Non-relativistic QFT Lagrangian for fermions

Question

Take the ordinary Hamiltonian from non-relativistic quantum mechanics expressed in terms of the fermi fields $\psi(\mathbf{x})$ and $\psi^\dagger(\mathbf{x})$ (as derived, for example, by A. L. Fetter and D. Walecka in Quantum Theory of Many-particle Systems, page 19):

$$\hat{H}~=~\int\hat\psi^\dagger(\mathbf{x})T(\mathbf{x})\hat\psi(\mathbf{x})d^3x$$ $$ + \frac{1}{2}\iint\hat\psi^\dagger(\mathbf{x})\hat\psi^\dagger(\mathbf{x'})V(\mathbf{x},\mathbf{x'})\hat\psi(\mathbf{x'})\hat\psi(\mathbf{x})d^3xd^3x' \tag{2.4}$$

The field $\psi(\mathbf{x})$ and $\Pi(\mathbf{x})=i\psi^\dagger(\mathbf{x})$ ($\hbar=1$) satisfy the usual canonical quantization relations, but if I try to build a Lagrangian as:

$$L=\int\Pi(\mathbf{x})d_t\psi(\mathbf{x})d\mathbf{x}-H.$$

It turns out that, because:

$$d_t\psi(\mathbf{x})=-iT(\mathbf{x})\psi(\mathbf{x}) - i\int\psi^\dagger(\mathbf{x})V(\mathbf{x},\mathbf{x'})\psi(\mathbf{x'})\psi(\mathbf{x})d\mathbf{x'}.$$

If I combine both expressions the Lagrangian turns out to be zero (a proof of the last equation can be found in Greiner's Field Quantization, chapter 3, it can be derived using $[a,bc]=[a,b]_\mp c\pm b[a,c]_\mp$).

My questions are:

1. What is wrong in this derivation?

(Greiner manages to get the Hamiltonian from the Lagrangian but he makes some integration by parts that he gives as obvious but that for me should have an extra term)

2. How can you derive $$\frac{\delta H}{\delta\psi}=-d_t\Pi$$ from the previous hamiltonian? From this expression, the Euler-Lagrange equations can be derived easily, but I can't seem to find the way to get it.

Answer 1

Comment to the question (v4):

1. Classically, the Lagrangian for a fermion system reads $$ L ~=~ \int\! d^3x~ i\psi^{\dagger}\dot{\psi}-H.\tag{A}$$

2. The Legendre transformation from the Lagrangian to the Hamiltonian formalism is tricky for at least three reasons:

   • The traditional Dirac-Bergmann analysis leads to constraints. See e.g. my Phys.SE answers here and here.
   • When differentiating wrt. Grassmann-odd fields, one must distinguish between differentiation $\frac{\delta_L}{\delta\psi}$ from left and differentiation $\frac{\delta_R}{\delta\psi}$ from right. See e.g. my Phys.SE answer here or B. De Witt, Supermanifolds.
   • How can one treat $\psi$ and $\psi^{\dagger}$ as independent variables, if they are Hermitian conjugate of each other? The resolution is similar to this Phys.SE post.

3. The equal-time super-Poisson brackets reads $$ \{\psi({\bf x},t), \psi^{\dagger}({\bf y},t)\}_{PB}~=~ -i \delta^3({\bf x}-{\bf y})~=~\{\psi^{\dagger}({\bf x},t), \psi({\bf y},t)\}_{PB},\tag{B} $$ and other fundamental super-Poisson brackets vanish.

4. Due to the QM correspondence principle, the canonical anticommutation relations (CARs) are the super-Poisson brackets (5) multiplied with $i\hbar$: $$ \{\hat{\psi}({\bf x},t), \hat{\psi}^{\dagger}({\bf y},t)\}_{+} ~=~ \hbar\delta^3({\bf x}-{\bf y})\hat{\bf 1} ~=~\{\hat{\psi}^{\dagger}({\bf x},t), \hat{\psi}({\bf y},t)\}_{+},\tag{C} $$ and other CARs vanish.

5. Hamilton's equations read $$\dot{\psi} ~\approx~\{ \psi, H\}_{PB}~\stackrel{(B)}{=}~-i\frac{\delta_L H}{\delta\psi^{\dagger}}, \qquad \dot{\psi}^{\dagger} ~\approx~\{ \psi^{\dagger}, H\}_{PB}~\stackrel{(B)}{=}~-i\frac{\delta_L H}{\delta\psi}.\tag{D} $$

6. Heisenberg equations of motion read $$i\hbar\partial_t\hat{\psi}~\approx~[\hat{\psi}, \hat{H}], \qquad i\hbar\partial_t\hat{\psi}^{\dagger}~\approx~[\hat\psi^{\dagger}, \hat{H}].\tag{E} $$

Answer 2

Regarding question 2, in Weinberg's The Quantum Theory of Fields page 295, the author defines the functional derivative of an arbitrary bosonic functional $F[q,p]$:

$$\frac{\delta F[q,p]}{\delta q}=i[p,F[q,p]]$$ $$\frac{\delta F[q,p]}{\delta p}=i[F[q,p],q]$$

where $q,p$ are a pair of operators that satisfy canonical conmutation relations. If that is indeed the case, then the result in question 2 follows directly from this. Weinberg says that this definition is motivated by the result that, if $F[q,p]$ is written "with all the $q$s to the left of all $p$s" then:

$$\delta F[q,p]=\int d^3 x \left( \delta q \frac{\delta F[q,p]}{\delta q} + \frac{\delta F[q,p]}{\delta p} \delta p\right) $$

I don't know much about functional differentiation and I haven't managed to go from this equation to the previous two. If anyone can explain that, it solves question two.