How Does the Dyson-Maleev Representation Benefit Solid State Physics?

[Petar Mali]

$$\hat{S}^+_i=\sqrt{2S}(\hat{a}_i-\frac{1}{2S}\hat{a}^+_i\hat{a}_i\hat{a}_i)$$

$$\hat{S}^-_i=\sqrt{2S}\hat{a}^+_i, \quad<br /> \hat{S}^z_i=S-\hat{a}^+_i\hat{a}_i$$

Why is in solid state physics often convenient to use this representation? It is obvious that

$$(\hat{S}^-_i)^{\dagger}\neq \hat{S}^+_i$$

And Hamiltonian of Heisenberg model is hermitian!

 

[Petar Mali]

Maybe this transformation has advantage that the Hamiltonian has finite number of terms? Is that only advantage?

 

[Petar Mali]

This representation is in use for $$S \geq \frac{1}{2}$$ like Holstein Primakoff representation

http://en.wikipedia.org/wiki/Holstein–Primakoff_transformation

in which we have square root of some function of operator which has infinitely terms when we expand it in Taylor series!

Dyson Maleev representation has finite number of terms but it isn't hermitian! So I suppose this is only adventage! Am I right?

 

[bjnartowt]

canonical transform

Hi Petar Mali

This is almost 3 years after the fact, so you may realize this by now. You are right that S+ and S- are not hermitian conjugates of one another. You are also right in realizing that Dyson/Maleev has finite number of terms in the Hamiltonian. The fact is that S+, S-, and Sz satisfy the spin-commutation relations (c.f., Milhaly/Martin's text on problems/solutions in solid state physics). The constraint that one must write down operators that are faithful to the spin-commutation relations is the only limit to your imagination, so to speak. All's far in love, war, and canonical transformations.