# How do I derive the Dirac Lagrangian?

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?

• Integrate by part the $\partial \bar{\psi }\gamma \psi$ term to get back the first Lagrangian Commented Nov 15, 2022 at 13:29
• These differ by $\frac12i\partial_\mu(\overline{\psi}\gamma^\mu\psi)$. What do you mean by "deriving" either Lagrangian?
– J.G.
Commented Nov 15, 2022 at 15:52
• Doesn’t the second have half the mass of the first? Commented Nov 15, 2022 at 18:16

Both are perfectly valid. I actually prefer the second.

• Well, because the second follows from the first one. But in general, we could multiply lagrangian by 2, for example. That would not affect the equation we derive, but it is not the same lagrangian. Commented Nov 15, 2022 at 13:33
• You get one from the other by adding a total derivative. That is the usual way of getting equivalent Lagrangians -both classical and quantum. Commented Nov 15, 2022 at 13:37
• I’m confused. How are they equivalent when the second has a mass term that is half the mass term in the first? Commented Nov 15, 2022 at 17:24
• You are right. The mass term should be outside the parentheses, I did not notice that. Commented Nov 15, 2022 at 17:29

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.