# Deriving adjoint Dirac equation from Lagrangian

I'm trying to derive the adjoint Dirac equation from the Lagrangian:

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

To start, I plugged it into the Euler-Lagrange equation with my variation variable being $$\psi$$:

$$\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)}\right)-\frac{\partial\mathcal{L}}{\partial\psi}=0$$

Which yields:

$$\partial_\mu(i\bar{\psi}\gamma^\mu)-m\bar{\psi}=0$$

Next, I factored out the $$i$$, then used the anticommutation relation $$\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$$ to swap $$\gamma^0$$ (from the adjoint wavefunction) and $$\gamma^\mu$$. However, when I do this I don't get something resembling the Dirac adjoint; I have an extra $$2\partial_\mu\psi^\dagger$$ term, so I must have done something wrong (I end up getting $$2\partial_\mu\psi^\dagger-\partial_\mu(\psi^\dagger\gamma^\mu\gamma^0)-m\bar{\psi}=0$$). Could I just have a pointer in the right direction; I'm unsure how to proceed from what I have from the Euler-Lagrange equation to $$\bar{\psi}(i\gamma^\mu\partial_\mu+m)=0.$$

If you want to derive the equation of motion for $$\bar \psi$$, then you already have it. It's commonly written as

$$\bar \psi \left(i\gamma^\mu \overleftarrow{\partial}_\mu+m\right)=0$$

Where it's understood that we should act with the derivative on $$\bar\psi$$ as normal, i.e. $$\partial_\mu \bar \psi$$, but that the $$\gamma^\mu$$ must still multiply on the right.

• I understand that's the equation, but how do I get from $\partial_\mu (i\bar{\psi}\gamma^\mu)-m\bar{\psi}=0$ to the equation you wrote. Does the derivative going to the right also conjugate $i$? Commented Jan 21, 2022 at 20:01
• @moboDawn_φ well your second term in the equation of motion $-\frac{\partial L}{\partial \psi}$ is off by a minus sign. Also your Lagrangian appears to be missing a factor of $i$ in the first term. Commented Jan 21, 2022 at 20:03
• Oh I see. Sorry, that was a stupid mistake on my part. Commented Jan 21, 2022 at 20:04
• @moboDawn_φ not stupid, just a small mistake. Happens all the time. :) Commented Jan 21, 2022 at 20:28
• Thanks for pointing it out to me! Commented Jan 21, 2022 at 20:29

With all the constants made explicit... \begin{align} \mathcal{L} &= iħ \bar{ψ} γ^μ ∂_μ ψ - mc \bar{ψ} ψ\\ Δ\mathcal{L} &= iħ Δ\bar{ψ} γ^μ ∂_μ ψ + iħ \bar{ψ} γ^μ ∂_μ Δψ - mc Δ\bar{ψ} ψ - mc \bar{ψ} Δψ\\ &= iħ Δ\bar{ψ} γ^μ ∂_μ ψ + \left(∂_μ (iħ \bar{ψ} γ^μ Δψ) - iħ ∂_μ\bar{ψ} γ^μ Δψ\right) - mc Δ\bar{ψ} ψ - mc \bar{ψ} Δψ\\ &= Δ\bar{ψ} \left(iħ γ^μ ∂_μ ψ - mc ψ\right) - \left(iħ ∂_μ\bar{ψ} γ^μ + mc \bar{ψ}\right) Δψ + ∂_μ \left(iħ \bar{ψ} γ^μ Δψ\right)\\ &⇒ iħ γ^μ ∂_μ ψ - mc ψ = 0, \quad iħ ∂_μ\bar{ψ} γ^μ + mc \bar{ψ} = 0, \quad Δ\mathcal{L} ≡ ∂_μ \left(iħ \bar{ψ} γ^μ Δψ\right) \end{align} where "≡" denotes "on-shell" equality, with $$Δ\mathcal{L}$$ reduced, on-shell, to the variational involving the involving the boundary term.

It should actually be made symmetric so that the boundary term gets contributions from $$Δ\bar{ψ}$$ as well as from $$Δψ$$: $$\mathcal{L} = \bar{ψ} \left(iħ \overleftrightarrow{γ^μ ∂_μ} - mc\right) ψ,$$ where $$\overleftrightarrow{γ^μ ∂_μ} = \frac{γ^μ \overrightarrow{∂_μ} - \overleftarrow{∂_μ} γ^μ}{2}.$$

Then \begin{align} \mathcal{L} &= \frac{1}{2}\overbrace{\bar{ψ} \left(iħ γ^μ \overrightarrow{∂_μ} - mc\right) ψ}^{\overrightarrow{\mathcal{L}}} - \frac{1}{2}\overbrace{\bar{ψ} \left(iħ \overleftarrow{∂_μ} γ^μ + mc\right) ψ}^{\overleftarrow{\mathcal{L}}}\\ Δ\overrightarrow{\mathcal{L}} &= Δ\left(\bar{ψ}\left(iħ γ^μ \overrightarrow{∂_μ} - mc\right) ψ\right)\\ &= Δ\bar{ψ} \left(iħ γ^μ \overrightarrow{∂_μ} - mc\right) ψ - \bar{ψ}\left(iħ \overleftarrow{∂_μ} γ^μ + mc\right) Δψ + ∂_μ \left(iħ \bar{ψ} γ^μ Δψ\right)\\ Δ\overleftarrow{\mathcal{L}} &=Δ\left(\bar{ψ}\left(iħ \overleftarrow{∂_μ} γ^μ + mc\right) ψ\right)\\ &= \bar{ψ}\left(iħ \overleftarrow{∂_μ} γ^μ + mc\right) Δψ - Δ\bar{ψ} \left(iħ γ^μ \overrightarrow{∂_μ} - mc\right) ψ + ∂_μ \left(iħ Δ\bar{ψ} γ^μ ψ\right)\\ &⇒\\ Δ\mathcal{L} &= Δ\bar{ψ} \left(iħ γ^μ \overrightarrow{∂_μ} - mc\right) ψ - \bar{ψ}\left(iħ \overleftarrow{∂_μ} γ^μ + mc\right) Δψ + ∂_μ \left(iħ \frac{\bar{ψ} γ^μ Δψ - Δ\bar{ψ} γ^μ ψ}{2}\right) \end{align}.

Defining $$\overleftrightarrow{γ^μ Δ} = \frac{γ^μ \overrightarrow{Δ} - \overleftarrow{Δ} γ^μ}{2},$$ the boundary term can be written: $$∂_μ \left(\bar{ψ} iħ\overleftrightarrow{γ^μ Δ} ψ\right).$$ Then, $$Δ\mathcal{L} = Δ\bar{ψ} \left(iħ γ^μ \overrightarrow{∂_μ} - mc\right) ψ - \bar{ψ}\left(iħ \overleftarrow{∂_μ} γ^μ + mc\right) Δψ + ∂_μ \left(\bar{ψ} iħ\overleftrightarrow{γ^μ Δ} ψ\right).$$