Quantum spin waves · magnons

Magnon Lab

The quantum counterpart of a classical spin wave is the magnon: a single spin quantum spread over the crystal. The simulations on this page build up the magnon from the simplest case. First a single magnon in a ferromagnet, then the interaction and bound states of two magnons, and finally the antiferromagnet, where quantum fluctuations melt the Néel state and magnons give way to spinons. Each simulation shows the exact numerical result side by side with an approximate theory. Choose a simulation with the selector below.

\[ H = -J\sum_{n}\mathbf S_n\cdot\mathbf S_{n+1} \;-\; h\sum_n S^z_n ,\qquad S=\tfrac12 \]

\(J>0\) ferromagnetic exchange · \(h\): field along \(\hat z\) · ring of \(N\) sites (periodic boundary) · \(\hbar=a=1\), time in units of \(\hbar/J\). The simulation solves the Schrödinger equation exactly in the one-magnon sector (an \(N\times N\) problem); no approximation is involved.

SimulationMagnon Lab

1 · Ground state. All spins ↑. This state is an exact eigenstate of the Hamiltonian: nothing moves and the magnon number is zero. Click a site to flip that spin.

periodic ring · N = 30

colour = phase of the amplitude arg c_n · saturation = |c_n|

Figure 1 · Magnon energy ω_k and magnon content of the state

Figure 1. The curve is \(\omega_k=h+J(1-\cos k)\). Grey dots are the allowed values \(k=2\pi m/N\) on the ring. The size of the coloured dots is the weight of the current state in the \(|k\rangle\) component, \(|\tilde c_k|^2\). For a single spin flip all \(k\) have equal weight; a magnon is a single dot; a packet is a narrow region around \(k_0\). Clicking a dot selects that magnon.

Figure 2

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Hamiltonian

Exchange J

Field h

h shifts all magnon energies by the same amount; it changes only the phases, not the probabilities.

State

Number of sites N

Wave number k = 2πm/N

Packet width σ

Superposition

One-magnon weight p

\(|\psi\rangle=\sqrt{1-p}\,|F\rangle+\sqrt p\,|k\rangle\)

Drawing gain

The transverse component of the arrows is magnified only in the drawing.

Cone angle θ

Time

Speed

1×: each real-time second is 2 ħ/J.

Method: the amplitudes are propagated exactly as \(\tilde c_k(t)=\tilde c_k(0)\,e^{-i\omega_k t}\) and transformed back to sites with \(c_n=N^{-1/2}\sum_k \tilde c_k e^{ikn}\). The numerical error is at floating-point level only.

1 · Two distant magnons. Two spins have been flipped at opposite points of the ring. Each spreads like a single magnon. The bars are the number of flipped spins on each site, \(\langle n_i\rangle\) (total 2). The circles are the prediction for two non-interacting magnons. The map on the right is the joint position probability of the two flipped spins, \(P(n_1,n_2)\). Until the magnons meet the result equals the non-interacting prediction; a difference starts only when they meet. Click the chain or the map to change the starting positions.

periodic ring

diagonal: same site (forbidden)

Figure 1 · Relative-distance distribution P(r)

Figure 2

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Hamiltonian

Exchange J

Anisotropy Δ

Δ = 1 Heisenberg, Δ = 0 XY (hard core only), Δ ≫ 1 Ising.

Field h

h shifts the two-magnon energies by 2h; it does not change the probabilities.

State

Number of sites N

Total momentum K = 2πM/N

State (index within the K block)

0: lowest energy. Clicking a dot in the spectrum also selects it.

Packet width σ

Time

Speed

1×: each real-time second is 2 ħ/J.

Method: for each \(K\) the real symmetric matrix of dimension \(\sim N/2\) on the relative coordinate \(r=n_2-n_1\) is diagonalized with the Jacobi method; the time evolution \(e^{-iE t}\) is exact. Total dimension \(N(N-1)/2\).

1 · The Néel state is not an eigenstate. The simulation starts from the classical ground state \(|\!\uparrow\downarrow\uparrow\downarrow\cdots\rangle\) and evolves the chain exactly in time. The bars are \(\langle S^z_i\rangle\); the dashed outlines are the Néel values \(\pm S\). Unlike \(|\!\uparrow\uparrow\cdots\rangle\) in the ferromagnet, the exchange term \(S^+_iS^-_{i+1}\) flips neighbouring ↑↓ pairs, so the order melts by itself. Figure 2 shows the bond energy flowing from its \(zz\) part into its \(xy\) part; their sum is constant.

S = 1/2 · N = 12

Figure 1

Figure 2

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Chain (exact diagonalization)

Spin S

Number of sites N

Linear spin-wave theory

Lattice

Spin S

Lattice size (k sum)

\(\Delta S\) is computed as a \(\mathbf k\) sum on a finite lattice; the \(\gamma_{\mathbf k}=\pm1\) modes (uniform rotation) are excluded.

Spectrum

Selected q = 2πn/N

Broadening η

The spectrum of a finite chain is discrete; η is only a Lorentzian broadening for display.

Time

Speed

1×: each real-time second is 2 ħ/J.

Reading guide

Bars, colours and arrows

Bar height

In steps 1–4 the bar on each site is \(P_n=|c_n|^2=\tfrac12-\langle S^z_n\rangle\): the probability of finding the flipped spin on that site. The right axis reads the same quantity as \(\langle S^z_n\rangle\). The vertical scale adjusts itself; values are on the left axis.

Colour

The colour is the phase of the complex amplitude, \(\arg c_n\); the colour wheel is the same as on the Spin Wave Lab page (0° red, 90° yellow-green, 180° cyan, 270° violet-blue). Because the energy is measured from the ground state, the phase rotates at rate \(\omega_k\). In step 5 the same colour is \(\arg\langle S^+_n\rangle\), the exact counterpart of the precession phase on the Spin Wave Lab page.

Meaning of the phase

The global phase of a single state cannot be measured; only phase differences between sites (\(k\)) and the relative phase in a superposition of two states are meaningful. The colour in steps 1–4 is the phase that appears when the state is superposed with \(|F\rangle\). This superposition is actually made in step 5(a).

Arrows (step 5 only)

The arrow is the expectation value \(\langle\mathbf S_n\rangle\). A quantum spin has no classical direction; the arrow is only an average. In 5(a) the arrow is shorter than \(1/2\). This shows that the reduced state of the site is mixed, i.e. the site is entangled with the rest of the chain.

Derivation

From the ground state to the magnon

Step 1

Ferromagnetic ground state

\(\mathbf S_n\cdot\mathbf S_{n+1}=S^z_nS^z_{n+1}+\tfrac12(S^+_nS^-_{n+1}+S^-_nS^+_{n+1})\). On \(|F\rangle=|\!\uparrow\uparrow\cdots\uparrow\rangle\) the \(S^+\) terms give zero and \(S^zS^z\) gives \(\tfrac14\):

\[ H|F\rangle=E_0|F\rangle,\qquad E_0=-\tfrac{NJ}{4}-\tfrac{Nh}{2}. \]

\(|F\rangle\) is an exact eigenstate and contains no quantum fluctuations. In the antiferromagnet the Néel state is not: the term \(S^+_nS^-_{n+1}\) turns a ↓↑ pair into ↑↓, so the AFM ground state differs from the Néel state.

Step 2

A flipped spin is a particle

In the state \(|n\rangle=S^-_n|F\rangle\) two bonds drop from \(\tfrac14\) to \(-\tfrac14\) (\(+J\)), the field term adds \(+h\), and the flip term of the exchange moves the flipped spin to a neighbour:

\[ \begin{aligned} H|n\rangle&=(E_0+J+h)\,|n\rangle\\ &\quad-\tfrac J2\big(|n+1\rangle+|n-1\rangle\big). \end{aligned} \]

This is the equation of a particle in a tight-binding model. Since \(S^z_{\rm tot}\) is conserved, the number of flipped spins stays exactly 1. The infinite-chain solution starting from site \(j\) is

\[ c_n(t)=i^{\,n-j}J_{n-j}(Jt)\,e^{-i(J+h)t}, \] \[ \langle (n-j)^2\rangle=\tfrac{(Jt)^2}{2}. \]

The spreading is ballistic (\(\propto t\)), not diffusive (\(\propto\sqrt t\)).

Step 3

Magnon: a momentum eigenstate

Because of translation symmetry the eigenstates are plane waves:

\[ |k\rangle=\frac{1}{\sqrt N}\sum_n e^{ikn}|n\rangle, \] \[ \omega_k=E_k-E_0=h+J(1-\cos k), \]

with \(k=2\pi m/N\). This is \(2JS(1-\cos k)\) for \(S=\tfrac12\), identical to the classical linear spin-wave frequency: for a single magnon in the FM the quantum and classical dispersions coincide. At small \(k\), \(\omega_k\simeq h+\tfrac J2k^2\): the magnon is a particle with effective mass \(1/J\).

\(k=0\) is special: \(|0\rangle\propto S^-_{\rm tot}|F\rangle\) is a rotated ground state, with zero energy at \(h=0\) (Goldstone mode). Magnons with \(k\neq0\) have total spin \(N/2-1\). A single spin flip is an equal superposition of all magnons: \(|j\rangle=N^{-1/2}\sum_k e^{-ikj}|k\rangle\).

Step 4

Wave packet and group velocity

In the one-magnon sector the position operator \(X\) evolves in the Heisenberg picture exactly as \(X(t)=X(0)+V\,t\), with velocity operator \(V=\partial\omega_k/\partial k=J\sin k\). For a real, symmetric envelope

\[ \langle x\rangle(t)=x_0+\langle v\rangle t,\qquad \sigma^2(t)=\sigma_0^2+\mathrm{Var}(v)\,t^2, \]

where the averages are taken over the distribution \(|\tilde c_k|^2\). For a narrow packet \(\langle v\rangle\simeq J\sin k_0\) and \(\mathrm{Var}(v)\simeq J^2\cos^2k_0\,\sigma_k^2\). The largest velocity is at \(k_0=\pi/2\); there the curvature of the dispersion vanishes, so the packet spreads most slowly.

Step 5

Where does the classical spin wave come from?

\(S^+_n\) lowers the magnon number by one. Therefore \(\langle S^\pm_n\rangle=0\) in any state with a definite magnon number. (a) For \(|\psi\rangle=\sqrt{1-p}\,|F\rangle+\sqrt p\,|k\rangle\)

\[ \langle S^+_n\rangle=\sqrt{\tfrac{p(1-p)}{N}}\;e^{i(kn-\omega_k t)},\qquad \langle S^z_n\rangle=\tfrac12-\tfrac pN . \]

This is a spin wave rotating at the classical precession frequency, but its amplitude is small, \(\propto1/\sqrt N\), and \(|\langle\mathbf S_n\rangle|<\tfrac12\). (b) The \(k=0\) coherent state \(\bigotimes_n\big(\cos\tfrac\theta2|\!\uparrow\rangle+e^{i\phi}\sin\tfrac\theta2|\!\downarrow\rangle\big)\) is a product state. Its magnon number is binomially distributed:

\[ \langle n\rangle=NS(1-\cos\theta),\qquad \frac{\Delta n}{\langle n\rangle}=\sqrt{\frac{1-s}{Ns}},\quad s=\sin^2\tfrac\theta2 . \]

As \(N\to\infty\) the relative fluctuation goes to zero: a classical wave is many magnons with a sharp phase and an uncertain number. For \(S=\tfrac12\) and \(k\neq0\) a many-magnon coherent state is not exact, because magnons interact. This is the subject of the Two magnons simulation.

Bridge

Comparison with the Spin Wave Lab page

Quantity	Classical (LLG, linear)	Quantum (single magnon)
Dispersion	\(\omega_k=h+2JS(1-\cos k)\)	The same (\(S=\tfrac12\)); exact for one magnon in the FM
Single-spin excitation	Transverse amplitude \(\psi_n\propto i^nJ_n(2JSt)\)	Probability amplitude \(c_n\propto i^nJ_n(Jt)\): same mathematics, different meaning
Excitation amplitude	Continuous: cone angle \(\theta_0\)	Discrete: magnon number \(n\), \(\langle n\rangle=NS(1-\cos\theta)\)
Transverse component	Always defined	\(\langle S^\perp\rangle=0\) in a number eigenstate; nonzero only in a superposition
Spin length	\(\|\mathbf S\|=S\)	\(\|\langle\mathbf S_n\rangle\|\le S\), \(\langle\mathbf S^2\rangle=S(S+1)\)
Thermal distribution	Equipartition \(k_BT/\omega_k\)	Bose–Einstein \(1/(e^{\omega_k/k_BT}-1)\) → Bloch \(T^{3/2}\)

Critical notes

Scope and limitations of this simulation

Why exact?

The FM ground state and the one-magnon states are exact eigenstates of the Heisenberg model, so no Holstein–Primakoff or linear spin-wave approximation is needed here. This is a convenience specific to the FM. In the AFM even the ground state is not known exactly, and magnons are defined only approximately (LSWT).

Is the magnon a boson?

At the one-magnon level \(b_k^\dagger=N^{-1/2}\sum_n e^{ikn}S^-_n\) behaves like a boson. For \(S=\tfrac12\) two flips do not fit on one site (\((S^-_n)^2=0\)); moreover two magnons interact through the \(S^zS^z\) term and form bound states in 1D (Bethe, 1931). The boson picture is valid in the dilute-magnon limit (\(\langle n\rangle/N\ll 2S\)).

Finite ring

\(k\) is quantized and the spectrum is discrete. The front of a single-spin excitation reaches the opposite point of the ring at \(t_s=N/2J\) and interferes with itself. After that the infinite-chain formulas (Bessel, \(\sigma(t)\)) no longer apply, but the simulation remains exact for the ring. Partial revivals appear at long times.

Quantum or classical?

The dynamics of a single magnon is mathematically identical to a classical linear spin wave. Its quantum nature shows in the discreteness of the excitation (\(\Delta S^z_{\rm tot}=-1\)), in \(\langle S^\perp\rangle=0\), and in entanglement. The “wave” of a single magnon is therefore a wave of probability amplitude, not a classical wave of arrows.

Suggested exercises

Ballistic spreading. Step 2, \(N=60\). Show in Figure 2 that \(\sigma(t)\) follows the line \(Jt/\sqrt2\). Double \(J\): what happens to the front velocity? How does this differ from diffusion (\(\sigma\propto\sqrt t\))?
What does the field change? In step 2 increase \(h\) from 0 to 1. Why does \(P_n\) not change at all? (Hint: \([H_{\rm Zeeman},H_{\rm ex}]=0\) and \(S^z_{\rm tot}\) is fixed in the sector.)
Goldstone mode. Step 3, \(h=0\), \(m=0\). Why does the colour not rotate? Show that \(|k=0\rangle\) is proportional to \(S^-_{\rm tot}|F\rangle\) and find the total spin of a \(k\neq0\) magnon.
Group velocity. Step 4, \(N=60\), \(\sigma=4\). Compare the measured velocity with \(J\sin k_0\) for \(k_0=\pi/4,\pi/2,3\pi/4\). At which \(k_0\) does the packet spread most slowly? Why?
Why is the arrow short? Step 5(a), \(p=0.5\). Read the transverse amplitude for \(N=20\) and \(N=60\) and verify that it scales as \(1/\sqrt N\). Explain \(|\langle\mathbf S_n\rangle|<\tfrac12\) using the reduced density matrix of a single site.
How many magnons are in a classical wave? How many magnons on average does the \(N=80\), \(\theta_0=5^\circ\) plane wave of the Spin Wave Lab page correspond to for \(S=\tfrac12\)? Discuss the result (\(\langle n\rangle\approx0.15\)) together with step 5(b): what does a “small-amplitude classical wave” correspond to quantum mechanically?

Reading guide

Bars, map and spectrum

Bars

\(\langle n_i\rangle=\tfrac12-\langle S^z_i\rangle\): the probability of finding a flipped spin on site \(i\). The total is exactly 2 at all times. The circles are the density computed assuming the two magnons do not interact at all: \(|\varphi_a(i,t)|^2+|\varphi_b(i,t)|^2\), where \(\varphi\) is the solution of the Single magnon simulation.

Two-particle map

The horizontal axis is the position of the first flipped spin, the vertical axis that of the second. The map is symmetric because magnons are identical. The diagonal (\(n_1=n_2\)) is grey: for \(S=\tfrac12\) two flips do not fit on one site. The cells right next to the diagonal are adjacent pairs. A bound state gives a stripe hugging the diagonal; free magnons give a pattern moving away from it. Because of the periodic boundary, the top-left and bottom-right corners are also adjacent to the diagonal.

Relative distance

\(P(r)\) is the probability of ring distance \(r=1,\dots,N/2\) between the two flipped spins. In a bound state \(P(r)\) decays exponentially; for free magnons the distance grows with time.

Spectrum

In step 3 each dot is a two-magnon eigenstate. The shaded band contains all possible sums of the energies of two free magnons. Most dots lie inside this band. Every dot outside it is a direct trace of the interaction.

Derivation

The two-magnon problem

Step 1

Two sources of interaction

A single flipped spin “breaks” two bonds and raises the energy by \(J\Delta\). Two distant flipped spins break four bonds: \(2J\Delta\). Two adjacent flipped spins break only two bonds:

\[ \underbrace{\uparrow\downarrow\uparrow\cdots\uparrow\downarrow\uparrow}_{2J\Delta}\qquad \underbrace{\uparrow\downarrow\downarrow\uparrow}_{J\Delta}. \]

Being neighbours lowers the energy by \(J\Delta\): an attractive interaction. The second source is kinematic: \((S^-_i)^2=0\), i.e. two flips do not fit on one site (hard core). Since the Hamiltonian is local, two distant magnons move exactly independently.

Step 2

Centre of mass and relative motion

The total momentum is conserved. Therefore, writing

\[ \psi(n_1,n_2)=e^{iK(n_1+n_2)/2}f(r),\qquad r=n_2-n_1 \]

turns the problem into a one-dimensional tight-binding problem in the relative coordinate:

\[ \begin{aligned} E f(r)=&\big(2h+2J\Delta-J\Delta\,\delta_{r,1}\big)f(r)\\ &-J\cos\tfrac K2\,\big[f(r+1)+f(r-1)\big],\quad f(0)=0. \end{aligned} \]

The relative particle sees a hard wall at \(r=0\) and a well of depth \(-J\Delta\) at \(r=1\). The hopping amplitude \(J\cos(K/2)\) depends on the total momentum. On the ring, the condition \(f(r)=e^{iKN/2}f(N-r)\) reduces each \(K\) block to dimension \(\sim N/2\).

Step 3

Continuum and bound state

Scattering states have the form \(f(r)\sim\sin(qr+\delta_q)\). Their energies are sums of two free-magnon energies:

\[ E=\omega_{K/2+q}+\omega_{K/2-q}=2h+2J\Delta-2J\cos\tfrac K2\cos q . \]

A bound state is sought as \(f(r)\propto x^{\,r-1}\). From the \(r=1\) equation

\[ x=\frac{\cos(K/2)}{\Delta},\qquad E_b(K)=2h+J\Delta-\frac{J\cos^2(K/2)}{\Delta}, \]

and the bound state exists when \(|x|<1\), i.e. \(|\cos(K/2)|<\Delta\). In the Heisenberg chain \(E_b=2h+J\sin^2(K/2)\): this is Bethe's 1931 result. The bound state then lies below the band for every \(K\neq0\). In the Ising limit the binding energy tends to \(J\Delta\), i.e. to “one bond saved”.

Step 4

The bound state is a particle

The bound state has its own dispersion relation, hence its own group velocity and mass:

\[ v_b=\frac{dE_b}{dK}=\frac{J}{2\Delta}\sin K,\qquad E_b\simeq 2h+\tfrac J4K^2\ (\Delta=1). \]

In the Heisenberg chain the mass of the bound pair is \(2/J\), exactly twice the single-magnon mass (\(1/J\)). Starting from two adjacent flipped spins, the total weight going into bound states is

\[ W_b=\frac1{2\pi}\int dK\,\big(1-x_K^2\big)=1-\frac{1}{2\Delta^2}\quad(\Delta\ge1). \]

At \(\Delta=1\), \(W_b=1/2\): half of the pair stays together and half breaks up. For \(\Delta<1\) the integral is taken only over the region \(|K|>2\arccos\Delta\) where the bound state exists.

Critical notes

To what extent are magnons bosons?

What does linear spin-wave theory miss?

In linear theory magnons are free bosons and the two-magnon spectrum consists only of the continuum. The bound state comes from the fourth-order (magnon–magnon) terms of the Holstein–Primakoff expansion. At large \(S\) these terms are reduced by a factor of order \(1/S\), but the bound state does not disappear (Wortis, 1963).

Δ = 0: where is the interaction?

At \(\Delta=0\) only the hard core remains. In 1D, hard-core bosons map onto free fermions through the Jordan–Wigner transformation. This is why in step 1 at \(\Delta=0\) the site density agrees with the non-interacting prediction even after the magnons have met and “passed through” each other. The trace of the hard core is visible in the two-particle correlations (\(P(r)\), the map). There is a subtle exception on the ring: for two fermions the Jordan–Wigner transformation produces an antiperiodic boundary condition. Therefore, when the wave fronts have travelled around the ring (around \(t\approx N/2J\)), a small difference appears in the density as well. This is a topological effect; on an infinite chain no difference ever arises.

Dimension and anisotropy

In the 1D Heisenberg chain the bound state exists for every \(K\neq0\). In 2D and 3D, in the isotropic case, it exists only in part of the Brillouin zone, near its boundary. Easy-axis anisotropy (\(\Delta>1\)) strengthens the binding; easy-plane anisotropy (\(\Delta<1\)) weakens it.

Δ < 1 and the ground state

For \(\Delta<1\) and \(h<J(1-\Delta)\) the single-magnon energy is negative at \(k=0\). Then \(|F\rangle\) is not the ground state. The dynamics in the simulation is still exact, because the sector is conserved, but the energies can no longer be read as excitations above the ground state (the readout warns about this).

Finite ring

The spectrum is discrete, and the bound state is exponentially weakly sensitive to the finite size. Near \(x\to1\) (\(K\to0\) or \(\Delta\to\cos(K/2)\)) the size of the bound state exceeds \(N\) and it can no longer be distinguished from the continuum. The part of the adjacent pair that flies apart returns from the other side of the ring at about \(t_s\approx N/2J\).

Bethe ansatz

The problem in this simulation is the first step Bethe took in 1931 when he solved the Heisenberg chain exactly. The two-magnon scattering phase shift \(\delta_q\) and the bound states (“string” solutions with complex momenta) are the building blocks of the exact many-magnon solution.

Suggested exercises

When do they interact? Step 1, \(\Delta=1\), \(N=48\). At what time does the deviation \(D(t)\) in Figure 2 start to grow? Compare with the meeting time of the two fronts, \(t\approx N/4J\).
Jordan–Wigner. Repeat the experiment with \(\Delta=0\). Why does the site density not deviate from the non-interacting prediction after the meeting? Look at the map and at \(P(r)\): where is the difference? Explain the small deviation that appears if you wait longer using the antiperiodic boundary condition on the ring.
Half of it stays. Step 2, \(\Delta=1\). Show that \(P(r\le3)\) approaches its asymptote and that the bound-state weight is \(W_b\simeq1/2\). Verify \(W_b=1-1/(2\Delta^2)=0.875\) for \(\Delta=2\).
Bethe bound state. Step 3, \(\Delta=1\), \(h=0\). Show that at \(K=\pi\) the bound-state energy is \(J\) and its wave function is \(f(r)=\delta_{r,1}\). Why is the relative hopping zero at \(K=\pi\)?
Disappearance of the bound state. At \(\Delta=0.5\) the bound state exists only for \(|K|>2\arccos\Delta=2\pi/3\). Find this in the spectrum and look at how the wave function spreads near the threshold.
A heavy particle. Step 4, \(K_0=\pi/2\). Compare the velocity of the bound packet with \(J\sin K_0\) (single magnon). How does it change at \(\Delta=2\)? Why does the bound pair hardly move in the Ising limit?

Reading guide

How to read the antiferromagnet simulation

Bars (step 1)

\(\langle S^z_i(t)\rangle\): up (cyan) and down (orange). The dashed outlines are the Néel values \(\pm S\). The order parameter \(m_s=\frac{1}{NS}\sum_i(-1)^i\langle S^z_i\rangle\) is 1 in the Néel state and 0 in the ground state.

Configuration list (step 2)

Each row is a state of the \(S^z\) basis (↑, ↓; ○ for \(m=0\) when \(S=1\)). The bar on the right is the probability \(|\langle c|\psi_0\rangle|^2\) of that configuration in the exact ground state. Configurations related by translation carry the same probability.

Bogoliubov curves (step 3)

Top: the magnon energy \(\varepsilon_{\mathbf k}\) of linear theory; bottom: the number of bosons in the ground state, \(v_{\mathbf k}^2\). \(v_{\mathbf k}^2\) diverges where \(\varepsilon_{\mathbf k}\to0\) (at \(\Gamma\) and at \(M\)/\(k=\pi\)). How much this divergence contributes to the integral depends on the dimension.

Spectrum map (step 4)

Horizontal axis: momentum transfer \(q\); vertical axis: energy \(\omega\). The colour is the intensity of \(S^{zz}(q,\omega)\) (square-root scale). On a finite chain each column consists of discrete poles. Lines: linear theory (dashed) and, for \(S=\tfrac12\), the lower and upper edges of the two-spinon continuum.

Derivation

From the Néel state to spinons

Step 1

Why does the Néel state melt?

\(\mathbf S_i\cdot\mathbf S_{i+1}=S^z_iS^z_{i+1}+\tfrac12(S^+_iS^-_{i+1}+S^-_iS^+_{i+1})\). In the FM state the flip terms give zero. In the Néel state, however, every ↑↓ bond can be flipped:

\[ H|{\rm N}\rangle=-NJS^2|{\rm N}\rangle+\tfrac J2\sum_i\big(S^+_iS^-_{i+1}+S^-_iS^+_{i+1}\big)|{\rm N}\rangle . \]

Since the second term is nonzero, \(|{\rm N}\rangle\) is not an eigenstate and changes in time. Its energy is conserved (\(-JS^2\) per bond) but is transferred from the \(zz\) correlation to the \(xy\) correlations. The Néel state is not an eigenstate of the total spin either, whereas the exact ground state is a singlet (\(S_{\rm tot}=0\), Lieb–Mattis).

Step 2

The exact ground state

The Néel state is a variational trial state and gives an upper bound to the true energy: \(E_{\rm N}/N=-JS^2\). For \(S=\tfrac12\) the exact result is known from the Bethe ansatz (Hulthén, 1938):

\[ \frac{E_0}{NJ}=\tfrac14-\ln2\simeq-0.4431\quad\text{(Néel: }-0.25\text{)} . \]

Quantum fluctuations lower the energy by almost a factor of two. The correlations decay as \(\langle\mathbf S_0\cdot\mathbf S_r\rangle\sim(-1)^r\sqrt{\ln r}/r\) for \(S=\tfrac12\) and exponentially (\(\xi\approx6\)) for \(S=1\). There is no long-range Néel order in 1D. Although the Néel states are the most probable configurations, their weight decreases exponentially with \(N\).

Step 3

Holstein–Primakoff and Bogoliubov

On sublattice A \(S^z=S-a^\dagger a\), on B \(S^z=-S+b^\dagger b\). To second order

\[ H_2=JzS\sum_{\mathbf k}\Big[a^\dagger_{\mathbf k}a_{\mathbf k}+b^\dagger_{\mathbf k}b_{\mathbf k}+\gamma_{\mathbf k}\big(a_{\mathbf k}b_{-\mathbf k}+{\rm h.c.}\big)\Big], \]

with \(\gamma_{\mathbf k}=\frac1z\sum_\delta e^{i\mathbf k\cdot\delta}\). The \(ab\) term does not conserve the boson number, so the transformation \(\alpha_{\mathbf k}=u_{\mathbf k}a_{\mathbf k}-v_{\mathbf k}b^\dagger_{-\mathbf k}\) is needed:

\[ \varepsilon_{\mathbf k}=JzS\sqrt{1-\gamma_{\mathbf k}^2},\qquad v_{\mathbf k}^2=\tfrac12\Big(\tfrac{1}{\sqrt{1-\gamma_{\mathbf k}^2}}-1\Big). \]

In the new vacuum \(\langle a^\dagger a\rangle=\Delta S=\overline{v_{\mathbf k}^2}\neq0\). \(\Delta S\) diverges logarithmically in 1D, and equals \(0.197\) in 2D and \(0.078\) in 3D. For the \(S=\tfrac12\) square lattice this gives \(m=0.303\) (QMC: \(0.307\)). The dispersion is the same as in classical linear theory; the difference lies in the zero-point fluctuations.

Step 4

Spinons and the Haldane gap

Neutron scattering measures the dynamical structure factor

\[ S^{zz}(q,\omega)=\sum_n|\langle n|S^z_q|0\rangle|^2\,\delta(\omega-E_n+E_0) . \]

LSWT gives a single sharp branch, \(\omega=2JS|\sin q|\). In the \(S=\tfrac12\) chain, however, a spin flip (\(\Delta S=1\)) fractionalizes into two \(S=\tfrac12\) spinons and the weight spreads over a continuum:

\[ \tfrac\pi2 J|\sin q|\;\le\;\omega\;\le\;\pi J|\sin\tfrac q2| . \]

The lower edge is the des Cloizeaux–Pearson result; LSWT underestimates it (\(J|\sin q|\)) by a factor \(\pi/2\). For \(S=1\) the spectrum is gapped (\(\Delta\simeq0.4105J\) at \(q=\pi\)) and there is a well-defined triplet mode around \(q=\pi\): the “magnon” returns, but massive (Haldane, 1983). On the 2D square lattice magnons survive; the LSWT velocity is corrected by only about 18%.

Critical notes

What it shows, and what it does not

Finite size

Exact diagonalization is limited to \(N\le16\) (\(S=\tfrac12\)) and \(N\le10\) (\(S=1\)). The spectrum is discrete, and for \(S=\tfrac12\) the lowest excitation at \(q=\pi\) goes to zero as \(\sim1/N\). On a finite chain the spinon continuum appears as individual poles spread between the edges of the continuum. At \(N=10\) the Haldane gap has not yet fully converged (it comes out somewhat too large).

Time evolution

Step 1 is a “quantum quench” experiment: the Néel state is not the ground state, so the system does not relax to it, and its energy stays constant. On the finite ring quasi-periodic revivals appear. Long-time behaviour (thermalization, integrability) is beyond the scope of this simulation.

When is LSWT reliable?

Linear theory is the first term of the \(1/S\) expansion and is consistent when \(\Delta S\ll S\). In 2D and 3D it is quantitatively good even for \(S=\tfrac12\). In 1D, however, \(\Delta S\) diverges and the theory is qualitatively wrong: there is no long-range order and there are no sharp magnons. Half-integer spin chains remain gapless (Lieb–Schultz–Mattis); integer spin chains are gapped (Haldane).

Connection to experiment

In quasi-1D \(S=\tfrac12\) antiferromagnets (e.g. KCuF\(_3\)) neutron scattering has shown the two-spinon continuum directly. The Haldane gap has been measured in \(S=1\) Ni-chain compounds. In \(S=\tfrac12\) square-lattice compounds (e.g. La\(_2\)CuO\(_4\)) magnons consistent with spin-wave theory are observed.

Suggested exercises

Comparison with the FM. Step 1. Using the action of \(H\) on the two states, show why the Néel state changes in time while the FM state does not. Explain why the sum of the \(zz\) and \(xy\) bond energies stays constant.
Variational bound. Step 2, \(S=\tfrac12\). Read \(E_0/N\) for \(N=8,12,16\) and compare its convergence to \(\tfrac14-\ln2\) with the Néel (\(-0.25\)) and LSWT (\(-0.4317\)) values.
Néel weight. Record the total weight of the two Néel states in the ground state as a function of \(N\). Why does it decrease exponentially?
Dimension and zero-point motion. Step 3. In 1D increase the lattice size and show that \(\Delta S\) grows as \(\ln N\). In 2D show that \(\Delta S\) converges. Compare \(m=S-\Delta S\) for the \(S=\tfrac12\) square lattice with the QMC value \(0.307\).
Spinon continuum. Step 4, \(S=\tfrac12\), \(N=16\). Show that at \(q=\pi/2\) the weight is not concentrated in a single pole, and that the lowest pole is much closer to the des Cloizeaux–Pearson curve than to LSWT.
Haldane gap. Switch to \(S=1\). Show that at \(q=\pi\) the lowest excitation stays away from zero and that most of the weight is concentrated in a single pole. Compare the gap for \(N=6,8,10\) with \(0.4105J\).

Roadmap

Simulations

Simulation	Content	Method	Status
Single magnon	FM ground state, flipped spin, \(\|k\rangle\), wave packet, classical limit	One-magnon sector, exact	ready
Two magnons	Origin of the interaction, scattering continuum, Bethe bound state, XXZ	Two-magnon sector, exact	ready
Antiferromagnet	Melting of the Néel state, exact ground state, Holstein–Primakoff/Bogoliubov, spinons and the Haldane gap	Exact diagonalization (Lanczos, Krylov) and LSWT	ready
Thermal magnons	Bose statistics, Bloch \(T^{3/2}\), comparison with classical equipartition	Analytic and numerical sums	planned

Magnon Lab · All computations run in the browser · Append ?sim=single, ?sim=two or ?sim=afm to the address to open a simulation directly, and #simulation for the full-page view. Classical counterpart: Spin Wave Lab.