# Connection between “classical” Grassmann variables and Heisenberg Equation of motion

I have been reading di Francesco et al's textbook on Conformal Field theory, and am confused by a particular statement they make on pg 22.

Let $\{\psi_i\}$ be a set of Grassmann variables. Starting with the Lagrangian $$L = \frac{i}{2} \psi_i T_{ij} \dot{\psi}_j - V(\psi) \tag{2.32}$$ they derive the equation $$\dot{\psi_i} = -i \, (T^{-1})_{ij} \frac{\partial V}{\partial \psi_j}.\tag{2.36}$$

They then claim that you get the same result if you use the Heisenberg equation of motion $$\dot{\psi} = i [H, \psi]\tag{2.36b}$$ with $$H = V(\psi)\quad\text{and}\quad \{\psi_i,\psi_j\}_{+} = (T^{-1})_{ij}.\tag{2.37}$$

I don't understand how they get from the Heisenberg equation of motion to the desired result. I tried setting $V = \psi_j$ for a particular $j$ and deriving the result in this particular case, but in trying to compute $[\psi_j,\psi_i]$ you'll end up getting extra terms of the form $\psi_i\psi_j$ which I don't know how to get rid of. Unfortunately the textbook doesn't work this out and leaves this as an exercise to the reader.

On a slightly deeper level, what exactly is meant when we say that Grassmann variables provide a "classical" description of Fermi fields?

Any help/insight would be much appreciated!

1. Grassmann-odd variables provide a classical description of Grassmann-odd quantum operators in the same way that Grassmann-even variables provide a classical description of Grassmann-even quantum operators. The classical super-Poisson bracket $$\{\psi^i,\psi^j\}_{PB} ~=~ -i (T^{-1})^{ij} \tag{A}$$ is related to the super-commutator$^1$ $$\hat{\psi}^i\hat{\psi}^j+\hat{\psi}^j\hat{\psi}^i ~=~\{\hat{\psi}^i,\hat{\psi}^j\}_{+} ~=~ [\hat{\psi}^i,\hat{\psi}^j]_{SC} ~=~ \hbar ~(T^{-1})^{ij}~{\bf 1} \tag{B}$$ in accordance with the correspondence principle between classical and quantum mechanics, cf. e.g. this Phys.SE post.

2. The EL equations (2.36) for the Lagrangian (2.32) are precisely the Hamilton's equations $$\dot{\psi}^i ~\approx~ \{ \psi^i, H\}_{PB} ~=~\{ \psi^i,\psi^j\}_{PB}\frac{\partial H}{\partial \psi^j} ~\stackrel{(A)}{=}~-i (T^{-1})^{ij} \frac{\partial H}{\partial \psi^j}.\tag{C}$$ Eq. (2.36b) (which uses units with $\hbar=1$) is the corresponding Heisenberg's EOM $$i\hbar\frac{d\hat{\psi}^i}{dt} ~\approx~ [ \hat{\psi}^i, \hat{H}]_{SC},\tag{D}$$ i.e. the quantum version of the classical Hamilton's eqs. (C).

3. Concerning the Legendre transformation between the Lagrangian and Hamiltonian formulation of Grassmann-odd variables, see e.g. this Phys.SE post and links therein.

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$^1$ The super-commutator $[\hat{A},\hat{B}]_{SC}$ of two operators $\hat{A}$, $\hat{B}$ (with Grassmann parities $|A|$, $|B|$) is defined as $$[\hat{A},\hat{B}]_{SC}~:=~\hat{A}\hat{B}-(-1)^{|A||B|}\hat{B}\hat{A} .\tag{E}$$

• Thanks for the answer! So, are you saying that the "commutator" in the Heisenberg EOM is supposed to be treated as an anticommutator? If so, isn't there a sign error? I am approaching this more simplistically, just trying to get the EOM to agree starting from the quantum picture. Sorry for a being a bit pedantic, and thanks again! – Aaron Jun 22 '16 at 17:38
• Actually, is there an implicit assumption that $V(\psi)$ is Grassmann even? I tried working out the details for a "order 2" grassmann-even $V$, and it gives me the right result. – Aaron Jun 22 '16 at 20:27
• Yes, $L$, $H$, $V$, $T_{ij}$ are implicitly assumed to be Grassmann-even. So if a $V$-term contains an odd number of $\psi$'s, the corresponding coefficient should be Grassmann-odd! – Qmechanic Jun 22 '16 at 20:31
• I'm sorry but I don't see how your equation D is correct. For example, section 6.2 of Arodz and Hadasz's "Lectures on Classical and Quantum Theory of Fields" uses a commutator in the Heisenberg equation (eq. (6.75)), not a super-commutator as your equation suggests. Could you please clarify what I am missing here? (Link to notes: link.springer.com/book/10.1007%2F978-3-319-55619-2) – Arkya Aug 8 '20 at 20:20
• No. I recommend you to review the notion of Grassmann parity. – Qmechanic Aug 8 '20 at 20:54