field theory - From Lagrangian to Hamiltonian in Fermionic Model

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 ?

Answer

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.