Canonical quantisation of conjugate Dirac field

Question

This may be a stupid question to ask. For Dirac field, we know the Lagrangian $${\cal L}=\bar{\psi}(i\gamma^{\mu}\partial_{\mu}-m)\psi \tag{1}$$ is not symmetric in $\psi$ and its conjugate field $$\bar{\psi}=\psi^\dagger\gamma^0,\tag{2}$$ so that the momentum of the conjugate field is $$\bar{\pi}=0,\tag{3}$$ with $$\pi=i\psi^\dagger.\tag{4}$$ Then would it be incorrect to impose the anticommutation relation $$\{\bar{\psi}_\alpha(t,\vec{x}),\bar{\pi}_\beta(t,\vec{x}')\}=i\hbar\delta(\vec{x}-\vec{x}')\delta_{\alpha\beta},\tag{5}$$ right? Why is this different compared to the complex scalar field where we treat $\phi$ and $\phi^\dagger$ as independent fields but not for Dirac field and its adjoint, which is also some sort of complex conjugation?

Answer 1

In this answer we sketch the Dirac-Bergmann analysis for Dirac fermions$^1$:

1. More generally, the Dirac Lagrangian density in 3+1D is$^2$ $$\begin{align} {\cal L}~=~&\overline{\psi}\left(\frac{i}{2}\sum_{\mu=0}^3 \gamma^{\mu}\stackrel{\leftrightarrow}{\partial}_{\!\mu}-m\right)\psi +\sum_{\mu=0}^3\partial_{\mu}(\alpha\overline{\psi} \gamma^{\mu}\psi) \cr ~=~&\frac{i}{2}\psi^{\dagger}\stackrel{\leftrightarrow}{\partial}_t\psi +\partial_t(\alpha\psi^{\dagger}\psi) -{\cal H}, \cr {\cal H}~=~& \overline{\psi}\left(m-\frac{i}{2}\sum_{j=1}^3 \gamma^j\stackrel{\leftrightarrow}{\partial}_{\!j}\right)\psi-\sum_{j=1}^3\partial_j(\alpha\overline{\psi} \gamma^j\psi),\cr \overline{\psi}~:=~&\psi^{\dagger}\gamma^0, \cr (\gamma^0)^2~=~&{\bf 1}_{4\times 4}, \cr \stackrel{\leftrightarrow}{\partial}_{\!\mu}~:=~&\stackrel{\rightarrow}{\partial}_{\mu}-\stackrel{\leftarrow}{\partial}_{\mu}, \cr \alpha~\in~&\mathbb{R}, \end{align}\tag{A}$$ up to possible total divergence terms.

2. It is in principle possible to perform the Dirac-Bergmann analysis in complex$^3$ variables, cf. OP's question. However, the counting of degrees of freedom (DOF) and the enforcing of the constraints then become quite subtle, so let us for simplicity switch to real variables, i.e. the real and imaginary parts, which are independent: $$\begin{align} \psi~=~&(\psi^1+i\psi^2)/\sqrt{2},\cr \psi^{\ast}~=~&(\psi^1-i\psi^2)/\sqrt{2},\cr {\cal L}~=~&\frac{i}{2}\sum_{a=1}^2\psi^{aT}\dot{\psi}^a +\partial_t(i\alpha\psi^{1T}\psi^2) -{\cal H} \cr ~=~&\sum_{a=1}^2\pi_{aR}^{T}\dot{\psi}^a -{\cal H} \cr ~=~&\sum_{a=1}^2\dot{\psi}^{aT}\pi_{aL} -{\cal H}. \end{align}\tag{B}$$

3. Due to the presence of left & right derivatives, we can define left & right canonical momenta $$ \begin{align} \pi^T_{1R}~=~&\frac{\partial_R{\cal L}}{\partial\dot{\psi}^1} ~=~ \frac{i}{2}\psi^{1T}-i\alpha\psi^{2T} ~=~-\frac{\partial_L{\cal L}}{\partial\dot{\psi}^1} ~=~-\pi^{1T}_L , \cr \pi^T_{2R}~=~&\frac{\partial_R{\cal L}}{\partial\dot{\psi}^2} ~=~ \frac{i}{2}\psi^{2T}+i\alpha\psi^{1T} ~=~-\frac{\partial_L{\cal L}}{\partial\dot{\psi}^2} ~=~-\pi^{2T}_L , \end{align}\tag{C}$$ which are Grassmann-odd and imaginary. It seems inconsistent to introduce a complex momentum $\pi_a$ for a real variable $\psi^a$, but an imaginary momentum $\pi_a$ is fine, as long as the degrees of freedom (DOF) match. (Here we use that the parameter $\alpha\in\mathbb{R}$ is real.)

4. Eq. (C) produces primary constraints $$ \begin{align} \chi_1~:=~i\pi_{1R}+\frac{1}{2}\psi^1-\alpha \psi^2 ~\approx~0, \cr \chi_2~:=~i\pi_{2R}+\frac{1}{2}\psi^2+\alpha \psi^1 ~\approx~0,\tag{D} \end{align}$$ which are Grassmann-odd and real.

5. The canonical super-Poisson brackets of fundamental phase space variables are $$\begin{align} \{\psi^a_{\alpha}({\bf x}), \pi^{\beta}_{bR}({\bf y})\}_{PB} ~=~&\delta^a_b \delta_{\alpha}^{\beta} \delta^3({\bf x}\!-\!{\bf y})\cr ~=~&-\{\pi^{\beta}_{bL}({\bf y}),\psi^a_{\alpha}({\bf x}) \}_{PB},\cr a,b~\in~&\{1,2\},\cr \alpha,\beta~\in~&\{1,2,3,4\}. \end{align}\tag{E} $$

6. It turns out that the primary constraints (D) are 2nd class constraints, and that there are no secondary constraints. The matrix elements of the 2nd class constraints (D) are $$ \begin{align} \Delta^{\alpha\beta}_{ab}({\bf x},{\bf y}) ~:=~&\{\chi^{\alpha}_a({\bf x}), \chi^{\beta}_b({\bf y})\}_{PB}\cr ~=~&i\delta_{ab}\delta^{\alpha\beta}\delta^3({\bf x}\!-\!{\bf y}), \end{align} \tag{F} $$ which are independent of the real parameter $\alpha\in\mathbb{R}$. The inverse matrix elements are $$ (\Delta^{-1})^{ab}_{\alpha\beta}({\bf x},{\bf y}) ~=~-i\delta^{ab}\delta_{\alpha\beta}\delta^3({\bf x}\!-\!{\bf y}). \tag{G} $$

7. The non-zero Dirac brackets of the fundamental fields are $$\begin{align} \{\psi^a_{\alpha}({\bf x}), \psi^{b}_{\beta}({\bf y})\}_{DB} ~=~&\{\psi^a_{\alpha}({\bf x}), \psi_{b}^{\beta}({\bf y})\}_{PB}\cr &-\sum_{c,d=1}^2\sum_{\gamma,\delta=1}^4\int_{\mathbb{R}^6}\! d^3{\bf u}~d^3{\bf v} \{\psi^a_{\alpha}({\bf x}), \chi^{\gamma}_c({\bf u})\}_{PB}\cr &\qquad\qquad (\Delta^{-1})^{cd}_{\gamma\delta}({\bf u},{\bf v})~\{\chi^{\delta}_d({\bf v}), \psi_{b}^{\beta}({\bf y})\}_{PB}\cr ~=~&-i\delta^{ab}\delta_{\alpha\beta}\delta^3({\bf x}\!-\!{\bf y}), \cr \{\psi_{\alpha}({\bf x}), \psi^{\ast}_{\beta}({\bf y})\}_{DB} ~=~&-i\delta_{\alpha\beta}\delta^3({\bf x}\!-\!{\bf y}). \end{align}\tag{H} $$

8. The non-zero equal-time anticommutators of the fundamental operator fields are multiplied with $i\hbar$: $$\begin{align} \{\hat{\psi}^a_{\alpha}({\bf x},t), \hat{\psi}^b_{\beta}({\bf y},t)\}_{+}~=~&\hbar\delta^{ab}\delta_{\alpha\beta}\hat{\bf 1}~\delta^3({\bf x}\!-\!{\bf y}),\cr \{\hat{\psi}_{\alpha}({\bf x},t), \hat{\psi}^{\dagger}_{\beta}({\bf y},t)\}_{+}~=~&\hbar\delta_{\alpha\beta}\hat{\bf 1}~\delta^3({\bf x}\!-\!{\bf y}). \end{align}\tag{I} $$

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$^1$ For more explanation and other methods, such as the Faddeev-Jackiw method, see e.g. this & this related posts.

$^2$ Here we assume that the Lagrangian density ${\cal L}$ is real, cf. e.g. this Phys.SE post. OP's Lagrangian density (1) corresponds to $\alpha=\frac{i}{2}$, which is not real.

The $\dagger$-symbol is a shorthand for the combined $T$- and $\ast$-symbol.

$^3$E.g. the left & right canonical momenta are $$ \begin{align} \pi^T_R~=~&\frac{\partial_R{\cal L}}{\partial\dot{\psi}} ~=~ \left(\frac{i}{2}+\alpha\right)\psi^{\dagger} ~=~-\frac{\partial_L{\cal L}}{\partial\dot{\psi}} ~=~-\pi^T_L , \cr \pi^{\dagger}_R~=~&\frac{\partial_R{\cal L}}{\partial\dot{\psi}^{\ast}} ~=~ \left(\frac{i}{2}-\alpha\right)\psi^T ~=~-\frac{\partial_L{\cal L}}{\partial\dot{\psi}^{\ast}} ~=~-\pi^{\dagger}_L . \end{align}\tag{J}$$ OP's momenta (3) & (4) correspond to $\alpha=\frac{i}{2}$.

Note that the corresponding momenta $\pi$ and $\pi^{\ast}$ are not necessarily complex conjugate fields, i.e. the $\ast$-symbol does not necessarily denote complex conjugation.

Also be aware that some authors exchange the meaning of $\pi$ and $\pi^{\ast}$.