Chapter 4: Q28P (page 177)

For the most general normalized spinor (Equation 4.139),
compute ${S_{x}}, {S_{y}}, {S_{z}}, {S_{x}^{2}}, {S_{y}^{2}}, and {S_{x}^{2}} . check that {S_{x}^{2}} + {S_{y}^{2}} + {S_{z}^{2}} = {S^{2}} .$
$X = (\begin{matrix} a \\ b \end{matrix}) = {a_{X}}_{+} + b_{X} (4.139) .$

Short Answer

Expert verified

By solving we find the above value to be $<S_{x}^{2}> + <S_{y}^{2}> + <S_{z}^{2}> = \frac{h^{2}}{4}$ .

Step by step solution

Calculating

Computing $<S_{x}>, <S_{y}>, <S_{z}>, <S_{x}^{2}>, <S_{y}^{2}> and <S_{z}^{2}>$

$<S_{x}> = \frac{h}{2} (a^{*} b^{*}) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) {(-)}_{b}^{a} = \frac{h}{2} (a^{*} b^{*}) {(-)}_{b}^{a} \frac{h}{2} (a^{*} b + b^{* a}) = h R e (a b^{*}) .$

$<S_{y}> = \frac{h}{2} (a^{*} b^{*}) (\begin{matrix} 0 & - i \\ i & 0 \end{matrix}) (\begin{matrix} a \\ b \end{matrix}) = \frac{h}{2} (a^{*} b^{*}) (\begin{matrix} - i b \\ i a \end{matrix})$

$= \frac{h}{2} (- i a^{*} b + i a b^{*}) = \frac{h}{2} i (a b^{*} - a^{*} b) = - h l m (a b^{*}) .$

$<S_{z}> = \frac{h}{2} (a^{*} b^{*}) (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) (\begin{matrix} a \\ b \end{matrix}) = \frac{h}{2} (a^{*} b^{*}) (\begin{matrix} a \\ - b \end{matrix}) = \frac{h}{2} (a^{*} a - b^{*} b) = \frac{h}{2} ({|a|}^{2} {|b|}^{2}) .$

$S_{x}^{2} = \frac{h^{2}}{4} (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) = \frac{h^{2}}{4} (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) = \frac{h^{2}}{4}$

$S_{y}^{2} = \frac{h^{2}}{4} (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) = \frac{h^{2}}{4}$

localid="1655968926137" $S_{z}^{2} = \frac{h^{2}}{4} (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) = \frac{h^{2}}{4} s o <S_{x}^{2}> = <S_{y}^{2}> = <S_{z}^{2}> = \frac{h^{2}}{4}$

Checking that<Sx2>+<Sy2>+<Sz2>+<S2>.

$<S_{x}^{2}> + <S_{y}^{2}> + <S_{z}^{2}> = \frac{3}{4} h^{2} \overset{?}{= s (s + 1) h^{2} = \frac{1}{2}} (\frac{1}{2} + 1) h^{2} = \frac{3}{4} h^{2} = <S^{2}>$

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For the most general normalized spinor (Equation 4.139),
compute ${S_{x}}, {S_{y}}, {S_{z}}, {S_{x}^{2}}, {S_{y}^{2}}, and {S_{x}^{2}} . check that {S_{x}^{2}} + {S_{y}^{2}} + {S_{z}^{2}} = {S^{2}} .$
$X = (\begin{matrix} a \\ b \end{matrix}) = {a_{X}}_{+} + b_{X} (4.139) .$

Short Answer

Step by step solution

Calculating

Checking that<Sx2>+<Sy2>+<Sz2>+<S2>.

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For the most general normalized spinor (Equation 4.139),compute{Sx},{Sy},{Sz},{Sx2},{Sy2},and{Sx2}.checkthat{Sx2}+{Sy2}+{Sz2}={S2}.X=(ab)=aX++bX(4.139).

Short Answer

Step by step solution

Calculating

Checking that<Sx2>+<Sy2>+<Sz2>+<S2>.

One App. One Place for Learning.

Most popular questions from this chapter

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Electricity and Magnetism

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Study anywhere. Anytime. Across all devices.

For the most general normalized spinor (Equation 4.139),
compute ${S_{x}}, {S_{y}}, {S_{z}}, {S_{x}^{2}}, {S_{y}^{2}}, and {S_{x}^{2}} . check that {S_{x}^{2}} + {S_{y}^{2}} + {S_{z}^{2}} = {S^{2}} .$
$X = (\begin{matrix} a \\ b \end{matrix}) = {a_{X}}_{+} + b_{X} (4.139) .$