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

Recommended explanations on Physics Textbooks

Modern Physics

Electromagnetism

Astrophysics

Circular Motion and Gravitation

Thermodynamics

Further Mechanics and Thermal Physics

Study anywhere. Anytime. Across all devices.

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