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 (or 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 $$\langle \chi_i, \bar{\chi}_i|0\rangle = \prod_{n=1}^{\infty}\chi_{-n} \bar{\chi}_{-n}.$$

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    $\begingroup$ There are many small issues here. When it's a disk, the area is $t\cdot dt\cdot ds$ and not without the $t$, right? Also, if you work with a Euclidean-signature disk, the kinetic terms for the fermions should use the complex $\partial_t \pm i \partial_s$ and not the "light-cone-like" sum and difference, right? But when you correct all these things, it should be straightforward to compute all such Gaussian integrals, by completing squares etc. You will need to find the classical solution with the boundary conditions. $\endgroup$ May 21, 2015 at 17:43
  • $\begingroup$ Thanks Lubos. Yes the issues you point out all need to be addressed (they mainly come from using the action on the cylinder rather than the disk), however they are not what's troubling me. I can't see what to plug in for the $\psi$'s and integrate over. The classical solution with boundary conditions would be simply that the $\psi's$ are either left of right moving. But plugging this into the action gives me zero. $\endgroup$ May 21, 2015 at 18:04
  • $\begingroup$ Dear @childofsaturn, if you had the idea to do "the same" for the fermions by yourself, you shouldn't be surprised by these "more trivial" results. This will be discussed in sections like 4.4, 5.3, 6.3 etc. especially for the $bc$ ghost system, which is really the same as your $\psi$'s on a flat world sheet. The simplest partition sum is zero, and you need certain insertions to get a nonzero result. Those will be generated automatically by the string formulae, and they may also be seen to depend on the Grassmann boundary conditions if you choose to have them. $\endgroup$ May 22, 2015 at 6:47
  • $\begingroup$ Dear Lubos, I'm not sure I follow. I have the wave function expression for the ground state which I got independently from the operator formalism, and which depends on infinitely many Grassmann variables. I'm now just trying to see whether the path integral gives me the same result. The classical action giving zero (I think) implies that the ground state doesn't depend on the boundary conditions, which isn't true once we look at the wave functional expression. $\endgroup$ May 22, 2015 at 7:59
  • $\begingroup$ Dear @LubošMotl. Please see the edits. $\endgroup$ May 22, 2015 at 8:15

1 Answer 1


There's a discussion of this problem in https://arxiv.org/abs/hep-th/9408089 where they overcome the issue of vanishing action by using a holomorphic factorization for the classical part: $$ \psi = \psi_\text{cl} + \delta \psi $$ where $\psi_\text{cl}$ is real and obeys Klein-Gordon with boundary conditions, while $\delta \psi=0$ on the boundary.

In this case the action factorizes as $A(\psi)= A(\psi_\text{cl}) + A(\delta \psi)$ (at least that's what they claim, I don't fully understand this step yet); if so, the second piece can be neglected, as it gives a contribution that is independent of boundary conditions.

Write explicitly $$ \psi_\text{cl} = \sum_{m>0} \psi_m e^{mz} + \psi_{-m} e^{m \bar z} $$ and then compute $$A(\psi_\text{cl}) \sim \sum_{m>0} \psi_m \psi_{-m}$$ by observing that $w=e^z$ lives on the disk $\{x^2+y^2=1\}$ in $R^2$, and finally the wave functional is $$\Psi[\psi_m]=e^{A(\psi_\text{cl})}\sim \prod_m \psi_{-m} \psi_m$$


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