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

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


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

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

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


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

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).

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.