# Holstein-Primakoff and Dyson-Maleev representation

In Holstein-Primakoff and Dyson-Maleev representation, spin operators are represented by bosonic operators. Roughly speaking, a state with $S^z=S-m$ corresponds to a state containing $m$ bosons. In the Dyson-Maleev representation, $(S^+)^\dagger\neq S^-$, so we have a non-unitary transformation.

Are eigenvalues and matrix elements invariant under such a transformation?

In matrix form, what exactly are the two transformations that will give rise to the two representations, say for spin $S=3/2$?

I) We are given an angular momentum operator $$\vec{S}$$ in an (unitary, finite-dimensional, irreducible) spin $$s$$-representation

$$\vec{S}^2~=~s(s+1){\bf 1}, \qquad s\in \frac{1}{2}\mathbb{N}_0, \tag{1}$$

\begin{align} [S_i,S_j]~=~&i\sum_{k=1}^3\epsilon_{ijk} S_k, \qquad i,j,k\in\{x,y,z\}, \cr S_i^{\dagger}~=~& S_i,\end{align}\tag{2}

or in terms of raising and lowering ladders operators

$$S_{\pm}~:=~S_x\pm i S_y, \qquad S_{\pm}^{\dagger}~=~ S_{\mp}, \tag{3}$$ $$[S_z, S_{\pm}]~=~\pm S_{\pm}, \qquad [S_+,S_-]~=~2S_z. \tag{4}$$

Here we have put the reduced Planck constant $$\hbar=1$$.

II) The Heisenberg algebra in terms of annihilation operator $$a_-\equiv a$$ and creation operator $$a_+\equiv a^{\dagger}$$ reads

$$[a_-, a_+]~\equiv~[a,a^{\dagger}]~=~{\bf 1}, \tag{5}$$

$$a_{\pm}^{\dagger}~=~ a_{\mp}.\tag{6}$$

The number operator is

$$n~:=~a_+a_-~\equiv~ a^{\dagger}a.\tag{7}$$

One has

$$[n,a_{\pm}]~=~\pm a_{\pm}, \qquad f(n)a_{\pm}~=~a_{\pm}f(n\pm 1), \tag{8}$$

where $$f$$ is an arbitrary function.

III) The Holstein-Primakoff unitary realization of the spin $$s$$-irrep is given as

$$S_+~=~ a_+h(n)~=~ h(n-1)a_+,\tag{9}$$ $$S_-~=~ h(n)a_-~=~ a_-h(n-1),\tag{10}$$ $$S_z~=~n-s, \tag{11}$$

where

$$h(n)~:=~\sqrt{2s-n}~=~\sqrt{2s} \sqrt{1-\frac{n}{2s}}.\tag{12}$$

It is straightforward to check that eqs. (9-11) yield the Lie algebra (4).

IV) The Dyson-Maleev non-unitary realization of the spin $$s$$-irrep is of the form

$$J_+~=~ S_+g(n)~=~g(n-1)S_+, \tag{13}$$ $$J_-~=~ g(n)^{-1}S_-~=~S_-g(n-1)^{-1},\tag{14}$$ $$J_z~=~S_z, \tag{15}$$

where

$$g(n)~:=~ \sqrt{1-\frac{n}{2s}}.\tag{16}$$

It is straightforward to check that eqs. (13-15) yield the Lie algebra (4) even without using the explicit form (16).

V) Let us define new creation and annihilation operators

$$A_+~:=~ a_+g(n)~=~g(n-1)a_+, \tag{17}$$ $$A_-~=~ g(n)^{-1}a_-~=~a_-g(n-1)^{-1}, \tag{18}$$

with the same number operator

$$N~:=~A_+A_-~=~a_+a_-~=~n,\tag{19}$$

and the same Heisenberg algebra (5).

VI) Note that the new creation and annihilation operators $$A_{\pm}$$ are not each others $$\dagger$$-conjugate a la eq. (6). But one can introduce another Hermitian involution $$\ddagger$$ as

$$a_-^{\ddagger}~:=~ a_+ g(n)^2~=~g(n-1)^2 a_+ \tag{20}$$ $$a_+^{\ddagger}~:=~ g(n)^{-2}a_-~=~a_- g(n-1)^{-2},\tag{21}$$ $$n^{\ddagger}~=~ n, \qquad (FG)^{\ddagger}~=~G^{\ddagger}F^{\ddagger}, \qquad F^{\ddagger\ddagger}~=~F, \tag{22}$$

where $$F$$ and $$G$$ are two arbitrary operators in the universal enveloping algebra. With the new Hermitian involution $$\ddagger$$, the creation and annihilation operators $$A_{\pm}$$ are each others $$\ddagger$$-conjugate

$$A_{\pm}^{\ddagger}~=~ A_{\mp}.\tag{23}$$

VII) Conclusion. The Dyson-Maleev realization (13-15) built with the annihilation and creation operators $$a_{\pm}$$ can be viewed as a Holstein-Primakoff realization built with the annihilation and creation operators $$A_{\pm}$$. Moreover, the Dyson-Maleev realization (13-15) is unitary wrt. the $$\ddagger$$-conjugation but not wrt. the original $$\dagger$$-conjugation, cf. eq. (23). We believe that these observations essentially answer OP's original questions(v1).

• The point about a new Hermitian involution is insightful. Do you have any references about this subject? – leongz Nov 4 '12 at 23:20
• @leongz: No, the answer was developed from scratch, but if I find a reference, I'll post it here. – Qmechanic Nov 4 '12 at 23:46