I am trying to prove a simple relation involving the Holstein-Primakoff Transformation. Here is the transformation: $$S_+ = \sqrt{2S-b^{\dagger}b} b$$ $$S_- = b^{\dagger}\sqrt{2S-b^{\dagger}b}$$ $$S_z = S - b^{\dagger}b$$ $$b^{\dagger}b = n$$ I am trying to plug these transformations into the formula $$S_x^2 + S_y^2 = S^2 - S_z^2$$ and see equality. However, when I plug the transformation in I get different answers on the two sides of the equation. First I transform the left hand side $$S_x^2+S_y^2= \frac{S_+S_- + S_- S_+}{2}$$ Using $$S_x = \frac{S_+ + S_-}{2}$$ $$S_y = \frac{S_+ - S_-}{2i}$$ Then I expand the right hand side $$S^2 - S_z^2 = S^2 - (S^2+n^2 -2Sn) = (2S-n)n$$ so that I have on the RHS: $$(2S-n)n$$ Then I go ahead and plug in the Holstein Primakoff transformation formulas into the left hand side. I have to apply some commutation relations to get things to look nice but in the end I have on the LHS: $$\frac{S_+S_- + S_- S_+}{2} = (2S-n)n +S$$

I get a strange, inconsistent, additional S on the left hand side when I apply the transformation.

$$(2S-n)n +S = (2S-n)n?$$

Can anyone else confirm that they see this or maybe point out the flaw in my algebra? I've decided not to include the operator algebra that I go through on the LHS since it would take a bit to write up, but if someone feels they need to see it to answer my question I can go ahead and add it in.

edit1: Is this just an artifact of the theory which implies it is necessary to have $S\gg1$ so that $S\ll S^2$ and we can just neglect the extra term for the transformation to work? My impression was that the transformation was in some sense exact but we sometimes make the approximation that $n\ll S$ so we can simplify things, but this is different than the first limit I mentioned.


In your equation

$$ S^2 - S_z^2 = S^2 - (S^2+n^2 -2Sn) = (2S-n)n$$

$ S^2$ in LHS is an operator.

$$ S^2 - S_z^2 = \hat{S}^2 - \hat{S_z}^2 = S(S+1) - (S^2+n^2 -2Sn) = (2S-n)n +S $$


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