How do I derive the Dirac Lagrangian?

Question

It's frequently said, that the Lagrangian of a Dirac field is

$$\mathcal{L}=i\bar{\psi}(\gamma^\mu\partial_\mu-m)\psi.$$

Applying the Euler-Lagrange equation we get the Dirac equation. Although, we can get a similar construction of Lagrangian, which leads to the same equation, e.g.

$$\mathcal{L}=\frac{i}{2}\left[\bar{\psi}\gamma^\mu\partial_\mu\psi-\left(\partial_\mu\bar{\psi}\right)\gamma^\mu\psi-m\bar{\psi}\psi\right].$$

Is there a way to derive the Lagrangian we normally use?

Answer 1

Both are perfectly valid. I actually prefer the second.

Answer 2

The second one is $𝔏 = \bar{ψ} \left(iħ∂^↔ - mc\right) ψ$, where $∂^↔ = ½(∂ - ∂^←)$, not $𝔏 = \bar{ψ} \left(iħ∂^↔ - ½ mc\right) ψ$, which is what you wrote. I'm using the notations $∂ = γ^μ ∂_μ$ (with the summation convention), with $∂^←$ being $∂$ acting to the left.

So, $$\bar{ψ} ∂^↔ ψ = ½\left(\bar{ψ} ∂ψ - ∂_μ\bar{ψ} γ^μ ψ\right) = ∂_μ\left(-½\bar{ψ} γ^μ ψ\right) + \bar{ψ} ∂ψ.$$ The two Lagrangians are equivalent up to a "total divergence" term.

Your question has the appearance of a much broader question: "How do you derive any Lagrangian for the Dirac equation?", when the reality of the matter is that writing down a Lagrangian is more of a situation of throwing mud at a wall to see what sticks.

In fact, we can throw entire mud cakes at the wall and make it all stick, turning the whole place into an adobe. Caking up the wall will also make the theory look normal (for a change): a Lagrangian for a second order field law that is not totally singular (i.e. 0 on-shell), but regular.

Rewrite $ψ = (iħ∂ + mc)χ$, where we will refer to $χ$ as the "fermionic potential" and $ψ$ as the "fermionic field strength". Then, the Dirac equation $(iħ∂ - mc)ψ = 0$ reduces to the Klein-Gordon equation $\left(-ħ^2∂^2 - (mc)^2\right)χ = 0$. The fermionic field strength is invariant with respect to a "fermionic gauge transform": $χ → χ + (iħ∂ - mc)ω$, provided also that $\left(-ħ^2∂^2 - (mc)^2\right)ω = 0$.

The Lagrangian density $𝔏 = ½\left(\overline{(iħ∂χ)} (iħ∂χ) - \overline{(mcχ)}(mcχ)\right)$ yields the Klein-Gordon equation for the fermionic potential $χ$. Under the above-mentioned "fermionic gauge transform" the Lagrangian density transforms to itself plus a residual term ... that happens to be a total divergence because of the condition placed on $ω$. So, the action is invariant with respect to the "fermionic gauge transform". The mud sticks to the wall and cakes it up into an adobe.

Things get really interesting if you try to use a similar trick with the Standard Model Lagrangian; but that's another matter that may have to be taken up at another time.