# The Holstein-Primakoff Representation (approximation)

I have a question regarding the Holstein-Primakoff representation.

In the HP-representation we define the spin operators in terms of bosonic creation and annihilation operators.

$$S_j^+ = \sqrt{2S - n_j} a_j \\ S_j^{^-} = a^\dagger_j\sqrt{2S - n_j} \\ S^z_j = S - n_j$$

Where $a_j$ and $n_j$ are operators. When we derive the magnon dispersion relation, we make the assumption that

$$\frac{\langle n_j \rangle}{\langle S^z \rangle}\ll 1$$

which is fine. As far as I understand this only means that we assume that most of the spins are pointing along the z-direction. However, when we go further in the derivation, we do a series expansion in $n_j/S$, so that i.e.

$$S_j^+ = \sqrt{2S}\sqrt{1- \frac{n_j}{2S}+...} \approx \sqrt{2S}\sqrt{1- \frac{n_j}{2S}}$$

This is where I don't understand. Say that we are in a spin $1/2$-system. In that case we can for one site have at most 1 magnon excitation and $S=\pm 1/2$, so for each individual site $n_j/2S$ is not much smaller than one.

My question is then, how can we justify that the series expansion makes sense in the operators at each individual site? Will the contributions from each site when summed up not contribute because

$$\frac{\langle n_j \rangle}{\langle S^z \rangle}\ll 1$$

or have I misunderstood something?

Any thoughts would be much appreciated.

The assumption is, that the spin $S$ is a large parameter. A conjecture that is apparently not valid for $S=1/2$. The expansion is in $1/S$, which is assumend to be close to zero.
$$S^+_j = \sqrt{2S-n_j}a^\dagger_j = \sqrt{2S}\sqrt{1-\frac{n_j}{2S}}a^\dagger_j\approx\sqrt{2S}\cdot\left(1-\frac{n_j}{4S}\right)a^\dagger_j$$ The second term, being of order $S^{-1/2}$, is neglected.
Assuming $S$ to be large, amounts to a semiclassical approximation. In this limit the relative uncertainty of the spin-operators becomes vanishingly small. (Use the spin-algebra to see this.) $$\frac{\Delta S^i\Delta S^j}{S^2} \longrightarrow 0$$ The working hypothesis is, that low-energy excitations are realized as small deviations around the fully aligned groundstate. For small spin, e.g $S=1/2$ there is no way to only slightly deviate from let's say $S^z_i=+1/2$ Which brings us back to your objection/observation, that the expansion is not valid in the 'small-spin' limit.
• @camzor00 $S$ is the spin on each site. For example if you consider a chain of spin 1/2 particles (electrons), then $S=1/2$ – Nephente Mar 1 '14 at 15:19