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  Fermionic path integral on the disk - Recovering the vacuum state

+ 4 like - 0 dislike

I'm trying to get a better feel for the operator to state map in quantum field theory. There is a general claim for 2d theories that doing the path integral on a disk with no operator insertions gives you (the wave function representation of) the ground state. Polchinski works this out explicitly for the free bosonic field on pages 66-68 in Volume 1 of his Superstring Theory. As a self-assigned exercise, I'm now trying to do the same for the fermionic case, though I'm having some trouble with evaluating the path integral.

The system I'm studying is the usual Dirac action given by $$S = \int dt ds [i\bar{\psi}_-(\partial_{t} + \partial_{s})\psi_- + i\bar{\psi}_+(\partial_{t} - \partial_{s})\psi_+].$$

Now I want to do the path integral on the semi-infinite cylinder (equivalently a  unit disk) with coordinates $t \in [0,\infty)$, and $s \in [0, 2\pi)$.

$$\int_{\psi_-(0,s) = f(s), \psi_+(0,s) = g(s)} \mathcal{D}\psi \mathcal{D} \bar{\psi} e^{-S}$$ where $$f(s) = \sum_{n \in \mathbb{z}}\chi_n e^{ins},\,\,\,\,\,g(s) = \sum_{n\in \mathbb{Z}} \phi_n e^{ins} $$ being the boundary conditions on the unit circle ($\chi_n$, $\phi_n$ being Grassmann numbers). Note that I'm imposing periodic boundary conditions on the fermions with respect to $s$.

I'm not sure how to proceed with the path integral at this point. The usual procedure (at least for bosonic fields) of writing your field as $\phi = \phi_{cl} + \phi_{q}$ where $\phi_{cl}$ obeys the classical equations on motion and obeys the right boundary conditions and $\phi_q$ being a fluctuation seems to give some weird stuff since the action evaluated at a classical solution gives $0$. Any help or hints would be appreciated.

Edit: The expression I'm trying to compare to is the following:

The Hilbert space from the wave function perspective is given by "square-integrable" functions of infinitely many Grassmann numbers $f(\chi_i, \bar{\chi}_i)$, $i$ being any integer. The expression for the Hamiltonian is then given by $$H = \sum_{n \in \mathbb{Z}} n(\chi_{n} \frac{\partial}{\partial \chi_n} + \bar{\chi}_{n} \frac{\partial}{\partial \bar{\chi}_n}).$$ From this one computes the ground state to be $$\langle \chi_i, \bar{\chi}_i|0\rangle = \prod_{n=1}^{\infty}\chi_{-n} \bar{\chi}_{-n}.$$

asked May 22, 2015 in Theoretical Physics by anonymous [ revision history ]
edited May 22, 2015

Let's consider a Dirac fermion first. Then $\mathcal{L} = \bar{\Psi} \hat{X} \Psi + $ (total derivative), where $\hat{X}$ is made of 0th and 1st order derivatives. Then the Euler-Lagrange eq is $\hat{X} \Psi = 0$, so classically $\mathcal{L} = 0$, if the total devirative can be neglected. This implies the vaninshing of $\mathcal{L}$ and the action is not so important.

I can't generalize it for compact spacetime. I don't know what's topological insulator, where edge currents matter. Let's wait for a lucid answer by someone else.

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