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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 $S=3/2$?

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1 Answer 1

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

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

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

or in terms of raising and lowering ladders operators

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

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

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

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

The number operator is

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

One has

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

where $f$ is an arbitrary function.

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

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

where

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

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

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

where

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

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

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

with the same number operator

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

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

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

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

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

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).

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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

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