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\begin{document}

\centerline{\bf Lecture~3}
\centerline{\bf Free massless fermion on the cylinder and on the plane}
\bigskip

Consider now a free massless Majorana fermion $\psi=\pMatrix{\psi_1\\\psi_2}$ on the cylinder:
\eq$$
S[\psi]={\i\over2\pi}\int d^2\xi\,\bpsi\gamma^\mu\d_\mu\psi
={\i\over\pi}\int d^2\xi(\psi_1\bd\psi_1-\psi_2\d\psi_2).
\label{S-freefermion}
$$
Here, as usual, $\bpsi=\psi^+\gamma^0$. The gamma-matrices should satisfy the conditions
\eq$$
\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2\eta^{\mu\nu},
\qquad
\gamma^0\gamma^\mu\gamma^0=(\gamma^\mu)^+.
\label{gamma-props}
$$
We choose them purely imaginary:
\eq$$
\gamma^0=\pMatrix{&-\i\\\i&}=\sigma^2,
\qquad
\gamma^1=\pMatrix{&\i\\\i&}=\i\sigma^1,
\qquad
\gamma^3=\gamma^0\gamma^1=\pMatrix{1&\\&-1}=\sigma^3.
\label{gamma-def}
$$
The Majorana condition in this case is
\eq$$
\psi^*=\psi.
\label{Majorana}
$$
The variables $\psi_\alpha(\xi)$ should be considered Grassmann (anticommuting): $\psi_\alpha(\xi)\psi_\beta(\xi')=-\psi_\beta(\xi')\psi_\alpha(\xi)$. The Hamiltonian reads%
%
\footnote{Accurate derivation of (\ref{H-freefermion}) and (\ref{psipsi-Poisson}) demands careful taking into account constraints, but we omit these subtleties.}
%
\eq$$
H=-{\i\over2\pi}\int^L_0d\xi^1\,\psi\gamma^3\d_1\psi
=-{\i\over2\pi}\int^L_0d\xi^1(\psi_1\d_1\psi_1-\psi_2\d_1\psi_2)
\label{H-freefermion}
$$
with the Poisson bracket
\eq$$
\{\psi_\alpha(\xi^0,\xi^1),\psi_\beta(\xi^0,\xi^{\prime1})\}=\i\pi\delta_{\alpha\beta}\delta(\xi^1-\xi^{\prime1}).
\label{psipsi-Poisson}
$$
Note that due to anticommutativity of the field the Poisson bracket is symmetric. The equation of motion looks like
\eq$$
\d_0\psi=-\sigma^3\d_1\psi
\label{psi-eqmo}
$$
or
\eq$$
\bd\psi_1=\d\psi_2=0,
\label{psipm-eqmo}
$$
where, as we remember, $\d={1\over2}(\d_1-\d_0)=\d_\zeta$, $\bd={1\over2}(\d_1+\d_0)=\d_\bzeta$. We see that $\psi_1=\psi_1(\zeta)$ is a right-moving Majorana\--Weyl fermion wave, while $\psi_2=\psi_2(\bzeta)$ is a left-moving Majorana\--Weyl fermion wave.

Let us expand the $\psi$ field into modes. Before doing it we have to fix the periodicity condition. There are two possibilities:
\begin{itemize}
\item Ramond (R) condition: $\psi(\xi^0,\xi^1+L)=\psi(\xi^0,\xi^1)$;
\item Neveu\--Schwarz (NS) condition: $\psi(\xi^0,\xi^1+L)=-\psi(\xi^0,\xi^1)$.
\end{itemize}
We have
\eq$$
\Aligned{
\psi_1(\xi)
&=\sqrt{\pi\over L}\sum_{k\in\Z+{\delta\over2}}b_k(\xi^0)\e^{2\pi\i k\xi^1/L},
\\
\psi_2(\xi)
&=\sqrt{\pi\over L}\sum_{k\in\Z+{\delta\over2}}\bb_k(\xi^0)\e^{-2\pi\i k\xi^1/L},
}\label{psi-decompose}
$$
Here
\eq$$
\Aligned{
\delta
&=0
&&\text{in the R case;}
\\
\delta
&=1
&&\text{in the NS case.}
}\label{delta01-def}
$$
Evidently, due to the Majorana condition we have
\eq$$
b_{-k}=\left(b_k\right)^*,
\qquad
\bb_{-k}=\left(\bb_k\right)^*.
\label{bk-Majoranacond}
$$
The Hamiltonian and momentum are
\eq$$
H={2\pi\over L}\sum_{k\in\Z_{\ge0}+{\delta\over2}}k(b_{-k}b_k+\bb_{-k}\bb_k),
\qquad
P={2\pi\over L}\sum_{k\in\Z_{\ge0}+{\delta\over2}}k(b_{-k}b_k-\bb_{-k}\bb_k).
\label{HPb-freeboson}
$$
The Poisson bracket is
\eq$$
\{b_k,b_l\}=\{\bb_k,\bb_l\}=\i\delta_{k+l,0},
\qquad
\{b_k,\bb_l\}=0.
\label{bb-Poisson}
$$
It is easy to check that the equations of motion for the modes $b_k,\bb_k$ in both cases read
\eq$$
\d_0b_k=-\i{2\pi\over L}kb_k,
\qquad
\d_0\bb_k=-\i{2\pi\over L}kb_k.
\label{bk-eqmo}
$$
and have the following solution
\eq$$
b^+_k=\beta_k\e^{-2\pi\i k\xi^0/L},
\qquad
b^-_k=\bbeta_k\e^{2\pi\i k\xi^0/L}.
\label{bpm-solution}
$$
We finally have
\eq$$
\Aligned{
\psi_1(\zeta)
&=\sqrt{\pi\over L}\sum_{k\in\Z+{\delta\over2}}\beta_k\e^{2\pi\i k\zeta/L},
\\
\psi_2(\bzeta)
&=\sqrt{\pi\over L}\sum_{k\in\Z+{\delta\over2}}\bbeta_k\e^{-2\pi\i k\bzeta/L}.
}\label{psipm-beta}
$$

Now let us quantize the system. The Poisson bracket is substituted by the anticommutator:
\eq$$
[\psi_\alpha(\xi^0,\xi^1),\psi_\beta(\xi^0,\xi^{\prime1})]_+=\pi\delta_{\alpha\beta}\delta(x-x')
\label{psipsi-commut}
$$
or
\eq$$
[\beta_k,\beta_l]_+=[\bbeta_k,\bbeta_l]_+=\delta_{k+l,0},
\qquad
[\beta_k,\bbeta_l]_+=0.
\label{bb-commut}
$$
We changed to constant operators $\beta_k,\bbeta_k$. Define the vacuums by the conditions $\beta_k|0\rangle_{\rm NS}=\bbeta_k|0\rangle_{\rm R}=\beta_k|0\rangle_{\rm R}=\bbeta_k|0\rangle_{\rm NS}=0$ for $k>$. But there will be some difference between the NS and R vacuums, which we specify later.

Let us write the Hamiltonian in terms of $\beta$-modes. As in the boson case we consider the products of operators as symmetrized ones:
$$
\beta_{-k}\beta_k\mapsto{1\over2}(\beta_{-k}\beta_k-\beta_k\beta_{-k})=\beta_{-k}\beta_k-{1\over2}.
$$
In the R sector we obtain for the vacuum energy
$$
E^{\rm R}_0=-2\pi\sum^\infty_{k=0}{k\over L}={\pi\over6L}.
$$
In the R sector we have
\Multline*$$
E^{\rm NS}_0=-2\pi\sum^\infty_{k=0}{k+1/2\over L}
=\left[2\pi{\d\over\d\varepsilon}\sum^\infty_{k=0}\e^{-\varepsilon(k+1/2)/L}+\const\cdot L\right]_{\varepsilon\to0}
\\
=\left[{\d\over\d\varepsilon}{\pi\over\sh{\varepsilon\over2L}}+\const\cdot L\right]_{\varepsilon\to0}
=\left[-{2\pi L\over\varepsilon^2}-{\pi\over12L}+\const\cdot L\right]_{\varepsilon\to0}
=-{\pi\over12L}.
$$
The Hamiltonians read
\eq$$
\Aligned{
H^{\rm R}
&={2\pi\over L}\sum_{k\in\Z_{>0}}k(\beta_{-k}\beta_k+\bbeta_{-k}\bbeta_k)+{\pi\over6L},
\\
H^{\rm NS}
&={2\pi\over L}\sum_{k\in\Z_{\ge0}+{1\over2}}k(\beta_{-k}\beta_k+\bbeta_{-k}\bbeta_k)-{\pi\over12L}.
}\label{H-freefermion-quantum}
$$
The expressions for the momentum remain unchanged.

Note that the energy of the NS vacuum is larger than that of the R vacuum:
\eq$$
E^{\rm R}_0-E^{\rm NS}_0={\pi\over4L}.
\label{ER-ENS}
$$
It means that the R vacuum is an excited state. Moreover, since $[b_0,H^{\rm R}]=[\bb_0,H^{\rm R}]=0$ and $b_0^2=\bb_0^2={1\over2}$, if $|v\rangle_{\rm R}$ is any eigenvector, we immediately obtain a quadruplet
\eq$$
|v\rangle_{\rm R},\quad\beta_0|v\rangle_{\rm R},\quad\bbeta_0|v\rangle_{\rm R},\quad\beta_0\bbeta_0|v\rangle_{\rm R}.
\label{R-4vacuums}
$$
Nevertheless, this representation is reducible. It splits into two equivalent irreducible representation, one of which we will associate with the fermion. Introduce two fermion operators
\eq$$
c_0={\i^{-1/2}\beta_0+\i^{1/2}\bbeta_0\over\sqrt2},
\qquad
c_0^+={\i^{1/2}\beta_0+\i^{-1/2}\bbeta_0\over\sqrt2}.
\label{cc+-def}
$$
They are mutually conjugate and satisfy the standard fermion relations
\eq$$
[c_0,c^+_0]_+=1,
\qquad
c_0^2=(c^+_0)^2=0.
\label{cc+-commut}
$$
It is natural to construct any representation from the $c_0$-vacuum $|u\rangle$: $c_0|u\rangle=0$. The representation is evidently two-dimensional: $\mathop{\rm span}\{|u\rangle,c_0^+|u\rangle\}$. It is easy to construct two such vectors in the representation (\ref{R-4vacuums}):
$$
|u_1\rangle=c_0\beta_0|v\rangle_{\rm R},
\qquad
|u_2\rangle=c_0|v\rangle_{\rm R}.
$$
We may remove doubling by choosing any of these representations.

Finally, define the vacuum vector $|0\rangle_{\rm R}$ by the conditions
\eq$$
\beta_k|0\rangle_{\rm R}=\bbeta_k|0\rangle_{\rm R}=0\quad(k>0),
\qquad
c_0|0\rangle_{\rm R}=0.
\label{R-vac}
$$
The second orthogonal vector is
\eq$$
|1\rangle_{\rm R}=c^+_0|0\rangle_{\rm R}
\label{R1-vac}
$$
satisfy the `dual' condition: $c^+_0|1\rangle_{\rm R}=0$.

The NS vacuum is nondegenerate and corresponds to the lowest energy in both sectors. The condition
\eq$$
\beta_k|0\rangle_{\rm NS}=\bbeta_k|0\rangle_{\rm NS}=0\quad(k>0)
\label{NS-vac}
$$
defines it uniquely.

To understand this picture better, let us make the transformation to the plane $z=\e^{-2\pi\i\zeta/L}$, $\bz=\e^{2\pi\i\bzeta/L}$. The action (\ref{S-freefermion}) is consistent with a conformal transformation, if the spinor field transforms as
\eq$$
\psi_1(\zeta,\bzeta)\to(f'(\zeta))^{1/2}\psi_1(f(\zeta),\bar f(\bzeta)),
\qquad
\psi_2(\zeta,\bzeta)\to(\bar f'(\bzeta))^{1/2}\psi_2(f(\zeta),\bar f(\bzeta)).
\label{psi-transform}
$$
Hence on the plane we have
\eq$$
\Aligned{
\psi_1(z)
&={\i^{1/2}\over\sqrt2}\Psi(z)={\i^{1/2}\over\sqrt2}\sum_{k\in\Z+{\delta\over2}}\beta_kz^{-1/2-k},
\\
\psi_2(\bz)
&={\i^{-1/2}\over\sqrt2}\bPsi(\bz)={\i^{-1/2}\over\sqrt2}\sum_{k\in\Z+{\delta\over2}}\bbeta_k\bz^{-1/2-k}.
}\label{psipm-beta-plane}
$$
Note that the periodicity conditions on the plane are opposite to those on the cylinder:
\eq$$
\Aligned{
&\text{R sector:}
&\psi(\e^{2\pi\i}z,\e^{-2\pi\i}\bz)
&=-\psi(z,\bz),
\\
&\text{NS sector:}
&\psi(\e^{2\pi\i}z,\e^{-2\pi\i}\bz)
&=\psi(z,\bz).
}\label{NSR-plane}
$$
In the NS sector the fermion is well-defined in the vicinity of the origin, while in the R sector there is a singularity. The singularity is described by a special operator depending on the R vacuum:
\eq$$
|0\rangle_{\rm R}=\sigma(0)|0\rangle_{\rm NS},
\qquad
|1\rangle_{\rm R}=\mu(0)|0\rangle_{\rm NS}.
$$
The operator $\sigma(x)$ is called \emph{spin operator} or \emph{order parameter}, while the operator $\mu(x)$ is called \emph{dual spin operator} or \emph{disorder parameter}.%
%
\footnote{Their names are related to their role in the Ising model as the operators corresponding to the spin variable in two dual representations of the Ising model. The parameter $\sigma$ becomes nonzero at low temperature $T<T_c$, while $\mu$ is only nonzero at high temperatures $T>T_c$. The free massless fermion describes the Ising model at the critical point, where there is no principal difference between these two objects.}

From (\ref{H-freefermion-quantum}) we may conclude that the Hamiltonian on the cylinder $H$ is related to the dilation operator~$D$ on the plane:
\eq$$
H={2\pi\over L}D-{\pi\over12L}.
\label{HD-rel-freefermion}
$$
The vacuum energy term is twice smaller than in the boson case. From (\ref{ER-ENS}) we conclude that
\eq$$
D\sigma(z)={1\over8}\sigma(z),
\qquad
D\mu(z)={1\over8}\mu(z),
\label{D-sigma-mu}
$$
which means that their scaling dimensions $d_\sigma=d_\mu={1\over8}$.
\bigbreak
\bigskip
\centerline{\bf Problems}
\medskip

{\bf 1.}~Compute the correlation functions:
\begin{enumerate}
\item $\langle\Psi(z')\Psi(z)\rangle\overset{\rm def}={}_{\rm NS}\langle0|\Psi(z')\Psi(z)|0\rangle_{\rm NS}$;
\item $\langle\mu(\infty)\Psi(z)\sigma(0)\rangle\overset{\rm def}={}_{\rm R}\langle1|\Psi(z)|0\rangle_{\rm R}$;
\item $\langle\sigma(\infty)\Psi(z')\Psi(z)\sigma(0)\rangle\overset{\rm def}={}_{\rm R}\langle0|\Psi(z')\Psi(z)|0\rangle_{\rm R}$.
\end{enumerate}

{\bf 2.}~The matrix element ${}_{\rm R}\langle0|0\rangle_{\rm R}$ defines a pair correlation function on the plane:
$$
{}_{\rm R}\langle0|0\rangle_{\rm R}=\langle\sigma(\infty)\sigma(0)\rangle
=\lim_{z,\bz\to\infty}z^{1/16}\bz^{1/16}\langle\sigma(z,\bz)\sigma(0)\rangle.
$$
By means of a Möbius transformation
$$
f(z)={az+b\over cz+d}
$$
calculate the correlation function
$$
\langle\sigma(x')\sigma(x)\rangle={}_{\rm NS}\langle0|\sigma(x')\sigma(x)|0\rangle_{\rm NS}.
$$

{\bf 3.}~By applying a Möbius transformation to the second matrix element of Problem~1 find the correlation function
$$
\langle\mu(z_1)\Psi(z_2)\sigma(z_3)\rangle.
$$



\end{document}
