Saturday, October 25, 2014

quantum field theory - 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!



Answer





  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 $$[\hat{A},\hat{B}]_{SC}~:=~\hat{A}\hat{B}-(-1)^{|A||B|}\hat{B}\hat{A} .\tag{E}$$


No comments:

Post a Comment

classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...