today’s outline - october 30, 2013 - iitphys.iit.edu/~segre/phys405/13f/lecture_18.pdf · 2013....
TRANSCRIPT
![Page 1: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/1.jpg)
Today’s Outline - October 30, 2013
• Problems 4.29 and 4.52
• Electrons in a magnetic field
• Adding angular momentum
• Total spin of hydrogen
• Adding angular momentum
• Problem 4.55
Midterm Exam 2: Monday, November 11, 2013 covers through Chapter4.4.2 (at least!)
Homework Assignment #09:Chapter 4: 20,23,27,31,43,55due Wednesday, November 6, 2013
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17
![Page 2: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/2.jpg)
Today’s Outline - October 30, 2013
• Problems 4.29 and 4.52
• Electrons in a magnetic field
• Adding angular momentum
• Total spin of hydrogen
• Adding angular momentum
• Problem 4.55
Midterm Exam 2: Monday, November 11, 2013 covers through Chapter4.4.2 (at least!)
Homework Assignment #09:Chapter 4: 20,23,27,31,43,55due Wednesday, November 6, 2013
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17
![Page 3: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/3.jpg)
Today’s Outline - October 30, 2013
• Problems 4.29 and 4.52
• Electrons in a magnetic field
• Adding angular momentum
• Total spin of hydrogen
• Adding angular momentum
• Problem 4.55
Midterm Exam 2: Monday, November 11, 2013 covers through Chapter4.4.2 (at least!)
Homework Assignment #09:Chapter 4: 20,23,27,31,43,55due Wednesday, November 6, 2013
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17
![Page 4: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/4.jpg)
Today’s Outline - October 30, 2013
• Problems 4.29 and 4.52
• Electrons in a magnetic field
• Adding angular momentum
• Total spin of hydrogen
• Adding angular momentum
• Problem 4.55
Midterm Exam 2: Monday, November 11, 2013 covers through Chapter4.4.2 (at least!)
Homework Assignment #09:Chapter 4: 20,23,27,31,43,55due Wednesday, November 6, 2013
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17
![Page 5: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/5.jpg)
Today’s Outline - October 30, 2013
• Problems 4.29 and 4.52
• Electrons in a magnetic field
• Adding angular momentum
• Total spin of hydrogen
• Adding angular momentum
• Problem 4.55
Midterm Exam 2: Monday, November 11, 2013 covers through Chapter4.4.2 (at least!)
Homework Assignment #09:Chapter 4: 20,23,27,31,43,55due Wednesday, November 6, 2013
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17
![Page 6: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/6.jpg)
Today’s Outline - October 30, 2013
• Problems 4.29 and 4.52
• Electrons in a magnetic field
• Adding angular momentum
• Total spin of hydrogen
• Adding angular momentum
• Problem 4.55
Midterm Exam 2: Monday, November 11, 2013 covers through Chapter4.4.2 (at least!)
Homework Assignment #09:Chapter 4: 20,23,27,31,43,55due Wednesday, November 6, 2013
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17
![Page 7: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/7.jpg)
Today’s Outline - October 30, 2013
• Problems 4.29 and 4.52
• Electrons in a magnetic field
• Adding angular momentum
• Total spin of hydrogen
• Adding angular momentum
• Problem 4.55
Midterm Exam 2: Monday, November 11, 2013 covers through Chapter4.4.2 (at least!)
Homework Assignment #09:Chapter 4: 20,23,27,31,43,55due Wednesday, November 6, 2013
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17
![Page 8: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/8.jpg)
Today’s Outline - October 30, 2013
• Problems 4.29 and 4.52
• Electrons in a magnetic field
• Adding angular momentum
• Total spin of hydrogen
• Adding angular momentum
• Problem 4.55
Midterm Exam 2: Monday, November 11, 2013 covers through Chapter4.4.2 (at least!)
Homework Assignment #09:Chapter 4: 20,23,27,31,43,55due Wednesday, November 6, 2013
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17
![Page 9: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/9.jpg)
Problem 4.29
(a) Find the eigenvalues and eigenspinors of Sy
(b) If you measured Sy on a particle in the general state
χ = aχ+ + bχ−
what values might you get, and what is theprobability of each? Check that the probabilitiesadd up to 1.
(c) If you measured S2y , what values might you get,
and with what probabilities?
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 2 / 17
![Page 10: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/10.jpg)
Problem 4.29a
Starting with the eigenvalue equation
Syχ =~2
(0 −ii 0
)= λχ
the eigenvectors are obtained by as-suming
χ =
(αβ
)0 = det
∣∣∣∣ −λ −i~/2i~/2 −λ
∣∣∣∣0 = λ2 −
(~2
)2
λ = ±~2
~2
(0 −ii 0
)(αβ
)= ±~
2
(αβ
)→
(−iβiα
)= ±
(αβ
)
β = ±iα → χ(y)+ =
1√2
(1i
), χ
(y)− =
1√2
(1−i
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 3 / 17
![Page 11: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/11.jpg)
Problem 4.29a
Starting with the eigenvalue equation
Syχ =~2
(0 −ii 0
)= λχ
the eigenvectors are obtained by as-suming
χ =
(αβ
)0 = det
∣∣∣∣ −λ −i~/2i~/2 −λ
∣∣∣∣0 = λ2 −
(~2
)2
λ = ±~2
~2
(0 −ii 0
)(αβ
)= ±~
2
(αβ
)→
(−iβiα
)= ±
(αβ
)
β = ±iα → χ(y)+ =
1√2
(1i
), χ
(y)− =
1√2
(1−i
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 3 / 17
![Page 12: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/12.jpg)
Problem 4.29a
Starting with the eigenvalue equation
Syχ =~2
(0 −ii 0
)= λχ
the eigenvectors are obtained by as-suming
χ =
(αβ
)0 = det
∣∣∣∣ −λ −i~/2i~/2 −λ
∣∣∣∣0 = λ2 −
(~2
)2
λ = ±~2
~2
(0 −ii 0
)(αβ
)= ±~
2
(αβ
)→
(−iβiα
)= ±
(αβ
)
β = ±iα → χ(y)+ =
1√2
(1i
), χ
(y)− =
1√2
(1−i
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 3 / 17
![Page 13: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/13.jpg)
Problem 4.29a
Starting with the eigenvalue equation
Syχ =~2
(0 −ii 0
)= λχ
the eigenvectors are obtained by as-suming
χ =
(αβ
)
0 = det
∣∣∣∣ −λ −i~/2i~/2 −λ
∣∣∣∣
0 = λ2 −(~2
)2
λ = ±~2
~2
(0 −ii 0
)(αβ
)= ±~
2
(αβ
)→
(−iβiα
)= ±
(αβ
)
β = ±iα → χ(y)+ =
1√2
(1i
), χ
(y)− =
1√2
(1−i
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 3 / 17
![Page 14: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/14.jpg)
Problem 4.29a
Starting with the eigenvalue equation
Syχ =~2
(0 −ii 0
)= λχ
the eigenvectors are obtained by as-suming
χ =
(αβ
)
0 = det
∣∣∣∣ −λ −i~/2i~/2 −λ
∣∣∣∣0 = λ2 −
(~2
)2
λ = ±~2
~2
(0 −ii 0
)(αβ
)= ±~
2
(αβ
)→
(−iβiα
)= ±
(αβ
)
β = ±iα → χ(y)+ =
1√2
(1i
), χ
(y)− =
1√2
(1−i
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 3 / 17
![Page 15: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/15.jpg)
Problem 4.29a
Starting with the eigenvalue equation
Syχ =~2
(0 −ii 0
)= λχ
the eigenvectors are obtained by as-suming
χ =
(αβ
)
0 = det
∣∣∣∣ −λ −i~/2i~/2 −λ
∣∣∣∣0 = λ2 −
(~2
)2
λ = ±~2
~2
(0 −ii 0
)(αβ
)= ±~
2
(αβ
)→
(−iβiα
)= ±
(αβ
)
β = ±iα → χ(y)+ =
1√2
(1i
), χ
(y)− =
1√2
(1−i
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 3 / 17
![Page 16: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/16.jpg)
Problem 4.29a
Starting with the eigenvalue equation
Syχ =~2
(0 −ii 0
)= λχ
the eigenvectors are obtained by as-suming
χ =
(αβ
)0 = det
∣∣∣∣ −λ −i~/2i~/2 −λ
∣∣∣∣0 = λ2 −
(~2
)2
λ = ±~2
~2
(0 −ii 0
)(αβ
)= ±~
2
(αβ
)→
(−iβiα
)= ±
(αβ
)
β = ±iα → χ(y)+ =
1√2
(1i
), χ
(y)− =
1√2
(1−i
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 3 / 17
![Page 17: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/17.jpg)
Problem 4.29a
Starting with the eigenvalue equation
Syχ =~2
(0 −ii 0
)= λχ
the eigenvectors are obtained by as-suming
χ =
(αβ
)0 = det
∣∣∣∣ −λ −i~/2i~/2 −λ
∣∣∣∣0 = λ2 −
(~2
)2
λ = ±~2
~2
(0 −ii 0
)(αβ
)= ±~
2
(αβ
)
→(−iβiα
)= ±
(αβ
)
β = ±iα → χ(y)+ =
1√2
(1i
), χ
(y)− =
1√2
(1−i
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 3 / 17
![Page 18: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/18.jpg)
Problem 4.29a
Starting with the eigenvalue equation
Syχ =~2
(0 −ii 0
)= λχ
the eigenvectors are obtained by as-suming
χ =
(αβ
)0 = det
∣∣∣∣ −λ −i~/2i~/2 −λ
∣∣∣∣0 = λ2 −
(~2
)2
λ = ±~2
~2
(0 −ii 0
)(αβ
)= ±~
2
(αβ
)→
(−iβiα
)= ±
(αβ
)
β = ±iα → χ(y)+ =
1√2
(1i
), χ
(y)− =
1√2
(1−i
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 3 / 17
![Page 19: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/19.jpg)
Problem 4.29a
Starting with the eigenvalue equation
Syχ =~2
(0 −ii 0
)= λχ
the eigenvectors are obtained by as-suming
χ =
(αβ
)0 = det
∣∣∣∣ −λ −i~/2i~/2 −λ
∣∣∣∣0 = λ2 −
(~2
)2
λ = ±~2
~2
(0 −ii 0
)(αβ
)= ±~
2
(αβ
)→
(−iβiα
)= ±
(αβ
)
β = ±iα
→ χ(y)+ =
1√2
(1i
), χ
(y)− =
1√2
(1−i
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 3 / 17
![Page 20: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/20.jpg)
Problem 4.29a
Starting with the eigenvalue equation
Syχ =~2
(0 −ii 0
)= λχ
the eigenvectors are obtained by as-suming
χ =
(αβ
)0 = det
∣∣∣∣ −λ −i~/2i~/2 −λ
∣∣∣∣0 = λ2 −
(~2
)2
λ = ±~2
~2
(0 −ii 0
)(αβ
)= ±~
2
(αβ
)→
(−iβiα
)= ±
(αβ
)
β = ±iα → χ(y)+ =
1√2
(1i
)
, χ(y)− =
1√2
(1−i
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 3 / 17
![Page 21: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/21.jpg)
Problem 4.29a
Starting with the eigenvalue equation
Syχ =~2
(0 −ii 0
)= λχ
the eigenvectors are obtained by as-suming
χ =
(αβ
)0 = det
∣∣∣∣ −λ −i~/2i~/2 −λ
∣∣∣∣0 = λ2 −
(~2
)2
λ = ±~2
~2
(0 −ii 0
)(αβ
)= ±~
2
(αβ
)→
(−iβiα
)= ±
(αβ
)
β = ±iα → χ(y)+ =
1√2
(1i
), χ
(y)− =
1√2
(1−i
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 3 / 17
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Problem 4.29b
The general spinor can be written in terms of the Sy eigenspinors as wellas the usual ones
χ = aχ+ + bχ− = c+χ(y)+ + c−χ
(y)−
By measuring Sy , the possible answers are the magnitudes of thesecoefficients squared. The coefficients may be extracted using the Fouriertrick
c+ = χ(y)†+ χ
=1√2
( 1 −i )
(ab
)=
1√2
(a− ib)
c− = χ(y)†− χ
=1√2
( 1 i )
(ab
)=
1√2
(a + ib)
Thus we will measure the eigenvalues with probabilites as follows, whichsum to 1 ir χ is normalized.
+~2, P+ =
1
2|a− ib|2 − ~
2, P− =
1
2|a + ib|2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 4 / 17
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Problem 4.29b
The general spinor can be written in terms of the Sy eigenspinors as wellas the usual ones
χ = aχ+ + bχ− = c+χ(y)+ + c−χ
(y)−
By measuring Sy , the possible answers are the magnitudes of thesecoefficients squared. The coefficients may be extracted using the Fouriertrick
c+ = χ(y)†+ χ
=1√2
( 1 −i )
(ab
)=
1√2
(a− ib)
c− = χ(y)†− χ
=1√2
( 1 i )
(ab
)=
1√2
(a + ib)
Thus we will measure the eigenvalues with probabilites as follows, whichsum to 1 ir χ is normalized.
+~2, P+ =
1
2|a− ib|2 − ~
2, P− =
1
2|a + ib|2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 4 / 17
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Problem 4.29b
The general spinor can be written in terms of the Sy eigenspinors as wellas the usual ones
χ = aχ+ + bχ− = c+χ(y)+ + c−χ
(y)−
By measuring Sy , the possible answers are the magnitudes of thesecoefficients squared. The coefficients may be extracted using the Fouriertrick
c+ = χ(y)†+ χ
=1√2
( 1 −i )
(ab
)=
1√2
(a− ib)
c− = χ(y)†− χ
=1√2
( 1 i )
(ab
)=
1√2
(a + ib)
Thus we will measure the eigenvalues with probabilites as follows, whichsum to 1 ir χ is normalized.
+~2, P+ =
1
2|a− ib|2 − ~
2, P− =
1
2|a + ib|2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 4 / 17
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Problem 4.29b
The general spinor can be written in terms of the Sy eigenspinors as wellas the usual ones
χ = aχ+ + bχ− = c+χ(y)+ + c−χ
(y)−
By measuring Sy , the possible answers are the magnitudes of thesecoefficients squared. The coefficients may be extracted using the Fouriertrick
c+ = χ(y)†+ χ
=1√2
( 1 −i )
(ab
)=
1√2
(a− ib)
c− = χ(y)†− χ
=1√2
( 1 i )
(ab
)=
1√2
(a + ib)
Thus we will measure the eigenvalues with probabilites as follows, whichsum to 1 ir χ is normalized.
+~2, P+ =
1
2|a− ib|2 − ~
2, P− =
1
2|a + ib|2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 4 / 17
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Problem 4.29b
The general spinor can be written in terms of the Sy eigenspinors as wellas the usual ones
χ = aχ+ + bχ− = c+χ(y)+ + c−χ
(y)−
By measuring Sy , the possible answers are the magnitudes of thesecoefficients squared. The coefficients may be extracted using the Fouriertrick
c+ = χ(y)†+ χ =
1√2
( 1 −i )
(ab
)
=1√2
(a− ib)
c− = χ(y)†− χ
=1√2
( 1 i )
(ab
)=
1√2
(a + ib)
Thus we will measure the eigenvalues with probabilites as follows, whichsum to 1 ir χ is normalized.
+~2, P+ =
1
2|a− ib|2 − ~
2, P− =
1
2|a + ib|2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 4 / 17
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Problem 4.29b
The general spinor can be written in terms of the Sy eigenspinors as wellas the usual ones
χ = aχ+ + bχ− = c+χ(y)+ + c−χ
(y)−
By measuring Sy , the possible answers are the magnitudes of thesecoefficients squared. The coefficients may be extracted using the Fouriertrick
c+ = χ(y)†+ χ =
1√2
( 1 −i )
(ab
)=
1√2
(a− ib)
c− = χ(y)†− χ
=1√2
( 1 i )
(ab
)=
1√2
(a + ib)
Thus we will measure the eigenvalues with probabilites as follows, whichsum to 1 ir χ is normalized.
+~2, P+ =
1
2|a− ib|2 − ~
2, P− =
1
2|a + ib|2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 4 / 17
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Problem 4.29b
The general spinor can be written in terms of the Sy eigenspinors as wellas the usual ones
χ = aχ+ + bχ− = c+χ(y)+ + c−χ
(y)−
By measuring Sy , the possible answers are the magnitudes of thesecoefficients squared. The coefficients may be extracted using the Fouriertrick
c+ = χ(y)†+ χ =
1√2
( 1 −i )
(ab
)=
1√2
(a− ib)
c− = χ(y)†− χ
=1√2
( 1 i )
(ab
)=
1√2
(a + ib)
Thus we will measure the eigenvalues with probabilites as follows, whichsum to 1 ir χ is normalized.
+~2, P+ =
1
2|a− ib|2 − ~
2, P− =
1
2|a + ib|2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 4 / 17
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Problem 4.29b
The general spinor can be written in terms of the Sy eigenspinors as wellas the usual ones
χ = aχ+ + bχ− = c+χ(y)+ + c−χ
(y)−
By measuring Sy , the possible answers are the magnitudes of thesecoefficients squared. The coefficients may be extracted using the Fouriertrick
c+ = χ(y)†+ χ =
1√2
( 1 −i )
(ab
)=
1√2
(a− ib)
c− = χ(y)†− χ =
1√2
( 1 i )
(ab
)
=1√2
(a + ib)
Thus we will measure the eigenvalues with probabilites as follows, whichsum to 1 ir χ is normalized.
+~2, P+ =
1
2|a− ib|2 − ~
2, P− =
1
2|a + ib|2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 4 / 17
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Problem 4.29b
The general spinor can be written in terms of the Sy eigenspinors as wellas the usual ones
χ = aχ+ + bχ− = c+χ(y)+ + c−χ
(y)−
By measuring Sy , the possible answers are the magnitudes of thesecoefficients squared. The coefficients may be extracted using the Fouriertrick
c+ = χ(y)†+ χ =
1√2
( 1 −i )
(ab
)=
1√2
(a− ib)
c− = χ(y)†− χ =
1√2
( 1 i )
(ab
)=
1√2
(a + ib)
Thus we will measure the eigenvalues with probabilites as follows, whichsum to 1 ir χ is normalized.
+~2, P+ =
1
2|a− ib|2 − ~
2, P− =
1
2|a + ib|2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 4 / 17
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Problem 4.29b
The general spinor can be written in terms of the Sy eigenspinors as wellas the usual ones
χ = aχ+ + bχ− = c+χ(y)+ + c−χ
(y)−
By measuring Sy , the possible answers are the magnitudes of thesecoefficients squared. The coefficients may be extracted using the Fouriertrick
c+ = χ(y)†+ χ =
1√2
( 1 −i )
(ab
)=
1√2
(a− ib)
c− = χ(y)†− χ =
1√2
( 1 i )
(ab
)=
1√2
(a + ib)
Thus we will measure the eigenvalues with probabilites as follows, whichsum to 1 ir χ is normalized.
+~2, P+ =
1
2|a− ib|2 − ~
2, P− =
1
2|a + ib|2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 4 / 17
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Problem 4.29b
The general spinor can be written in terms of the Sy eigenspinors as wellas the usual ones
χ = aχ+ + bχ− = c+χ(y)+ + c−χ
(y)−
By measuring Sy , the possible answers are the magnitudes of thesecoefficients squared. The coefficients may be extracted using the Fouriertrick
c+ = χ(y)†+ χ =
1√2
( 1 −i )
(ab
)=
1√2
(a− ib)
c− = χ(y)†− χ =
1√2
( 1 i )
(ab
)=
1√2
(a + ib)
Thus we will measure the eigenvalues with probabilites as follows, whichsum to 1 ir χ is normalized.
+~2, P+ =
1
2|a− ib|2
− ~2, P− =
1
2|a + ib|2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 4 / 17
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Problem 4.29b
The general spinor can be written in terms of the Sy eigenspinors as wellas the usual ones
χ = aχ+ + bχ− = c+χ(y)+ + c−χ
(y)−
By measuring Sy , the possible answers are the magnitudes of thesecoefficients squared. The coefficients may be extracted using the Fouriertrick
c+ = χ(y)†+ χ =
1√2
( 1 −i )
(ab
)=
1√2
(a− ib)
c− = χ(y)†− χ =
1√2
( 1 i )
(ab
)=
1√2
(a + ib)
Thus we will measure the eigenvalues with probabilites as follows, whichsum to 1 ir χ is normalized.
+~2, P+ =
1
2|a− ib|2 − ~
2, P− =
1
2|a + ib|2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 4 / 17
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Problem 4.29c
Measuring S2y is simply a question of applying the
same operator twice, thus pulling out the same eigen-value twice.
χ†SySyχ = χ†Sy (+~2c+χ
(y)+ −
~2c−χ
(y)− )
= χ†(~2
4c+χ
(y)+ +
~2
4c−χ
(y)− )
= χ†~2
4χ =
~2
4
with a probability of 1
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 5 / 17
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Problem 4.29c
Measuring S2y is simply a question of applying the
same operator twice, thus pulling out the same eigen-value twice.
χ†SySyχ = χ†Sy (+~2c+χ
(y)+ −
~2c−χ
(y)− )
= χ†(~2
4c+χ
(y)+ +
~2
4c−χ
(y)− )
= χ†~2
4χ =
~2
4
with a probability of 1
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 5 / 17
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Problem 4.29c
Measuring S2y is simply a question of applying the
same operator twice, thus pulling out the same eigen-value twice.
χ†SySyχ = χ†Sy (+~2c+χ
(y)+ −
~2c−χ
(y)− )
= χ†(~2
4c+χ
(y)+ +
~2
4c−χ
(y)− )
= χ†~2
4χ =
~2
4
with a probability of 1
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 5 / 17
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Problem 4.29c
Measuring S2y is simply a question of applying the
same operator twice, thus pulling out the same eigen-value twice.
χ†SySyχ = χ†Sy (+~2c+χ
(y)+ −
~2c−χ
(y)− )
= χ†(~2
4c+χ
(y)+ +
~2
4c−χ
(y)− )
= χ†~2
4χ
=~2
4
with a probability of 1
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 5 / 17
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Problem 4.29c
Measuring S2y is simply a question of applying the
same operator twice, thus pulling out the same eigen-value twice.
χ†SySyχ = χ†Sy (+~2c+χ
(y)+ −
~2c−χ
(y)− )
= χ†(~2
4c+χ
(y)+ +
~2
4c−χ
(y)− )
= χ†~2
4χ =
~2
4
with a probability of 1
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 5 / 17
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Problem 4.29c
Measuring S2y is simply a question of applying the
same operator twice, thus pulling out the same eigen-value twice.
χ†SySyχ = χ†Sy (+~2c+χ
(y)+ −
~2c−χ
(y)− )
= χ†(~2
4c+χ
(y)+ +
~2
4c−χ
(y)− )
= χ†~2
4χ =
~2
4
with a probability of 1
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 5 / 17
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Problem 4.52
Find the matrix representing Sx for a particle of spin 3/2 (using, as always,the basis of eigenstates of Sz . Solve the characteristic equation todetermine the eigenvalues of Sx .
First we write down the eigenstates of Sz in the S = 3/2 system.
| 32
32〉 =
1000
| 32
12〉 =
0100
| 32− 1
2〉 =
0010
| 32− 3
2〉 =
0001
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 6 / 17
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Problem 4.52
Find the matrix representing Sx for a particle of spin 3/2 (using, as always,the basis of eigenstates of Sz . Solve the characteristic equation todetermine the eigenvalues of Sx .
First we write down the eigenstates of Sz in the S = 3/2 system.
| 32
32〉 =
1000
| 32
12〉 =
0100
| 32− 1
2〉 =
0010
| 32− 3
2〉 =
0001
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 6 / 17
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Problem 4.52
Find the matrix representing Sx for a particle of spin 3/2 (using, as always,the basis of eigenstates of Sz . Solve the characteristic equation todetermine the eigenvalues of Sx .
First we write down the eigenstates of Sz in the S = 3/2 system.
| 32
32〉 =
1000
| 32
12〉 =
0100
| 32− 1
2〉 =
0010
| 32− 3
2〉 =
0001
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 6 / 17
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Problem 4.52
Find the matrix representing Sx for a particle of spin 3/2 (using, as always,the basis of eigenstates of Sz . Solve the characteristic equation todetermine the eigenvalues of Sx .
First we write down the eigenstates of Sz in the S = 3/2 system.
| 32
32〉 =
1000
| 32
12〉 =
0100
| 32− 1
2〉 =
0010
| 32− 3
2〉 =
0001
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 6 / 17
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Problem 4.52
Find the matrix representing Sx for a particle of spin 3/2 (using, as always,the basis of eigenstates of Sz . Solve the characteristic equation todetermine the eigenvalues of Sx .
First we write down the eigenstates of Sz in the S = 3/2 system.
| 32
32〉 =
1000
| 32
12〉 =
0100
| 32− 1
2〉 =
0010
| 32− 3
2〉 =
0001
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 6 / 17
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Problem 4.52
Find the matrix representing Sx for a particle of spin 3/2 (using, as always,the basis of eigenstates of Sz . Solve the characteristic equation todetermine the eigenvalues of Sx .
First we write down the eigenstates of Sz in the S = 3/2 system.
| 32
32〉 =
1000
| 32
12〉 =
0100
| 32− 1
2〉 =
0010
| 32− 3
2〉 =
0001
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 6 / 17
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Problem 4.52 (cont.)
The raising and lowering operators are
S±|sm〉 = ~√
s(s + 1)−m(m ± 1) |s(m ± 1)〉
we build the matrix for S+ by applying the raising operator to each of theeigenstates
S+ = ~
0√
3 0 00 0 2 0
0 0 0√
30 0 0 0
S+| 32 3
2〉 = 0
S+| 32 12〉 =
√32( 5
2)− 12( 3
2)~ | 32 32〉 =√
3~ | 32
32〉
S+| 32 − 12〉 =
√32( 5
2)+ 12( 1
2)~ | 32 12〉 = 2~ | 3
212〉
S+| 32 − 32〉 =
√32( 5
2)+ 32(− 1
2)~ | 32 − 12〉 =√
3~ | 32− 1
2〉
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 7 / 17
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Problem 4.52 (cont.)
The raising and lowering operators are
S±|sm〉 = ~√
s(s + 1)−m(m ± 1) |s(m ± 1)〉
we build the matrix for S+ by applying the raising operator to each of theeigenstates
S+ = ~
0√
3 0 00 0 2 0
0 0 0√
30 0 0 0
S+| 32 3
2〉 = 0
S+| 32 12〉 =
√32( 5
2)− 12( 3
2)~ | 32 32〉 =√
3~ | 32
32〉
S+| 32 − 12〉 =
√32( 5
2)+ 12( 1
2)~ | 32 12〉 = 2~ | 3
212〉
S+| 32 − 32〉 =
√32( 5
2)+ 32(− 1
2)~ | 32 − 12〉 =√
3~ | 32− 1
2〉
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 7 / 17
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Problem 4.52 (cont.)
The raising and lowering operators are
S±|sm〉 = ~√
s(s + 1)−m(m ± 1) |s(m ± 1)〉
we build the matrix for S+ by applying the raising operator to each of theeigenstates
S+ = ~
0√
3 0 00 0 2 0
0 0 0√
30 0 0 0
S+| 32 32〉 = 0
S+| 32 12〉 =
√32( 5
2)− 12( 3
2)~ | 32 32〉 =√
3~ | 32
32〉
S+| 32 − 12〉 =
√32( 5
2)+ 12( 1
2)~ | 32 12〉 = 2~ | 3
212〉
S+| 32 − 32〉 =
√32( 5
2)+ 32(− 1
2)~ | 32 − 12〉 =√
3~ | 32− 1
2〉
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 7 / 17
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Problem 4.52 (cont.)
The raising and lowering operators are
S±|sm〉 = ~√
s(s + 1)−m(m ± 1) |s(m ± 1)〉
we build the matrix for S+ by applying the raising operator to each of theeigenstates
S+ = ~
0√
3 0 00 0 2 0
0 0 0√
30 0 0 0
S+| 32 3
2〉 = 0
S+| 32 12〉 =
√32( 5
2)− 12( 3
2)~ | 32 32〉 =√
3~ | 32
32〉
S+| 32 − 12〉 =
√32( 5
2)+ 12( 1
2)~ | 32 12〉 = 2~ | 3
212〉
S+| 32 − 32〉 =
√32( 5
2)+ 32(− 1
2)~ | 32 − 12〉 =√
3~ | 32− 1
2〉
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 7 / 17
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Problem 4.52 (cont.)
The raising and lowering operators are
S±|sm〉 = ~√
s(s + 1)−m(m ± 1) |s(m ± 1)〉
we build the matrix for S+ by applying the raising operator to each of theeigenstates
S+ = ~
0
√3 0 0
0
0 2 0
0
0 0√
3
0
0 0 0
S+| 32 3
2〉 = 0
S+| 32 12〉 =
√32( 5
2)− 12( 3
2)~ | 32 32〉 =√
3~ | 32
32〉
S+| 32 − 12〉 =
√32( 5
2)+ 12( 1
2)~ | 32 12〉 = 2~ | 3
212〉
S+| 32 − 32〉 =
√32( 5
2)+ 32(− 1
2)~ | 32 − 12〉 =√
3~ | 32− 1
2〉
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Problem 4.52 (cont.)
The raising and lowering operators are
S±|sm〉 = ~√
s(s + 1)−m(m ± 1) |s(m ± 1)〉
we build the matrix for S+ by applying the raising operator to each of theeigenstates
S+ = ~
0
√3 0 0
0
0 2 0
0
0 0√
3
0
0 0 0
S+| 32 3
2〉 = 0
S+| 32 12〉 =
√32( 5
2)− 12( 3
2)~ | 32 32〉
=√
3~ | 32
32〉
S+| 32 − 12〉 =
√32( 5
2)+ 12( 1
2)~ | 32 12〉 = 2~ | 3
212〉
S+| 32 − 32〉 =
√32( 5
2)+ 32(− 1
2)~ | 32 − 12〉 =√
3~ | 32− 1
2〉
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Problem 4.52 (cont.)
The raising and lowering operators are
S±|sm〉 = ~√
s(s + 1)−m(m ± 1) |s(m ± 1)〉
we build the matrix for S+ by applying the raising operator to each of theeigenstates
S+ = ~
0
√3 0 0
0
0 2 0
0
0 0√
3
0
0 0 0
S+| 32 3
2〉 = 0
S+| 32 12〉 =
√32( 5
2)− 12( 3
2)~ | 32 32〉 =√
3~ | 32
32〉
S+| 32 − 12〉 =
√32( 5
2)+ 12( 1
2)~ | 32 12〉 = 2~ | 3
212〉
S+| 32 − 32〉 =
√32( 5
2)+ 32(− 1
2)~ | 32 − 12〉 =√
3~ | 32− 1
2〉
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Problem 4.52 (cont.)
The raising and lowering operators are
S±|sm〉 = ~√
s(s + 1)−m(m ± 1) |s(m ± 1)〉
we build the matrix for S+ by applying the raising operator to each of theeigenstates
S+ = ~
0√
3
0 0
0 0
2 0
0 0
0√
3
0 0
0 0
S+| 32 3
2〉 = 0
S+| 32 12〉 =
√32( 5
2)− 12( 3
2)~ | 32 32〉 =√
3~ | 32
32〉
S+| 32 − 12〉 =
√32( 5
2)+ 12( 1
2)~ | 32 12〉 = 2~ | 3
212〉
S+| 32 − 32〉 =
√32( 5
2)+ 32(− 1
2)~ | 32 − 12〉 =√
3~ | 32− 1
2〉
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Problem 4.52 (cont.)
The raising and lowering operators are
S±|sm〉 = ~√
s(s + 1)−m(m ± 1) |s(m ± 1)〉
we build the matrix for S+ by applying the raising operator to each of theeigenstates
S+ = ~
0√
3
0 0
0 0
2 0
0 0
0√
3
0 0
0 0
S+| 32 3
2〉 = 0
S+| 32 12〉 =
√32( 5
2)− 12( 3
2)~ | 32 32〉 =√
3~ | 32
32〉
S+| 32 − 12〉 =
√32( 5
2)+ 12( 1
2)~ | 32 12〉
= 2~ | 32
12〉
S+| 32 − 32〉 =
√32( 5
2)+ 32(− 1
2)~ | 32 − 12〉 =√
3~ | 32− 1
2〉
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Problem 4.52 (cont.)
The raising and lowering operators are
S±|sm〉 = ~√
s(s + 1)−m(m ± 1) |s(m ± 1)〉
we build the matrix for S+ by applying the raising operator to each of theeigenstates
S+ = ~
0√
3
0 0
0 0
2 0
0 0
0√
3
0 0
0 0
S+| 32 3
2〉 = 0
S+| 32 12〉 =
√32( 5
2)− 12( 3
2)~ | 32 32〉 =√
3~ | 32
32〉
S+| 32 − 12〉 =
√32( 5
2)+ 12( 1
2)~ | 32 12〉 = 2~ | 3
212〉
S+| 32 − 32〉 =
√32( 5
2)+ 32(− 1
2)~ | 32 − 12〉 =√
3~ | 32− 1
2〉
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Problem 4.52 (cont.)
The raising and lowering operators are
S±|sm〉 = ~√
s(s + 1)−m(m ± 1) |s(m ± 1)〉
we build the matrix for S+ by applying the raising operator to each of theeigenstates
S+ = ~
0√
3 0
0
0 0 2
0
0 0 0
√3
0 0 0
0
S+| 32 3
2〉 = 0
S+| 32 12〉 =
√32( 5
2)− 12( 3
2)~ | 32 32〉 =√
3~ | 32
32〉
S+| 32 − 12〉 =
√32( 5
2)+ 12( 1
2)~ | 32 12〉 = 2~ | 3
212〉
S+| 32 − 32〉 =
√32( 5
2)+ 32(− 1
2)~ | 32 − 12〉 =√
3~ | 32− 1
2〉
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Problem 4.52 (cont.)
The raising and lowering operators are
S±|sm〉 = ~√
s(s + 1)−m(m ± 1) |s(m ± 1)〉
we build the matrix for S+ by applying the raising operator to each of theeigenstates
S+ = ~
0√
3 0
0
0 0 2
0
0 0 0
√3
0 0 0
0
S+| 32 3
2〉 = 0
S+| 32 12〉 =
√32( 5
2)− 12( 3
2)~ | 32 32〉 =√
3~ | 32
32〉
S+| 32 − 12〉 =
√32( 5
2)+ 12( 1
2)~ | 32 12〉 = 2~ | 3
212〉
S+| 32 − 32〉 =
√32( 5
2)+ 32(− 1
2)~ | 32 − 12〉
=√
3~ | 32− 1
2〉
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Problem 4.52 (cont.)
The raising and lowering operators are
S±|sm〉 = ~√
s(s + 1)−m(m ± 1) |s(m ± 1)〉
we build the matrix for S+ by applying the raising operator to each of theeigenstates
S+ = ~
0√
3 0
0
0 0 2
0
0 0 0
√3
0 0 0
0
S+| 32 3
2〉 = 0
S+| 32 12〉 =
√32( 5
2)− 12( 3
2)~ | 32 32〉 =√
3~ | 32
32〉
S+| 32 − 12〉 =
√32( 5
2)+ 12( 1
2)~ | 32 12〉 = 2~ | 3
212〉
S+| 32 − 32〉 =
√32( 5
2)+ 32(− 1
2)~ | 32 − 12〉 =√
3~ | 32− 1
2〉
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Problem 4.52 (cont.)
The raising and lowering operators are
S±|sm〉 = ~√
s(s + 1)−m(m ± 1) |s(m ± 1)〉
we build the matrix for S+ by applying the raising operator to each of theeigenstates
S+ = ~
0√
3 0 00 0 2 0
0 0 0√
30 0 0 0
S+| 32 3
2〉 = 0
S+| 32 12〉 =
√32( 5
2)− 12( 3
2)~ | 32 32〉 =√
3~ | 32
32〉
S+| 32 − 12〉 =
√32( 5
2)+ 12( 1
2)~ | 32 12〉 = 2~ | 3
212〉
S+| 32 − 32〉 =
√32( 5
2)+ 32(− 1
2)~ | 32 − 12〉 =√
3~ | 32− 1
2〉
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Problem 4.52 (cont.)
Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates
S− = ~
0 0 0 0√3 0 0 0
0 2 0 0
0 0√
3 0
S−| 32 32〉 =
√32( 5
2)− 32( 1
2)~ | 32 12〉 =√
3~ | 32
12〉
S−| 32 12〉 =
√32( 5
2)− 12(− 1
2)~ | 32 −12〉 = 2~ | 3
2− 1
2〉
S−| 32 − 12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32− 3
2〉
S−| 32 − 32〉 = 0
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Problem 4.52 (cont.)
Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates
S− = ~
0 0 0 0√3 0 0 0
0 2 0 0
0 0√
3 0
S−| 32 32〉 =
√32( 5
2)− 32( 1
2)~ | 32 12〉 =√
3~ | 32
12〉
S−| 32 12〉 =
√32( 5
2)− 12(− 1
2)~ | 32 −12〉 = 2~ | 3
2− 1
2〉
S−| 32 − 12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32− 3
2〉
S−| 32 − 32〉 = 0
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Problem 4.52 (cont.)
Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates
S− = ~
0 0 0 0√3 0 0 0
0 2 0 0
0 0√
3 0
S−| 32 32〉 =
√32( 5
2)− 32( 1
2)~ | 32 12〉
=√
3~ | 32
12〉
S−| 32 12〉 =
√32( 5
2)− 12(− 1
2)~ | 32 −12〉 = 2~ | 3
2− 1
2〉
S−| 32 − 12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32− 3
2〉
S−| 32 − 32〉 = 0
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Problem 4.52 (cont.)
Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates
S− = ~
0 0 0 0√3 0 0 0
0 2 0 0
0 0√
3 0
S−| 32 32〉 =
√32( 5
2)− 32( 1
2)~ | 32 12〉 =√
3~ | 32
12〉
S−| 32 12〉 =
√32( 5
2)− 12(− 1
2)~ | 32 −12〉 = 2~ | 3
2− 1
2〉
S−| 32 − 12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32− 3
2〉
S−| 32 − 32〉 = 0
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Problem 4.52 (cont.)
Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates
S− = ~
0
0 0 0
√3
0 0 0
0
2 0 0
0
0√
3 0
S−| 32 32〉 =
√32( 5
2)− 32( 1
2)~ | 32 12〉 =√
3~ | 32
12〉
S−| 32 12〉 =
√32( 5
2)− 12(− 1
2)~ | 32 −12〉 = 2~ | 3
2− 1
2〉
S−| 32 − 12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32− 3
2〉
S−| 32 − 32〉 = 0
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Problem 4.52 (cont.)
Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates
S− = ~
0
0 0 0
√3
0 0 0
0
2 0 0
0
0√
3 0
S−| 32 32〉 =
√32( 5
2)− 32( 1
2)~ | 32 12〉 =√
3~ | 32
12〉
S−| 32 12〉 =
√32( 5
2)− 12(− 1
2)~ | 32 −12〉
= 2~ | 32− 1
2〉
S−| 32 − 12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32− 3
2〉
S−| 32 − 32〉 = 0
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Problem 4.52 (cont.)
Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates
S− = ~
0
0 0 0
√3
0 0 0
0
2 0 0
0
0√
3 0
S−| 32 32〉 =
√32( 5
2)− 32( 1
2)~ | 32 12〉 =√
3~ | 32
12〉
S−| 32 12〉 =
√32( 5
2)− 12(− 1
2)~ | 32 −12〉 = 2~ | 3
2− 1
2〉
S−| 32 − 12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32− 3
2〉
S−| 32 − 32〉 = 0
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Problem 4.52 (cont.)
Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates
S− = ~
0 0
0 0
√3 0
0 0
0 2
0 0
0 0
√3 0
S−| 32 32〉 =
√32( 5
2)− 32( 1
2)~ | 32 12〉 =√
3~ | 32
12〉
S−| 32 12〉 =
√32( 5
2)− 12(− 1
2)~ | 32 −12〉 = 2~ | 3
2− 1
2〉
S−| 32 − 12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32− 3
2〉
S−| 32 − 32〉 = 0
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Problem 4.52 (cont.)
Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates
S− = ~
0 0
0 0
√3 0
0 0
0 2
0 0
0 0
√3 0
S−| 32 32〉 =
√32( 5
2)− 32( 1
2)~ | 32 12〉 =√
3~ | 32
12〉
S−| 32 12〉 =
√32( 5
2)− 12(− 1
2)~ | 32 −12〉 = 2~ | 3
2− 1
2〉
S−| 32 − 12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉
=√
3~ | 32− 3
2〉
S−| 32 − 32〉 = 0
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Problem 4.52 (cont.)
Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates
S− = ~
0 0
0 0
√3 0
0 0
0 2
0 0
0 0
√3 0
S−| 32 32〉 =
√32( 5
2)− 32( 1
2)~ | 32 12〉 =√
3~ | 32
12〉
S−| 32 12〉 =
√32( 5
2)− 12(− 1
2)~ | 32 −12〉 = 2~ | 3
2− 1
2〉
S−| 32 − 12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32− 3
2〉
S−| 32 − 32〉 = 0
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Problem 4.52 (cont.)
Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates
S− = ~
0 0 0
0
√3 0 0
0
0 2 0
0
0 0√
3
0
S−| 32 32〉 =
√32( 5
2)− 32( 1
2)~ | 32 12〉 =√
3~ | 32
12〉
S−| 32 12〉 =
√32( 5
2)− 12(− 1
2)~ | 32 −12〉 = 2~ | 3
2− 1
2〉
S−| 32 − 12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32− 3
2〉
S−| 32 − 32〉 = 0
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Problem 4.52 (cont.)
Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates
S− = ~
0 0 0
0
√3 0 0
0
0 2 0
0
0 0√
3
0
S−| 32 32〉 =
√32( 5
2)− 32( 1
2)~ | 32 12〉 =√
3~ | 32
12〉
S−| 32 12〉 =
√32( 5
2)− 12(− 1
2)~ | 32 −12〉 = 2~ | 3
2− 1
2〉
S−| 32 − 12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32− 3
2〉
S−| 32 − 32〉 = 0
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Problem 4.52 (cont.)
Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates
S− = ~
0 0 0 0√3 0 0 0
0 2 0 0
0 0√
3 0
S−| 32 32〉 =
√32( 5
2)− 32( 1
2)~ | 32 12〉 =√
3~ | 32
12〉
S−| 32 12〉 =
√32( 5
2)− 12(− 1
2)~ | 32 −12〉 = 2~ | 3
2− 1
2〉
S−| 32 − 12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32− 3
2〉
S−| 32 − 32〉 = 0
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Problem 4.52 (cont.)
Sx =1
2(S+ + S−)
=~2
0√
3 0 0√3 0 2 0
0 2 0√
3
0 0√
3 0
The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)
0 =
∣∣∣∣∣∣∣∣− λ
√3 0 0√
3 −λ 2 0
0 2 −λ√
3
0 0√
3 −λ
∣∣∣∣∣∣∣∣ = −λ
∣∣∣∣∣∣−λ 2 0
2 −λ√
3
0√
3 −λ
∣∣∣∣∣∣−√3
∣∣∣∣∣∣√
3 2 0
0 −λ√
3
0√
3 −λ
∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−
√3[√
3(λ2 − 3)]
= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9
0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)
the eigenvalues of Sx are : 32~, 1
2~,− 1
2~,− 3
2~
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 9 / 17
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Problem 4.52 (cont.)
Sx =1
2(S+ + S−) =
~2
0√
3 0 0√3 0 2 0
0 2 0√
3
0 0√
3 0
The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)
0 =
∣∣∣∣∣∣∣∣− λ
√3 0 0√
3 −λ 2 0
0 2 −λ√
3
0 0√
3 −λ
∣∣∣∣∣∣∣∣ = −λ
∣∣∣∣∣∣−λ 2 0
2 −λ√
3
0√
3 −λ
∣∣∣∣∣∣−√3
∣∣∣∣∣∣√
3 2 0
0 −λ√
3
0√
3 −λ
∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−
√3[√
3(λ2 − 3)]
= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9
0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)
the eigenvalues of Sx are : 32~, 1
2~,− 1
2~,− 3
2~
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 9 / 17
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Problem 4.52 (cont.)
Sx =1
2(S+ + S−) =
~2
0√
3 0 0√3 0 2 0
0 2 0√
3
0 0√
3 0
The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)
0 =
∣∣∣∣∣∣∣∣− λ
√3 0 0√
3 −λ 2 0
0 2 −λ√
3
0 0√
3 −λ
∣∣∣∣∣∣∣∣ = −λ
∣∣∣∣∣∣−λ 2 0
2 −λ√
3
0√
3 −λ
∣∣∣∣∣∣−√3
∣∣∣∣∣∣√
3 2 0
0 −λ√
3
0√
3 −λ
∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−
√3[√
3(λ2 − 3)]
= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9
0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)
the eigenvalues of Sx are : 32~, 1
2~,− 1
2~,− 3
2~
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 9 / 17
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Problem 4.52 (cont.)
Sx =1
2(S+ + S−) =
~2
0√
3 0 0√3 0 2 0
0 2 0√
3
0 0√
3 0
The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)
0 =
∣∣∣∣∣∣∣∣− λ
√3 0 0√
3 −λ 2 0
0 2 −λ√
3
0 0√
3 −λ
∣∣∣∣∣∣∣∣
= −λ
∣∣∣∣∣∣−λ 2 0
2 −λ√
3
0√
3 −λ
∣∣∣∣∣∣−√3
∣∣∣∣∣∣√
3 2 0
0 −λ√
3
0√
3 −λ
∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−
√3[√
3(λ2 − 3)]
= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9
0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)
the eigenvalues of Sx are : 32~, 1
2~,− 1
2~,− 3
2~
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 9 / 17
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Problem 4.52 (cont.)
Sx =1
2(S+ + S−) =
~2
0√
3 0 0√3 0 2 0
0 2 0√
3
0 0√
3 0
The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)
0 =
∣∣∣∣∣∣∣∣− λ
√3 0 0√
3 −λ 2 0
0 2 −λ√
3
0 0√
3 −λ
∣∣∣∣∣∣∣∣ = −λ
∣∣∣∣∣∣−λ 2 0
2 −λ√
3
0√
3 −λ
∣∣∣∣∣∣
−√
3
∣∣∣∣∣∣√
3 2 0
0 −λ√
3
0√
3 −λ
∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−
√3[√
3(λ2 − 3)]
= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9
0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)
the eigenvalues of Sx are : 32~, 1
2~,− 1
2~,− 3
2~
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 9 / 17
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Problem 4.52 (cont.)
Sx =1
2(S+ + S−) =
~2
0√
3 0 0√3 0 2 0
0 2 0√
3
0 0√
3 0
The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)
0 =
∣∣∣∣∣∣∣∣−λ
√3 0 0√
3 −λ 2 0
0 2 −λ√
3
0 0√
3 −λ
∣∣∣∣∣∣∣∣ = −λ
∣∣∣∣∣∣−λ 2 0
2 −λ√
3
0√
3 −λ
∣∣∣∣∣∣−√3
∣∣∣∣∣∣√
3 2 0
0 −λ√
3
0√
3 −λ
∣∣∣∣∣∣
= −λ[−λ(λ2 − 3)− 2(−2λ)]−√
3[√
3(λ2 − 3)]
= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9
0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)
the eigenvalues of Sx are : 32~, 1
2~,− 1
2~,− 3
2~
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 9 / 17
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Problem 4.52 (cont.)
Sx =1
2(S+ + S−) =
~2
0√
3 0 0√3 0 2 0
0 2 0√
3
0 0√
3 0
The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)
0 =
∣∣∣∣∣∣∣∣−λ
√3 0 0√
3 −λ 2 0
0 2 −λ√
3
0 0√
3 −λ
∣∣∣∣∣∣∣∣ = −λ
∣∣∣∣∣∣−λ 2 0
2 −λ√
3
0√
3 −λ
∣∣∣∣∣∣−√3
∣∣∣∣∣∣√
3 2 0
0 −λ√
3
0√
3 −λ
∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−
√3[√
3(λ2 − 3)]
= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9
0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)
the eigenvalues of Sx are : 32~, 1
2~,− 1
2~,− 3
2~
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 9 / 17
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Problem 4.52 (cont.)
Sx =1
2(S+ + S−) =
~2
0√
3 0 0√3 0 2 0
0 2 0√
3
0 0√
3 0
The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)
0 =
∣∣∣∣∣∣∣∣−λ
√3 0 0√
3 −λ 2 0
0 2 −λ√
3
0 0√
3 −λ
∣∣∣∣∣∣∣∣ = −λ
∣∣∣∣∣∣−λ 2 0
2 −λ√
3
0√
3 −λ
∣∣∣∣∣∣−√3
∣∣∣∣∣∣√
3 2 0
0 −λ√
3
0√
3 −λ
∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−
√3[√
3(λ2 − 3)]
= −λ[−λ3 + 7λ]− [3λ2 − 9]
= λ4 − 10λ2 + 9
0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)
the eigenvalues of Sx are : 32~, 1
2~,− 1
2~,− 3
2~
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 9 / 17
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Problem 4.52 (cont.)
Sx =1
2(S+ + S−) =
~2
0√
3 0 0√3 0 2 0
0 2 0√
3
0 0√
3 0
The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)
0 =
∣∣∣∣∣∣∣∣−λ
√3 0 0√
3 −λ 2 0
0 2 −λ√
3
0 0√
3 −λ
∣∣∣∣∣∣∣∣ = −λ
∣∣∣∣∣∣−λ 2 0
2 −λ√
3
0√
3 −λ
∣∣∣∣∣∣−√3
∣∣∣∣∣∣√
3 2 0
0 −λ√
3
0√
3 −λ
∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−
√3[√
3(λ2 − 3)]
= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9
0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)
the eigenvalues of Sx are : 32~, 1
2~,− 1
2~,− 3
2~
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 9 / 17
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Problem 4.52 (cont.)
Sx =1
2(S+ + S−) =
~2
0√
3 0 0√3 0 2 0
0 2 0√
3
0 0√
3 0
The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)
0 =
∣∣∣∣∣∣∣∣−λ
√3 0 0√
3 −λ 2 0
0 2 −λ√
3
0 0√
3 −λ
∣∣∣∣∣∣∣∣ = −λ
∣∣∣∣∣∣−λ 2 0
2 −λ√
3
0√
3 −λ
∣∣∣∣∣∣−√3
∣∣∣∣∣∣√
3 2 0
0 −λ√
3
0√
3 −λ
∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−
√3[√
3(λ2 − 3)]
= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9
0 = λ4 − 10λ2 + 9
= (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)
the eigenvalues of Sx are : 32~, 1
2~,− 1
2~,− 3
2~
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 9 / 17
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Problem 4.52 (cont.)
Sx =1
2(S+ + S−) =
~2
0√
3 0 0√3 0 2 0
0 2 0√
3
0 0√
3 0
The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)
0 =
∣∣∣∣∣∣∣∣−λ
√3 0 0√
3 −λ 2 0
0 2 −λ√
3
0 0√
3 −λ
∣∣∣∣∣∣∣∣ = −λ
∣∣∣∣∣∣−λ 2 0
2 −λ√
3
0√
3 −λ
∣∣∣∣∣∣−√3
∣∣∣∣∣∣√
3 2 0
0 −λ√
3
0√
3 −λ
∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−
√3[√
3(λ2 − 3)]
= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9
0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1)
= (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)
the eigenvalues of Sx are : 32~, 1
2~,− 1
2~,− 3
2~
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 9 / 17
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Problem 4.52 (cont.)
Sx =1
2(S+ + S−) =
~2
0√
3 0 0√3 0 2 0
0 2 0√
3
0 0√
3 0
The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)
0 =
∣∣∣∣∣∣∣∣−λ
√3 0 0√
3 −λ 2 0
0 2 −λ√
3
0 0√
3 −λ
∣∣∣∣∣∣∣∣ = −λ
∣∣∣∣∣∣−λ 2 0
2 −λ√
3
0√
3 −λ
∣∣∣∣∣∣−√3
∣∣∣∣∣∣√
3 2 0
0 −λ√
3
0√
3 −λ
∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−
√3[√
3(λ2 − 3)]
= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9
0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)
the eigenvalues of Sx are : 32~, 1
2~,− 1
2~,− 3
2~
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 9 / 17
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Problem 4.52 (cont.)
Sx =1
2(S+ + S−) =
~2
0√
3 0 0√3 0 2 0
0 2 0√
3
0 0√
3 0
The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)
0 =
∣∣∣∣∣∣∣∣−λ
√3 0 0√
3 −λ 2 0
0 2 −λ√
3
0 0√
3 −λ
∣∣∣∣∣∣∣∣ = −λ
∣∣∣∣∣∣−λ 2 0
2 −λ√
3
0√
3 −λ
∣∣∣∣∣∣−√3
∣∣∣∣∣∣√
3 2 0
0 −λ√
3
0√
3 −λ
∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−
√3[√
3(λ2 − 3)]
= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9
0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)
the eigenvalues of Sx are : 32~, 1
2~,− 1
2~,− 3
2~
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 9 / 17
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Electron in a magnetic field
Orbital angular momentum implies a circulating charge which leads to amagnetic moment
because spin behaves in all ways like orbital angular momentum, it mustalso be associated with a magnetic dipole moment
~µ = γ~S
in the presence of a magnetic field,the dipole feels a torque
and has total energy
the Hamiltonian for a particle withspin in the presence of a magneticfield, is thus
γ is the gyromagnetic ratio
~τ = ~µ× ~B
H = −~µ · ~B
H = −γ~B · ~S
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 10 / 17
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Electron in a magnetic field
Orbital angular momentum implies a circulating charge which leads to amagnetic moment
because spin behaves in all ways like orbital angular momentum, it mustalso be associated with a magnetic dipole moment
~µ = γ~S
in the presence of a magnetic field,the dipole feels a torque
and has total energy
the Hamiltonian for a particle withspin in the presence of a magneticfield, is thus
γ is the gyromagnetic ratio
~τ = ~µ× ~B
H = −~µ · ~B
H = −γ~B · ~S
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 10 / 17
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Electron in a magnetic field
Orbital angular momentum implies a circulating charge which leads to amagnetic moment
because spin behaves in all ways like orbital angular momentum, it mustalso be associated with a magnetic dipole moment
~µ = γ~S
in the presence of a magnetic field,the dipole feels a torque
and has total energy
the Hamiltonian for a particle withspin in the presence of a magneticfield, is thus
γ is the gyromagnetic ratio
~τ = ~µ× ~B
H = −~µ · ~B
H = −γ~B · ~S
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 10 / 17
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Electron in a magnetic field
Orbital angular momentum implies a circulating charge which leads to amagnetic moment
because spin behaves in all ways like orbital angular momentum, it mustalso be associated with a magnetic dipole moment
~µ = γ~S
in the presence of a magnetic field,the dipole feels a torque
and has total energy
the Hamiltonian for a particle withspin in the presence of a magneticfield, is thus
γ is the gyromagnetic ratio
~τ = ~µ× ~B
H = −~µ · ~B
H = −γ~B · ~S
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 10 / 17
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Electron in a magnetic field
Orbital angular momentum implies a circulating charge which leads to amagnetic moment
because spin behaves in all ways like orbital angular momentum, it mustalso be associated with a magnetic dipole moment
~µ = γ~S
in the presence of a magnetic field,the dipole feels a torque
and has total energy
the Hamiltonian for a particle withspin in the presence of a magneticfield, is thus
γ is the gyromagnetic ratio
~τ = ~µ× ~B
H = −~µ · ~B
H = −γ~B · ~S
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 10 / 17
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Electron in a magnetic field
Orbital angular momentum implies a circulating charge which leads to amagnetic moment
because spin behaves in all ways like orbital angular momentum, it mustalso be associated with a magnetic dipole moment
~µ = γ~S
in the presence of a magnetic field,the dipole feels a torque
and has total energy
the Hamiltonian for a particle withspin in the presence of a magneticfield, is thus
γ is the gyromagnetic ratio
~τ = ~µ× ~B
H = −~µ · ~B
H = −γ~B · ~S
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 10 / 17
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Electron in a magnetic field
Orbital angular momentum implies a circulating charge which leads to amagnetic moment
because spin behaves in all ways like orbital angular momentum, it mustalso be associated with a magnetic dipole moment
~µ = γ~S
in the presence of a magnetic field,the dipole feels a torque
and has total energy
the Hamiltonian for a particle withspin in the presence of a magneticfield, is thus
γ is the gyromagnetic ratio
~τ = ~µ× ~B
H = −~µ · ~B
H = −γ~B · ~S
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 10 / 17
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Electron in a magnetic field
Orbital angular momentum implies a circulating charge which leads to amagnetic moment
because spin behaves in all ways like orbital angular momentum, it mustalso be associated with a magnetic dipole moment
~µ = γ~S
in the presence of a magnetic field,the dipole feels a torque
and has total energy
the Hamiltonian for a particle withspin in the presence of a magneticfield, is thus
γ is the gyromagnetic ratio
~τ = ~µ× ~B
H = −~µ · ~B
H = −γ~B · ~S
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 10 / 17
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Electron in a magnetic field
Orbital angular momentum implies a circulating charge which leads to amagnetic moment
because spin behaves in all ways like orbital angular momentum, it mustalso be associated with a magnetic dipole moment
~µ = γ~S
in the presence of a magnetic field,the dipole feels a torque
and has total energy
the Hamiltonian for a particle withspin in the presence of a magneticfield, is thus
γ is the gyromagnetic ratio
~τ = ~µ× ~B
H = −~µ · ~B
H = −γ~B · ~S
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 10 / 17
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Electron in a magnetic field
Orbital angular momentum implies a circulating charge which leads to amagnetic moment
because spin behaves in all ways like orbital angular momentum, it mustalso be associated with a magnetic dipole moment
~µ = γ~S
in the presence of a magnetic field,the dipole feels a torque
and has total energy
the Hamiltonian for a particle withspin in the presence of a magneticfield, is thus
γ is the gyromagnetic ratio
~τ = ~µ× ~B
H = −~µ · ~B
H = −γ~B · ~S
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 10 / 17
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Dipole at rest in a magnetic field
consider a particle of spin 1/2 atrest in the presence of a magneticfield in the z-direction
the Hamiltonian is
and its eigenstates are those of Szwith eigenvalues E±
the time dependent Schrodingerequation can be solved in terms ofthe stationary states
~B = B0z
H = −γB0Sz = −γB0~2
(1 00 −1
){χ+, E+ = −γB0~
2
χ−, E− = +γB0~2
i~∂χ
∂t= Hχ
χ(t) = aχ+e−E+t/~ + bχ−e
−E−t/~ =
(ae iγB0t/2
be−iγB0t/2
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 11 / 17
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Dipole at rest in a magnetic field
consider a particle of spin 1/2 atrest in the presence of a magneticfield in the z-direction
the Hamiltonian is
and its eigenstates are those of Szwith eigenvalues E±
the time dependent Schrodingerequation can be solved in terms ofthe stationary states
~B = B0z
H = −γB0Sz = −γB0~2
(1 00 −1
){χ+, E+ = −γB0~
2
χ−, E− = +γB0~2
i~∂χ
∂t= Hχ
χ(t) = aχ+e−E+t/~ + bχ−e
−E−t/~ =
(ae iγB0t/2
be−iγB0t/2
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 11 / 17
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Dipole at rest in a magnetic field
consider a particle of spin 1/2 atrest in the presence of a magneticfield in the z-direction
the Hamiltonian is
and its eigenstates are those of Szwith eigenvalues E±
the time dependent Schrodingerequation can be solved in terms ofthe stationary states
~B = B0z
H = −γB0Sz = −γB0~2
(1 00 −1
){χ+, E+ = −γB0~
2
χ−, E− = +γB0~2
i~∂χ
∂t= Hχ
χ(t) = aχ+e−E+t/~ + bχ−e
−E−t/~ =
(ae iγB0t/2
be−iγB0t/2
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 11 / 17
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Dipole at rest in a magnetic field
consider a particle of spin 1/2 atrest in the presence of a magneticfield in the z-direction
the Hamiltonian is
and its eigenstates are those of Szwith eigenvalues E±
the time dependent Schrodingerequation can be solved in terms ofthe stationary states
~B = B0z
H = −γB0Sz
= −γB0~2
(1 00 −1
){χ+, E+ = −γB0~
2
χ−, E− = +γB0~2
i~∂χ
∂t= Hχ
χ(t) = aχ+e−E+t/~ + bχ−e
−E−t/~ =
(ae iγB0t/2
be−iγB0t/2
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 11 / 17
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Dipole at rest in a magnetic field
consider a particle of spin 1/2 atrest in the presence of a magneticfield in the z-direction
the Hamiltonian is
and its eigenstates are those of Szwith eigenvalues E±
the time dependent Schrodingerequation can be solved in terms ofthe stationary states
~B = B0z
H = −γB0Sz = −γB0~2
(1 00 −1
)
{χ+, E+ = −γB0~
2
χ−, E− = +γB0~2
i~∂χ
∂t= Hχ
χ(t) = aχ+e−E+t/~ + bχ−e
−E−t/~ =
(ae iγB0t/2
be−iγB0t/2
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 11 / 17
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Dipole at rest in a magnetic field
consider a particle of spin 1/2 atrest in the presence of a magneticfield in the z-direction
the Hamiltonian is
and its eigenstates are those of Szwith eigenvalues E±
the time dependent Schrodingerequation can be solved in terms ofthe stationary states
~B = B0z
H = −γB0Sz = −γB0~2
(1 00 −1
)
{χ+, E+ = −γB0~
2
χ−, E− = +γB0~2
i~∂χ
∂t= Hχ
χ(t) = aχ+e−E+t/~ + bχ−e
−E−t/~ =
(ae iγB0t/2
be−iγB0t/2
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 11 / 17
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Dipole at rest in a magnetic field
consider a particle of spin 1/2 atrest in the presence of a magneticfield in the z-direction
the Hamiltonian is
and its eigenstates are those of Szwith eigenvalues E±
the time dependent Schrodingerequation can be solved in terms ofthe stationary states
~B = B0z
H = −γB0Sz = −γB0~2
(1 00 −1
){χ+, E+ = −γB0~
2
χ−, E− = +γB0~2
i~∂χ
∂t= Hχ
χ(t) = aχ+e−E+t/~ + bχ−e
−E−t/~ =
(ae iγB0t/2
be−iγB0t/2
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 11 / 17
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Dipole at rest in a magnetic field
consider a particle of spin 1/2 atrest in the presence of a magneticfield in the z-direction
the Hamiltonian is
and its eigenstates are those of Szwith eigenvalues E±
the time dependent Schrodingerequation
can be solved in terms ofthe stationary states
~B = B0z
H = −γB0Sz = −γB0~2
(1 00 −1
){χ+, E+ = −γB0~
2
χ−, E− = +γB0~2
i~∂χ
∂t= Hχ
χ(t) = aχ+e−E+t/~ + bχ−e
−E−t/~ =
(ae iγB0t/2
be−iγB0t/2
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 11 / 17
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Dipole at rest in a magnetic field
consider a particle of spin 1/2 atrest in the presence of a magneticfield in the z-direction
the Hamiltonian is
and its eigenstates are those of Szwith eigenvalues E±
the time dependent Schrodingerequation
can be solved in terms ofthe stationary states
~B = B0z
H = −γB0Sz = −γB0~2
(1 00 −1
){χ+, E+ = −γB0~
2
χ−, E− = +γB0~2
i~∂χ
∂t= Hχ
χ(t) = aχ+e−E+t/~ + bχ−e
−E−t/~ =
(ae iγB0t/2
be−iγB0t/2
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 11 / 17
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Dipole at rest in a magnetic field
consider a particle of spin 1/2 atrest in the presence of a magneticfield in the z-direction
the Hamiltonian is
and its eigenstates are those of Szwith eigenvalues E±
the time dependent Schrodingerequation can be solved in terms ofthe stationary states
~B = B0z
H = −γB0Sz = −γB0~2
(1 00 −1
){χ+, E+ = −γB0~
2
χ−, E− = +γB0~2
i~∂χ
∂t= Hχ
χ(t) = aχ+e−E+t/~ + bχ−e
−E−t/~ =
(ae iγB0t/2
be−iγB0t/2
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 11 / 17
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Dipole at rest in a magnetic field
consider a particle of spin 1/2 atrest in the presence of a magneticfield in the z-direction
the Hamiltonian is
and its eigenstates are those of Szwith eigenvalues E±
the time dependent Schrodingerequation can be solved in terms ofthe stationary states
~B = B0z
H = −γB0Sz = −γB0~2
(1 00 −1
){χ+, E+ = −γB0~
2
χ−, E− = +γB0~2
i~∂χ
∂t= Hχ
χ(t) = aχ+e−E+t/~ + bχ−e
−E−t/~
=
(ae iγB0t/2
be−iγB0t/2
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 11 / 17
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Dipole at rest in a magnetic field
consider a particle of spin 1/2 atrest in the presence of a magneticfield in the z-direction
the Hamiltonian is
and its eigenstates are those of Szwith eigenvalues E±
the time dependent Schrodingerequation can be solved in terms ofthe stationary states
~B = B0z
H = −γB0Sz = −γB0~2
(1 00 −1
){χ+, E+ = −γB0~
2
χ−, E− = +γB0~2
i~∂χ
∂t= Hχ
χ(t) = aχ+e−E+t/~ + bχ−e
−E−t/~ =
(ae iγB0t/2
be−iγB0t/2
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 11 / 17
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Time-dependent expectation value
The constants a and b can be de-termined from the initial conditions
giving |a|2 + |b|2 = 1 and suggest-ing the relationship
the time-dependent solution is thus
χ(0) =
(ab
)a = cos(
α
2), b = sin(
α
2)
χ(t) =
(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)the expectation value of S as a function of time can now be calculated
〈Sx〉 = χ†(t)Sxχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 11 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(sin(α2 )e−iγB0t/2
cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(e−iγB0t + e iγB0t
)=
~2
sinα cos(γB0t)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 12 / 17
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Time-dependent expectation value
The constants a and b can be de-termined from the initial conditions
giving |a|2 + |b|2 = 1 and suggest-ing the relationship
the time-dependent solution is thus
χ(0) =
(ab
)
a = cos(α
2), b = sin(
α
2)
χ(t) =
(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)the expectation value of S as a function of time can now be calculated
〈Sx〉 = χ†(t)Sxχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 11 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(sin(α2 )e−iγB0t/2
cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(e−iγB0t + e iγB0t
)=
~2
sinα cos(γB0t)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 12 / 17
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Time-dependent expectation value
The constants a and b can be de-termined from the initial conditions
giving |a|2 + |b|2 = 1 and suggest-ing the relationship
the time-dependent solution is thus
χ(0) =
(ab
)
a = cos(α
2), b = sin(
α
2)
χ(t) =
(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)the expectation value of S as a function of time can now be calculated
〈Sx〉 = χ†(t)Sxχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 11 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(sin(α2 )e−iγB0t/2
cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(e−iγB0t + e iγB0t
)=
~2
sinα cos(γB0t)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 12 / 17
![Page 111: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/111.jpg)
Time-dependent expectation value
The constants a and b can be de-termined from the initial conditions
giving |a|2 + |b|2 = 1 and suggest-ing the relationship
the time-dependent solution is thus
χ(0) =
(ab
)a = cos(
α
2), b = sin(
α
2)
χ(t) =
(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)the expectation value of S as a function of time can now be calculated
〈Sx〉 = χ†(t)Sxχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 11 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(sin(α2 )e−iγB0t/2
cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(e−iγB0t + e iγB0t
)=
~2
sinα cos(γB0t)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 12 / 17
![Page 112: Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s](https://reader033.vdocuments.mx/reader033/viewer/2022061003/60b204f980fab33fea531f2a/html5/thumbnails/112.jpg)
Time-dependent expectation value
The constants a and b can be de-termined from the initial conditions
giving |a|2 + |b|2 = 1 and suggest-ing the relationship
the time-dependent solution is thus
χ(0) =
(ab
)a = cos(
α
2), b = sin(
α
2)
χ(t) =
(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)the expectation value of S as a function of time can now be calculated
〈Sx〉 = χ†(t)Sxχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 11 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(sin(α2 )e−iγB0t/2
cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(e−iγB0t + e iγB0t
)=
~2
sinα cos(γB0t)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 12 / 17
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Time-dependent expectation value
The constants a and b can be de-termined from the initial conditions
giving |a|2 + |b|2 = 1 and suggest-ing the relationship
the time-dependent solution is thus
χ(0) =
(ab
)a = cos(
α
2), b = sin(
α
2)
χ(t) =
(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)
the expectation value of S as a function of time can now be calculated
〈Sx〉 = χ†(t)Sxχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 11 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(sin(α2 )e−iγB0t/2
cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(e−iγB0t + e iγB0t
)=
~2
sinα cos(γB0t)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 12 / 17
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Time-dependent expectation value
The constants a and b can be de-termined from the initial conditions
giving |a|2 + |b|2 = 1 and suggest-ing the relationship
the time-dependent solution is thus
χ(0) =
(ab
)a = cos(
α
2), b = sin(
α
2)
χ(t) =
(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)the expectation value of S as a function of time can now be calculated
〈Sx〉 = χ†(t)Sxχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 11 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(sin(α2 )e−iγB0t/2
cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(e−iγB0t + e iγB0t
)=
~2
sinα cos(γB0t)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 12 / 17
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Time-dependent expectation value
The constants a and b can be de-termined from the initial conditions
giving |a|2 + |b|2 = 1 and suggest-ing the relationship
the time-dependent solution is thus
χ(0) =
(ab
)a = cos(
α
2), b = sin(
α
2)
χ(t) =
(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)the expectation value of S as a function of time can now be calculated
〈Sx〉 = χ†(t)Sxχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 11 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(sin(α2 )e−iγB0t/2
cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(e−iγB0t + e iγB0t
)=
~2
sinα cos(γB0t)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 12 / 17
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Time-dependent expectation value
The constants a and b can be de-termined from the initial conditions
giving |a|2 + |b|2 = 1 and suggest-ing the relationship
the time-dependent solution is thus
χ(0) =
(ab
)a = cos(
α
2), b = sin(
α
2)
χ(t) =
(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)the expectation value of S as a function of time can now be calculated
〈Sx〉 = χ†(t)Sxχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 11 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)
=~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(sin(α2 )e−iγB0t/2
cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(e−iγB0t + e iγB0t
)=
~2
sinα cos(γB0t)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 12 / 17
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Time-dependent expectation value
The constants a and b can be de-termined from the initial conditions
giving |a|2 + |b|2 = 1 and suggest-ing the relationship
the time-dependent solution is thus
χ(0) =
(ab
)a = cos(
α
2), b = sin(
α
2)
χ(t) =
(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)the expectation value of S as a function of time can now be calculated
〈Sx〉 = χ†(t)Sxχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 11 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(sin(α2 )e−iγB0t/2
cos(α2 )e iγB0t/2
)
=~2
cos(α/2) sin(α/2)(e−iγB0t + e iγB0t
)=
~2
sinα cos(γB0t)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 12 / 17
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Time-dependent expectation value
The constants a and b can be de-termined from the initial conditions
giving |a|2 + |b|2 = 1 and suggest-ing the relationship
the time-dependent solution is thus
χ(0) =
(ab
)a = cos(
α
2), b = sin(
α
2)
χ(t) =
(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)the expectation value of S as a function of time can now be calculated
〈Sx〉 = χ†(t)Sxχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 11 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(sin(α2 )e−iγB0t/2
cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(e−iγB0t + e iγB0t
)
=~2
sinα cos(γB0t)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 12 / 17
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Time-dependent expectation value
The constants a and b can be de-termined from the initial conditions
giving |a|2 + |b|2 = 1 and suggest-ing the relationship
the time-dependent solution is thus
χ(0) =
(ab
)a = cos(
α
2), b = sin(
α
2)
χ(t) =
(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)the expectation value of S as a function of time can now be calculated
〈Sx〉 = χ†(t)Sxχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 11 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(sin(α2 )e−iγB0t/2
cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(e−iγB0t + e iγB0t
)=
~2
sinα cos(γB0t)
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 12 / 17
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More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(−i sin(α2 )e−iγB0t/2
i cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(−ie−iγB0t + ie iγB0t
)= −~
2sinα sin(γB0t)
〈Sz〉 = χ†(t)Szχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 13 / 17
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More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)
=~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(−i sin(α2 )e−iγB0t/2
i cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(−ie−iγB0t + ie iγB0t
)= −~
2sinα sin(γB0t)
〈Sz〉 = χ†(t)Szχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 13 / 17
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More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(−i sin(α2 )e−iγB0t/2
i cos(α2 )e iγB0t/2
)
=~2
cos(α/2) sin(α/2)(−ie−iγB0t + ie iγB0t
)= −~
2sinα sin(γB0t)
〈Sz〉 = χ†(t)Szχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 13 / 17
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More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(−i sin(α2 )e−iγB0t/2
i cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(−ie−iγB0t + ie iγB0t
)
= −~2
sinα sin(γB0t)
〈Sz〉 = χ†(t)Szχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 13 / 17
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More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(−i sin(α2 )e−iγB0t/2
i cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(−ie−iγB0t + ie iγB0t
)= −~
2sinα sin(γB0t)
〈Sz〉 = χ†(t)Szχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 13 / 17
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More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(−i sin(α2 )e−iγB0t/2
i cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(−ie−iγB0t + ie iγB0t
)= −~
2sinα sin(γB0t)
〈Sz〉 = χ†(t)Szχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
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More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(−i sin(α2 )e−iγB0t/2
i cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(−ie−iγB0t + ie iγB0t
)= −~
2sinα sin(γB0t)
〈Sz〉 = χ†(t)Szχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)
=~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
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More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(−i sin(α2 )e−iγB0t/2
i cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(−ie−iγB0t + ie iγB0t
)= −~
2sinα sin(γB0t)
〈Sz〉 = χ†(t)Szχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)
=~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 13 / 17
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More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(−i sin(α2 )e−iγB0t/2
i cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(−ie−iγB0t + ie iγB0t
)= −~
2sinα sin(γB0t)
〈Sz〉 = χ†(t)Szχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)
=~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 13 / 17
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More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(−i sin(α2 )e−iγB0t/2
i cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(−ie−iγB0t + ie iγB0t
)= −~
2sinα sin(γB0t)
〈Sz〉 = χ†(t)Szχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 13 / 17
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Larmor precession
The expectation values for the compo-nents of S are thus
〈Sx〉 =~2
sinα cos(γB0t)
〈Sy 〉 = −~2
sinα sin(γB0t)
〈Sz〉 =~2
cosα
S precesses about the z-axis, at a fixedangle α so 〈Sz〉 is constant and the othertwo are constantly changing
the precession frequency ω = γB0 iscalled the Larmor frequency
z
y
x
S
ω
α
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Larmor precession
The expectation values for the compo-nents of S are thus
〈Sx〉 =~2
sinα cos(γB0t)
〈Sy 〉 = −~2
sinα sin(γB0t)
〈Sz〉 =~2
cosα
S precesses about the z-axis, at a fixedangle α so 〈Sz〉 is constant and the othertwo are constantly changing
the precession frequency ω = γB0 iscalled the Larmor frequency
z
y
x
S
ω
α
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 14 / 17
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Larmor precession
The expectation values for the compo-nents of S are thus
〈Sx〉 =~2
sinα cos(γB0t)
〈Sy 〉 = −~2
sinα sin(γB0t)
〈Sz〉 =~2
cosα
S precesses about the z-axis, at a fixedangle α so 〈Sz〉 is constant and the othertwo are constantly changing
the precession frequency ω = γB0 iscalled the Larmor frequency
z
y
x
S
ω
α
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 14 / 17
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Larmor precession
The expectation values for the compo-nents of S are thus
〈Sx〉 =~2
sinα cos(γB0t)
〈Sy 〉 = −~2
sinα sin(γB0t)
〈Sz〉 =~2
cosα
S precesses about the z-axis, at a fixedangle α so 〈Sz〉 is constant and the othertwo are constantly changing
the precession frequency ω = γB0 iscalled the Larmor frequency
z
y
x
S
ω
α
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 14 / 17
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Larmor precession
The expectation values for the compo-nents of S are thus
〈Sx〉 =~2
sinα cos(γB0t)
〈Sy 〉 = −~2
sinα sin(γB0t)
〈Sz〉 =~2
cosα
S precesses about the z-axis, at a fixedangle α so 〈Sz〉 is constant and the othertwo are constantly changing
the precession frequency ω = γB0 iscalled the Larmor frequency
z
y
x
S
ω
α
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 14 / 17
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Larmor precession
The expectation values for the compo-nents of S are thus
〈Sx〉 =~2
sinα cos(γB0t)
〈Sy 〉 = −~2
sinα sin(γB0t)
〈Sz〉 =~2
cosα
S precesses about the z-axis, at a fixedangle α so 〈Sz〉 is constant and the othertwo are constantly changing
the precession frequency ω = γB0 iscalled the Larmor frequency
z
y
x
S
ω
α
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 14 / 17
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Larmor precession
The expectation values for the compo-nents of S are thus
〈Sx〉 =~2
sinα cos(γB0t)
〈Sy 〉 = −~2
sinα sin(γB0t)
〈Sz〉 =~2
cosα
S precesses about the z-axis, at a fixedangle α so 〈Sz〉 is constant and the othertwo are constantly changing
the precession frequency ω = γB0 iscalled the Larmor frequency
z
y
x
S
ω
α
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 14 / 17
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Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0).
Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
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Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2.
What is the total angularmomentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 15 / 17
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Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 15 / 17
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Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 15 / 17
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Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑
↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 15 / 17
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Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓
↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 15 / 17
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Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑
↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 15 / 17
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Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
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Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 15 / 17
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Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
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Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2
= (S(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
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Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
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Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2)
= ~(m1 + m2)χ1χ2
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Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1.
Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2))
= (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑
=~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)
=~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)
=~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑
=~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)
= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)
= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 16 / 17
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑
=~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)
=~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)
=~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ =
~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑
+ ~21
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑
+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
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Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
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Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑
= (S(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
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Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
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Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓)
= ~(↓↑+ ↑↓) =√
2~1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
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Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓)
=√
2~1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
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Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
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Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
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Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
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Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]
=1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
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Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
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Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
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Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
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Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 17 / 17