Antiferromagnets, Spin Chains, and Theta Terms

The previous page solved the two-dimensional $O(N)$ nonlinear sigma model at large $N$ and found a mass gap. That result is the right generic expectation for the $O(3)$ model too: the coupling is asymptotically free, grows in the infrared, and usually produces a finite correlation length. But the $O(3)$ model has a special extra ingredient in two Euclidean dimensions. Its fields are maps from spacetime into a sphere, and such maps have an integer winding number. The action can therefore contain a theta term.

This page explains why theta terms are not optional decorations in antiferromagnets. They are forced on us by the Berry phases of microscopic spins. The low-energy theory of a one-dimensional antiferromagnetic Heisenberg chain is not merely

$$S_{\rm kin}={1\over 2g}\int d^2x\,\partial_\mu\mathbf n\cdot\partial_\mu\mathbf n, \qquad \mathbf n^2=1,$$

but rather

$$S_E[\mathbf n] ={1\over 2g}\int d^2x\,\partial_\mu\mathbf n\cdot\partial_\mu\mathbf n -i\theta Q[\mathbf n].$$

Here $Q\in\mathbb Z$ is the degree of the map $\mathbf n:\text{spacetime}\to S^2$ when Euclidean spacetime is closed. For a spin-$S$ antiferromagnetic chain,

$$\boxed{\theta=2\pi S.}$$

Thus integer and half-integer spin chains land at different points on the theta circle:

$$S\in\mathbb Z\quad\Rightarrow\quad \theta=0\pmod {2\pi},$$

whereas

$$S\in\mathbb Z+{1\over2}\quad\Rightarrow\quad \theta=\pi\pmod {2\pi}.$$

This is the continuum-field-theory origin of Haldane’s distinction between integer and half-integer antiferromagnetic spin chains.

For readers coming from relativistic QFT, the important conceptual point is that $\theta$ is not chosen by hand in the spin-chain problem. It is fixed by the microscopic spin representation. Changing $S$ by $1/2$ changes the interference among topological sectors by a sign.

Required background. Nonlinear sigma models and constraints supplies the $O(3)$ kinetic theory, and the large-N saddle explains why the model is generically massive without a topological term. Only the coherent-state Berry phase is new here.

Theta-term normalization and spin-chain dictionary

Normalization of the theta term

We use Euclidean coordinates $(\tau,x)$ and orient them so that

$$Q[\mathbf n] ={1\over4\pi}\int d\tau\,dx\, \mathbf n\cdot(\partial_\tau\mathbf n\times\partial_x\mathbf n) \in\mathbb Z.$$

Equivalently,

$$Q[\mathbf n] ={1\over8\pi}\int d^2x\,\epsilon_{\mu\nu}\, \mathbf n\cdot(\partial_\mu\mathbf n\times\partial_\nu\mathbf n), \qquad \epsilon_{\tau x}=+1.$$

The Euclidean path-integral weight is

$$e^{-S_E}=\exp\left[-S_{\rm kin}+i\theta Q\right].$$

On a closed oriented spacetime, $Q$ is an integer. Changing the orientation convention sends $Q\mapsto -Q$ and therefore sends $\theta\mapsto-\theta$. The physics is unchanged after this convention is kept consistently. Boundaries require an additional qualification developed below.

Spin-chain dictionary

For the nearest-neighbor antiferromagnetic chain, the long-wavelength dictionary is convention-dependent in details but has a stable core:

Lattice quantity	Continuum meaning
coherent-state spin direction	$(-1)^j\mathbf n\sqrt{1-a^2\boldsymbol\ell^2/S^2}+(a/S)\boldsymbol\ell$
$\mathbf n^2=1$	staggered Néel direction
$\boldsymbol\ell\perp\mathbf n$	smooth uniform magnetization density
$c\sim 2JSa$	spin-wave velocity
$g\sim 2/S$	sigma-model coupling at large $S$
$\theta=2\pi S$	Berry-phase/topological angle

The precise normalization of $c$ and $g$ depends on the definition of $\boldsymbol\ell$ and on rescaling Euclidean time. The quantized statement $\theta=2\pi S$ is the robust one.

From a Néel chain to a slowly varying field

Consider the antiferromagnetic Heisenberg chain

$$H=J\sum_j \mathbf S_j\cdot\mathbf S_{j+1}, \qquad J>0,$$

with spin length

$$\mathbf S_j^2=S(S+1).$$

Classically, the lowest-energy pattern alternates:

$$\uparrow\downarrow\uparrow\downarrow\cdots.$$

The correct slow variable is therefore not the uniform magnetization, but the staggered Néel field $\mathbf n(x,\tau)$. For a coherent-state unit vector $\mathbf N_j$, a length-preserving decomposition is

$$\mathbf N_j(\tau) =(-1)^j\mathbf n(x_j,\tau) \sqrt{1-{a^2\boldsymbol\ell^2\over S^2}} +{a\over S}\boldsymbol\ell(x_j,\tau), \qquad x_j=ja,$$

where

$$\mathbf n^2=1, \qquad \mathbf n\cdot\boldsymbol\ell=0.$$

The coherent-state expectation value is $\langle\mathbf S_j\rangle=S\mathbf N_j$. Expanding the square root gives the familiar leading shorthand

$$\langle\mathbf S_j\rangle =S(-1)^j\mathbf n+a\boldsymbol\ell+O(a^2\boldsymbol\ell^2/S).$$

The field $\mathbf n$ records the local staggered direction. The smaller field $\boldsymbol\ell$ records the smooth ferromagnetic canting, or uniform magnetization density. The two constraints ensure $\mathbf N_j^2=1$; the square-root correction matters when deriving coefficients consistently beyond leading order.

A one-dimensional antiferromagnet has alternating microscopic spins. The low-energy continuum variables are a unit staggered field $\mathbf n(x)$ and a small smooth magnetization density $\boldsymbol\ell(x)$ orthogonal to $\mathbf n$.

Expanding the exchange energy in slowly varying fields gives the continuum Hamiltonian density

$$\mathcal H =2Ja\,\boldsymbol\ell^2+{JS^2a\over2}(\partial_x\mathbf n)^2+\text{irrelevant terms}.$$

The first term says that uniform canting costs energy. The second term says that spatial gradients of the Néel field cost stiffness. The antiferromagnet is special because the time derivative of $\mathbf n$ does not come from the microscopic Hamiltonian alone. It comes from the Berry phase of spin coherent states.

Spin coherent states and the Berry phase

A spin-$S$ coherent state $|\mathbf N\rangle$ is labeled by a unit vector $\mathbf N\in S^2$ and satisfies

$$\langle \mathbf N|\mathbf S|\mathbf N\rangle=S\mathbf N.$$

In the coherent-state path integral, a spin trajectory $\mathbf N(\tau)$ contributes the phase

$$\exp\left[iS\Omega[\mathbf N]\right],$$

where $\Omega[\mathbf N]$ is the oriented solid angle swept out by the path on the unit sphere. Because the path-integral weight is $e^{-S_E}$, the corresponding term in the Euclidean action is

$$S_{E,B}[\mathbf N] =-iS\Omega[\mathbf N] =-iS\int d\tau\,\mathbf A(\mathbf N)\cdot\partial_\tau\mathbf N,$$

where $\mathbf A$ is a monopole vector potential on $S^2$ satisfying

$$\nabla_{\mathbf N}\times\mathbf A=\mathbf N.$$

This formula is local only in patches on the sphere. The solid angle is defined modulo $4\pi$, so the phase is well-defined precisely because $2S$ is an integer:

$$e^{iS(\Omega+4\pi)}=e^{iS\Omega}e^{i4\pi S}=e^{iS\Omega}.$$

A spin coherent-state path $\mathbf N(\tau)$ on $S^2$ contributes the factor $e^{iS\Omega[\mathbf N]}$. Equivalently, its Euclidean action contains $-iS\Omega[\mathbf N]$. The ambiguity $\Omega\sim\Omega+4\pi$ is harmless because $2S\in\mathbb Z$.

For a whole spin chain, the coherent-state path integral contains

$$\exp\left[iS\sum_j\Omega[\mathbf N_j]\right],$$

where $\mathbf N_j(\tau)$ is the direction of the spin at site $j$. In an antiferromagnet,

$$\mathbf N_j(\tau)\approx(-1)^j\mathbf n(x_j,\tau)+{a\over S}\boldsymbol\ell(x_j,\tau).$$

The smooth part of the Berry factor gives the following term in the Euclidean action:

$$S_{E,B}^{\rm smooth}=-i\int d\tau\,dx\, \boldsymbol\ell\cdot(\mathbf n\times\partial_\tau\mathbf n).$$

Combining this with the continuum Hamiltonian density gives, to leading order,

$$S_E[\mathbf n,\boldsymbol\ell] =\int d\tau\,dx\, \left[ 2Ja\,\boldsymbol\ell^2 +{JS^2a\over2}(\partial_x\mathbf n)^2 -i\boldsymbol\ell\cdot(\mathbf n\times\partial_\tau\mathbf n) \right] -i\theta Q[\mathbf n].$$

The field $\boldsymbol\ell$ is Gaussian. Integrating it out gives

$$S_{\rm kin} =\int d\tau\,dx\, \left[{1\over8Ja}(\partial_\tau\mathbf n)^2 +{JS^2a\over2}(\partial_x\mathbf n)^2 \right].$$

Although the stationary value of $\boldsymbol\ell$ is imaginary in this Euclidean representation, the Gaussian integral is well defined and produces the positive time-derivative term shown above. This is the same harmless complex saddle that appears whenever a real canonical momentum is integrated out of a Euclidean phase-space path integral.

This can be written in relativistic-looking form

$$S_{\rm kin} ={1\over2g}\int d\tau\,dx\, \left[{1\over c}(\partial_\tau\mathbf n)^2+c(\partial_x\mathbf n)^2\right],$$

with

$$\boxed{ g={2\over S}, \qquad c=2JSa. }$$

After rescaling Euclidean time by the spin-wave velocity $c$, this becomes the ordinary two-dimensional $O(3)$ nonlinear sigma model,

$$S_{\rm kin}={1\over2g}\int d^2x\,(\partial_\mu\mathbf n)^2.$$

The remaining, alternating part of the Berry phase is the topological term. It is here that the microscopic spin length survives in a way no local gradient expansion could guess.

Ferromagnets and antiferromagnets use Berry phases differently

For a ferromagnet, neighboring spins point in nearly the same direction, so their Berry phases add. The continuum Euclidean action begins schematically as

$$S_E^{\rm ferro} =-i{S\over a}\int d\tau\,dx\, \mathbf A(\mathbf n)\cdot\partial_\tau\mathbf n +{\rho_s\over2}\int d\tau\,dx\,(\partial_x\mathbf n)^2.$$

The time derivative is first order. After continuation to real time, it pairs the two transverse spin-wave coordinates as conjugate variables and gives the quadratic magnon dispersion $\omega\propto k^2$.

For an antiferromagnet, the leading Berry phases alternate and nearly cancel. Their smooth remainder couples $\boldsymbol\ell$ to $\mathbf n\times\partial_\tau\mathbf n$; integrating out the costly canting field $\boldsymbol\ell$ produces $(\partial_\tau\mathbf n)^2$. The semiclassical dispersion is therefore linear, $\omega\simeq c|k|$. The cancellation is not complete: its quantized remainder is the theta term derived next.

How the theta term appears

The Berry phase is sensitive to the fact that neighboring antiferromagnetic spins live near opposite points on the sphere. Pair neighboring sites. The Berry phases of the two spins nearly cancel, but their small mismatch is a total derivative in field space. Summed over the chain, these mismatches become the winding number of the map $\mathbf n(\tau,x)$.

A useful identity is the variation of the solid angle:

$$\delta\Omega[\mathbf n] =\int d\tau\, \delta\mathbf n\cdot(\mathbf n\times\partial_\tau\mathbf n).$$

Taking $\delta\mathbf n=a\partial_x\mathbf n$ gives

$$\Omega[\mathbf n(x+a)]-\Omega[\mathbf n(x)] =a\int d\tau\, \partial_x\mathbf n\cdot(\mathbf n\times\partial_\tau\mathbf n)+O(a^2).$$

Using cyclic symmetry of the scalar triple product,

$$\partial_x\mathbf n\cdot(\mathbf n\times\partial_\tau\mathbf n) = \mathbf n\cdot(\partial_\tau\mathbf n\times\partial_x\mathbf n).$$

After summing pairs, the alternating part of the exponent in the path-integral weight gives

$$iS\sum_j(-1)^j\Omega[\mathbf n(x_j)] = i(2\pi S)Q[\mathbf n] \quad \text{mod }2\pi i.$$

Equivalently, the Euclidean action contains $-i\theta Q$. Thus

$$\boxed{\theta=2\pi S.}$$

The coefficient is quantized because the microscopic spin representation is quantized. This is the conceptual punchline: the continuum theta angle remembers whether the microscopic spin is integer or half-integer.

Topological charge as degree

Before using the theta term, it is worth seeing why $Q$ is an integer. Start with the simpler $O(2)$ case. A path on a circle may be written as

$$n(t)=e^{i\varphi(t)}, \qquad 0\le t\le 2\pi,$$

with periodicity of the physical point

$$n(2\pi)=n(0).$$

The angle itself may wind:

$$\varphi(2\pi)=\varphi(0)+2\pi k, \qquad k\in\mathbb Z.$$

The winding number is

$$q={1\over2\pi}\int_0^{2\pi}dt\,{d\varphi\over dt}=k.$$

The $O(3)$ topological charge is the two-dimensional version of this statement. Finite-action configurations on the Euclidean plane approach a constant at infinity, so the domain may be compactified to a sphere:

$$\mathbb R^2\cup\{\infty\}\simeq S^2.$$

Thus a field configuration is a map

$$\mathbf n:S^2_{\rm spacetime}\to S^2_{\rm target}.$$

Such maps have an integer degree. In angular coordinates on the target sphere,

$$\mathbf n=(\sin\Theta\cos\Phi,\sin\Theta\sin\Phi,\cos\Theta),$$

one finds

$$\mathbf n\cdot(\partial_\tau\mathbf n\times\partial_x\mathbf n) =\sin\Theta\, {\partial(\Theta,\Phi)\over\partial(\tau,x)}.$$

Therefore

$$Q={1\over4\pi}\int d\tau\,dx\, \sin\Theta\,{\partial(\Theta,\Phi)\over\partial(\tau,x)}.$$

This is the signed target-sphere area swept out by the map, divided by the area $4\pi$ of the unit sphere. A configuration that covers the target sphere once has $Q=1$; one that covers it $k$ times has $Q=k$.

For finite-action configurations, Euclidean spacetime may be compactified to $S^2$. The sigma-model field is then a map $S^2\to S^2$, and $Q$ counts the signed number of times the domain wraps the target sphere.

On a closed spacetime, the theta term does not change the local classical equations of motion for smooth variations within a fixed topological sector. It multiplies sectors by phases:

$$Z(\theta)=\sum_{Q\in\mathbb Z}e^{i\theta Q}Z_Q,$$

where $Z_Q$ is the contribution from configurations of topological charge $Q$. Since $Q$ is integer on a closed oriented spacetime,

$$Z(\theta+2\pi)=Z(\theta).$$

Boundaries and edge Berry phases

The statements “$Q$ is an integer,” “$\theta$ is $2\pi$-periodic,” and “the theta term does not affect the local variation” all require a qualification when spacetime has a boundary. On an open strip, the field values at the two spatial endpoints trace curves on $S^2$. After choosing caps for those curves, the strip integral obeys, up to the orientation convention,

$$Q_{\rm strip} =k+{\Omega_R-\Omega_L\over4\pi}, \qquad k\in\mathbb Z.$$

The integer $k$ changes when a different cap is chosen, while the full microscopic Berry factor remains unambiguous. Substituting $\theta=2\pi S$ shows that the boundary-dependent part of the Euclidean action is

$$S_{\rm edge} =-i{S\over2}\left(\Omega_R-\Omega_L\right).$$

Semiclassically, the two endpoints therefore carry Berry phases of spins $S/2$ with opposite boundary orientations. For an integer-spin chain, the closed-bulk phase $e^{i2\pi S k}$ is trivial, but the edge term need not be. In particular, an open spin-1 Haldane chain can carry spin-$1/2$ endpoint degrees of freedom.

This resolves an apparent contradiction. The bulk values $\theta=0$ and $\theta=2\pi$ give the same partition function on a closed spacetime, yet they can encode different boundary physics when the protecting symmetries are retained. Theta periodicity should therefore never be used to erase an open chain’s boundary Berry phases.

Integer and half-integer spin

Combining

$$\theta=2\pi S$$

with theta periodicity gives two basic universality classes:

$$\begin{array}{ccl} S\in\mathbb Z &\Rightarrow& \theta=0\pmod{2\pi},\\ S\in\mathbb Z+{1\over2} &\Rightarrow& \theta=\pi\pmod{2\pi}. \end{array}$$

At $\theta=0$, the closed-bulk $O(3)$ model behaves like the ordinary asymptotically free sigma model. The coupling grows in the infrared and the theory produces a mass gap,

$$M\sim \Lambda e^{-2\pi/g},$$

up to prefactors and scheme-dependent definitions of the ultraviolet scale. For the uniform nearest-neighbor chain and the phase continuously connected to it, this is the field-theory explanation of the Haldane gap for integer spin. The statement concerns the bulk gap; it does not remove the edge degrees of freedom of an open Haldane chain.

At $\theta=\pi$, the topological sectors enter with the sign

$$e^{i\pi Q}=(-1)^Q.$$

Even and odd topological sectors interfere destructively. This changes the infrared theory. For the nearest-neighbor spin-$1/2$ Heisenberg antiferromagnet, the infrared fixed point is the $SU(2)_1$ Wess–Zumino–Witten conformal field theory, with logarithmic corrections from a marginally irrelevant operator. More broadly, the Lieb–Schultz–Mattis constraint says that a half-odd-integer spin per unit cell, together with translation and spin-rotation symmetry, cannot have a completely trivial, unique, symmetry-preserving gapped ground state. The alternatives include a gapless phase or ground-state degeneracy from symmetry breaking.

On closed spacetime the theta angle is periodic with period $2\pi$. Integer spin gives $\theta=0$ modulo $2\pi$ and a massive bulk sigma-model phase. Half-integer spin gives $\theta=\pi$ modulo $2\pi$; a symmetry-preserving infrared theory must be gapless or otherwise avoid a trivial unique ground state.

This is a striking lesson. The local Lagrangian density

$${1\over2g}(\partial\mathbf n)^2$$

knows nothing about whether $S=1$ or $S=1/2$. The distinction is entirely in a topological phase invisible in ordinary perturbation theory around a smooth configuration.

Symmetry constraints on the theta angle

The antiferromagnetic chain has microscopic symmetries that act nontrivially on the continuum field. A one-site translation reverses the Néel field:

$$T_a:\quad \mathbf n\mapsto -\mathbf n.$$

The topological density changes sign under this map because

$$(-\mathbf n)\cdot[\partial_\tau(-\mathbf n)\times\partial_x(-\mathbf n)] =-\mathbf n\cdot(\partial_\tau\mathbf n\times\partial_x\mathbf n),$$

so

$$Q\mapsto -Q.$$

Because translation is unitary, the path-integral phase can be invariant under this transformation only when

$$e^{i\theta Q}=e^{-i\theta Q} \qquad \text{for all }Q\in\mathbb Z,$$

which requires

$$\theta=0\quad\text{or}\quad \theta=\pi\pmod{2\pi}.$$

These two symmetry-invariant values are precisely the two values realized by uniform integer and half-integer antiferromagnetic spin chains.

Physical time reversal requires a slightly different bookkeeping. It acts schematically as

$$\mathcal T:\quad \mathbf n(\tau,x)\mapsto-\mathbf n(-\tau,x)$$

and is antiunitary, so it also complex-conjugates the Berry phase. Tracking the field transformation together with complex conjugation again sends the theta weight to its $\theta\mapsto-\theta$ partner. Thus $\theta=0$ and $\theta=\pi$ are the time-reversal-invariant values modulo $2\pi$; assigning a sign to $Q$ without also tracking antiunitarity is incomplete.

This also explains why perturbations matter. If the microscopic chain is dimerized, translation by one site is no longer a symmetry. In the continuum theory this allows the effective theta angle to move away from $0$ or $\pi$. Then the special interference at $\theta=\pi$ can be destroyed and a gap may open without violating the Lieb–Schultz–Mattis constraint, because the doubled unit cell contains an integer total spin.

A local coordinate check

Locally, the $O(3)$ field has only two independent components. Choose a patch near the north pole and write

$$\mathbf n=(\sqrt{1-\boldsymbol\pi^2},\pi_1,\pi_2), \qquad \boldsymbol\pi^2=\pi_1^2+\pi_2^2.$$

Then

$$(\partial_\mu\mathbf n)^2 =(\partial_\mu\boldsymbol\pi)^2 +{(\boldsymbol\pi\cdot\partial_\mu\boldsymbol\pi)^2\over1-\boldsymbol\pi^2}.$$

So the two local fields $\pi_1,\pi_2$ are the spin-wave coordinates. At weak coupling, after rescaling $\boldsymbol\pi=\sqrt g\,\boldsymbol\varphi$, the leading interaction is order $g$. This is the same perturbative sigma-model expansion used on the previous pages.

The topological term is different. In a single coordinate patch, it looks like a total derivative or a curl. On a closed spacetime it does not change the perturbative beta function of $g$ at any finite order. With a boundary, the same total derivative is precisely why a boundary term remains. Globally, no single smooth coordinate patch covers all configurations of nonzero $Q$. The theta term can therefore be invisible in bulk perturbation theory and still decide the infrared and edge physics.

Relation to the next step

The continuum antiferromagnet raises a question that sounds paradoxical at first. The microscopic classical picture has a Néel vector, but the one-dimensional quantum chain has no ordinary long-range staggered magnetization in its symmetric ground state. In field-theory language, low-dimensional fluctuations restore the continuous symmetry.

The next page studies this symmetry restoration more directly. It returns to the sigma model as a quantum field theory and explains why the order parameter vanishes even when the classical field wants to choose a point on the sphere.

Summary

A one-dimensional antiferromagnetic Heisenberg spin chain has low-energy variables

$$\mathbf S_j\approx S(-1)^j\mathbf n(x_j)+a\boldsymbol\ell(x_j), \qquad \mathbf n^2=1, \qquad \mathbf n\cdot\boldsymbol\ell=0.$$

The exchange interaction gives spatial stiffness and a cost for uniform canting. The spin coherent-state Berry phase gives the time derivative term and, more importantly, the theta term. After integrating out $\boldsymbol\ell$, the low-energy Euclidean action is

$$S_E[\mathbf n] ={1\over2g}\int d\tau\,dx\, \left[{1\over v}(\partial_\tau\mathbf n)^2+v(\partial_x\mathbf n)^2\right] -i\theta Q[\mathbf n],$$

with

$$g={2\over S}, \qquad v=2JSa, \qquad \theta=2\pi S.$$

The topological charge is

$$Q={1\over4\pi}\int d\tau\,dx\, \mathbf n\cdot(\partial_\tau\mathbf n\times\partial_x\mathbf n) \in\mathbb Z \qquad\text{on closed spacetime}.$$

Thus integer spin gives $\theta=0$ modulo $2\pi$ and the ordinary massive bulk sigma-model behavior, while half-integer spin gives $\theta=\pi$ modulo $2\pi$. The nearest-neighbor spin-$1/2$ chain is critical; more generally, the half-odd-integer chain cannot be a trivial unique symmetric gapped state. On an open chain, the bulk periodicity must be supplemented by the endpoint Berry phases, which retain information that a closed-spacetime reduction $\theta\sim\theta+2\pi$ would discard.

Common pitfalls

Forgetting the Berry phase. The ordinary gradient energy only produces the sigma-model kinetic term. The integer versus half-integer distinction comes from the spin coherent-state Berry phase.

Mixing the Berry factor with the Euclidean action. A coherent spin contributes $e^{iS\Omega}$ to the path-integral weight, so the Euclidean action contains $-iS\Omega$. Switching that sign midway also flips the smooth $\boldsymbol\ell$ coupling and the theta term.

Treating the theta term as a small local perturbation. It weights entire topological sectors by phases $e^{i\theta Q}$. Its leading density can be a total derivative in one patch while its global effect remains nonperturbative.

Losing the factor of $2\pi$. The spin-chain result is $\theta=2\pi S$, not $\theta=S$. Since $\theta$ is periodic modulo $2\pi$ on closed spacetime, this factor is exactly what separates integer from half-integer bulk theories.

Using closed-spacetime periodicity on an open chain. When there is a boundary, $Q$ need not be an integer by itself and the theta term leaves endpoint Berry phases. In particular, reducing $\theta=2\pi$ to zero before retaining the boundary term erases the spin-$1/2$ edges of the spin-1 Haldane chain.

Overstating the gapless claim. The uniform nearest-neighbor spin-$1/2$ Heisenberg antiferromagnet is critical, but the general half-odd-integer constraint allows either gaplessness or degeneracy. Explicit dimerization can open a gap because it doubles the unit cell and breaks one-site translation.

Exercises

Exercise 1: integrating out the uniform magnetization

Starting from

$$S_E[\mathbf n,\boldsymbol\ell] =\int d\tau\,dx\, \left[ 2Ja\,\boldsymbol\ell^2 +{JS^2a\over2}(\partial_x\mathbf n)^2 -i\boldsymbol\ell\cdot(\mathbf n\times\partial_\tau\mathbf n) \right],$$

with $\mathbf n^2=1$ and $\mathbf n\cdot\boldsymbol\ell=0$, integrate out $\boldsymbol\ell$ classically and show that the time-derivative term is

$${1\over8Ja}(\partial_\tau\mathbf n)^2.$$

Solution

The $\boldsymbol\ell$-dependent part is

$$2Ja\,\boldsymbol\ell^2 -i\boldsymbol\ell\cdot\mathbf A, \qquad \mathbf A=\mathbf n\times\partial_\tau\mathbf n.$$

The stationary point satisfies

$$4Ja\,\boldsymbol\ell-i\mathbf A=0,$$

so

$$\boldsymbol\ell_\star={i\over4Ja}\mathbf A.$$

Completing the square,

$$2Ja\left(\boldsymbol\ell-{i\mathbf A\over4Ja}\right)^2 +{1\over8Ja}\mathbf A^2.$$

Since $\mathbf n^2=1$, we have

$$\mathbf n\cdot\partial_\tau\mathbf n=0,$$

and therefore

$$\mathbf A^2=(\mathbf n\times\partial_\tau\mathbf n)^2 =(\partial_\tau\mathbf n)^2.$$

Thus integrating out $\boldsymbol\ell$ gives

$${1\over8Ja}(\partial_\tau\mathbf n)^2.$$

Exercise 2: topological charge of the identity map

Let the domain sphere have coordinates $(\vartheta,\varphi)$ with $0\le\vartheta\le\pi$ and $0\le\varphi<2\pi$. Consider the identity map to the target sphere,

$$\mathbf n(\vartheta,\varphi)=(\sin\vartheta\cos\varphi,\sin\vartheta\sin\varphi,\cos\vartheta).$$

Show that $Q=1$.

Solution

For this parameterization,

$$\mathbf n\cdot(\partial_\vartheta\mathbf n\times\partial_\varphi\mathbf n)=\sin\vartheta.$$

Therefore

$$Q={1\over4\pi}\int_0^\pi d\vartheta\int_0^{2\pi}d\varphi\,\sin\vartheta.$$

The integrals give

$$\int_0^\pi d\vartheta\,\sin\vartheta=2, \qquad \int_0^{2\pi}d\varphi=2\pi.$$

Thus

$$Q={1\over4\pi}(2)(2\pi)=1.$$

The identity map covers the target sphere exactly once with positive orientation.

Exercise 3: the boundary term at θ = 2πS

For a field on an open strip, suppose the capped topological charge is

$$Q_{\rm strip}=k+{\Omega_R-\Omega_L\over4\pi}, \qquad k\in\mathbb Z.$$

Insert $\theta=2\pi S$ into $S_\theta=-i\theta Q_{\rm strip}$. Show that the boundary-dependent part is the difference of Berry actions for spins of magnitude $S/2$. What does this predict for an open spin-1 chain?

Solution

Substitution gives

$$S_\theta =-i2\pi S k -i{S\over2}(\Omega_R-\Omega_L).$$

The second term is

$$-i{S\over2}\Omega_R+i{S\over2}\Omega_L.$$

The opposite signs reflect the opposite orientations of the two ends. Each has the coherent-state Berry phase of an effective spin $S/2$. For $S=1$, the bulk integer-sector factor $e^{i2\pi k}$ is trivial, but the two endpoints carry spin-$1/2$ Berry phases. This is the continuum signature of the edge degrees of freedom of the open Haldane chain.

Exercise 4: translation and the topological charge

A one-site translation acts on the Néel field as $\mathbf n\mapsto-\mathbf n$. Show that this sends $Q\mapsto -Q$. For which theta angles is the phase $e^{i\theta Q}$ invariant under this transformation for all $Q\in\mathbb Z$?

Solution

Under $\mathbf n\mapsto-\mathbf n$,

$$\partial_\mu\mathbf n\mapsto-\partial_\mu\mathbf n.$$

The cross product of two derivatives is unchanged:

$$\partial_\tau(-\mathbf n)\times\partial_x(-\mathbf n) =\partial_\tau\mathbf n\times\partial_x\mathbf n.$$

But the remaining factor of $\mathbf n$ changes sign, so

$$(-\mathbf n)\cdot[\partial_\tau(-\mathbf n)\times\partial_x(-\mathbf n)] =-\mathbf n\cdot(\partial_\tau\mathbf n\times\partial_x\mathbf n).$$

Therefore

$$Q\mapsto -Q.$$

The phase is invariant if

$$e^{i\theta Q}=e^{-i\theta Q}$$

for all integers $Q$. This requires

$$e^{i2\theta Q}=1$$

for all $Q$, hence

$$2\theta=2\pi k, \qquad k\in\mathbb Z.$$

Modulo $2\pi$, the solutions are

$$\theta=0, \qquad \theta=\pi.$$

Exercise 5: why the theta term is invisible in small-field perturbation theory

In a local patch write

$$\mathbf n=(\sqrt{1-\pi_1^2-\pi_2^2},\pi_1,\pi_2).$$

Show that, to leading order in $\pi$, the topological density is a total derivative:

$$\mathbf n\cdot(\partial_\tau\mathbf n\times\partial_x\mathbf n) =\partial_\tau\pi_1\partial_x\pi_2- \partial_\tau\pi_2\partial_x\pi_1+O(\pi^3).$$

Then rewrite the leading term as a total derivative and explain what changes when spacetime has a boundary.

Solution

To leading order,

$$\mathbf n=(1,\pi_1,\pi_2)+O(\pi^2),$$

so

$$\partial_\tau\mathbf n=(0,\partial_\tau\pi_1,\partial_\tau\pi_2)+O(\pi),$$

and

$$\partial_x\mathbf n=(0,\partial_x\pi_1,\partial_x\pi_2)+O(\pi).$$

The leading cross product points in the first internal direction:

$$\partial_\tau\mathbf n\times\partial_x\mathbf n =\left(\partial_\tau\pi_1\partial_x\pi_2- \partial_\tau\pi_2\partial_x\pi_1,0,0\right)+O(\pi^3).$$

Dotting with $\mathbf n=(1,\pi_1,\pi_2)+\cdots$ gives

$$\mathbf n\cdot(\partial_\tau\mathbf n\times\partial_x\mathbf n) =\partial_\tau\pi_1\partial_x\pi_2- \partial_\tau\pi_2\partial_x\pi_1+O(\pi^3).$$

The leading term is

$$\partial_\tau\pi_1\partial_x\pi_2- \partial_\tau\pi_2\partial_x\pi_1 =\partial_\tau(\pi_1\partial_x\pi_2)-\partial_x(\pi_1\partial_\tau\pi_2),$$

because mixed derivatives commute. On a closed spacetime, or for fluctuations that vanish at infinity, the integral of this term vanishes. With a boundary it instead leaves a boundary contribution, consistent with the endpoint Berry phases derived above. This is why the theta term does not modify ordinary bulk perturbation theory around the trivial sector while remaining physically important.

References

• I. Affleck, T. Kennedy, E. H. Lieb, and H. Tasaki, “Rigorous Results on Valence-Bond Ground States in Antiferromagnets,” Physical Review Letters 59 (1987) 799–802, doi:10.1103/PhysRevLett.59.799.
• F. D. M. Haldane, “Continuum Dynamics of the 1-D Heisenberg Antiferromagnet: Identification with the O(3) Nonlinear Sigma Model,” Physics Letters A 93 (1983) 464–468, doi:10.1016/0375-9601(83)90631-X.
• F. D. M. Haldane, “Nonlinear Field Theory of Large-Spin Heisenberg Antiferromagnets: Semiclassically Quantized Solitons of the One-Dimensional Easy-Axis Néel State,” Physical Review Letters 50 (1983) 1153–1156, doi:10.1103/PhysRevLett.50.1153.
• E. H. Lieb, T. Schultz, and D. Mattis, “Two Soluble Models of an Antiferromagnetic Chain,” Annals of Physics 16 (1961) 407–466, doi:10.1016/0003-4916(61)90115-4.

Further reading

• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, Cambridge, 2010.
• E. Fradkin, Field Theories of Condensed Matter Physics, 2nd ed., Cambridge University Press, Cambridge, 2013.
• A. M. Polyakov, Gauge Fields and Strings, Contemporary Concepts in Physics, Vol. 3, Harwood Academic Publishers, Chur, 1987.
• A. Zee, Quantum Field Theory in a Nutshell, 2nd ed., Princeton University Press, Princeton, 2010.