not give you the Lagrangian density you want. 

That being said, you can usually guess the correct Lagrangian density and then just check whether it gives the correct field equations. If you look at the form of the Euler–Lagrange equations:$\frac{\partial \mathcal{L}}{\partial \psi} - \partial_\mu \frac{\partial L}{\partial(\partial_\mu \psi)} = 0$ you can see what general kind of transformation they induce in going from the Lagrangian density to the field equations. Where you have a term of degree $$n$$ in $$\psi$$ in the Lagrangian density you'll get one of degree $$n-1$$ in the equations of motion, so if you see a term of degree $$n$$ in the EOM then put in a term of degree $$n+1$$ in the Lagrangian density. We can similarly work backward with the other term.

For example, here's the Dirac equation: $(i\hbar \gamma^\mu\partial_\mu - m)\psi = 0$ We see a term $$-m\psi$$ so we can guess that it probably comes from a term in the Lagrangian proportional to $$-m\psi^2$$. As for the term $$i\hbar \gamma^\mu \partial_\mu \psi$$, if it came from the $$\partial\mathcal{L}/\partial\psi$$ part then the corresponding term in the Lagrangian must have been something like $$i \hbar \psi \gamma^\mu \partial_\mu \psi$$. If it came from the other term in the E–L equations, then we can work backward by undoing a derivative, giving $$i\hbar \gamma^\mu \psi$$, then inserting a factor of $$\partial_\mu \psi$$. Either way the conclusion is the same: there's probably a term that contains the field multiplied by its derivative (and the gamma matrices).

Now, we want our Lagrangian density to be a manifest Lorentz scalar. The only scalar quadratic in the field is $$\overline \psi \psi$$.  We also want a term that's linear in the field and linear in the derivative. The only way to construct such a term and have it be Lorentz invariant is:$\overline\psi \gamma^\mu \partial_\mu \psi$ or its conjugate.  And we had a factor of $$i\hbar$$, don't forget.

So we can write down the Lagrangian density$\mathcal{L} = i\hbar \overline \psi \gamma^\mu \partial_\mu \psi - m \overline \psi \psi$ and this is actually correct. But you can always determine the constant factors easily, anyway.

So the general procedure was:
1. For each term in the EOM, work backward using the E–L equations to determine the degree of the corresponding term in the Lagrangian density
2. Use Lorentz invariance (or whatever symmetry applies) to narrow down the possible range of terms that can actually appear in the Lagrangian density
3. Figure out the constant factors (a piece of cake as you can just evaluate the Euler–Lagrange operator on each term)

 Example: The standard Lagrangian density for the free classical electromagnetic field is $$\mathcal{L} = -\frac{1}{4\mu_0} F^{\mu\nu} F_{\mu\nu}$$. But you can add any multiple of $$G^{\mu\nu} G_{\mu\nu}$$ and get another valid Lagrangian density for the field, because this is a four-divergence.
 There's also a pseudoscalar, $$\overline \psi \gamma^5 \psi$$.
 Again, you can insert a $$\gamma^5$$, which would make this into a pseudoscalar.
 I set $$c = 1$$ for convenience.