Product of two spin operators | Quantum Science Philippines

## Product of two spin operators

By: Argielou A. Flores, MSU-IIT

Show that $\overrightarrow{S_1} \cdot \overrightarrow{S_1} = \frac{1}{2}(S_{1+}S_{2-} + S_{1-}S_{2+}) + S_{1z}S_{2z}$

where \overrightarrow{S_1} and \overrightarrow{S_2} are spin operators of particle 1 and 2, respectively.

Solution:

We define the following: $\overrightarrow{S_1} = S_{1x}\hat{i} + S_{1y}\hat{j} + S_{1z}\hat{k}\\*\overrightarrow{S_2} = S_{2x}\hat{i} + S_{2y}\hat{j} + S_{2z}\hat{k}$

Taking the product of the two operators $\overrightarrow{S_1} \cdot \overrightarrow{S_2} = S_{1x}S_{2x} + S_{1y}S_{2y} + S_{1z}S_{2z}$ $(a)$

Introducing the ladder operators: $S_{1\pm} = S_{1x} \pm iS_{1y}\\*S_{2\pm} = S_{2x}\pm iS_{2y}$

Multiplying the following, $\begin{array}{lll} S_{1+}S_{2-}& = &(S_{1x}+S_{1y})(S_{2x}-S_{2y})\\&= &S_{1x}S_{2x}-iS_{1x}S_{2y}+iS_{1y}S_{2x}+S_{1y}S_{2y}\\&\end{array}$ $\begin{array}{lll} S_{1-}S_{2+}& = &(S_{1x} - S_{1y})(S_{2x} +S_{2y})\\& = &S_{1x}S_{2x} +iS_{1x}S_{2y} - iS_{1y}S_{2x}+S_{1y}S_{2y}\end{array}$ $\begin{array}{lll} S_{1+}S_{2-}+ S_{1-}S_{2+}& = &S_{1x}S_{2x}-iS_{1x}S_{2y}+iS_{1y}S_{2x}+S_{1y}S_{2y}+S_{1x}S_{2x}+iS_{1x}S_{2y}-iS_{1y}S_{2x}+S_{1y}S_{2y}\\&=&2(S_{1x}S_{2x}+S_{1y}S_{2y})\end{array}$ $S_{1x}S_{2x}+S_{1y}S_{2y} = \frac{1}{2}( S_{1+}S_{2-}+ S_{1-}S_{2+})$ $(b)$

Substituting $(b)$ to $(a)$, we then have: $\overrightarrow{S_1} \cdot \overrightarrow{S_2} = \frac{1}{2}( S_{1+}S_{2-}+ S_{1-}S_{2+})+ S_{1z}S_{2z}$

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