Question

Currently, I am puzzled by the formula 2.8 in the paper arXiv;1907.03363, where the author states that Z_R (β) = 0 for N being odd. I am having difficulty understanding the discussion in the paragraph on page 13.

"Now let us consider the Ramond sector. For odd N, the path integral of the SYK model in the Ramond sector (with no operator insertions) is actually identically zero. To see this, first note that if H=0, so that the action (2.4) consists only of the kinetic energy, then each of the ψₖ has a zero-mode in the Ramond sector, so in all there are an odd number of zero-modes. Now including the Hamiltonian, a term that is proportional to Hʳ for some r has an insertion of qr fermions, which is an even number since q is even. An even number of these fermions can be paired up by propagators, and the remaining ones – also an even number – can be used to soak up zero-modes. Since we started with an odd number of fermion zero-modes, there is always an odd number left over. In particular, we can never soak up all of the zero-modes, and therefore the Ramond sector path integral of the odd N SYK model is identically zero."

Is there a more detailed discussion regarding this formula? From the perspective of the path integral, there are also zero modes present when N is even. I am wondering why these zero modes, when N is even, do not cause the path integral to vanish. Specifically, when considering the case H=0, the zero mode always appears in the determinant ∏ᵢωᵢ, right?

Answer 1

Main Answer

In the paper arXiv:1907.03363, the author states that for odd N, the path integral in the Ramond sector of the SYK model is identically zero. This result is due to the fact that for H=0, each of the ψₖ fields has a zero mode in the Ramond sector, resulting in an odd number of zero modes in total.

Step-by-step explanation

The discussion in the paper regarding this formula is quite detailed, but it may still be unclear why zero modes do not cause the path integral to vanish when N is even.

When considering the case H=0, zero modes always appear in the determinant product ∏ᵢωᵢ. However, when N is even, there are an even number of zero modes present.

This means that we can pair up these zero modes using propagators and still have an even number left over. Therefore, there are always enough propagators to soak up all of the zero modes, and the path integral does not vanish in this case.

In contrast, when N is odd, there is always an odd number of zero modes left over after pairing them up with propagators. This is because there is one more zero mode than there are propagators available to soak them up. As a result, there is no way to completely eliminate all of the zero modes in this case, causing the path integral to vanish identically.

Overall, the key difference between these two cases is the parity of N. When N is even, there are enough propagators available to soak up all of the zero modes present in the determinant product ∏ᵢωᵢ. However, when N is odd, there are more zero modes than propagators available to soak them up, causing the path integral to vanish identically.