# Some questions about the free Fermionic partition function on a circle (Ginsparg's CFT lectures)

The following questions are based on these lectures, http://arxiv.org/abs/hep-th/9108028

• I would like to know what is the relationship between the last equation on page 82 ($(L_0)_{cyl} = L_0 - \frac{c}{24}$) and the last 2 sets of equations on page 84 (where one has derived the zero-point energies of $\pm 1/24$, $\pm 1/48$ of periodic and anti-periodic Fermions on the cylinder)

Can the equations below 7.6 (on page 84) be derived from the last equation on page 82?

• I can understand the derivation of the partition function in equation 7.5 at the top of page 84 where one gets that $Z = q^{-\frac{c}{24}}\bar{q}^{-\frac{\bar{c}}{24} } Tr q^{L_0}\bar{q}^{\bar{L}_0}$

But from there how does one derive the statements in the first paragraph of page 92 where he states that (1) when the Fermion is anti-periodic along both time and space the partition function is $q^{-1/48}Tr_A q^{L_0}$, (2) when its periodic in time and anti-periodic in space the answer is $q^{-1/48}Tr_A (-1)^Fq^{L_0}$ (3) when its anti-periodic in time and periodic in space the answer is $(1/\sqrt{2})q^{-1/48}Tr_P q^{L_0}$ and (4) when its periodic along both its $(1/\sqrt{2})q^{-1/48}Tr_P (-1)^Fq^{L_0}$

These 4 answers above look very mysterious to me at various levels - (1) I don't see how they can be read off from the general answer of $Z = q^{-\frac{c}{24}}\bar{q}^{-\frac{\bar{c}}{24} } Tr q^{L_0}\bar{q}^{\bar{L}_0}$, (2) how is the (-1)^F operator seeing the boundary conditions? (3) why do these answers have only the $q^{L_0}$ part and not the full $q^{L_0}\bar{q}^{\bar{L}_0}$

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• Partly yes. The equations below (7.6) use the zeta-regularization to compute the regulated values of the sums $$1+ 2+3+4+\dots = -\frac{1}{12}$$ and $$\frac{1}{2}+ \frac{3}{2} + \frac{5}{2}+\dots = \frac 12\left( 1+2+3+4+\dots \right)-(1+2+3+\dots) = -\frac 12(1+2+3+\dots) = +\frac{1}{24}$$ So the zero-point energy for a single periodic or antiperiodic boson is $-1/24$ and $+1/48$ – the extra $1/2$ comes from the $\hbar\omega/2$ in the harmonic oscillator zero-point energy. For fermions on a cylinder, the signs are reverted.
However, at least some of these values may be also derived from the last equation on page 82, the cylinder-plane translation. It's because for a single boson CFT, the identity operator in the plane has to be translated to the ground state of a single periodic boson CFT on a cylinder. The dimension of 1 in the plane is 0 but the energy of a single boson on a cylinder is $c/24=-1/24$ which implies that at least for $c$ periodic bosons, the equivalence holds. One may also argue that the sign of the ground state energy is reverted for the fermions. However, the equation on page 82 is more general because it holds for any CFT. On the other hand, the ground state energies for the different periodicities of free bosons and free fermions produce many specific examples what to calculate.
• On page 92, a single holomorphic fermion is explicitly mentioned to be debated. It means that the antiholomorphic part of the CFT under discussion is empty and we have $\bar c =0$ and $\tilde L_0=0$. We just drop this whole part of the expression. Also, we substitute $c=1/2$ for a single real fermion, thus getting $-1/48$ at various places.
The operator $(-1)^F$ is what switches the sign of all the fermions included in $F$. You may verify that this operator $(-1)^F$ anticommutes with all these fermions. This anticommutativity means that if you first measure the fermion and then act by $(-1)^F$, you get the opposite sign of the result than if you first act by $(-1)^F$ and then measure the fermion. It means that $(-1)^F$ is literally the operator that flips the fermions – much like $\exp(i\phi J_z)$ is the operator that literally rotates the degrees of freedom around the $z$-axis. If you insert it at $\tau=0$, the signs of the fermions are flipped at $\tau=0$ so it's equivalent to changing the periodicity of the fermions in the $\tau$ direction – from periodic to antiperiodic or vice versa.
Motl Thanks for your reply. Let me try to understand your arguments in detail - (1) so in the first place itself, isn't there a reason why a free Fermion $1+1$ CFT makes sense only if the Fermion is antiperiodic (in space? in time?) (2) And I thought when one is given a CFT one gets the holomorphic and the anti-holomorphic parts together. I thought one can't just choose to have one and not the other. So you are saying that one can choose? – user6818 Aug 19 '13 at 17:13