Semiclassical spin wave theory of general single-Q structures

"This.. is the business.. we've chosen" - Lee Strasberg in 'The Godfather I and The Godfather II'
Spin Hamiltonian

Now lets tie everything together...

General spin Hamiltonian


$$ H = \sum_{mi,nj} \mathbf{S}_{mi}^T \cdot J_{mi} \cdot \mathbf{S}_{nj} + \sum_{mi}\mathbf{S}_{mi}^T \cdot A_i \cdot \mathbf{S}_{mi} + \mu_B\mathbf{H}^T\sum_{mi}g_i\mathbf{S}_{mi} $$

Where:


  • $\mathbf{S}_{mi}$ column vector of spin operators at position $\mathbf{r}_{mi} = \mathbf{r}_m + t$
    • $m$ index of the unit cell
    • $i$ index of the magnetic atom iin the unit cell
  • $J_{mi,nj}$ 3x3 exchange matrix
  • $A_i$ single ion anisotropy
  • $g_i$ g-tensor
  • $\mathbf{H}$ column vector of external magnetic field
  • $\mu_B$ Bohr magneton (0.05788 meV/T)
Rotation matrices

As we've seen, we need a lot of rotation matrices to convert vectors and reference frames

Rotation, inversion and reflection operations can be described by a multiplication with an orthogonal matrix $R$.

The $R$ transformation can be applied to the basis vectors of the coordinate system or to the objects themselves using $R^{−1}=R^T$.

Transforming the basis vectors:

  • New components of polar vectors (e.g. position vector $\mathbf{r}_{mi}$ ):
$$\mathbf{v} = R\mathbf{v}'$$
  • New components of axis vectors (e.g. spin vector $\mathbf{S}_{mi}$ ):
$$\mathbf{v} = \det (R)\cdot \mathbf{v}'$$
  • New components of tensors (e.g. exchange matrix $J^{\alpha,\beta}$, correlation function $S^{\alpha,\beta}$):
$$T' = R^T T R$$

To exactly (in an LSWT sense) solve the general Hamiltonian the magnetic structure has to be transformed into a ferromagnet. This is possible for single-k magnetic structures (with additional $k=0$ component). For multi-k magnetic structures an approximate magnetic supercell has to be defined.

LSWT solution of the general spin Hamiltonian
Magnetic ground states

So we have the following 'exactly' solvable magnetic structures:

  • Incommensurate spirals with ordering wave vector $Q$
    • Rotation of the n-th cell

$$\varphi_n = \mathbf{Q}\cdot\mathbf{r}_n$$
    • Rotating coordinate system

$$\mathbf{S}_{nj} = R(\mathbf{Q} \cdot \mathbf{r}_n)S_{nj}' R_n$$
  • Magnetic moment direction within the unit cell
    • Additional local rotation independent of $n$

$$S_{nj}' = R_j'S_{nj}''$$
Magnetic ground states

The local spin rotation is now:

$$ S_{nj}^{\alpha\prime} = \sum_\mu R_j^{\prime\alpha\mu} S_{nj}^{\prime\prime\mu} $$

Which can be rewritten as a multiplication with a complex and a real unit vector:

$$ \begin{align} \mathbf{u}_j^\alpha & = R_j^{\prime\alpha 1} + i R_j^{\prime\alpha 2} \\ \mathbf{v}_j^\alpha & = R_j^{\prime\alpha 3} \end{align}$$

Where:

  • $\mathbf{v}_j$ is a unit vector parallel to $\mathbf{S}_{nj}$
  • $\mathbf{u}_j$ is a complex vector, $\Re \, u_j$ and $\Im \, u_j$ spanning the perpendicular space
Symmetries of the Hamiltonian

Now lets look at the symmetries of the Hamiltonian.

Lattice periodicity:

$$J_{mi,nj} = J_{ij}(\mathbf{d})$$

Invariance under exchange of two spins:

$$J_{ij}(\mathbf{d}) = J_{ji}^T(-\mathbf{d})$$

Invariance under the $R_n$ rotations:

$$J_{ij}(\mathbf{d}) = R^T_n J_{ij}(\mathbf{d}) R_n, \; n = 1,2,3,...$$

Fourier-transform of $J$:

$$J_{ij}(\mathbf{k}) = \sum_\mathbf{d} J_{ij}(\mathbf{d}) e^{-i \mathbf{k} \cdot \mathbf{d}}$$

And the identities:

$$\begin{align} J_{ij}(\mathbf{k}) & = \overline{J_{ij}(-\mathbf{k})} \\ J_{ij}(\mathbf{k}) & = \overline{J_{ji}^T(\mathbf{k})} \end{align}$$
Holstein-Primakoff approximation

How does this effect the Boson expansion of the spin ladder operators:

$$ \begin{align} {S_{nj}^{\prime\prime}}^+ & = \sqrt{2S_j} b_{nj},\\ {S_{nj}^{\prime\prime}}^- & = \sqrt{2S_j} b^{\dagger}_{nj},\\ {S_{nj}^{\prime\prime}}^Z & = S_j - b_{nj}^{\dagger}b_{nj} \end{align}$$

Bosonic commutator

$$ \left[b_{mi}, b^{\dagger}_{nj} \right] = \delta_{mn}\delta_{ij} $$

Spin vector operator:

$$ \mathbf{S}^{\prime}_{nj} = \sqrt{\frac{S_j}{2}}\left( \bar{\mathbf{u}}b_{nj} + \mathbf{u}_jb_{nj}^{\dagger} \right) + \mathbf{v}_j \left( S_j - b_{nj}^{\dagger} b_{nj}\right) $$
Quadratic spin Hamiltonian

Now let's apply all of this to the Hamiltonian.

Hamiltonian in the rotating coordinate system:


$$ H = \sum_{mi,nj} \mathbf{S}_{mi}^{\prime T}R_m^TJ_{mi,nj}R_n\mathbf{S}^{\prime}_{mn} $$

Note that using the rotational symmetry of the Hamiltonian:


$$ J_{mi,nj}^\prime = R_m^TJ_{mi,nj}R_n = J_{mi,nj}R_{n-m} $$
Quadratic spin Hamiltonian

After transforming the operators from real space to reciprocal space

$$ b_{mi} = \frac{1}{\sqrt{N}}\sum_{\mathbf{k} \in \text{B.Z}} b_i (\mathbf{k} ) e^{i\mathbf{k}\cdot\mathbf{r}_m} $$

The leading term for excitations is quadratic:

$$ H = \sum_{\mathbf{k} \in \text{B.Z}} \mathbf{x}^\dagger(\mathbf{k})h(\mathbf{k})\mathbf{x}(\mathbf{k}) $$

Where $\mathbf{x}(\mathbf{k})$ is a vector of spin operators:

$$ \mathbf{x}(\mathbf{k}) = \left[ b_1(\mathbf{k}),...b_N(\mathbf{k}),b_1^\dagger(-\mathbf{k}),...,b_N^\dagger(-\mathbf{k}) \right]^T $$

Note! the length of $\mathbf{x}(\mathbf{k})$ is $2N$ where N is number of magnetic atoms in the unit cell
Quadratic spin Hamiltonian

Which is in matrix form:

$$ h(\mathbf{k}) = \begin{bmatrix} A(\mathbf{k})-C & B(\mathbf{k}) \\ B^\dagger(\mathbf{k}) & \bar{A}(-\mathbf{k})-C \end{bmatrix} $$

Where:

$$ \begin{align} A(\mathbf{k})^{i,j} & = \frac{\sqrt{S_i S_j}}{2}\mathbf{u}_i^T J_{ij}^\prime (-\mathbf{k})\bar{\mathbf{u}}_j , \\ B(\mathbf{k})^{i,j} & = \frac{\sqrt{S_i S_j}}{2}\mathbf{u}_i^T J_{ij}^\prime (-\mathbf{k})\mathbf{u}_j , \\ C(\mathbf{k})^{i,j} & = \delta_{ij}\sum_l S_l \mathbf{v}^T J_{il}^\prime(0)\mathbf{v}_l \end{align}$$

Note: $A(\mathbf{k})$ is Hermitian and C is real
Quadratic spin Hamiltonian - Zeeman Term

The Zeeman term is given by:

  • Operator form:
$$H^Z = -\mu_B\mathbf{H}^T \sum_{\mathbf{k}j} g_j\mathbf{v}_j b^\dagger_j(\mathbf{k})b_j(\mathbf{k})$$
  • Matrix form:
$$A^Z(\mathbf{k})^{i,j} = - \frac{1}{2} \mu_B \delta_{ij} \mathbf{H}^T g_i \mathbf{v}_i$$
Diagonalization of the spin Hamiltonian

To diagonalize the Hamiltonian we need a $T$ linear transformation $x_i = \sum_j T_{ij} x_j^\prime $ which fulfills:

  • $h(\mathbf{k})$ is diagomnal in the new basis
  • The new boson operators also fulfill the commutation relations:
$$ \left[\mathbf{x}^\prime, {\mathbf{x}^\prime}^\dagger \right] = \mathbf{x}^\prime ({\mathbf{x}^\prime}^\star)^T - (\mathbf{x}^{\prime\star} \mathbf{x}^{\prime T})^T = g $$

Where:

$$ g = \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix} $$

We will use the method of Colpa which include the Cholesky decomposition of $h(\mathbf{k})$ and solving an eigenvalue problem of a Hermitian matrix.

Result:

  • $2N$ spin wave energies $\mathbf{E}_i$ with both positive and negative sign
  • $T$ transformation from the atomic modes to normal modes $\rightarrow$ can be used to calculate spin-spin correlation function
Spin-spin correlation function in rotating frame

General formula:

$$ S^{\prime\alpha\beta}(\mathbf{k},\omega) = \frac{1}{2N}\sum_{i=1}^{2N} S^{\prime\alpha\beta}_i \delta(\omega \pm \omega_i) \left( n(\omega) + \frac{1}{2}(1\pm 1) \right) $$

Boson population is described by the Bose statistics:

$$ n(\omega_i) = \frac{1}{e^{\hbar\omega_i / k_B T} - 1} $$

The dimensions of $S^{\prime\alpha\beta}$ are $3 \times 3 \times 2N$ for each $\mathbf{k}$ point
Spin-spin correlation function in laboratory frame

Transforming from rotating frame to the laboratory frame:

$$ \left< \mathbf{S}_{mi} \mathbf{S}_{nj}^T (\tau) \right> = \left< R_m \mathbf{S}_{mi}^\prime \mathbf{S}_{nj}^{\prime T} (\tau) R_n^T \right> = \left< \mathbf{S}_{mi}^\prime \mathbf{S}_{nj}^{\prime T} (\tau) \right> R^T_{n-m} $$

Using Rodriguez’s formula:

$$ R(\mathbf{Q}\cdot\mathbf{r}_n) = e^{i\mathbf{Q}\cdot\mathbf{r}_n}R_1 + e^{-i\mathbf{Q}\cdot\mathbf{r}_n}\bar{R}_1 + R_2 $$

Correlation function:

$$ S(\mathbf{k}, \omega) = S^{\prime}(\mathbf{k},\omega)R_2 + S^{\prime}(\mathbf{k} + \mathbf{Q},\omega)R_1 + S^{\prime}(\mathbf{k} - \mathbf{Q},\omega)\bar{R}_1 $$

In $\mathbf{Q} \neq 0 $ magnetic structure generally $3N$ spin wave modes can be observed: $\omega (\mathbf{k})$ and $\omega (\mathbf{k}\pm \mathbf{Q})$. Dimensions of $S^{\alpha\beta}$ is $3 \times 3 \times 3 \times 2N$ for each $\mathbf{k}$ point.