Spin Models

  • Up to now our degrees of freedom have been particles

  • In this lecture: systems of interacting spins

$$ \nonumber \newcommand{\cN}{\mathcal{N}} \newcommand{\br}{\mathbf{r}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bk}{\mathbf{k}} \newcommand{\bq}{\mathbf{q}} \newcommand{\bv}{\mathbf{v}} \newcommand{\pop}{\psi^{\vphantom{\dagger}}} \newcommand{\pdop}{\psi^\dagger} \newcommand{\Pop}{\Psi^{\vphantom{\dagger}}} \newcommand{\Pdop}{\Psi^\dagger} \newcommand{\Phop}{\Phi^{\vphantom{\dagger}}} \newcommand{\Phdop}{\Phi^\dagger} \newcommand{\phop}{\phi^{\vphantom{\dagger}}} \newcommand{\phdop}{\phi^\dagger} \newcommand{\aop}{a^{\vphantom{\dagger}}} \newcommand{\adop}{a^\dagger} \newcommand{\bop}{b^{\vphantom{\dagger}}} \newcommand{\bdop}{b^\dagger} \newcommand{\cop}{c^{\vphantom{\dagger}}} \newcommand{\cdop}{c^\dagger} \newcommand{\bra}[1]{\langle{#1}\rvert} \newcommand{\ket}[1]{\lvert{#1}\rangle} \newcommand{\inner}[2]{\langle{#1}\rvert #2 \rangle} \newcommand{\braket}[3]{\langle{#1}\rvert #2 \lvert #3 \rangle} \newcommand{\sgn}{\mathrm{sgn}} \newcommand{\brN}{\br_1, \ldots, \br_N} \newcommand{\xN}{x_1, \ldots, x_N} \newcommand{\zN}{z_1, \ldots, z_N} $$

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


  • 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

  • What about?

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

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

  • 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 NĂ©el 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?