today’s outline - november 05, 2012 - iitphys.iit.edu/~segre/phys405/12f/lecture_20.pdf · 2012....
TRANSCRIPT
![Page 1: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/1.jpg)
Today’s Outline - November 05, 2012
• Problem 4.52
• Electrons in a magnetic field
• Adding angular momentum
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 1 / 10
![Page 2: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/2.jpg)
Today’s Outline - November 05, 2012
• Problem 4.52
• Electrons in a magnetic field
• Adding angular momentum
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 1 / 10
![Page 3: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/3.jpg)
Today’s Outline - November 05, 2012
• Problem 4.52
• Electrons in a magnetic field
• Adding angular momentum
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 1 / 10
![Page 4: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/4.jpg)
Today’s Outline - November 05, 2012
• Problem 4.52
• Electrons in a magnetic field
• Adding angular momentum
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 1 / 10
![Page 5: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/5.jpg)
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
−12〉 =
0010
| 32
−32〉 =
0001
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 2 / 10
![Page 6: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/6.jpg)
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
−12〉 =
0010
| 32
−32〉 =
0001
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 2 / 10
![Page 7: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/7.jpg)
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
−12〉 =
0010
| 32
−32〉 =
0001
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 2 / 10
![Page 8: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/8.jpg)
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
−12〉 =
0010
| 32
−32〉 =
0001
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 2 / 10
![Page 9: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/9.jpg)
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
−12〉 =
0010
| 32
−32〉 =
0001
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 2 / 10
![Page 10: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/10.jpg)
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
−12〉 =
0010
| 32
−32〉 =
0001
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 2 / 10
![Page 11: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/11.jpg)
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
−12〉
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 3 / 10
![Page 12: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/12.jpg)
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
−12〉
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 3 / 10
![Page 13: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/13.jpg)
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
−12〉
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 3 / 10
![Page 14: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/14.jpg)
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
−12〉
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 3 / 10
![Page 15: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/15.jpg)
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
−12〉
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 3 / 10
![Page 16: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/16.jpg)
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
−12〉
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 3 / 10
![Page 17: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/17.jpg)
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
−12〉
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 3 / 10
![Page 18: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/18.jpg)
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
−12〉
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 3 / 10
![Page 19: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/19.jpg)
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
−12〉
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 3 / 10
![Page 20: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/20.jpg)
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
−12〉
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 3 / 10
![Page 21: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/21.jpg)
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
−12〉
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 3 / 10
![Page 22: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/22.jpg)
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
−12〉
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 3 / 10
![Page 23: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/23.jpg)
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
−12〉
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 3 / 10
![Page 24: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/24.jpg)
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
−12〉
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 3 / 10
![Page 25: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/25.jpg)
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
2−12〉
S−| 32 −12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32
−32〉
S−| 32 −32〉 = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 4 / 10
![Page 26: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/26.jpg)
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
2−12〉
S−| 32 −12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32
−32〉
S−| 32 −32〉 = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 4 / 10
![Page 27: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/27.jpg)
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
2−12〉
S−| 32 −12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32
−32〉
S−| 32 −32〉 = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 4 / 10
![Page 28: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/28.jpg)
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
2−12〉
S−| 32 −12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32
−32〉
S−| 32 −32〉 = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 4 / 10
![Page 29: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/29.jpg)
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
2−12〉
S−| 32 −12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32
−32〉
S−| 32 −32〉 = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 4 / 10
![Page 30: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/30.jpg)
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
−12〉
S−| 32 −12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32
−32〉
S−| 32 −32〉 = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 4 / 10
![Page 31: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/31.jpg)
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
2−12〉
S−| 32 −12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32
−32〉
S−| 32 −32〉 = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 4 / 10
![Page 32: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/32.jpg)
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
2−12〉
S−| 32 −12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32
−32〉
S−| 32 −32〉 = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 4 / 10
![Page 33: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/33.jpg)
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
2−12〉
S−| 32 −12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉
=√
3~ | 32
−32〉
S−| 32 −32〉 = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 4 / 10
![Page 34: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/34.jpg)
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
2−12〉
S−| 32 −12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32
−32〉
S−| 32 −32〉 = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 4 / 10
![Page 35: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/35.jpg)
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
2−12〉
S−| 32 −12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32
−32〉
S−| 32 −32〉 = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 4 / 10
![Page 36: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/36.jpg)
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
2−12〉
S−| 32 −12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32
−32〉
S−| 32 −32〉 = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 4 / 10
![Page 37: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/37.jpg)
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
2−12〉
S−| 32 −12〉 =
√32( 5
2)+ 12(− 3
2)~ | 32 −32〉 =√
3~ | 32
−32〉
S−| 32 −32〉 = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 4 / 10
![Page 38: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/38.jpg)
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
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 2012 November 05, 2012 5 / 10
![Page 39: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/39.jpg)
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
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 2012 November 05, 2012 5 / 10
![Page 40: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/40.jpg)
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
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 2012 November 05, 2012 5 / 10
![Page 41: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/41.jpg)
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
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 2012 November 05, 2012 5 / 10
![Page 42: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/42.jpg)
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
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 2012 November 05, 2012 5 / 10
![Page 43: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/43.jpg)
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
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 2012 November 05, 2012 5 / 10
![Page 44: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/44.jpg)
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
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 2012 November 05, 2012 5 / 10
![Page 45: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/45.jpg)
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
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 2012 November 05, 2012 5 / 10
![Page 46: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/46.jpg)
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
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 2012 November 05, 2012 5 / 10
![Page 47: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/47.jpg)
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
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 2012 November 05, 2012 5 / 10
![Page 48: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/48.jpg)
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
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 2012 November 05, 2012 5 / 10
![Page 49: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/49.jpg)
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
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 2012 November 05, 2012 5 / 10
![Page 50: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/50.jpg)
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
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 2012 November 05, 2012 5 / 10
![Page 51: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/51.jpg)
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 2012 November 05, 2012 6 / 10
![Page 52: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/52.jpg)
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 2012 November 05, 2012 6 / 10
![Page 53: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/53.jpg)
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 2012 November 05, 2012 6 / 10
![Page 54: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/54.jpg)
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 2012 November 05, 2012 6 / 10
![Page 55: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/55.jpg)
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 2012 November 05, 2012 6 / 10
![Page 56: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/56.jpg)
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 2012 November 05, 2012 6 / 10
![Page 57: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/57.jpg)
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 2012 November 05, 2012 6 / 10
![Page 58: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/58.jpg)
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 2012 November 05, 2012 6 / 10
![Page 59: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/59.jpg)
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 2012 November 05, 2012 6 / 10
![Page 60: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/60.jpg)
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 2012 November 05, 2012 6 / 10
![Page 61: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/61.jpg)
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 2012 November 05, 2012 7 / 10
![Page 62: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/62.jpg)
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 2012 November 05, 2012 7 / 10
![Page 63: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/63.jpg)
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 2012 November 05, 2012 7 / 10
![Page 64: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/64.jpg)
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 2012 November 05, 2012 7 / 10
![Page 65: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/65.jpg)
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 2012 November 05, 2012 7 / 10
![Page 66: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/66.jpg)
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 2012 November 05, 2012 7 / 10
![Page 67: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/67.jpg)
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 2012 November 05, 2012 7 / 10
![Page 68: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/68.jpg)
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 2012 November 05, 2012 7 / 10
![Page 69: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/69.jpg)
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 2012 November 05, 2012 7 / 10
![Page 70: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/70.jpg)
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 2012 November 05, 2012 7 / 10
![Page 71: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/71.jpg)
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 2012 November 05, 2012 7 / 10
![Page 72: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/72.jpg)
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 2012 November 05, 2012 7 / 10
![Page 73: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/73.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 2012 November 05, 2012 8 / 10
![Page 74: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/74.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 2012 November 05, 2012 8 / 10
![Page 75: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/75.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 2012 November 05, 2012 8 / 10
![Page 76: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/76.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 2012 November 05, 2012 8 / 10
![Page 77: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/77.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 2012 November 05, 2012 8 / 10
![Page 78: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/78.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 2012 November 05, 2012 8 / 10
![Page 79: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/79.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 2012 November 05, 2012 8 / 10
![Page 80: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/80.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 2012 November 05, 2012 8 / 10
![Page 81: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/81.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 2012 November 05, 2012 8 / 10
![Page 82: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/82.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 2012 November 05, 2012 8 / 10
![Page 83: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/83.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 2012 November 05, 2012 8 / 10
![Page 84: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/84.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 2012 November 05, 2012 8 / 10
![Page 85: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/85.jpg)
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 2012 November 05, 2012 9 / 10
![Page 86: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/86.jpg)
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 2012 November 05, 2012 9 / 10
![Page 87: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/87.jpg)
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 2012 November 05, 2012 9 / 10
![Page 88: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/88.jpg)
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 2012 November 05, 2012 9 / 10
![Page 89: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/89.jpg)
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 2012 November 05, 2012 9 / 10
![Page 90: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/90.jpg)
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 2012 November 05, 2012 9 / 10
![Page 91: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/91.jpg)
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 2012 November 05, 2012 9 / 10
![Page 92: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/92.jpg)
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 2012 November 05, 2012 9 / 10
![Page 93: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/93.jpg)
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 2012 November 05, 2012 9 / 10
![Page 94: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/94.jpg)
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 2012 November 05, 2012 9 / 10
![Page 95: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/95.jpg)
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 2012 November 05, 2012 10 / 10
![Page 96: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/96.jpg)
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 2012 November 05, 2012 10 / 10
![Page 97: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/97.jpg)
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 2012 November 05, 2012 10 / 10
![Page 98: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/98.jpg)
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 2012 November 05, 2012 10 / 10
![Page 99: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/99.jpg)
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 2012 November 05, 2012 10 / 10
![Page 100: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/100.jpg)
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 2012 November 05, 2012 10 / 10
![Page 101: Today’s Outline - November 05, 2012 - IITphys.iit.edu/~segre/phys405/12F/lecture_20.pdf · 2012. 11. 6. · Today’s Outline - November 05, 2012 Problem 4.52 Electrons in a magnetic](https://reader033.vdocuments.net/reader033/viewer/2022061003/60b204fa80fab33fea531f2b/html5/thumbnails/101.jpg)
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 2012 November 05, 2012 10 / 10