Q28P For the most general normalized ... [FREE SOLUTION]

91影视

Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

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>.

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

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

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>.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Scientific Method Physics

Magnetism and Electromagnetic Induction

Quantum Physics

Space Physics

Fundamentals of Physics

Rotational Dynamics

Study anywhere. Anytime. Across all devices.

91影视

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

Scientific Method Physics

Magnetism and Electromagnetic Induction

Quantum Physics

Space Physics

Fundamentals of Physics

Rotational Dynamics

Study anywhere. Anytime. Across all devices.

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