I am trying to build the fermionic coherent state formalism in conformance with the grassmann conventions used in the book "Mirror Symmetry", relation (9.20), where the fermionic integration is defined to be:$$\int\psi d\psi=1\space(1)$$It is immensely desirable to have:$$\frac{\partial\psi}{\partial\psi}=1\space(2)$$And consequently we can follow the convention that a $\psi$ differentiation as in (2) means a right integration as in (1). This works perfectly fine as all proofs of supersymmetry in chapters 9 and 10 come out spot on with these conventions. Following these conventions above, and building the fermionic coherent state formalism, the ladder operators acquire the following representations in the coherent state basis:$$\Psi^\dagger\rightarrow\bar{\psi},\space\Psi\rightarrow-\frac{\partial}{\partial\bar{\psi}}\space(3)$$Now when I evaluate the anti-commutator of the ladder operators in the coherent state basis I get:$$[\Psi,\Psi^\dagger]_+=-1\space(4)$$This should have come out +1 since this what has been utilized in the book when it comes to quantization, see e.g. (10.83) and (10.209):$$[\Psi,\Psi^\dagger]_+=+1\space(5)$$$$[\Psi^I,\Psi^{\dagger J}]_+=g^{IJ}\space(6)$$ I have checked the computations leading to (4) and they seem correct, should the details be required I can try and write them. Can some one please help out with the unwanted minus sign appearing in (4)?

  • $\begingroup$ In deriving the above, I assumed commutativity of the fermions with the oscillator ground state. Something that fixes the problem is the assumption that fermions anticommute with the oscillator ground state, meaning that they commute with the first excited state, and anticommute with the coherent states. But this seems nonsensical: to have fermions anticommute with the ground state: a state which has fermion number zero and is consequently bosonic! $\endgroup$ – Kong Jun 13 '18 at 18:35
  • $\begingroup$ Looks like fermionic ladder operators in coherent state representation ($|\zeta\rangle=e_{}^{-\zeta\Psi_{}^{\dagger}}|0\rangle$, $\langle\zeta_{}^{*}|=\langle 0|e_{}^{-\Psi_{}^{}\zeta_{}^{*}}|0\rangle$) are represented as : $\Psi_{}^{\dagger}\rightarrow\zeta_{}^{*}$, $\Psi_{}^{}\rightarrow\frac{\partial}{\partial\zeta_{}^{*}}$. (Note: Here $\zeta_{}^{}$ and $\zeta_{}^{*}$ are assumed to commute with vaccum state.) $\endgroup$ – Sunyam Jun 13 '18 at 20:49
  • $\begingroup$ Agreed under usual conventions, but the sign flip in (3) comes from the conventions (1) and (2), which work so remarkably that I am finding it hard to doubt them: they put fermions and bosons on exactly the same footing. $\endgroup$ – Kong Jun 13 '18 at 23:57
  • $\begingroup$ I don't see how sign flip follows just from the defining rules of grassman calculus. You need grassman extension of fermionic fock space to represent fermionic operators aka Grassman-Bergmann representation. $\endgroup$ – Sunyam Jun 14 '18 at 0:58

The sign flip in (4), occurs (as it should indeed) when one keeps track of the order flipping while doing the adjoint operation:$$\Psi^\dagger|\psi>=\frac{\partial}{\partial\psi}|\psi>\Rightarrow<\bar{\psi}|\Psi=<\bar{\psi}|\biggl(\frac{\partial}{\partial\psi}\biggr)^\dagger$$Now the derivative operator hits from the right, and in conformance with the conventions, corresponds to a left integration, this produces the required sign flip.


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.