Spin Models

Ising Model

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

  • ‘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 \lvert{\sigma_1}\rangle\lvert{\sigma_2}\rangle\cdots \lvert{\sigma_N}\rangle form a basis

\lvert{\Psi}\rangle=\sum_{\{\sigma_j=\pm\}}\Psi_{\sigma_1\cdots \sigma_N}\lvert{\sigma_1}\rangle\lvert{\sigma_2}\rangle\cdots \lvert{\sigma_N}\rangle

Tensor product

  • For (distinguishable) particles product states were

\Psi_{\alpha_{1}\alpha_{2}\cdots \alpha_{N}}(\mathbf{r}_1,\ldots \mathbf{r}_N)=\varphi_{\alpha_{1}}(\mathbf{r_{1}})\varphi_{\alpha_{2}}(\mathbf{r_{2}})\cdots\varphi_{\alpha_{N}}(\mathbf{r_{N}})

  • Abstract form

\lvert{\Psi_{\alpha_{1}\alpha_{2}\cdots \alpha_{N}}}\rangle=\lvert{\varphi_{\alpha_{1}}}\rangle\lvert{\varphi_{\alpha_{2}}}\rangle\cdots\lvert{\varphi_{\alpha_{N}}}\rangle

\lvert{\Psi_{\alpha_{1}\alpha_{2}\cdots \alpha_{N}}}\rangle=\lvert{\varphi_{\alpha_{1}}}\rangle\otimes\lvert{\varphi_{\alpha_{2}}}\rangle\cdots\otimes\lvert{\varphi_{\alpha_{N}}}\rangle

  • 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\lvert{\sigma_1}\rangle\lvert{\sigma_2}\rangle\cdots \lvert{\sigma_N}\rangle = \lvert{\sigma_1}\rangle\lvert{\sigma_2}\rangle\cdots \lvert{\sigma_{j-1}}\rangle (s^a \lvert{\sigma_j}\rangle) \lvert{\sigma_{j+1}}\rangle\cdots\lvert{\sigma_N}\rangle

  • 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}

Check

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 \lvert{\sigma_1}\rangle\lvert{\sigma_2}\rangle\cdots \lvert{\sigma_N}\rangle 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

  • \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

\lvert{\text{FM}}\rangle \equiv \lvert{+}\rangle_1 \lvert{+}\rangle_2 \cdots \lvert{+}\rangle_N

Check

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 \lvert{\text{FM}}\rangle 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^-\lvert{\text{FM}}\rangle = \sum_{j=1}^N s^-_j\lvert{\text{FM}}\rangle = \sum_{j=1}^N \lvert{+}\rangle_1\lvert{+}\rangle_2\cdots \lvert{+}\rangle_{j-1} \lvert{-}\rangle_j\lvert{+}\rangle_{j+1}\cdots \lvert{+}\rangle_N.

Check

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

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

First Excited States

  • What about?

\lvert{j}\rangle = \lvert{+}\rangle_1\lvert{+}\rangle_2\cdots \lvert{+}\rangle_{j-1} \lvert{-}\rangle_j\lvert{+}\rangle_{j+1}\cdots \lvert{+}\rangle_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)\lvert{+}\rangle_j\lvert{-}\rangle_{j+1} = \lvert{-}\rangle_j\lvert{+}\rangle_{j+1}

  • Action of H on \lvert{j}\rangle is

H\lvert{j}\rangle = (1-N/4) \lvert{j}\rangle - \frac{1}{2}\left(\lvert{j-1}\rangle+\lvert{j+1}\rangle\right) (set J=-1 from now)

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

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

  • Eigenstates are plane waves

H\lvert{j}\rangle = (1-N/4) \lvert{j}\rangle - \frac{1}{2}\left(\lvert{j-1}\rangle+\lvert{j+1}\rangle\right).

\lvert{\eta}\rangle = \frac{1}{\sqrt{N}}\sum_{j=1}^N e^{i\eta j}\lvert{j}\rangle, \qquad \eta = \frac{2\pi n}{N}

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

\omega(\eta) = 2\sin^2\eta/2

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

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

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

\lvert{\text{AFM}}\rangle \equiv \lvert{+}\rangle_1\lvert{-}\rangle_{2}\cdots \lvert{+}\rangle_{N-1}\lvert{-}\rangle_{N}

  • 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 }\lvert{+}\rangle_j\lvert{-}\rangle_{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 \lvert{\text{AFM}}\rangle 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{aligned} s^+ &=\sqrt{2s}\sqrt{1-\frac{a^\dagger a^{\vphantom{\dagger}}}{2s}}a^{\vphantom{\dagger}}\\ s^- &= \sqrt{2s}a^\dagger\sqrt{1-\frac{a^\dagger a^{\vphantom{\dagger}}}{2s}} \\ s^z &= \left(s - a^\dagger a^{\vphantom{\dagger}}\right) \end{aligned}

Check

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

One way to think of it…

  • s^{\pm} and a^{\vphantom{\dagger}}, a^\dagger both shift us up and down a ladder of states. s^\pm\lvert{s,m}\rangle = \sqrt{s(s+1)-m(m\pm 1)}\lvert{s,m\pm 1}\rangle 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 a^{\vphantom{\dagger}}, s^-\propto a^\dagger doesn’t work. Something needed to stop us lowering beyond s^z=-s s^- = \sqrt{2s}a^\dagger\sqrt{1-\frac{a^\dagger a^{\vphantom{\dagger}}}{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 a^{\vphantom{\dagger}}, a^\dagger

Harmonic Spin Waves

  • Large s approximation

\begin{aligned} s^+ &\sim \sqrt{2s}a^{\vphantom{\dagger}}\qquad s^- \sim \sqrt{2s}a^\dagger\qquad s_z = \left(s - a^\dagger a^{\vphantom{\dagger}}\right). \end{aligned}

neglecting terms of order s^{-1/2}.

  • Heisenberg Hamiltonian becomes quadratic oscillator Hamiltonian

\begin{aligned} 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{aligned}

where x = \frac{1}{\sqrt{2}}(a^{\vphantom{\dagger}}+a^\dagger) and p = \frac{i}{\sqrt{2}}(a^\dagger-a^{\vphantom{\dagger}})

\begin{aligned} 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{aligned}

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

  • Use Fourier expansion of the position and momentum

\begin{aligned} 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{aligned} `

  • 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 \lvert{FM}\rangle

AFM case

  • Close to \lvert{\text{FM}}\rangle 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], corresponding to a dispersion relation

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

\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|, Same functional form, but with a different overall scale

Ground state fluctuations

  • Crudest approximation

\langle{\text{FM}}\rvert s_j^z \lvert{\text{FM}}\rangle = s, \qquad \langle{\text{AFM}}\rvert s_j^z \lvert{\text{AFM}}\rangle = s(-1)^j

  • However, in Holstein–Primakoff representation

s^z_j = s - a^\dagger_ja^{\vphantom{\dagger}}_j.

  • How does second term effect \langle{0}\rvert s^z_j \lvert{0}\rangle in ground state of H^{(2)}?

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

s^z_j = s - a^\dagger_ja^{\vphantom{\dagger}}_j.

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

\begin{aligned} \langle{0}\rvert a^\dagger_j a^{\vphantom{\dagger}}_j\lvert{0}\rangle &= \langle{0}\rvert\frac{1}{N}\sum_{j=1}^N a^\dagger_j a^{\vphantom{\dagger}}_j\lvert{0}\rangle\\ \sum_{j=1}^N a^\dagger_j a^{\vphantom{\dagger}}_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{aligned}

  • 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-a^\dagger_ja^{\vphantom{\dagger}}_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{aligned} &a^{\vphantom{\dagger}}_\eta = \sqrt{\frac{|\tan(\eta /2)|}{2}}\left(q_n + \frac{i}{|\tan(\eta /2)|}\pi_{-n}\right)\nonumber\\ &a^\dagger_\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[a^\dagger_\eta a^{\vphantom{\dagger}}_\eta+a^{\vphantom{\dagger}}_\eta a^\dagger_\eta\right]. \end{aligned}

  • To evaluate \Delta s = \langle{0}\rvert a^\dagger_ja^{\vphantom{\dagger}}_j\lvert{0}\rangle write in terms of a^\dagger_\eta, a_\eta

\begin{aligned} \frac{1}{N}\sum_{j=1}^N a^\dagger_j a^{\vphantom{\dagger}}_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{aligned} \begin{aligned} \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{aligned}

  • 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

Check

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