# Spin Models

• Up to now our degrees of freedom have been particles

• In this lecture: systems of interacting spins


## Ising Model

$$H = J\sum_{\langle j,k\rangle} \sigma_j \sigma_k, \label{spin_ising}$$

• ‘spin’ variables $\sigma_j=\pm 1$

• ${\langle j,k\rangle}$ indicates sum over nearest neighbour pairs

• $J<0$ favours aligned spins, leading to ferromagnetism

But… not really a quantum model

## $N$ spin-1/2: states and operators

• 1 spin-1/2: $\psi_{\pm}$ in $s^z$ basis (say). $N$-spins: $\Psi_{\sigma_1,\ldots \sigma_N}$ has $2^N$ components

• Product states $\ket{\sigma_1}\ket{\sigma_2}\cdots \ket{\sigma_N}$ form a basis

$$\ket{\Psi}=\sum_{{\sigma_j=\pm}}\Psi_{\sigma_1\cdots \sigma_N}\ket{\sigma_1}\ket{\sigma_2}\cdots \ket{\sigma_N}$$

## Tensor product

• For (distinguishable) particles product states were

$$\label{quantum_statistics_disting} \Psi_{\alpha_{1}\alpha_{2}\cdots \alpha_{N}}(\br_1,\ldots \br_N)=\varphi_{\alpha_{1}}(\mathbf{r_{1}})\varphi_{\alpha_{2}}(\mathbf{r_{2}})\cdots\varphi_{\alpha_{N}}(\mathbf{r_{N}})$$

• Abstract form

$$\ket{\Psi_{\alpha_{1}\alpha_{2}\cdots \alpha_{N}}}=\ket{\varphi_{\alpha_{1}}}\ket{\varphi_{\alpha_{2}}}\cdots\ket{\varphi_{\alpha_{N}}}$$

$$\ket{\Psi_{\alpha_{1}\alpha_{2}\cdots \alpha_{N}}}=\ket{\varphi_{\alpha_{1}}}\otimes\ket{\varphi_{\alpha_{2}}}\cdots\otimes\ket{\varphi_{\alpha_{N}}}$$

• We drop the $\otimes$ for brevity

## Operators

• Spin operators obey $[s^a,s^b]=i\epsilon_{abc}s^c$

• A spin operator $s^a_j$ acts on $j^\text{th}$ spin

$$s^a_j\ket{\sigma_1}\ket{\sigma_2}\cdots \ket{\sigma_N} = \ket{\sigma_1}\ket{\sigma_2}\cdots \ket{\sigma_{j-1}} (s^a \ket{\sigma_j}) \ket{\sigma_{j+1}}\cdots\ket{\sigma_N}$$

• Which is the same as

$$s^a_j = \overbrace{\mathbb{1}\otimes \cdots \mathbb{1}}^{j-1}\otimes s^a \otimes \overbrace{\mathbb{1} \otimes\cdots \mathbb{1}}^{N-j}$$

What is $[s^a_j,s^b_k]$ for $j\neq k$?

• Ising model Hamiltonian

$$H = 4J\sum_{\langle j,k\rangle} s^z_j s^z_k$$

• Eigenstates are product states $\ket{\sigma_1}\ket{\sigma_2}\cdots \ket{\sigma_N}$ with energy

$$E_{\sigma_1\cdots \sigma_N} = J\sum_{\langle j,k\rangle} \sigma_j \sigma_k$$

• Fine for statistical mechanics but a bit boring for QM!

## Heisenberg Model

• More realistic

$$H = J \sum_{\langle j,k\rangle} \mathbf{s}_j \cdot \mathbf{s}_k. \label{spin_Hberg}$$

• $\left[s^a_j,s^b_k\right]=i\delta_{jk}\epsilon_{abc}s^c_j$

• Unlike the elastic lattice, not possible to solve in general

• Captures many of the dynamical features (e.g. spin waves) of real magnetic materials.

## Heisenberg Ferromagnetic Chain

• Begin with the 1D case

$$H = J \sum_{j=1}^N \mathbf{s}_j \cdot \mathbf{s}_{j+1},$$

• As usual $\mathbf{s}_j=\mathbf{s}_{j+N}$ (periodic boundary conditions).

• Some anisotropic materials have magnetic atoms arranged in weakly coupled chains

• It’s often useful to write $$\mathbf{s}_j \cdot \mathbf{s}_{j+1} = s^z_js^z_{j+1} + \frac{1}{2}\left(s^+_js^-_{j+1} +s^-_js^+_{j+1}\right),$$ where $s^\pm = s^x\pm i s^y$ $$s^+ = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\qquad s^- = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$$ are the spin raising and lowering operators

## Ground state for $J<0$

$$\ket{\text{FM}} \equiv \ket{+}_1 \ket{+}_2 \cdots \ket{+}_N$$

Show that this is eigenstate of $H$ with $E_0\equiv JN/4$

• Also eigenstate of $S^z$ and $\mathbf{S}^2$, where $\mathbf{S}$ is total spin

$$\mathbf{S} = \sum_{j=1}^N \mathbf{s}_j=(S^x, S^y, S^z)$$

• Eigenvalues are $S^z = N/2$ and $\mathbf{S}^2 = \frac{N}{2}\left(\frac{N}{2}+1\right)$.

## Ground state multiplet

• Rotational invariance implies that $\ket{\text{FM}}$ is member of multiplet of $N+1$ degenerate eigenstates related by rotations

• These states can be generated by acting with $S^-=S^x-iS^y$ $$S^-\ket{\text{FM}} = \sum_{j=1}^N s^-_j\ket{\text{FM}} = \sum_{j=1}^N \ket{+}_1\ket{+}_2\cdots \ket{+}_{j-1} \ket{-}_j\ket{+}_{j+1}\cdots \ket{+}_N. \label{spin_Lowered}$$

$S^z = N/2-1$, but $\mathbf{S}^2$ and $H$ unchanged.

• $\left(S^-\right)^2\ket{\text{FM}}$ is constant amplitude superposition of states with two spins flipped, etc.

## First Excited States

$$\ket{j} = \ket{+}_1\ket{+}_2\cdots \ket{+}_{j-1} \ket{-}_j\ket{+}_{j+1}\cdots \ket{+}_N.$$

• Is this an eigenstate? Act with Hamiltonian, using

$$\mathbf{s}_j \cdot \mathbf{s}_{j+1} = s^z_js^z_{j+1} + \frac{1}{2}\left(s^+_js^-_{j+1} +s^-_js^+_{j+1}\right),$$

• Now note that

$$\left(s^+_j s^-_{j+1} +s^-_js^+_{j+1}\right)\ket{+}_j\ket{-}_{j+1} = \ket{-}_j\ket{+}_{j+1}.$$

• Action of $H$ on $\ket{j}$ is $$H\ket{j} = (1-N/4) \ket{j} - \frac{1}{2}\left(\ket{j-1}+\ket{j+1}\right).$$ (set $J=-1$ from now)

• Leaves us in subspace spanned by states $\ket{j}$: this is subspace with $S_z=N/2-1$

• Flips are like particles (magnons), with Hamiltonian conserving number

• Eigenstates are plane waves $$H\ket{j} = (1-N/4) \ket{j} - \frac{1}{2}\left(\ket{j-1}+\ket{j+1}\right).$$ $$$$\ket{\eta} = \frac{1}{\sqrt{N}}\sum_{j=1}^N e^{i\eta j}\ket{j}, \qquad \eta = \frac{2\pi n}{N}$$$$

• Eigenvalues $E = E_0 + \omega(\eta)$

$$\omega(\eta) = 2\sin^2\eta/2. \label{spin_dispersion}$$

• Dispersion is periodic, as for elastic chain, but quadratic, rather than linear, at small $\eta$

• $\eta=0$ corresponds to the state $S^-\ket{\text{FM}}$

## $N$-Magnon States

• A magnon has energy $\propto J$

• System with extensive energy / finite temperature must have many magnons

• Dimension of subspace of $n$ flipped spins is $\binom{N}{n}$

• Magnons can’t sit on the same site. Things get difficult!

## Antiferromagnets Are Different!

• Let’s try and guess the ground state for $J>0$

• Since anti-aligning spins should be favoured, we might try $$\ket{\text{AFM}} \equiv \ket{+}_1\ket{-}_{2}\cdots \ket{+}_{N-1}\ket{-}_{N}, \label{spin_AFM}$$

• What does $H$ do? Remember $$\mathbf{s}_j \cdot \mathbf{s}_{j+1} = s^z_js^z_{j+1} + \frac{1}{2}\overbrace{\left(s^+_js^-_{j+1} +s^-_js^+_{j+1}\right)}^{\text{swaps }\ket{+}_j\ket{-}_{j+1}},$$ spin flip terms cause spins to move about. Ground state is more complicated!

• For the AFM chain, quantum fluctuations too strong for AFM order

• Antiferromagnets do exist in higher dimensions, and Néel state $\ket{\text{AFM}}$ is good starting approximation

## Large $s$ Expansion

• Generalize model to $s>1/2$ (magnetic ions can have higher spin)

• Develop approximations that work for $s\gg 1/2$

• Hope that the qualitative behaviour we find holds for $s=1/2$

## Holstein–Primakoff Representation

• Represent spins as oscillators!

• Coupled spins becomes coupled oscillators

• Representation not linear, so we get anharmonic chain

• Harmonic approximation justified when spin large

\begin{align} s^+ &=\sqrt{2s}\sqrt{1-\frac{\adop\aop}{2s}}\aop \\ s^- &= \sqrt{2s}\adop\sqrt{1-\frac{\adop\aop}{2s}} \\ s^z &= \left(s - \adop \aop\right). \end{align}

Show that $[\aop,\adop]=1$ reproduces the spin commutation relations $[s^a,s^b]=i\epsilon_{abc}s^c$

## One way to think of it…

• $s^{\pm}$ and $\aop$, $\adop$ both shift us up and down a ladder of states. $$s^\pm\ket{s,m} = \sqrt{s(s+1)-m(m\pm 1)}\ket{s,m\pm 1}$$ Relation between $s^z$ and number of quanta $n$ is simple: $s^z = s - n$

• Difference: $2s+1$ spin states, but infinite oscillator states

• $s^+\propto \aop$, $s^-\propto \adop$ doesn’t work. Something needed to stop us lowering beyond $s^z=-s$ $$s^- = \sqrt{2s}\adop\sqrt{1-\frac{\adop\aop}{2s}}$$

## Another way…

• Classical spin described by point on sphere of radius $\sim s$

• Large $s$: approximate locally by plane

• Near north pole $[s^x,s^y]=is^z\sim is$ resembles $[x,p]=i$

• Therefore $s^\pm$ resemble $\aop$, $\adop$

## Harmonic Spin Waves

• Large $s$ approximation \begin{align} s^+ &\sim \sqrt{2s}\aop \qquad s^- \sim \sqrt{2s}\adop \qquad s_z = \left(s - \adop \aop\right). \label{spin_HPapprox} \end{align} neglecting terms of order $s^{-1/2}$.

• Heisenberg Hamiltonian becomes quadratic oscillator Hamiltonian \begin{align} s^x &\sim \sqrt{s}x \nonumber\\ s^y &\sim \sqrt{s}p\nonumber\\ s_z &= \left(s - \frac{1}{2}[x^2 + p^2 - 1] \right), \end{align} where $x = \frac{1}{\sqrt{2}}(\aop+\adop)$ and $p = \frac{i}{\sqrt{2}}(\adop-\aop)$

\begin{align} s^x &\sim \sqrt{s}x \nonumber\\ s^y &\sim \sqrt{s}p\nonumber\\ s_z &= \left(s - \frac{1}{2}[x^2 + p^2 - 1] \right), \end{align}

$$H = J \sum_{j=1}^N \mathbf{s}_j \cdot \mathbf{s}_{j+1}$$

$$H\sim NJ s^2 + sNJ+ \overbrace{sJ \sum_{j=1}^N \left[x_j x_{j+1} + p_j p_{j+1}-x_j^2 - p_j^2\right]}^{\equiv H^{(2)}} + \ldots, \label{spin_Harmonic}$$

• Use Fourier expansion of the position and momentum \begin{align} x_j(t) &= \frac{1}{\sqrt{N}}\sum_{|n| \leq (N-1)/2} q_n(t) e^{i\eta_n j},\nonumber\\ p_j(t) &= \frac{1}{\sqrt{N}}\sum_{|n| \leq (N-1)/2} \pi_n(t) e^{-i\eta_n j}\\ H^{(2)} &= -2sJ \sum_{|n| \leq (N-1)/2} \sin^2(\eta_n/2)\left[q_n q_{-n} + \pi_n\pi_{-n}\right] \end{align}

• We can read off dispersion $$\omega_{\text{FM}}(\eta) = 4s\left|J\right|\sin^2(\eta/2)$$ c.f. $\omega(\eta) = 2\sin^2\eta/2$ that we found for $s=1/2$

$$H\sim NJ s^2 + sNJ+ \overbrace{ -2sJ \sum_{|n| \leq (N-1)/2} \sin^2(\eta_n/2)\left[q_n q_{-n} + \pi_n\pi_{-n}\right]}^{\equiv H^{(2)}} + \ldots,$$

• Ground state energy of $H^{(2)}$ is

$$-2sJ\sum_{|n| \leq (N-1)/2} \frac{1}{2}\sin^2(\eta_n/2) = -sJN$$

• Overall $E_0 = NJs^2$, which is exact energy of $\ket{FM}$

## AFM case

• Close to $\ket{\text{FM}}$ few oscillator quanta; harmonic approximation OK

• Classically, small amplitude nonlinear oscillations treated as linear

• What about AFM case? Make it look like FM

• Rotate every other spin through $\pi$ about the $y$ axis, so that

$$(s^x_j,s^y_j,s^z_j)\longrightarrow (-s^x_j,s^y_j,-s^z_j),\quad j\text{ odd}.$$

• The Heisenberg chain Hamiltonian becomes $$H = -J \sum_{j=1}^N \left[s^x_j s^x_{j+1} - s^y_j s^y_{j+1} + s^z_j s^z_{j+1}\right].$$
• Harmonic approximation means: close to AFM in original variables

• Oscillator Hamiltonian is now $$H^{(2)} = 2sJ \sum_{|n| \leq (N-1)/2} \left[\sin^2(\eta/2)q_n q_{-n} + \cos^2(\eta/2)\pi_n\pi_{-n}\right], \label{spin_H2AFM}$$ corresponding to a dispersion relation

$$\omega_{\text{AFM}}(\eta) = 2sJ\left|\sin(\eta)\right|. \label{spin_AFMDispersion}$$

$$\omega_{\text{AFM}}(\eta) = 2sJ\left|\sin(\eta)\right|.$$

• Vanishes at both $\eta=0$ and Brillouin zone boundary $\eta=\pi$

• Linear near both points, c.f. quadratic for FM

• Compare $$H_\text{FM}^{(2)} = -2sJ \sum_{|n| \leq (N-1)/2} \sin^2(\eta_n/2)\left[q_n q_{-n} + \pi_n\pi_{-n}\right]$$ In FM both position and momentum terms vanish at $\eta=0$. This is the origin of quadratic dispersion at small $\eta$

• In AFM position term vanishes at $\eta=0$, with momentum term vanishing at $\eta=\pi$. This gives linear dispersion at these points

$$\omega_{\text{AFM}}(\eta) = 2sJ\left|\sin(\eta)\right|.$$

• We know (by other means) exact dispersion relation for lowest excited state of momentum $\eta$ (des Cloiseaux–Pearson mode) $$\omega_{\text{dCP}}(\eta) = \frac{\pi J}{2}\left|\sin(\eta)\right|, \label{spin_dCP}$$ Same functional form, but with a different overall scale

## Ground state fluctuations

• Crudest approximation

$$\bra{\text{FM}}s_j^z \ket{\text{FM}} = s, \qquad \bra{\text{AFM}}s_j^z \ket{\text{AFM}} = s(-1)^j$$

• However, in Holstein–Primakoff representation

$$s^z_j = s - \adop_j\aop_j.$$

• How does second term effect $\bra{0}s^z_j \ket{0}$ in ground state of $H^{(2)}$?

• We know it doesn’t for FM, because $\ket{\text{FM}}=\ket{0}$

$$s^z_j = s - \adop_j\aop_j.$$

• Why doesn’t second term contribute? By translational invariance

\begin{align} \bra{0}\adop_j \aop_j\ket{0} &= \bra{0}\frac{1}{N}\sum_{j=1}^N \adop_j \aop_j\ket{0}\\ \sum_{j=1}^N \adop_j \aop_j &= \frac{1}{2} \sum_{j=1}^N \left(x_j^2 + p_j^2 - 1\right) = -\frac{N}{2} + \frac{1}{2}\sum_n \left(q_n q_{-n} + \pi_n\pi_{-n}\right). \end{align}

• commutes with $$H_\text{FM}^{(2)} = -2sJ \sum_{|n| \leq (N-1)/2} \sin^2(\eta_n/2)\left[q_n q_{-n} + \pi_n\pi_{-n}\right]$$ and is zero in the ground state (c.f. ground state energy)

## AFM case

$$s^z_j = (-1)^j(s-\adop_j\aop_j)$$

• $q_n q_{-n} + \pi_n\pi_{-n}$ doesn’t commute with $\sin^2(\eta/2)q_n q_{-n} + \cos^2(\eta/2)\pi_n\pi_{-n}$

• Express both in oscillator variables

\begin{align} &\aop_\eta = \sqrt{\frac{|\tan(\eta /2)|}{2}}\left(q_n + \frac{i}{|\tan(\eta /2)|}\pi_{-n}\right)\nonumber\\ &\adop_\eta = \sqrt{\frac{|\tan(\eta /2)|}{2}}\left(q_{-n} - \frac{i}{|\tan(\eta /2)|}\pi_{n}\right),\qquad \eta=2\pi n/N \\ &\sin^2(\eta/2)q_n q_{-n} + \cos^2(\eta/2)\pi_n\pi_{-n}=\frac{\omega(\eta)}{2}\left[\adop_\eta\aop_\eta+\aop_\eta\adop_\eta\right]. \end{align}

• To evaluate $\Delta s = \bra{0}\adop_j\aop_j\ket{0}$ write in terms of $\adop_\eta$, $a_\eta$

\begin{align} \frac{1}{N}\sum_{j=1}^N \adop_j \aop_j &= \frac{1}{2N} \sum_{j=1}^N \left(x_j^2 + p_j^2 - 1\right) \\ &= -\frac{1}{2} + \frac{1}{2N}\sum_n \left(q_n q_{-n} + \pi_n\pi_{-n}\right) \end{align} \begin{align} \Delta s &= -\frac{1}{2}+\frac{1}{4N}\sum_n \left[|\tan(\eta_n/2)| + |\cot(\eta_n/2)|\right].\\ &= -\frac{1}{2}+\frac{1}{4}\int_{-\pi}^\pi \frac{d\eta}{2\pi} \left[|\tan(\eta_n/2)| + |\cot(\eta_n/2)|\right]. \end{align}

• Integral diverges logarithmically at $\eta=0$ and $\eta=\pi$.
• What went wrong? Our replacement $$\sum_n (\ldots) \longrightarrow \frac{N}{2\pi}\int_{-\pi}^\pi (\ldots)d\eta$$ failed us because the summand is singular (c.f. $\langle (u_i-u_j)^2\rangle$ in the elastic chain)

• At finite $N$ the sums are all finite if $\eta=0, \pi$ are excluded

$$\Delta s \propto \log N$$

• No AFM in 1D at zero temperature in $N\to\infty$ limit

Repeat the analysis on a 2D square lattice. You should find an integral over the two-dimensional Brillouin zone. Do you find divergences?