# Quantum ferromagnet

I came across the following exercise.

The quantum Heisenberg ferromagnet, which is specified by the Hamiltonian:

$$\hat{H} = - J\sum_{\langle m n \rangle} \hat{\vec{S}}_m \cdot \hat{\vec{S}}_n,$$

where $J>0, \hat{\vec{S}}_m$ represents the quantum mechanical spin operator at lattice site m, $\langle mn \rangle$ denotes the summation over neighboring sites and $\vec{S}_m ^2 = S(S+1)$. Now we introduce the following transformation:

$$\hat{S}_m^- = a^{\dagger}_m (2S - a^{\dagger}_m a_m )^{1/2}, \ \ \ \hat{S}_m^+ = (2S - a^{\dagger}_m a_m )^{1/2} a_m, \ \ \ \hat{S}_m^z = S- a_m^{\dagger}a_m.$$

Now we look at the one dimensional problem and put the lattice constant to unity. At low temperatures, for $S<<1/2$ we expect the deviations of the magnetization from its value to be very small, i.e. $S - \langle S_m^z \rangle = \langle a_m^{\dagger}a_m \rangle <<S$. In this case we may expand $(2S - a^{\dagger}_m a_m )^{1/2}$ in powers of $a^{\dagger}_m a_m$.

Show that to first order in $a_m^{\dagger}a_m /S$ the Heisenberg Hamiltonian takes the form

$$\hat{H} = - J N S^2 + J S \sum_m \left( a_{m+1}^{\dagger} - a_{m}^{\dagger} \right) \left( a_{m+1} - a_{m} \right) + \text{higher order terms.}$$

Now also show that if we Fourier transform the operators, the Hamiltonian takes the form

$$\hat{H}= - J N S^2 + \sum_k \hbar \omega_k \alpha_k^{\dagger} \alpha_k + \text{higher order terms}$$

with $\hbar \omega_k = 4 J S \sin^2(k/2)$.

Now I don't understand for the first Hamiltonian what I need to do. My first guess is to do a mean field approximation, since we have the expectation value. But this approach doesn't give me the desired cross terms. Furthermore I have a hard time on what to do with the dot product between $S_m$ and $S_n$. Would we need to write out the inner product, invert our operators to get $\hat{S}_x$ and $\hat{S}_y$? Since this seems really over complicated.

Then for the Fourier transform I get $2 \sin^2(k/a)$ and I wonder if the exercise is incorrect or if I made a small error.

$\newcommand{\Sz}[1]{\hat {S^{z}_\text{$#1$}} }$ $\newcommand{\Sx}[1]{\hat {S^{x}_\text{$#1$}} }$ $\newcommand{\Sy}[1]{\hat {S^{y}_\text{$#1$}} }$ $\newcommand{\Splus}[1]{\hat {S}^{+}_\text{$#1$} }$ $\newcommand{\Smin}[1]{\hat {S}^{-}_\text{$#1$} }$ $\newcommand{\ad}[1]{ a^{\dagger}_\text{$#1$} }$ $\newcommand{\a}[1]{ a_\text{$#1$} }$

Allright I have finished your problem.

Firstly write the quantum mechanical spin operator in terms of the ladder operators. $\hat{\textbf{S}}_m = ( \Sx{m},\Sy{m},\Sz{m} )$. Now we use the known relations of the ladder operators and these spatial operators to rewrite in terms of ladder operators. $\Sx{m} = \frac{1}{2}(\Splus{m} + \Smin{m}), \Sy{m} = \frac{1}{2i}(\Splus{m} - \Smin{m})$ and it follows from the commutation relations that $\Sz{m} = S - a^{\dagger}_m a_m$ as is given in your exercise.

Now let's calculate this inner product that you have in the Hamiltonian for some $l,m$:

\begin{align} \hat{\textbf{S}}_l \cdot \hat{\textbf{S}}_m &= \Sx{l} \Sx{m} + \Sy{l} \Sy{m} + \Sz{l} \Sz{m} \\ &= \frac{1}{2}(\Splus{l} + \Smin{l})\frac{1}{2}(\Splus{m} + \Smin{m}) + \frac{1}{2i}(\Splus{l} - \Smin{l}) \frac{1}{2i}(\Splus{m} - \Smin{m}) + S^2 - Sa^{\dagger}_l a_l - Sa^{\dagger}_m a_m + a^{\dagger}_l a_la^{\dagger}_m a_m \\ &= \frac{1}{4} ( \Splus{l} \Splus{m} + \Smin{l} \Splus{m} + \Splus{l} \Smin{m} + \Smin{l}\Smin{m}) - \frac{1}{4} ( \Splus{l} \Splus{m} - \Smin{l} \Splus{m} - \Splus{l} \Smin{m} + \Smin{l}\Smin{m}) + S^2 - Sa^{\dagger}_l a_l - Sa^{\dagger}_m a_m + a^{\dagger}_l a_la^{\dagger}_m a_m \\ &= \frac{1}{2}(\Smin{l} \Splus{m} + \Smin{l}\Smin{m}) + S^2 - Sa^{\dagger}_l a_l - Sa^{\dagger}_m a_m \end{align}

Where we have omitted the higher order term $a^{\dagger}_l a_la^{\dagger}_m a_m$. We now rewrite the ladder operators in terms of creation operators as is suggested in your question, but first let's rewrite these expressions a bit using $S - \langle S_m^z \rangle = \langle a_m^{\dagger}a_m \rangle <<S$ so $(1 - \frac{a_m^{\dagger}a_m}{2S})^{1/2} = 1$.

\begin{align} \hat{S}_m^- &= a^{\dagger}_m (2S - a^{\dagger}_m a_m )^{1/2} = a^{\dagger}_m \sqrt{2S} (1 - \frac{a^{\dagger}_m a_m}{2S} )^{1/2} = \sqrt{2S} \ a^{\dagger}_m \\ \hat{S}_m^+ &= (2S - a^{\dagger}_m a_m )^{1/2} a_m = \sqrt{2S} \ a_m\\ \end{align}

Your question simplifies the solution by only checking the nearest neighbour. That reduces the Hamiltonian to just:

$$\hat{H} = - J\sum_{m=1}^N \hat{\textbf{S}}_m \cdot \hat{\textbf{S}}_{m+1}$$

Plugging all our derivations and rewriting a bit gives:

\begin{align} \hat{H} &= - J\sum_{m=1}^N \frac{1}{2}(\Smin{m} \Splus{m+1} + \Smin{m+1}\Smin{m}) + S^2 - Sa^{\dagger}_m a_m - Sa^{\dagger}_{m+1} a_{m+1} \\ &= - J\sum_{m=1}^N S(\ad{m}\a{m+1} + \ad{m+1}\ad{m} ) + S^2 - S\ad{m}\a{m} - S\ad{m+1}\a{m+1} \\ &= - J\sum_{m=1}^N S^2 + S(\ad{m}(\a{m+1} - \a{m}) + \ad{m+1}(\a{m} - \a{m+1})) \\ &= -JNS^2 - JS \sum_{m=1}^N(\ad{m} - \ad{m+1})(\a{m+1} - \a{m}) \\ &= -JNS^2 + JS \sum_{m=1}^N(\ad{m+1} - \ad{m})(\a{m+1} - \a{m}) \end{align}

Since this is the one - dimensional problem, one can rewrite the Hamiltonian as follows: $$H=-J\sum_{m=1}^{N}\vec{S}_m\vec{S}_{m+1}$$ with $S_{N+1} = S_1$ (periodic boundary condition,for example)

$S^+$ and $S^-$ are ladder operators of the spin angular momentum $\vec{S}$ and have the following form: $$S^{\pm} = S^x\pm iS^y$$ So: $$S^x = \frac{S^+ + S^-}{2}, S^y = \frac{S^+ - S^-}{2i}$$ Now, we calculate $\vec{S}_m\vec{S}_{m+1}$.

$$\vec{S}_m\vec{S}_{m+1} = S^x_mS^x_{m+1} + S^y_mS^y_{m+1} + S^z_mS^z_{m+1} = \frac{1}{2}\left(S^+_mS^-_{m+1}+S^-_mS^+_{m+1}\right)+S^z_mS^z_{m+1}$$ We also have (to the first order): $$(2S - a^\dagger a)^{1/2}\approx (2S)^{1/2}\left(1 - \frac{a^\dagger a}{4S}\right)$$ Subsituting this into the expressions of $S^+$ and $S^-$, then expanding the previous equation (neglecting terms with second order and higher), you will get the desired answer (note that $[a_m,a^\dagger_n]=\delta_{mn}$)