Deriving adjoint Dirac equation from Lagrangian

Question

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

Answer 1

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.

Answer 2

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, $$ Δ\mathcal{L} = Δ\bar{ψ} \left(iħ γ^μ \overrightarrow{∂_μ} - mc\right) ψ - \bar{ψ}\left(iħ \overleftarrow{∂_μ} γ^μ + mc\right) Δψ + ∂_μ \left(\bar{ψ} iħ\overleftrightarrow{γ^μ Δ} ψ\right). $$