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:


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)

  1. 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.

2 Answers 2


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 $$i\hbar\partial_t\hat{\psi}~\approx~[\hat{\psi}, \hat{H}], \qquad i\hbar\partial_t\hat{\psi}^{\dagger}~\approx~[\hat\psi^{\dagger}, \hat{H}].\tag{E} $$

  • $\begingroup$ Thanks for the clarification. However I have two questions about your answer: 1) Why is the lagrangian as you expressed it and not the way I did? The field and its transpose don't satisfy canonical conmutation relations (they lack an i somewhere). 2) Can you give a reference for point 2. I am interested in a formal definition of differentiation from left and differentiation from right, as that appeared to be a problematic concept when dealing with this subject. $\endgroup$
    – recicle
    Apr 13, 2015 at 21:52
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Apr 14, 2015 at 13:27
  • $\begingroup$ Thanks, I'll look at the references to see if I can get something clear. By the way, the title has been edited to say that this is for fermions, but I can't see why my question would not apply to bosons too. $\endgroup$
    – recicle
    Apr 14, 2015 at 13:53
  • $\begingroup$ @recicle: It seems that for various reasons the post would become too broad if it were to also include bosons. $\endgroup$
    – Qmechanic
    Apr 14, 2015 at 13:57

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.


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