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.


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

  • $\begingroup$ 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! $\endgroup$ – Aaron Jun 22 '16 at 17:38
  • $\begingroup$ 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. $\endgroup$ – Aaron Jun 22 '16 at 20:27
  • $\begingroup$ 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! $\endgroup$ – Qmechanic Jun 22 '16 at 20:31
  • $\begingroup$ 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) $\endgroup$ – Arkya Aug 8 '20 at 20:20
  • 1
    $\begingroup$ No. I recommend you to review the notion of Grassmann parity. $\endgroup$ – Qmechanic Aug 8 '20 at 20:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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