Euler-Lagrange for Dirac Lagrangian - is $\bar \psi$ independent of $\psi$?

Question

In Peskin and Schroeder (3.34) they write the Dirac Lagrangian:

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

where $\bar \psi = \psi^\dagger \gamma^0$.

Then, they write: "The Euler-Lagrange equation for $\bar\psi$ immediately yields the Dirac Equation" $(i \gamma^\mu \partial_mu - m) \psi = 0$. This is just saying

$$ \frac{\partial L }{\partial \bar \psi} = 0 $$

because there is no term dependent on $\partial_\mu \bar \psi$. However, to me this is a very strange statement, because $\bar \psi$ does not appear to be independent of $\psi$, and therefore it's not clear to me that you can do the calculus of variations on $\psi$ and $\bar \psi$ independently. Can someone help justify this to me?

Answer 1

We generally treat $\psi$ and $\bar\psi$ as independent variables. This follows from the fact that the most general form of $\psi$ (and consequently the implied form of $\psi^\dagger$, where $\bar\psi=\psi^\dagger\gamma^0$) can be written in terms of two independent real fields $\psi_1$ and $\psi_2$ as $$\psi=\psi_1+\mathrm{i}\psi_2;\quad\psi^\dagger=\psi_1-\mathrm{i}\psi_2.$$

If you instead consider the Euler-Lagrange equation for a real field $\begin{bmatrix}\psi_1\\\psi_2\end{bmatrix}$, you will recover the same equations of motion.

I suspect there are a few typos (or perhaps an incomplete quotation) in the question's version of the Dirac equation for $\bar\psi$: the complex field and the conjugate thereof should (respectively) give EL equations $$\quad-\mathrm{i}\partial_\mu\gamma^\mu\bar\psi-m\bar\psi=0;\quad\mathrm{i}\gamma^\mu\partial_\mu\psi-m\psi=0.$$

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An alternative, and perhaps more compelling way to think about this is by realizing that $\psi$ and $\psi^\dagger$ are a conjugate pair. This follows from the expression for the canonical momentum implied by the Lagrangian: $$\mathcal{L}=\bar\psi(\mathrm{i}\gamma^\mu\partial_\mu-m)\psi\implies\Pi=\frac{\delta\mathcal{L}}{\delta\partial_0\psi}=\mathrm{i}\psi^\dagger.$$ By analogy to classical mechanics, where we treat the position and momentum independently, we should do the same for $\psi$ and $\Pi$.

Answer 2

While $\psi$ and $\bar{\psi}$ do not appear to be independent, the key point is that they are linearly independent. This is because the operation of complex conjugation (in the hermitian conjugate) is not actually a linear operation.

For example, try to write the complex number $z$ in terms of $z^*$ using only linear operations (addition and scalar multiplication). You'll see that it's impossible. Similarly, you cannot write $\psi$ in terms of $\bar{\psi}$ linearly, so they must be independent.

The reason why there is an additional "degree of freedom" created here is because a complex field necessarily has double the degrees of freedom of the corresponding real field. So if you demand that $\psi$ is real, then $\psi$ and $\bar{\psi}$ wouldn't be independent anymore.