Matrix representation of the square of the spin angular momentum | Quantum Science Philippines
Quantum Science Philippines

Matrix representation of the square of the spin angular momentum

By: Maria Christine L. Lugo, MS Physics I, MSU-IIT

 

Show that

 

a.) (S^{2} = S_{1}^{2}+S_{2}^{2}+ 2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+})

 

b.) S^{2} =\hbar^{2}\left( \begin{array}{cccc} 2& 0&0&0 \\ 0 &1&1&0 \\0&1&1&0\\0&0&0&2 \end{array} \right)

 

Solution:

a.) We can expand S^{2} by its components S_{1}^{2} and S_{2}^{2},

S^{2} = S_{1}^{2}+S_{2}^{2} + 2(S_{1}\cdot S_{2})

 

But we know that,

S_{1}\cdot S_{2} = \frac{1}{2} (S_{1+}S_{2-}+S_{1-}S_{2+}) +S_{1z}S_{2z}

(See link https://www.quantumsciencephilippines.com/?p=5698 )

 

Therefore,

S^{2} = S_{1}^{2} + S_{2}^{2} + 2[\frac{1}{2}(S_{1+}S_{2-} + S_{1-}S_{2+})+S_{1z}S_{2z}]

 

S^{2} = S_{1}^{2} + S_{2}^{2} +2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+}

 

b.) Using the eigenvalue equation of S^{2} and substituting the basis (|++>, |+->, |-+>, |- ->)

S^{2}|sm> = (S_{1}^{2}+S_{2}^{2} +2S_{1z}S_{2z}+ S_{1+}S_{2-}+S_{1-}S_{2+})|sm>

 

We have,

 

\begin{array}{lll} S^{2}|++>&=& S_{1}^{2}|++>+ S_{2}^{2}|++>+2S_{1z}S_{2z}|++> +S_{1+}S_{2-}|++>+S_{1-}S_{2+}|++>\\ \\&=&\frac{3}{4}\hbar^{2}|++>+\frac{3}{4}\hbar^{2}|++>+2(\frac{1}{2}\hbar)(\frac{1}{2}\hbar)|++>\\ \\&=&2\hbar^{2}|++>\end{array}

 

\begin{array}{lll} S^{2}|+->&=& S_{1}^{2}|+->+ S_{2}^{2}|+->+2S_{1z}S_{2z}|+-> +S_{1+}S_{2-}|+->+S_{1-}S_{2+}|+->\\ \\&=&\frac{3}{4}\hbar^{2}|+->+\frac{3}{4}\hbar^{2}|+->-\frac{1}{2}|+->+\hbar^{2}|-+>\\ \\&=&\hbar^{2}(|+->+|-+>)\end{array}

 

\begin{array}{lll} S^{2}|-+>&=& S_{1}^{2}|-+>+ S_{2}^{2}|-+>+2S_{1z}S_{2z}|-+> +S_{1+}S_{2-}|-+>+S_{1-}S_{2+}|-+>\\ \\&=&\frac{3}{4}\hbar^{2}|-+>+\frac{3}{4}\hbar^{2}|-+>-\frac{1}{2}|-+>+\hbar^{2}|+->\\ \\&=&\hbar^{2}(|-+>+|+->)\end{array}

 

\begin{array}{lll} S^{2}|->&=& S_{1}^{2}|- ->+ S_{2}^{2}|- ->+2S_{1z}S_{2z}|- -> +S_{1+}S_{2-}|- ->+S_{1-}S_{2+}|- ->\\ \\&=&\frac{3}{4}\hbar^{2}|- ->+\frac{3}{4}\hbar^{2}|- ->+2(\frac{1}{2}\hbar)(\frac{1}{2}\hbar)|- ->\\ \\&=&2\hbar^{2}|- ->\end{array}

 

Therefore, the matrix representation will give us

S^{2} =\hbar^{2}\left( \begin{array}{cccc} 2& 0&0&0 \\ 0 &1&1&0 \\0&1&1&0\\0&0&0&2 \end{array} \right)

 

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