Relativistic Fields, Spinors, and the Ising Continuum Limit

The previous page produced a surprise: a two-dimensional Ising model written in terms of ordinary commuting spins contains a natural fermionic field. The field is not inserted by hand. It is the point-split product of an order operator and a disorder operator, and its fermionic sign comes from transporting the disorder cut past the order insertion.

This page translates that observation into continuum field theory. We first classify scalar, vector and spinor fields by their rotation laws. Then we explicitly project the square-lattice corner equation onto its two critical modes, obtaining a Majorana equation with $m=4(K-K_c)/a$ to leading scaling order in the conventions below. This limit assumes $a|k|\ll1$ and $|K-K_c|\ll1$ in the bulk, away from additional coincident insertions. The same calculation identifies the critical part of the Ising energy with the fermion mass operator and explains the logarithmic specific heat.

Required background. Order, disorder, and the Ising fermion supplies the four corner directions, transported cuts and exact local propagation relation.

Helpful background. Critical propagators and the upper critical dimension supplies the scaling interpretation of a correlation length; two-dimensional disorder lines supplies the underlying mixed-correlator sign proof. The rotation review below is self-contained.

Fields as rotation representations

Euclidean spinor basis and Ising mass sign

Most of this page is Euclidean. Spatial indices $i,j$ are summed with $\delta_{ij}$, and the two-dimensional coordinates are

$$z=x^1+i x^2, \qquad \partial={1\over2}(\partial_1-i\partial_2), \qquad \bar\partial={1\over2}(\partial_1+i\partial_2).$$

For the two-dimensional Dirac equation we use Hermitian Euclidean gamma matrices

$$\gamma_1=\begin{pmatrix}0&1\\1&0\end{pmatrix}, \qquad \gamma_2=\begin{pmatrix}0&-i\\ i&0\end{pmatrix}, \qquad \{\gamma_i,\gamma_j\}=2\delta_{ij}.$$

We use the equation $(\gamma_i\partial_i-m)\Psi=0$. The projection below fixes $m>0$ on the ordered side $K>K_c$ in this basis; a different field convention can reverse both this labeling and the displayed mass sign. The scaling mass changes sign under Kramers–Wannier duality.

The word Majorana below counts one Lorentzian real fermionic degree of freedom. After Euclidean continuation the two chiral Grassmann fields are independent integration variables; no pointwise condition such as $\bar\psi=\psi^*$ is imposed. The Majorana structure is instead encoded by the antisymmetric fermion operator and its Pfaffian.

A continuum field is not defined only by its equation of motion. It is also defined by how it transforms under spacetime symmetries. In Euclidean field theory, the local rotation group is $\mathrm{SO}(d)$, or more precisely its double cover $\operatorname{Spin}(d)$ when spinors are present. A field with components $\Phi_A(x)$ transforms as

$$\Phi'_A(x')=D(R)_{AB}\Phi_B(x), \qquad x'=Rx,$$

where $D(R)$ is a finite-dimensional representation carried by the field indices. A scalar has $D(R)=1$. A vector has $D(R)=R$. A spinor has a matrix $S(R)$ satisfying

$$S(R)^{-1}\gamma_i S(R)=R_{ij}\gamma_j.$$

This is the first difference between writing down a field and writing down a collection of components. A two-component object is not automatically a spinor; it is a spinor only if rotations act by the spin representation.

A free scalar field is the simplest example. In Euclidean signature one may write

$$S_0[\phi] ={1\over2}\int d^d x\, \left[(\partial_i\phi)^2+m^2\phi^2\right],$$

so the classical equation is

$$(-\partial^2+m^2)\phi=0.$$

After analytic continuation to Lorentzian signature this becomes the Klein–Gordon equation. The field carries no internal rotation index; all angular momentum of a scalar excitation is orbital.

A vector field carries one rotation index:

$$A_i'(x')=R_{ij}A_j(x).$$

A tensor field carries several such indices. For example,

$$B_{ij}'(x')=R_{ik}R_{j\ell}B_{k\ell}(x).$$

These transformation laws are local. They do not yet say which field equations are correct or how many physical polarizations propagate. Those are dynamical questions.

Vector fields and transversality

A massive vector field can be described by the Euclidean Proca action

$$S[A]=\int d^d x\, \left[ {1\over4}F_{ij}F_{ij}+{m^2\over2}A_iA_i \right], \qquad F_{ij}=\partial_iA_j-\partial_jA_i.$$

Varying $A_i$ gives

$$-\partial_jF_{ji}+m^2A_i=0.$$

Expanding $F_{ji}$,

$$\partial_jF_{ji} =\partial^2A_i-\partial_i\partial_jA_j,$$

so the equation is

$$(-\partial^2+m^2)A_i+\partial_i\partial_jA_j=0.$$

Taking a divergence gives

$$m^2\partial_iA_i=0.$$

Therefore, for $m\ne0$,

$$\boxed{\partial_iA_i=0.}$$

The vector equation plus this constraint is equivalent to a Klein–Gordon equation for the transverse components:

$$(-\partial^2+m^2)A_i=0, \qquad \partial_iA_i=0.$$

In momentum space, the decomposition is especially transparent:

$$A_i(k)=A_i^{\perp}(k)+A_i^{\parallel}(k), \qquad k_iA_i^{\perp}(k)=0,$$

with transverse projector for nonzero Euclidean momentum $k$,

$$P_{ij}^{\perp}(k)=\delta_{ij}-{k_i k_j\over k^2}.$$

A vector field decomposes into components transverse and longitudinal to the momentum. The massive Proca equation imposes $\partial_iA_i=0$ on shell, leaving the spin-one polarizations.

This is a useful warning for later gauge theory. A vector field has a vector index, but not every component of that index represents an independent physical polarization. Gauge redundancy, constraints, and equations of motion all matter. In the Ising problem the analogous lesson is that the four corner labels of the lattice fermion are not four independent particles. They are a convenient lattice package whose long-distance content is a smaller spinor field.

Spinors and half-angle rotations

Spinors are not ordinary tensors. For an active counterclockwise rotation, fix

$$R_\theta=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}, \qquad x'=R_\theta x.$$

In the chosen gamma basis, the spin transformation is

$$\begin{gathered} \Psi=\begin{pmatrix}u\\v\end{pmatrix},\qquad \Psi'(x')=S_\theta\Psi(x),\\ S_\theta=\exp\!\left(-\frac\theta2\gamma_1\gamma_2\right) =\begin{pmatrix}e^{-i\theta/2}&0\\0&e^{i\theta/2}\end{pmatrix}. \end{gathered}$$

Indeed, $\gamma_1\gamma_2=i\,\mathrm{diag}(1,-1)$ and direct multiplication gives $S_\theta^{-1}\gamma_iS_\theta=(R_\theta)_{ij}\gamma_j$. Together with $\partial'_i=(R_\theta)_{ij}\partial_j$, this makes the Dirac equation covariant. The same rotor is derived in the one-plane Spin example.

Thus $u'(x')=e^{-i\theta/2}u(x)$ and $v'(x')=e^{i\theta/2}v(x)$. These signs depend on the complete coordinate-and-field convention; checking only a full turn would not detect their reversal. In particular,

$$S_{2\pi}(u)=-u, \qquad S_{2\pi}(v)=-v.$$

A spinor changes sign under a full $2\pi$ rotation. Only after a $4\pi$ rotation does it return to itself.

For $x'=R_\theta x$ and $\gamma_1=\sigma_1$, $\gamma_2=\sigma_2$, the upper component acquires $e^{-i\theta/2}$ and the lower component $e^{i\theta/2}$. Both acquire $-1$ under a full turn. The coordinate rotation is schematic; its orientation and the component phases use the same convention.

The Euclidean two-dimensional Dirac equation is a first-order equation,

$$(\gamma_i\partial_i-m)\Psi=0.$$

With the gamma matrices stated above and $\Psi=(u,v)^T$, this is equivalent to

$$\boxed{ (\partial_1+i\partial_2)u=m v, \qquad (\partial_1-i\partial_2)v=m u. }$$

At $m=0$ these equations split into holomorphic and antiholomorphic pieces:

$$\bar\partial u=0, \qquad \partial v=0.$$

Thus one component depends only on $z$, and the other depends only on $\bar z$. This is the continuum seed of the Ising conformal field theory: the critical fermion separates into left-moving and right-moving Majorana components.

There is no contradiction between the first-order Dirac equation and the second-order Klein–Gordon equation. Squaring the operator gives

$$(\gamma_i\partial_i+m)(\gamma_j\partial_j-m)\Psi =(\partial^2-m^2)\Psi,$$

With these fixed conventions, this identity is exact. Multiplying by an overall minus sign gives the positive Euclidean operator $-\partial^2+m^2$. Each component obeys the associated second-order equation, while the first-order equation also relates the components.

The Ising corner field as a spinor

Recall the point-split order–disorder composite from the previous page. Put a spin at an original-lattice site $x$ and a disorder endpoint at a neighboring dual site $x+e_a$, where the four half-lattice vectors have angles

$$\theta_a={\pi\over4}+{\pi\over2}(a-1), \qquad a=1,2,3,4.$$

The four corner fields are

$$\chi_a(x)=\sigma_x\mu_{x+e_a}.$$

Because moving the disorder endpoint once around the spin crosses the branch cut once, the corner field is antiperiodic:

$$\chi_{a+4}(x)=-\chi_a(x).$$

This antiperiodicity is exactly the lattice form of spinorial behavior. A scalar function of a continuous direction would have integer angular harmonics, whereas an antiperiodic function has half-integer harmonics. Here, however, the direction is sampled at only four corners. The corner-label space is therefore four-dimensional, with four independent Fourier characters. A convenient set of representatives is

$$s\in\left\{+{1\over2},-{1\over2},+{3\over2},-{3\over2}\right\} \qquad (s\ \mathrm{mod}\ 4).$$

A complete finite Fourier expansion is

$$\begin{gathered} f_s(x)=\frac14\sum_{a=1}^4e^{-is\theta_a}\chi_a(x),\\ \chi_a(x)=\sum_{s=\pm1/2,\,\pm3/2}e^{is\theta_a}f_s(x). \end{gathered}$$

Here $f_s$ is the coefficient called $u_s$ on the previous page; the temporary name keeps it distinct from the continuum components $u,v$. The angles are lifted continuously, so $\theta_{a+4}=\theta_a+2\pi$. Advancing a corner frame multiplies its character by $e^{is\theta}$, whereas rotating the field and its argument actively transforms the coefficient by $e^{-is\theta}$. For example, a lattice quarter-turn gives $\chi'_a(R_{\pi/2}x)=\chi_{a-1}(x)$ with the cut transported, and hence $f'_s(R_{\pi/2}x)=e^{-is\pi/2}f_s(x)$. This is the same convention as $S_\theta$ above.

The critical projection retains $f_{+1/2},f_{-1/2}$. The other two characters have a nonzero lattice kernel at criticality and can be eliminated in a long-wavelength expansion. They are not extra independent continuum particles or the beginning of an unlimited corner-harmonic tower. Inspect the figure’s original sites, dual face centers and the two successive reductions before following the matrix calculation.

The four Ising corner fields split into a critical $s=\pm1/2$ pair and a gapped $s=\pm3/2$ pair. With the lifted angles and active rotation convention stated here, the critical coefficients give $(u,v)$ up to one common normalization, and the leading mass is $m=4(K-K_c)/a$. In the figure, $j$ labels the corners and $a$ is lattice spacing. The geometry is schematic; the continuum step requires $a|k|\ll1$ and $|K-K_c|\ll1$.

Projecting the four-component lattice equation

Let $\ell$ denote the lattice spacing in this calculation ($\ell=a$ in the surrounding text), to distinguish it from the corner index $a$. Define the antiperiodic shift $T$ by $(T\chi)_a=\chi_{a+1}$, including $\chi_5=-\chi_1$. The exact previous-page relation is

$$\chi_a(x)-\cosh(2K)\chi_{a+1}(x) +\sinh(2K)\chi_{a+2}(x+\delta_a)=0.$$

This is a correlator equation away from spectator contacts. With inverse Fourier phase $e^{ik\cdot x}$, write it as $A(k,K)\chi(k)=0$, where

$$\begin{gathered} A(k,K)=I-\cosh(2K)T+\sinh(2K)E(k)T^2,\\ E(k)=\operatorname{diag}(e^{ik\cdot\delta_1},\ldots,e^{ik\cdot\delta_4}),\\ (\delta_1,\delta_2,\delta_3,\delta_4) =\ell\big((0,1),(-1,0),(0,-1),(1,0)\big). \end{gathered}$$

Use the orthonormal matrix $B_{as}=\tfrac12e^{is\theta_a}$ with columns ordered $s=(1/2,-1/2,3/2,-3/2)$. Then $B^\dagger B=I$ and $B^\dagger\chi=2f$. At zero momentum and critical coupling,

$$B^\dagger A(0,K_c)B =\operatorname{diag}(0,0,2-2i,2+2i).$$

Write $\delta K=K-K_c$, $k_\pm=k_1\pm ik_2$, and keep first order in $\delta K$ and $\ell k$. The critical $2\times2$ block is

$$\begin{aligned} A_{\rm L}(k,K) &=\begin{pmatrix} 2e^{3i\pi/4}\delta K & \tfrac{i\ell}{2}e^{-i\pi/4}k_-\\ \tfrac{i\ell}{2}e^{i\pi/4}k_+ & 2e^{-3i\pi/4}\delta K \end{pmatrix}\\ &\quad+O(\delta K^2,\,|\delta K|\ell|k|,\,\ell^2|k|^2). \end{aligned}$$

These coefficients can be checked directly from $B^\dagger T B=\operatorname{diag}(e^{is\pi/2})$, $\cosh(2K)=\sqrt2+2\delta K+O(\delta K^2)$ and $\sinh(2K)=1+2\sqrt2\delta K+O(\delta K^2)$. The derivative term comes from $E(k)=I+i\operatorname{diag}(k\cdot\delta_a)+O(\ell^2|k|^2)$.

Eliminating the heavy block gives the Schur complement $A_{\rm LL}-A_{\rm LH}A_{\rm HH}^{-1}A_{\rm HL}$. The heavy eigenvalues remain nonzero, and both mixed blocks vanish at $k=0$. Their correction therefore starts at $O(\ell^2|k|^2)$ and cannot alter the displayed first-order coefficients.

Replace $ik_i$ by $\partial_i$ and multiply the two equations, respectively, by $(2/\ell)e^{i\pi/4}$ and $(2/\ell)e^{-i\pi/4}$. The phases cancel:

$$\begin{gathered} \begin{pmatrix} -m & \partial_1-i\partial_2\\ \partial_1+i\partial_2 & -m \end{pmatrix} \begin{pmatrix}f_{+1/2}\\f_{-1/2}\end{pmatrix}=0,\\ m=\frac{4\delta K}{\ell}. \end{gathered}$$

The omitted operator terms have order $\delta K^2/\ell+|\delta K||k|+\ell|k|^2$. In the scaling limit $\ell\to0$ with $m$ and physical momentum fixed, $\delta K=m\ell/4$ and this remainder vanishes. The lattice spacing is now again denoted by $a$.

The continuum components may be normalized as

$$\begin{pmatrix}u\\v\end{pmatrix} =Z_\psi a^{-1/2} \begin{pmatrix}f_{+1/2}\\f_{-1/2}\end{pmatrix}.$$

The common dimensionless factor $Z_\psi$ fixes the chosen two-point amplitude; the homogeneous propagation equation does not determine it. No relative field rephasing is needed with these lifted angles. This completes the promised differential and mass normalization. Polyakov 1987, § 10.3.1, pp. 276–278, Eqs. (10.60)–(10.69) gives the lattice equation and soft-mode argument; the explicit matrix calculation above fixes our spacing and phase dictionary. His same-argument transformation $u(x)\mapsto e^{i\theta/2}u(R_\theta x)$ corresponds to the inverse of the active transformation used here.

The Majorana action

The two low modes describe one Lorentzian Majorana field, with two real spinor components, rather than a charged complex Dirac field. In the Euclidean functional integral use independent Grassmann variables $u,v$ and

$$\begin{gathered} S_{\rm M}=\frac12\int d^2x\, \big[u(\partial_1+i\partial_2)u -v(\partial_1-i\partial_2)v-2muv\big],\\ \mathcal A=C(\gamma_i\partial_i-m),\qquad C=\begin{pmatrix}0&1\\-1&0\end{pmatrix}. \end{gathered}$$

Varying this action gives the two displayed component equations. Since $C^T=-C$ and $C\gamma_i$ is symmetric, integration by parts makes $\mathcal A$ antisymmetric with boundary conditions that remove its surface term. With a fixed Grassmann measure orientation the Gaussian integral is proportional to $\operatorname{Pf}\mathcal A$; its sign is not specified by taking an arbitrary square root of a determinant. No pointwise condition $v=u^*$ is imposed. This action describes the local bulk scaling theory; a global partition function also requires the spin structure and sector combination inherited from the lattice boundaries.

Mass, duality, and the two phases

The Kramers–Wannier relation is

$$\sinh 2K\,\sinh 2K^*=1.$$

At the self-dual point,

$$K^*=K=K_c, \qquad \sinh 2K_c=1.$$

Expanding around criticality gives

$$K^*-K_c=-(K-K_c)+O((K-K_c)^2).$$

Therefore the continuum mass changes sign under duality:

$$m(K^*)=-m(K)+O(a m^2).$$

With the field phases fixed above, $m>0$ is the ordered side $K>K_c$ and $m<0$ is the disordered side. Duality exchanges long-range order of the spin and disorder variables in the thermodynamic limit. The free-fermion correlation length is $\xi_{\rm f}=1/|m|$ in the continuum normalization. Other Ising channels can have different correlation-length amplitudes, but their critical divergence has the same exponent:

$$\frac{\xi}{a}\propto\frac1{|K-K_c|}.$$

Thus the two-dimensional Ising correlation-length exponent is

$$\nu=1.$$

This agrees with the exact solution and is much sharper than mean-field theory. The deeper reason is that the critical Ising model is not an interacting scalar theory at long distances; it is a free massless Majorana fermion plus nontrivial spin and disorder fields.

Energy operator and the specific heat

The lattice energy is the operator conjugate to the coupling. For the nearest-neighbor square-lattice model,

$$Z(K)=\sum_{\{\sigma\}}\exp\!\left(K\sum_{\langle ij\rangle}\sigma_i\sigma_j\right),$$

define the bond energy

$$\epsilon_{ij}=\sigma_i\sigma_j.$$

The dimensionless free energy $F=-\log Z$ satisfies

$$-{\partial F\over\partial K} =\left\langle\sum_{\langle ij\rangle}\epsilon_{ij}\right\rangle,$$

and

$$-{\partial^2 F\over\partial K^2} =\sum_{\langle ij\rangle,\langle k\ell\rangle} \left\langle \epsilon_{ij}\epsilon_{k\ell}\right\rangle_c.$$

Up to conventional powers of the temperature, this is the specific heat. Translational invariance turns it into an integral of the connected energy–energy correlator:

$$C\propto \sum_x \langle\epsilon(0)\epsilon(x)\rangle_c.$$

The lattice bond has a nonzero identity contribution. Its critical part has an expansion $\epsilon_{ij}=c_0+c_1a\,\epsilon_{\rm cont}(x)+\cdots$, with nonuniversal coefficients; the connected correlator removes $c_0$. In the action convention above,

$$\epsilon_{\rm cont}(x) =\frac{\partial\mathcal L_{\rm M}}{\partial m} =-u(x)v(x)=i\bar\psi(x)\psi(x), \qquad \psi=u,\quad\bar\psi=-iv.$$

Here $\bar\psi$ is the independent opposite-chirality Grassmann field, not the complex conjugate of $\psi$. Anticommutation gives $i(-iv)u=vu=-uv$, so the factor of $i$ is consistent with the already fixed $u,v$ phases. The sign of $c_1$ relates this mass derivative to the particular lattice bond convention; it does not affect the connected two-point scaling.

Equivalently, the energy operator is the operator multiplying the fermion mass in the action. Since the Majorana field has scaling dimension $1/2$, the bilinear has scaling dimension

$$\Delta_\epsilon=1.$$

At criticality,

$$\langle\epsilon(x)\epsilon(0)\rangle_c\sim {1\over |x|^{2\Delta_\epsilon}} ={1\over |x|^2}.$$

Therefore the specific heat diverges logarithmically:

$$C\sim \int_a^\xi 2\pi r\,dr\,{1\over r^2} \sim \log {\xi\over a} \sim \log {1\over |K-K_c|}.$$

The same result appears in momentum space as the one-loop fermion bubble

$$C\sim \int_{|m|}^{1/a}{d^2k\over k^2} \sim \log {1\over |m|a}.$$

After subtracting its identity contribution and fixing the field amplitude, the critical part of the Ising bond energy is proportional to the fermion mass operator $i\bar\psi\psi$. Its two-point function scales as $1/|x|^2$, so the integrated energy–energy correlator gives the logarithmic specific-heat singularity. The diagram represents the scaling argument; it does not fix its lattice amplitude.

Once the correct scaling field is known, the thermodynamic singularity follows from its connected two-point function. The exact lattice solution fixes the nonuniversal amplitude and regular terms; the continuum argument fixes the logarithm. The equivalent fermion-integral calculation appears in Polyakov 1987, § 10.3.1, p. 278, Eq. (10.70) and the following specific-heat formula. These estimates assume $a\ll\xi$ and a sample much larger than $\xi$; at criticality in a finite sample the sample size supplies the infrared cutoff.

From lattice transport to critical thermodynamics

A continuum field is classified not only by its equation of motion but also by its rotation representation. Scalars transform trivially, vectors transform by the ordinary rotation matrix, and spinors transform by the double cover. The essential spinor fact is $S(2\pi)=-1$.

The Ising order–disorder composite has precisely this sign. Its four corner components are antiperiodic under a full rotation of the disorder endpoint around the order insertion. Their four independent corner characters may be represented by spins $\pm1/2$ and $\pm3/2$ modulo $4$. In the scaling limit the leading $\pm1/2$ combinations form a real two-component fermion. The local lattice transfer relation becomes the massive two-dimensional Dirac equation

$$(\partial_1+i\partial_2)u=m v, \qquad (\partial_1-i\partial_2)v=m u, \qquad m=\frac{4(K-K_c)}a\quad\text{to leading scaling order}.$$

At $K=K_c$, the mass vanishes and the two components become holomorphic and antiholomorphic Majorana fields. The energy operator is the fermion mass bilinear $i\bar\psi\psi$, whose scaling dimension is $1$. Hence the specific heat is logarithmically singular:

$$C\sim \log {1\over |K-K_c|}.$$

Common pitfalls

The four Ising corner fields are not four independent continuum fermions, nor do four samples support an unlimited independent harmonic tower. They form a four-dimensional lattice-label space; the long-distance spin $\pm1/2$ projection gives one Majorana fermion, while derivative descendants account for higher continuum-spin corrections.

The statement $\chi_{a+4}=-\chi_a$ is not ordinary anticommutation of microscopic spins. It is monodromy of mixed order–disorder correlation functions. The original Ising variables are still commuting classical spins.

The sign of the continuum mass depends on phase conventions, but its oddness under Kramers–Wannier duality is physical. The two signs encode the ordered and disordered phases.

The Feynman-bubble picture of the specific heat should be understood as a continuum scaling argument. The lattice cutoff $a$ and the normalization of $\epsilon$ determine nonuniversal constants, while the logarithmic divergence is universal.

Exercises

Exercise 1: Proca transversality

Derive the transversality condition for a massive vector field from the Proca equation

$$-\partial_jF_{ji}+m^2A_i=0, \qquad F_{ij}=\partial_iA_j-\partial_jA_i.$$

Then show that the transverse momentum-space projector is

$$P_{ij}^{\perp}(k)=\delta_{ij}-{k_i k_j\over k^2}.$$

Solution

Taking a divergence of the Proca equation gives

$$-\partial_i\partial_jF_{ji}+m^2\partial_iA_i=0.$$

The first term vanishes because $\partial_i\partial_j$ is symmetric in $i,j$, while $F_{ji}$ is antisymmetric:

$$\partial_i\partial_jF_{ji}=0.$$

For $m\ne0$, this implies

$$\partial_iA_i=0.$$

In momentum space, the longitudinal part of a vector is proportional to $k_i$. The projector onto this direction is

$$P_{ij}^{\parallel}={k_i k_j\over k^2}.$$

Therefore the projector onto the orthogonal subspace is

$$P_{ij}^{\perp}=\delta_{ij}-P_{ij}^{\parallel} =\delta_{ij}-{k_i k_j\over k^2}.$$

It obeys

$$k_iP_{ij}^{\perp}=k_j-{k^2k_j\over k^2}=0,$$

so $P^{\perp}A$ is transverse.

Exercise 2: Half-angle rotation and corner characters

Let a two-dimensional spinor transform under rotations by

$$S_\theta(u)=e^{-i\theta/2}u, \qquad S_\theta(v)=e^{i\theta/2}v.$$

Use the active convention $\Psi'(R_\theta x)=S_\theta\Psi(x)$. Check $S_\theta^{-1}\gamma_1S_\theta=\cos\theta\,\gamma_1-\sin\theta\,\gamma_2$, then show that a $2\pi$ rotation gives a minus sign. Explain why a continuously defined antiperiodic angular field has half-integer harmonics, and why four corner directions contain only four independent characters, labeled by $s=\pm1/2,\pm3/2$ modulo $4$.

Solution

Direct multiplication gives

$$S_\theta^{-1}\gamma_1S_\theta =\begin{pmatrix}0&e^{i\theta}\\e^{-i\theta}&0\end{pmatrix} =\cos\theta\,\gamma_1-\sin\theta\,\gamma_2.$$

For $\theta=2\pi$,

$$e^{i\theta/2}=e^{i\pi}=-1, \qquad e^{-i\theta/2}=e^{-i\pi}=-1.$$

Thus

$$S_{2\pi}(u)=-u, \qquad S_{2\pi}(v)=-v.$$

Now suppose a function of the corner angle has an angular harmonic $e^{is\theta}$. Under $\theta\mapsto\theta+2\pi$ it transforms as

$$e^{is(\theta+2\pi)}=e^{2\pi i s}e^{is\theta}.$$

Antiperiodicity requires

$$e^{2\pi i s}=-1.$$

Therefore

$$s\in {1\over2}+\mathbb Z.$$

Thus a continuously defined antiperiodic angular field has half-integer harmonics. On the four corner angles $\theta_a=\theta_0+\pi a/2$, however,

$$e^{i(s+4)\theta_a}=e^{4i\theta_0}e^{is\theta_a},$$

so $s$ and $s+4$ give the same corner character up to an $a$-independent phase. There are only four independent characters. One may choose the representatives

$$s=+{1\over2},\ -{1\over2},\ +{3\over2},\ -{3\over2} \qquad (\mathrm{mod}\ 4).$$

Exercise 3: Euclidean Dirac components

Using

$$\gamma_1=\begin{pmatrix}0&1\\1&0\end{pmatrix}, \qquad \gamma_2=\begin{pmatrix}0&-i\\i&0\end{pmatrix}, \qquad \Psi=\begin{pmatrix}u\\v\end{pmatrix},$$

show that

$$(\gamma_i\partial_i-m)\Psi=0$$

is equivalent to

$$(\partial_1+i\partial_2)u=m v, \qquad (\partial_1-i\partial_2)v=m u.$$

What happens at $m=0$?

Solution

Compute

$$\gamma_1\partial_1\Psi =\begin{pmatrix}\partial_1 v\\ \partial_1 u\end{pmatrix}, \qquad \gamma_2\partial_2\Psi =\begin{pmatrix}-i\partial_2 v\\ i\partial_2 u\end{pmatrix}.$$

Therefore

$$\gamma_i\partial_i\Psi =\begin{pmatrix} (\partial_1-i\partial_2)v\\ (\partial_1+i\partial_2)u \end{pmatrix}.$$

The equation $\gamma_i\partial_i\Psi=m\Psi$ gives

$$(\partial_1-i\partial_2)v=m u, \qquad (\partial_1+i\partial_2)u=m v.$$

These are the desired equations, just written in the opposite order.

At $m=0$ they become

$$(\partial_1+i\partial_2)u=0, \qquad (\partial_1-i\partial_2)v=0.$$

Since

$$\bar\partial={1\over2}(\partial_1+i\partial_2), \qquad \partial={1\over2}(\partial_1-i\partial_2),$$

we get

$$\bar\partial u=0, \qquad \partial v=0.$$

Thus $u$ is holomorphic and $v$ is antiholomorphic.

Exercise 4: The duality-odd Ising mass

Use the Kramers–Wannier relation

$$\sinh 2K\,\sinh 2K^*=1$$

to show that near the self-dual point,

$$K^*-K_c=-(K-K_c)+O((K-K_c)^2).$$

Why does this imply that the Ising fermion mass is duality-odd?

Solution

Set

$$K=K_c+\delta K, \qquad K^*=K_c+\delta K^*.$$

At criticality,

$$\sinh 2K_c=1.$$

Expanding to first order,

$$\sinh 2K =1+2\cosh 2K_c\,\delta K+O(\delta K^2),$$

and similarly

$$\sinh 2K^* =1+2\cosh 2K_c\,\delta K^*+O((\delta K^*)^2).$$

Multiplying and using $\sinh 2K\sinh 2K^*=1$ gives

$$1+2\cosh 2K_c(\delta K+\delta K^*)+O(\delta K^2)=1.$$

Therefore

$$\delta K^*=-\delta K+O(\delta K^2),$$

or

$$K^*-K_c=-(K-K_c)+O((K-K_c)^2).$$

The continuum mass is proportional to the relevant perturbation away from criticality:

$$m\propto K-K_c.$$

Since duality reverses $K-K_c$, it reverses $m$. Thus the mass term is duality-odd.

Exercise 5: The logarithmic specific heat

Assume the energy operator at the critical Ising point has scaling dimension $\Delta_\epsilon=1$. Show that the integrated energy–energy correlator diverges logarithmically. Then express the answer in terms of the correlation length $\xi$ and the mass $m$.

Solution

At criticality, an operator of scaling dimension $\Delta_\epsilon$ has two-point function

$$\langle\epsilon(x)\epsilon(0)\rangle_c\sim {1\over |x|^{2\Delta_\epsilon}}.$$

For $\Delta_\epsilon=1$,

$$\langle\epsilon(x)\epsilon(0)\rangle_c\sim {1\over |x|^2}.$$

The specific heat is proportional to the integral of this correlator. With a UV cutoff $a$ and an IR cutoff $\xi$,

$$C\sim \int_a^\xi 2\pi r\,dr\,{1\over r^2} =2\pi\int_a^\xi {dr\over r} =2\pi\log {\xi\over a},$$

up to the normalization of $\epsilon$.

The mass gap cuts off correlations at

$$\xi\sim {1\over |m|}.$$

Since $m\propto K-K_c$,

$$C\sim \log {1\over |m|a} \sim \log {1\over |K-K_c|}$$

plus nonuniversal regular terms.

References

• A. M. Polyakov, Gauge Fields and Strings. Contemporary Concepts in Physics, Vol. 3. Harwood Academic Publishers, 1987, § 10.3.1, doi:10.1201/9780203755082.

Further reading

• P. Di Francesco, P. Mathieu, and D. Sénéchal, Conformal Field Theory. Graduate Texts in Contemporary Physics. Springer, 1997, Chapters 4 and 12.
• B. M. McCoy and T. T. Wu, The Two-Dimensional Ising Model. Harvard University Press, 1973.
• T. D. Schultz, D. C. Mattis, and E. H. Lieb, “Two-Dimensional Ising Model as a Soluble Problem of Many Fermions,” Reviews of Modern Physics 36, 856–871 (1964), doi:10.1103/RevModPhys.36.856.