Product of two spin operators | Quantum Science Philippines
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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}

 

Adding these two,

\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}

 

One Response to “Product of two spin operators”

  1. Matrix representation of the square of the spin angular momentum | Quantum Science Philippines Says:

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