In the previous section we looked at the case of a single Ferromagnetic $J$. But what happens when we have $J_{i,j}$ on a non-trivial lattice?

Frustration from geometrical competing interactions
Degenerate ground states
Weak interactions (e.g. order by disorder) select state from manifold

Heisenberg Hamiltonian (Bravais lattice):

$$ \mathcal{H} = \sum_{i,j} J_{i,j} \mathbf{S}_i \cdot \mathbf{S}_j$$

Introduce the fourier transform the spin:

$$ \mathbf{S}_i = \frac{1}{\sqrt{N}}\sum_k S_k e^{-i \mathbf{k} \cdot \mathbf{r_i}} $$

So.....

Over the first Brillouin zone.

$$\begin{align} \mathcal{H} & = -\frac{1}{2N} \sum_{k, \acute{k}} \sum_{i,j} \mathbf{F}_k \cdot \mathbf{F}_\acute{k} J(\mathbf{r}_i - \mathbf{r}_j) e^{i \mathbf{k} \cdot \mathbf{r}_i} e^{i \mathbf{\acute{k}} \cdot \mathbf{r}_j}\\ & = -\frac{1}{2N} \sum_{k, \acute{k}} \sum_{d} \mathbf{F}_k \cdot \mathbf{F}_\acute{k} J(\mathbf{d}) \sum_i e^{i (\mathbf{k} + \mathbf{\acute{k}}) \cdot \mathbf{r}_i} e^{i \mathbf{k} \cdot \mathbf{d}}\\ \end{align}$$

Since $\sum_i e^{i (\mathbf{k} + \mathbf{\acute{k}}) \cdot \mathbf{r}_i} = N \delta_{(k+\acute{k},0)} $

$$\begin{align} \mathcal{H} & = -\frac{1}{2} \sum_{k} \mathbf{F}_k \cdot \mathbf{F}_{-k} \sum_{d} e^{-i \mathbf{k} \cdot \mathbf{d}} J(\mathbf{d}) \\ & = -\frac{1}{2} \sum_{k} \mathbf{F}_k \cdot \mathbf{F}_{-k} J(\mathbf{k}) \end{align}$$

So $J(k)$ is real and even since $J(\mathbf{d})$ is real and even. Forcing $\mathbf{F}_k$ to respect the normalisation condition of $\mathbf{S}_i$:

$$ S^2 = \mathbf{S}_i \cdot \mathbf{S}_i = \frac{1}{N} \sum_{k,\acute{k}} \mathbf{F}_k \cdot \mathbf{F}_\acute{k} e^{i (\mathbf{k} + \mathbf{\acute{k}})\cdot \mathbf{r}_i} = \frac{1}{N} \sum_{k,\acute{k}} \mathbf{F}_k \cdot \mathbf{F}_{-\acute{k}} e^{i (\mathbf{k} - \mathbf{\acute{k}})\cdot \mathbf{r}_i}$$

Let $\mathbf{Q}$ be a single wavevector which maximises $J(k)$. Minimising $\mathcal{H}$ to get the ground-state we have 2 cases:

$\mathbf{Q} = 0$: So $\mathbf{S}_i = N^{-\frac{1}{2}} \mathbf{S}_{\mathbf{Q} = 0}$ which gives us homogeneous spin polarisation or a ferromagnet as shown before.
$\mathbf{Q} \neq 0$: Our condition is now:

Since the R.H.S must be independent of $\mathbf{r}_i$ we get:

$$ \mathbf{S}_Q \cdot \mathbf{S}_Q = \mathbf{S}_{-Q} \cdot \mathbf{S}_{-Q} $$

That looks simple......

But

remember $\mathbf{S}_Q$ is a Fourier transform

$$ \mathbf{S}_Q = \mathbf{u} + i\mathbf{v}$$

After some replacement, we find that $\mathbf{u}$ and $\mathbf{v}$ are Orthoganal vectors of equal magnitude. So

$$ S^2 = \frac{1}{N} 2 \mathbf{S}_Q \cdot \mathbf{S}_{-Q} = \frac{2}{N} (\mathbf{u}^2 + \mathbf{v}^2)$$

Giving:

$$ \mathbf{u}^2 = \mathbf{v}^2 = \frac{NS^2}{4}$$

And a ground state of:

$$\mathcal{H}_0 = -\frac{1}{2} J(\mathbf{Q}) \mathbf{S}_Q \cdot \mathbf{S}_{-Q} -\frac{1}{2} J(\mathbf{-Q}) \mathbf{S}_{-Q} \cdot \mathbf{S}_{Q} = -J(\mathbf{Q})\frac{NS^2}{2}$$

Reversing the Fourier transform

$$ \begin{align} \mathbf{S}_i & = \frac{1}{\sqrt{N}} \left( e^{i \mathbf{Q} \cdot \mathbf{r}_i} \mathbf{S}_{Q} e^{i \mathbf{-Q} \cdot \mathbf{r}_i} \mathbf{S}_{-Q} \right) \\ & = \frac{1}{\sqrt{N}} \left[ e^{i \mathbf{Q} \cdot \mathbf{r}_i} (\mathbf{u} + i\mathbf{v}) + e^{i \mathbf{-Q} \cdot \mathbf{r}_i} (\mathbf{u} - i\mathbf{v})\right] \\ & = -\frac{2}{\sqrt{N}} \left( \mathbf{u}\cos (\mathbf{Q} \cdot \mathbf{r}) - \mathbf{v}\sin (\mathbf{Q} \cdot \mathbf{r}) \right) \end{align}$$

Normalising $\mathbf{u}$ and $\mathbf{v}$ we get:

$$\mathbf{S}_i = S\left( \mathbf{\hat{u}}\cos (\mathbf{Q} \cdot \mathbf{r}) - \mathbf{\hat{v}}\sin (\mathbf{Q} \cdot \mathbf{r}) \right)$$

So:

$\mathbf{\hat{u}} \perp \mathbf{\hat{v}}$
$\mathbf{\hat{u}}$ and $\mathbf{\hat{v}}$ are not fixed by minimising $\mathcal{H}$
Anisotropy would break the degeneracy

In spin-space

Choosing the points $\mathbf{\hat{x}}=\mathbf{\hat{u}}$ and $\mathbf{\hat{y}}=-\mathbf{\hat{v}}$ we see:

$$ \begin{align} S_i^x & = S\cos(\mathbf{Q} \cdot \mathbf{r}) \\ S_i^y & = S\sin(\mathbf{Q} \cdot \mathbf{r}) \\ S_i^z & = 0 \end{align}$$

A helical state which is coplanar but not colinear.

$\mathbf{\hat{u}}$ and $\mathbf{\hat{y}}$ can be arbitrarily orientated relative to $\mathbf{Q}$
$\mathbf{S}_i = C$ on any plane orthogonal to $\mathbf{Q}$

We have proved Luttinger-Tisza theory

Quantum mechanical treatment of spin waves

Important points to note:

Excitations are quasiparticles: magnons
- They are Bosons
- Obey commutator relations: $\left[ a, a^+ \right] = 1$
- Obey Bose-Einstein statistics: $n(\omega ) = \frac{1}{\exp (\hbar \omega / k_B T) - 1}$
Equivalent to coupled quantum harmonic oscillators:
- Zero point energy
- Reduction of magnetic order
- Quantum order by disorder
Strong magnon-magnon interaction in non-collinear magnets

Rotating coordinate system

Rotating coordinate system in the lab frame

We introduced the concept of the rotating coordinate system in the helical subsection. Choosing a suitable $\mathbf{\hat{u}}$ and $\mathbf{\hat{v}}$:

$$\begin{align} S_i^x & = S_i^\eta \\ S_i^y & = S_i^\mu \cos (\varphi_i ) + S_i^\xi \sin (\varphi_i)\\ S_i^z & = -S_i^\mu \sin (\varphi_i ) + S_i^\xi \cos (\varphi_i) \end{align} $$

Where the rotation angle:

$$ \varphi_i = \mathbf{Q}\cdot\mathbf{r}_i $$

Heisenberg Hamiltonian on Bravais lattice:

$$ \mathcal{H} = \sum_{i,d} J(d) \mathbf{S}_i \cdot \mathbf{S}_{i+d}$$

We derived this previously and after substitution and a bit of trigonometry:

$$ \begin{align} \mathcal{H} = & \sum_{i,j} J(\mathbf{d}) \left( S_i^\eta S_j^\eta + \sin (\mathbf{Q} \cdot \mathbf{d}_{i,j})(S_i^\mu S_j^\xi - S_i^\xi S_j^\eta \right) + \\ & \cos(\mathbf{Q}\cdot \mathbf{d}_{i,j}\left( S_i^\mu S_j^\mu + S_i^\xi S_j^\xi\right) \end{align}$$

Spin state with a small field along the $z$ axis:

$$ \lvert n \rangle \equiv S, m = S -n $$

Ladder operators for a harmonic oscillator:

$$ \begin{align} a^+ \lvert n \rangle & = \sqrt{n+1} \lvert n + 1 \rangle \\ a \lvert n \rangle & = \sqrt{n} \lvert n -1 \rangle \\ \left[ a, a^+ \right] & = 1 \end{align}$$

Ladder operators for spin:

$$ \begin{align} S^- \lvert n \rangle & = \sqrt{2S}\sqrt{1 - \frac{n}{2S}}\sqrt{n+1} \lvert n + 1 \rangle \\ S^+ \lvert n \rangle & = \sqrt{2S}\sqrt{1 - \frac{n - 1}{2S}}\sqrt{n} \lvert n -1 \rangle \\ \end{align}$$

Holstein–Primakoff transformation

Ladder operators for spin:

$$ \begin{align} S^- & = \sqrt{2S} a^+ \hat{f} \\ S^+ & = \sqrt{2S} \hat{f} a \end{align}$$

Where:

$$ \hat{f} = \sqrt{1 - \frac{n}{2S}} $$

Applying the transormation, $S$ in boson creation and annihilation operators:

$$ \begin{align} S_i^\eta & = \frac{1}{2}(S^+ + S^- ) = \sqrt{S/2}\hat{f}(a^+_i + a_i) \\ S_i^\mu & = -\frac{i}{2}(S^+ - S^- ) = \sqrt{S/2}\hat{f}(a^+_i - a_i) \\ S_i^\xi & = S - n_i = S - a_i^+a_i \end{align}$$

So:

$$ S_i^\eta S_j^\eta = S/2 \hat{f}_i \hat{f}_j (a^+_i + a_i)(a^+_j + a_j) $$

etc...

We have a hamiltonian of boson operators which can be expanded in powers of $1/S$ to obtain:

$$ \mathcal{H} = E_0 + \mathcal{H}_1 + \mathcal{H}_2 + \mathcal{H}_3 + \mathcal{H}_4 ... $$

Where $E_0$ is the classical result which we derived from the spiral example:

$$ E_0 = S^2 \sum_{i, \mathbf{d}} J(\mathbf{d})\cos (\mathbf{Q} \cdot \mathbf{d}) $$

1 operator term:

$$ \mathcal{H}_1 = S^{3/2} \sum_{i,j} \frac{i}{2} J(\mathbf{d}_{i,j}) \sin (\mathbf{Q} \cdot \mathbf{d}_{i,j})(a_i^+ - a_i + a_j - a_j^+) $$

2 operator term:

$$\begin{align} \mathcal{H}_2 = S \sum_{i,j} \frac{1}{2} J(\mathbf{d}_{i,j}) &( (1 - \cos (\mathbf{Q} \cdot \mathbf{d}_{i,j}))(a_ia_j + a_ia_j) \\ & + (1 + \cos (\mathbf{Q} \cdot \mathbf{d}_{i,j}))(a_ia_j^+ + a_i^+a_j) \\ & - 2\cos (\mathbf{Q} \cdot \mathbf{d}_{i,j})(a_i^+a_i + a_j^+a_j) ) \end{align}$$

3 operator term:

Non-zero in non-collinear structures
Can change the ground state

4 operator term:

Renormalizes the magnon dispersion
Gives finite magnon lifetime

We have a mix of operators on the different sites and we need to separate them.

Bring on more Fourier transforms:

$$\begin{align} a_i = \frac{1}{\sqrt{N}}\sum_{k} a_k e^{i \mathbf{k}\cdot\mathbf{r}_i} & , a_i^+ = \frac{1}{\sqrt{N}}\sum_{q} a_k^+ e^{-i \mathbf{k}\cdot\mathbf{r}_i}\\ a_k = \frac{1}{\sqrt{N}}\sum_{i} a_i e^{-i \mathbf{k}\cdot\mathbf{r}_i} & , a_k^+ = \frac{1}{\sqrt{N}}\sum_{i} a_i^+ e^{i \mathbf{k}\cdot\mathbf{r}_i} \end{align}$$

Where the usual Boson commutator relations are obeyed.

After substitution:

$$ \begin{align} \mathcal{H}_2 & = \frac{S}{2} \sum_\mathbf{k} \left( J(\mathbf{k}) - \frac{J(\mathbf{k} + \mathbf{Q}) + J(\mathbf{k} - \mathbf{Q})}{2} \right) \left( a_\mathbf{k}a_\mathbf{-k} + a_\mathbf{k}^+a_\mathbf{-k}^+\right) \\ & + \left( J(\mathbf{k}) + \frac{J(\mathbf{k} + \mathbf{Q}) + J(\mathbf{k} - \mathbf{Q})}{2} \right) \left( a_\mathbf{k}a_\mathbf{k}^+ + a_\mathbf{k}^+a_\mathbf{k}\right) - 4J(\mathbf{Q})a_\mathbf{k}^+a_\mathbf{k} \end{align}$$

Luckily, $\mathcal{H}_2$ can now be written in matrix form:

$$\mathcal{H}_2 = \sum_\mathbf{k} \mathbf{x}^\dagger H(\mathbf{k}) \mathbf{x} $$

Where $\mathbf{x}$ is a vector of Boson operators:

$$ \mathbf{x} = \begin{bmatrix} a_\mathbf{k} \\ a_{-\mathbf{k}}^+\end{bmatrix}$$

And the matrix of the Hamiltonian has the form:

$$ H = \begin{bmatrix} A & B \\ B & A \end{bmatrix} $$
$$\begin{align} A & = J(\mathbf{k}) + J(\mathbf{k} + \mathbf{Q})/2 + J(\mathbf{k} - \mathbf{Q})/2 - 2J(\mathbf{Q}) \\ B & = J(\mathbf{k}) - J(\mathbf{k} + \mathbf{Q})/2 - J(\mathbf{k} - \mathbf{Q})/2 \end{align}$$

In Bogoliubov method, we define new operator $b$ with the following transformation:

$$ \begin{align} b & = ua + va^+ \\ b^+ & = ua^+ +va \end{align}$$

The new operator has to fulfill the commutation relations:

$$ \begin{align} [b, b^+] & = 1 \\ u^2 + v^2 & = 1 \end{align}$$

With the right parameter choice:

$$ \mathcal{H}_2 = \sum_\mathbf{k} \omega_\mathbf{k} \left( b^+b + \frac{1}{2} \right) $$

And the spin wave dispersion:

$$ \omega_\mathbf{k} = \sqrt{A^2 - B^2} $$

The above calculation is equivalent to solving the eigenvalue problem of $gH$, where $g = [x, x^\dagger]$ commutator matrix.

Now we have $\omega_\mathbf{k}$ and half the story. Remember for neutron scattering:

$$ \frac{d^2\sigma}{d\Omega dE} = C \cdot F^2(\mathbf{K}) \sum_{\alpha , \beta}(\delta_{\alpha , \beta} - \hat{k}_\alpha\hat{k}_\beta ) (gSg^T)^{\alpha , \beta} (\mathbf{k}, \omega) $$

With the correlation function:

$$ S^{\alpha \beta}(\mathbf{k},\omega) = \frac{1}{2\pi\hbar}\int dt e^{-i \omega t} \left< S^\alpha (\mathbf{k}, 0) S^\beta (-\mathbf{k}, t) \right> $$

Rotating coordinate system

Rotating coordinate system in the lab frame

Correlation function:

$$\begin{align} S^{x,x}(t, \mathbf{k}) & = \frac{1}{N^2} \sum_{i,j} \left< S_i^x S_j^x(t) \right> e^{i \mathbf{k}\cdot \mathbf{d}_{i,j}} \\ & = \frac{S}{2} \sum_\mathbf{k} \left< a^+_{\mathbf{-k}} a^+_{\mathbf{k}}(t) + a_{\mathbf{-k}} a_{\mathbf{k}}(t) + a^+_{\mathbf{k}} a_{\mathbf{k}}(t) + a_{\mathbf{k}} a^+_{\mathbf{k}}(t) \right> \end{align} $$

Converting into $b$ and $b^+$ operators, taking the Fourier transform in time:

$$ S^{xx}(\omega,\mathbf{k}) = S \frac{A_\mathbf{k} - B_\mathbf{k}}{\omega_\mathbf{k}} \cdot \delta (\omega - \omega_\mathbf{k}) \cdot (n_\omega + 1) $$

Using

$$ \left< b^+_\mathbf{k} b_\mathbf{k} (t) \right> = n_\omega e^{-i \omega_\mathbf{k} t} $$

Rotating coordinate system

Rotating coordinate system in the lab frame

Correlation function:

$$ S^{y,y}(t) = -\frac{S}{2} \sum_\mathbf{i,j} \cos (\mathbf{Q} \cdot \mathbf{d}_{i,j}) \left< a^+_{i} a^+_{j}(t) + a_{i} a_{j}(t) - a^+_{i} a_{j}(t) - a_{i} a^+_{j}(t) \right> $$

Converting into $b$ and $b^+$ operators, taking the Fourier transform in time. The $\cos (\mathbf{Q} \cdot \mathbf{d}_{i,j})$ brings in a $\pm \mathbf{Q}$ shift in reciprocal space:

$$ S^{yy}(\omega,\mathbf{k}) = S^{zz}(\omega,\mathbf{k}) = S \frac{1}{4} \left( S^{\xi\xi} (\omega , \mathbf{k} - \mathbf{Q}) + S^{\xi\xi} (\omega , \mathbf{k} + \mathbf{Q}) \right) $$

Where the correlation function in the rotating frame:

$$ S^{\xi\xi} (\omega , \mathbf{k} ) = S \frac{A_\mathbf{k} + B_\mathbf{k}}{\omega_\mathbf{k}} \cdot \delta (\omega - \omega_\mathbf{k}) \cdot (n_\omega + 1) $$

Worked Example: Triangular lattice antiferromagnet

Spin configuration.

Spins in a triangular antiferromagnet.

There are 3 bond vectors given by:

$$\begin{align} \mathbf{e}_1 & = (1, 0) \\ \mathbf{e}_2 & = (-1/2, \sqrt(3)/2) \\ \mathbf{e}_3 & = (-1/2, -\sqrt(3)/2) \end{align}$$

The exchange couplings are given by:

$$ J(\mathbf{k}) = \sum_{i=1}^3 J\cos (\mathbf{k} \cdot \mathbf{e}_i ) $$

Ground State energy.

Energy manifold for a triangular antiferromagnet.

The ground-state energy is given by:

$$ \omega_\mathbf{k} = \sqrt{A^2 - B^2} = \sqrt{(1-J(\mathbf{k}))(1 + 2J(\mathbf{k}))} $$

Where $A$ and $B$ have been simplified for this case.

The classical ground state is given by minimising $J(\mathbf{k})$. (Note, by notation in some books it's maximising).

$$ \mathbf{Q} = (1/3, 1/3) $$

Spin-wave spectra.

Spin-wave dispersion in a triangular antiferromagnet.

By looking at the definition of $A$ and $B$ we can see that there are 3 spinwave modes.

The three spin wave modes are: $S(\mathbf{k},\omega)$ and $S(\mathbf{k}\pm\mathbf{Q},\omega)$ due to the incommensurate ordering wave vector.

Using SpinW we can plot the spin polarisation of each mode

In red is a phason mode polarised along the $c^*$-axis.
The blue Goldstone modes have polarisation in the easy-plane.

Goldstone modes appear due to spontaneous breaking of continuous symmetry.

Semiclassical spin wave theory on Bravais lattices

The spiral ground state

Heisenberg Hamiltonian (Bravais lattice):

Over the first Brillouin zone.

But

Reversing the Fourier transform

In spin-space

Quantum mechanical treatment of spin waves

Heisenberg Hamiltonian on Bravais lattice:

Holstein–Primakoff transformation

Worked Example: Triangular lattice antiferromagnet