While going from a given Lagrangian to Hamiltonian for a fermionic field, we use the following formula. $$ H = \Sigma_{i} \pi_i \dot{\phi_i} - L$$ where $\pi_i = \dfrac{\partial L}{\partial \dot{\phi_i}} $ In a Lagrangian involving fermionic fields given by, $$ L = \dfrac{1}{2}(\bar{\psi_i} \dot{\psi_j} - \dot{\bar{\psi_i}} \psi_j)$$ a direct computation gives $\pi_{\psi_j} = -\dfrac{1}{2}\bar{\psi_i}$ and $\pi_{\bar{\psi_i}} = -\dfrac{1}{2}\psi_j$. But on adding a total derivative $\dfrac{1}{2} \dfrac{d}{dt} (\bar{\psi_i} \psi_j)$ to the Lagrangian (which can always be done as the action won't change) but $\pi$'s become different. So the Hamiltonian as well changes. How do we resolve the issue ?


2 Answers 2


The canonical momenta don't change if you add a total derivative to the Lagrangian.

The particular total derivative you wanted to add to the Lagrangian as well as the Lagrangian itself has free $i,j$ indices. You surely meant something else because the Lagrangian should have no free indices like that. Let me assume that you meant both expressions to be summed with the sum and prefactor $\sum_{ij} c_{ij}$. Of maybe you really meant the Lagrangian to be a monomial for fixed values of $i,j$.

But that's not the issue here. The error relevant for your question is that you considered a phase space that has coordinates $\psi_j$, $\bar\psi_i$, $\pi_{\psi_i}$, and $\pi_{\bar\psi_j}$, and you think they're independent coordinates on the phase space. That would be too many phase space coordinates for such a limited system.

Well, they're not independent. The right derivation, using any form of the Lagrangian you want, will give you $\pi_{\psi_i}=-\bar \psi_i$ (without one-half; and equations that may be obtained by simple conjugations from this one!) so it means that the "same" non-differentiated $\psi$'s are their own momenta, too.

If you rewrite the Lagrangian in such a way that the redundant notation is eliminated, i.e. you don't think that coordinates that are dependent are actually independent (this is the error that made you end up with the canonical momenta being 1/2 of their right value; for example, you incorrectly used $\partial\dot{\bar\psi_i} / \partial \psi_j = 0$, which is not true, in the first momentum you mentioned), you will see that $$\frac{\partial L}{\partial \dot\psi_j }=-\bar\psi_i$$ if I use your confusing non-summation over $i,j$. There's no factor of 1/2. Indeed, to derive this thing without problems, it's helpful to first rewrite the Lagrangian as $\bar\psi_i\dot\psi_j$ by adding the appropriate total derivative. This form is unique because it contains no $\dot{\bar\psi_i}$ and no $\psi_j$, so it's only expressed as a function of the independent 1/2 of the degrees of freedom.

Needless to say, the Hamiltonian is zero if the fermionic Lagrangian only contains the kinetic term with the time derivative.

  • $\begingroup$ +1 Nice answer, What is the correct value of the partial derivative that the OP made a mistake with? $\endgroup$
    – Prathyush
    Commented Mar 26, 2013 at 14:33
  • $\begingroup$ Sorry for the sloppy way of posting the question. I think I was in a haste while doing so, actually Lagrangian does have a summation over all the indices. $\endgroup$
    – Jaswin
    Commented Mar 27, 2013 at 16:42
  • $\begingroup$ Dear @Prathyush, I don't know how to answer this question. The Lagrangian must first be rewritten to a form in which only the predecided independent coordinates and velocities appear... $\endgroup$ Commented Mar 28, 2013 at 5:55

It is not completely clear what Lagrangian OP has in mind. Here we will assume that the Lagrangian reads

$$ L~=~\frac{i}{2} g_{IJ} \left(\overline{\psi}^I \dot{\psi}^J-\dot{\overline{\psi}}^I \psi^J \right) + \frac{1}{2} h_{IJ} \left(\overline{\psi}^I \dot{\psi}^J+\dot{\overline{\psi}}^I \psi^J \right), \tag{1}$$

where $\psi^{I}$ is a complex Grassmann-odd scalar field, and $\overline{\psi}^I$ is the complex conjugate field. (This choice is partly inspired by one of OP's other Phys.SE questions.) The metrics are constant

$$ g_{JI}~=~g_{IJ}~=~\overline{g}_{JI}, \qquad h_{JI}~=~h_{IJ}~=~\overline{h}_{JI}. \tag{2}$$

The second term in the Lagrangian (1) is a total derivative term. This is just included for fun to see how this does not affect the quantization procedure. To derive the Hamiltonian formalism, we will use a Grassmann-odd version of this Phys.SE answer. (We recommend that the reader familiarize himself with the Grassmann-even model in that answer before trying to understand the Grassmann-odd model in this answer.)

The canonical anticommutation relations (CAR) read

$$ \{\psi^I, \pi_J \}_{PB}~=~\delta^I_J~=~\{\overline{\psi}^I, \overline{\pi}_J \}_{PB} ,\tag{3}$$

$$ \{\overline{\psi}^I, \pi_J \}_{PB}~=~0~=~\{\psi^I, \overline{\pi}_J \}_{PB} .\tag{4}$$

The Grassmann-odd momenta are given by right derivatives of the Lagrangian

$$ \pi_I~:=~L\frac{\stackrel{\leftarrow}{\partial^r}}{\partial \dot{\psi}^I}~=~\frac{1}{2}\overline{\psi}^J(i g_{JI}+h_{JI}), \tag{5}$$

$$ \overline{\pi}_I~:=~L\frac{\stackrel{\leftarrow}{\partial^r}}{\partial \dot{\overline{\psi}}^I} ~=~\frac{1}{2}(i g_{IJ}-h_{IJ})\psi^J.\tag{6}$$

The Hamiltonian is identically zero,

$$ H~:= ~ \pi_I\dot{\psi}^I+\overline{\pi}_I\dot{\overline{\psi}}^I - L~=~0.\tag{7} $$

Equations (5) and (6) yield two primary constraints

$$ 0~\approx~\chi_I~:=~\pi_I-\frac{1}{2}\overline{\psi}^J(i g_{JI}+h_{JI}), \tag{8}$$

$$ 0~\approx~\overline{\chi}_I~:=~\overline{\pi}_I-\frac{1}{2}(i g_{IJ}-h_{IJ})\psi^J.\tag{9}$$

They are, in turn, second-class constraints,

$$ \{\chi_I, \overline{\chi}_J \}_{PB}~=~-ig_{IJ}~=~\{\overline{\chi}_I, \chi_J \}_{PB} ,\tag{10}$$

$$ \{\chi_I, \chi_J \}_{PB}~=~0~=~\{\overline{\chi}_I, \overline{\chi}_J \}_{PB} ,\tag{11}$$

independent of the $h_{IJ}$ metric.

The Dirac bracket becomes

$$\begin{align}\{f, g \}_{DB}~:=~& \{f, g \}_{PB}- i\{f, \chi_I\}_{PB}g^{IJ}\{ \overline{\chi}_J,g\}_{PB}\cr &- i\{f, \overline{\chi}_I\}_{PB}g^{IJ}\{ \chi_J,g\}_{PB}.\end{align}\tag{12}$$

In other words, the Dirac anticommutation relations become

$$ \{\psi^I, \overline{\psi}^J \}_{DB}~=~-ig^{IJ}~=~\{\overline{\psi}^I, \psi^J \}_{DB} ,\tag{13}$$

$$ \{\psi^I, \psi^J \}_{DB}~=~0~=~\{\overline{\psi}^I, \overline{\psi}^J \}_{DB} ,\tag{14}$$

in agreement with the Faddeev-Jackiw method. The corresponding operator anticommutation relations read

$$ \{\hat{\psi}^I, \hat{\overline{\psi}}^J \}_{+}~=~\hbar g^{IJ}~=~\{\hat{\overline{\psi}}^I, \hat{\psi}^J \}_{+} ,\tag{15}$$

$$ \{\hat{\psi}^I, \hat{\psi}^J \}_{+}~=~0~=~\{\hat{\overline{\psi}}^I, \hat{\overline{\psi}}^J \}_{+} .\tag{16}$$

  • $\begingroup$ Comment to the answer (v1): Note that $\overline{\psi}^I$ denotes the complex conjugate of $\psi^I$. In particular, there is no hidden Dirac index or $\gamma^0$-matrix in this model. $\endgroup$
    – Qmechanic
    Commented May 31, 2015 at 13:31
  • $\begingroup$ Hi Sir can you tell me why Dirac gamma matrices (clifford algebra) can appear in classical Lagrangian of spinors (Grassmann algebra)? The former is the quantization of the latter. $\endgroup$
    – Valac
    Commented Oct 5, 2023 at 2:51
  • $\begingroup$ @Libertarian Feudalist Bot. Thanks for the feedback. Well, the canonical anticommutation relations (CARs) of the $\psi$ fields and their hermitian conjugate fields do not depend on $\gamma^{\mu}$, so one can argue that $\gamma^{\mu}$ (and the Clifford algebra) are not part of canonical quantization per se. $\endgroup$
    – Qmechanic
    Commented Oct 5, 2023 at 7:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.