Wrong sign anticommutation relation for the Dirac field?

Consider the Dirac Lagrangian $$\mathcal{L}=\psi ^{\dagger }\gamma ^{0}\left( \mathrm{i}\gamma ^{\rho }\partial _{\rho }-m\right) \psi .$$ The conjugate momenta to $\psi ^{a}$ are defined, as usual, by $$\pi _{a}=\partial \mathcal{% L}/\partial \dot{\psi}^{a}.$$ All references I have consulted claims that this implies that $\pi _{a}=\mathrm{i}\psi _{a}^{\dagger }$, which seems quite obvious. But why not $$\pi _{a}=-\mathrm{i}\psi _{a}^{\dagger }$$ in view of fermion anticommutation, the minus sign arising from having to pass $% \partial /\partial \dot{\psi}^{a}$ through $\psi ^{\dagger }$? Using such a minus sign, though, leads of course to a nonsensical minus sign in the anticommutation relation, $$\left\{ \psi ^{a}\left( \mathbf{x}\right) ,\psi _{b}^{\dagger }\left( \mathbf{y}\right) \right\} ~=~-\delta _{b}^{a}\delta ^{\left( 3\right) }\left( \mathbf{x}-\mathbf{y}\right) .$$ What am I missing?

Update: Following the note 'Dirac Brackets' by Steven Avery, define the following Poisson bracket (note the exact order of $\pi _{a}$ and $\psi ^{a}$): $$\left\{ f,g\right\} _{\text{P.B.}}=\int d^{3}x\left[ -\frac{\delta f}{\delta \pi _{a}\left( \mathbf{x}\right) }\frac{\delta g}{\delta \psi ^{a}\left( \mathbf{x}\right) }+\left( -1\right) ^{\varepsilon _{f}\varepsilon _{g}}% \frac{\delta g}{\delta \pi _{a}\left( \mathbf{x}\right) }\frac{\delta f}{% \delta \psi ^{a}\left( \mathbf{x}\right) }\right] ,$$ where $\varepsilon _{f}$ and $\varepsilon _{g}$ are the Grassmann parities of $f$ and $g$, respectively. For Grassmann-even $f$ and $g$, this bracket correctly reduces to the usual Poisson bracket. For the 'classical' Dirac Lagrangian, formulated in terms of Grassmann-odd $\psi^{a}$ and $\pi_{a}$, it becomes \begin{eqnarray*} \left\{ \psi ^{a}\left( \mathbf{x}\right) ,\pi _{b}\left( \mathbf{y}\right) \right\} _{\text{P.B.}} &=&-\int d^{3}z\left[ \frac{\delta \psi ^{a}\left( \mathbf{x}\right) }{\delta \pi _{c}\left( \mathbf{z}\right) }\frac{\delta \pi _{b}\left( \mathbf{y}\right) }{\delta \psi ^{c}\left( \mathbf{z}\right) } +\frac{\delta \pi _{b}\left( \mathbf{y}\right) }{\delta \pi _{c}\left( \mathbf{z}\right) }\frac{\delta \psi ^{a}\left( \mathbf{x}\right) }{\delta \psi ^{c}\left( \mathbf{z}\right) }\right] \\ &=&-\int d^{3}z\left[ 0+\delta _{b}^{c}\delta ^{\left( 3\right) }\left( \mathbf{y}-\mathbf{z}\right) \delta _{c}^{a}\delta ^{\left( 3\right) }\left( \mathbf{x}-\mathbf{z}\right) \right] \\ &=&-\delta _{b}^{a}\delta ^{\left( 3\right) }\left( \mathbf{x}-\mathbf{y} \right) . \end{eqnarray*} If $\pi _{a}=-\mathrm{i}\psi _{a}^{\dagger }$, note minus sign, then $$\left\{ \psi ^{a}\left( \mathbf{x}\right) ,\psi_{b}^{\dagger}\left( \mathbf{y}\right) \right\} _{\text{P.B.}}=-\mathrm{i}\delta _{b}^{a}\delta ^{\left( 3\right) }\left( \mathbf{x}-\mathbf{y}\right) ,$$ which using the 'quantum bracket = $\mathrm{i} \times$ Poisson bracket'-rule yields the correct quantum anticommutation relation: $$\left\{ \psi ^{a}\left( \mathbf{x}\right) ,\psi_{b}^{\dagger}\left( \mathbf{y}\right) \right\} =\delta _{b}^{a}\delta ^{\left( 3\right) }\left( \mathbf{x}-\mathbf{% y}\right) .$$

• What do you mean "why not" - what is the actual result of calculating $\partial\mathcal{L}/\partial \dot{\psi}$? – ACuriousMind May 31 '15 at 12:53
• The "why not" is meant rhetorically, as I think there should be a minus sign coming from, as I write, passing the variational derivative through from the left to right. – John Fredsted Jun 1 '15 at 12:02

OP asks about the Legendre transformation from the Lagrangian to the Hamiltonian formulation of fermions. This question has already been asked and answered in e.g. this, this and this Phys.SE posts.

Let us here just list a number of subtle points in this important and interesting calculation:

1. More generally, let $\phi^{\alpha}$ denote a field of Grassmann-parity $\epsilon_{\alpha}$. When defining Grassmann-odd canonical momentum, should we use derivatives $$\tag{1} \pi^R_{\alpha}~:=~\frac{\partial_R {\cal L}}{\partial \dot{\phi}^{\alpha}}~=~(-1)^{\epsilon_{\alpha}}\frac{\partial_L {\cal L}}{\partial \dot{\phi}^{\alpha}}~=:~(-1)^{\epsilon_{\alpha}}\pi^L_{\alpha}$$ that act from the right or from the left? Short answer: It should be correlated with choice of kinetic term $$\tag{2} {\cal L}_H~=~\pi^R_{\alpha}\dot{\phi}^{\alpha}-{\cal H}~=~\dot{\phi}^{\alpha}\pi^L_{\alpha}-{\cal H}$$ in the Hamiltonian Lagrangian density ${\cal L}_H$. This seems to resolves OP's question about sign conventions.$^1$

2. Be careful to use a consistent sign convention for the Poisson bracket$^2$ (PB) $$\{F,G\}_{PB}~=~\int d^3x\left( \frac{\delta_R F}{\delta \phi^{\alpha}({\bf x})} \frac{\delta_L G}{\delta \pi^R_{\alpha}({\bf x})}-\frac{\delta_R F}{\delta \pi^L_{\alpha}({\bf x})} \frac{\delta_L G}{\delta \phi^{\alpha}({\bf x})}\right)$$
$$~=~-(-1)^{\varepsilon_F\varepsilon_G}\{G,F\}_{PB}. \tag{3}$$ The fundamental PBs read $$\tag{4} \{\phi^{\alpha}({\bf x}), \pi^R_{\beta}({\bf x}^{\prime})\}_{PB} ~=~\delta^{\alpha}_{\beta}~\delta^3({\bf x}-{\bf x}^{\prime}) ~=~-\{\pi^L_{\beta}({\bf x}),\phi^{\alpha}({\bf x}^{\prime})\}_{PB} ,$$ and the rest are zero. Note that the above PB (3) is consistent with super-canonical transformations, and satisfies super-skew/anti-symmetry, a super-Jacobi identity $$\tag{5}\sum_{\text{cycl. }F,G,H} (-1)^{\varepsilon_F\varepsilon_H}\{\{F,G\}_{PB},H\}_{PB}~=~0,$$ and a super-Leibniz rule. So does the super-commutator $$\tag{6} [\hat{F}, \hat{G}\} ~:=~\hat{F}\hat{G}-(-1)^{\varepsilon_F\varepsilon_G}\hat{G}\hat{F} ~\longleftrightarrow~i\hbar\{F,G\}_{PB} +{\cal O}(\hbar^2) ,$$ which is consistent with the correspondence principle.

3. In the fermionic case, be careful not to confuse the classical PB $\{\cdot,\cdot\}_{PB}$ and the quantum anti-commutator $\{\cdot,\cdot\}_+$.

4. Going back to OP's example, can we treat $\psi$ and $\psi^{\dagger}$ as independent variables? If so, is the momentum for $\psi^{\dagger}$ zero?

5. Is the Legendre transformation singular? Are there constraints?

The answers to the last points 4 and 5 are given in the linked Phys.SE posts.

--

$^1$ Conventionally, one uses $\pi^R_{\alpha}$ rather than $\pi^L_{\alpha}$, cf. e.g. a comment between eqs. (44.6) and (44.7) in Srednicki's QFT book. A prepublication draft PDF file is available here.

$^2$ Here we are ignoring a discussion of the existence of functional derivatives, which rely on a consistent choice of boundary conditions, cf. e.g. this Phys.SE post.

• Earlier today I checkmarked your answer. But later, after having discovered the material by Steven Avery, compare my Update above, I had second thoughts. It seems to me that acting from the right, as you suggest, is not necessary, at least not at the 'classical' level. But perhaps I am just messing together Poisson brackets and quantum brackets, or what seems to be equivalent, the 'classical' formalism and the quantum formalism. But my point is, I guess, that I would like the usual 'quantum bracket = i times Poisson bracket'-rule to be applicable in the fermionic/Grassmann-odd case as well. – John Fredsted Jun 1 '15 at 12:32
• Thanks a lot. Eq. (3) is particularly interesting to me. I will look more into the details in the days to come. I would have liked to give you a Vote Up point, but my low reputation score does not presently allow me to do so. – John Fredsted Jun 1 '15 at 15:45