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Today’s Outline - November 11, 2015
• Chapter 3 problems
• Chapter 4 problems
• Stern-Gerlach problem
• Additional Chapter 4 problems:5,7,9,15,18,29,31,38
• Solid state systems
Midterm Exam 2: Monday, November 16, 2015, Room 204 Stuart Buildingcovers through Chapter 4
Homework Assignment #10:Chapter 4:33,47,49; Chapter 5:1,2,5due Monday, November 23, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 1 / 19
Today’s Outline - November 11, 2015
• Chapter 3 problems
• Chapter 4 problems
• Stern-Gerlach problem
• Additional Chapter 4 problems:5,7,9,15,18,29,31,38
• Solid state systems
Midterm Exam 2: Monday, November 16, 2015, Room 204 Stuart Buildingcovers through Chapter 4
Homework Assignment #10:Chapter 4:33,47,49; Chapter 5:1,2,5due Monday, November 23, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 1 / 19
Today’s Outline - November 11, 2015
• Chapter 3 problems
• Chapter 4 problems
• Stern-Gerlach problem
• Additional Chapter 4 problems:5,7,9,15,18,29,31,38
• Solid state systems
Midterm Exam 2: Monday, November 16, 2015, Room 204 Stuart Buildingcovers through Chapter 4
Homework Assignment #10:Chapter 4:33,47,49; Chapter 5:1,2,5due Monday, November 23, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 1 / 19
Today’s Outline - November 11, 2015
• Chapter 3 problems
• Chapter 4 problems
• Stern-Gerlach problem
• Additional Chapter 4 problems:5,7,9,15,18,29,31,38
• Solid state systems
Midterm Exam 2: Monday, November 16, 2015, Room 204 Stuart Buildingcovers through Chapter 4
Homework Assignment #10:Chapter 4:33,47,49; Chapter 5:1,2,5due Monday, November 23, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 1 / 19
Today’s Outline - November 11, 2015
• Chapter 3 problems
• Chapter 4 problems
• Stern-Gerlach problem
• Additional Chapter 4 problems:5,7,9,15,18,29,31,38
• Solid state systems
Midterm Exam 2: Monday, November 16, 2015, Room 204 Stuart Buildingcovers through Chapter 4
Homework Assignment #10:Chapter 4:33,47,49; Chapter 5:1,2,5due Monday, November 23, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 1 / 19
Today’s Outline - November 11, 2015
• Chapter 3 problems
• Chapter 4 problems
• Stern-Gerlach problem
• Additional Chapter 4 problems:5,7,9,15,18,29,31,38
• Solid state systems
Midterm Exam 2: Monday, November 16, 2015, Room 204 Stuart Buildingcovers through Chapter 4
Homework Assignment #10:Chapter 4:33,47,49; Chapter 5:1,2,5due Monday, November 23, 2015
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 1 / 19
Problem 3.6
Show that Q = d2/dφ2 is hermitian
〈f |Qg〉 =
∫ 2π
0f ∗
d2g
dφ2dφ = f ∗
dg
dφ
∣∣∣∣2π0
−∫ 2π
0
df ∗
dφ
dg
dφdφ
= f ∗dg
dφ
∣∣∣∣2π0
− df ∗
dφg
∣∣∣∣2π0
+
∫ 2π
0
d2f ∗
dφ2gdφ
=
∫ 2π
0
d2f ∗
dφ2gdφ = 〈Qf |g〉
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 2 / 19
Problem 3.6
Show that Q = d2/dφ2 is hermitian
〈f |Qg〉 =
∫ 2π
0f ∗
d2g
dφ2dφ
= f ∗dg
dφ
∣∣∣∣2π0
−∫ 2π
0
df ∗
dφ
dg
dφdφ
= f ∗dg
dφ
∣∣∣∣2π0
− df ∗
dφg
∣∣∣∣2π0
+
∫ 2π
0
d2f ∗
dφ2gdφ
=
∫ 2π
0
d2f ∗
dφ2gdφ = 〈Qf |g〉
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 2 / 19
Problem 3.6
Show that Q = d2/dφ2 is hermitian
〈f |Qg〉 =
∫ 2π
0f ∗
d2g
dφ2dφ = f ∗
dg
dφ
∣∣∣∣2π0
−∫ 2π
0
df ∗
dφ
dg
dφdφ
= f ∗dg
dφ
∣∣∣∣2π0
− df ∗
dφg
∣∣∣∣2π0
+
∫ 2π
0
d2f ∗
dφ2gdφ
=
∫ 2π
0
d2f ∗
dφ2gdφ = 〈Qf |g〉
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 2 / 19
Problem 3.6
Show that Q = d2/dφ2 is hermitian
〈f |Qg〉 =
∫ 2π
0f ∗
d2g
dφ2dφ = f ∗
dg
dφ
∣∣∣∣2π0
−∫ 2π
0
df ∗
dφ
dg
dφdφ
= f ∗dg
dφ
∣∣∣∣2π0
− df ∗
dφg
∣∣∣∣2π0
+
∫ 2π
0
d2f ∗
dφ2gdφ
=
∫ 2π
0
d2f ∗
dφ2gdφ = 〈Qf |g〉
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 2 / 19
Problem 3.6
Show that Q = d2/dφ2 is hermitian
〈f |Qg〉 =
∫ 2π
0f ∗
d2g
dφ2dφ = f ∗
dg
dφ
∣∣∣∣2π0
−∫ 2π
0
df ∗
dφ
dg
dφdφ
= f ∗dg
dφ
∣∣∣∣2π0
− df ∗
dφg
∣∣∣∣2π0
+
∫ 2π
0
d2f ∗
dφ2gdφ
=
∫ 2π
0
d2f ∗
dφ2gdφ
= 〈Qf |g〉
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 2 / 19
Problem 3.6
Show that Q = d2/dφ2 is hermitian
〈f |Qg〉 =
∫ 2π
0f ∗
d2g
dφ2dφ = f ∗
dg
dφ
∣∣∣∣2π0
−∫ 2π
0
df ∗
dφ
dg
dφdφ
= f ∗dg
dφ
∣∣∣∣2π0
− df ∗
dφg
∣∣∣∣2π0
+
∫ 2π
0
d2f ∗
dφ2gdφ
=
∫ 2π
0
d2f ∗
dφ2gdφ = 〈Qf |g〉
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 2 / 19
Problem 3.11
Find the momentum space wave function, Φ(p, t), for a particle in theground state of the harmonic oscillator. What is the probability that ameasurement of p on a particle in this state would yield a value outsidethe classical range?
The ground state of the harmonic oscillator is
Ψ0(x , t) =(mωπ~
)14e−mωx
2/2~e−iωt/2
Φ(p, t) =1√2π~
(mωπ~
)14e−iωt/2
∫ ∞−∞
e−ipx/~e−mωx2/2~ dx
=1√2π~
(mωπ~
)14e−iωt/2
√2π~mω
e−p2/2mω~
=
(1
πmω~
)14
e−p2/2mω~e−iωt/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 3 / 19
Problem 3.11
Find the momentum space wave function, Φ(p, t), for a particle in theground state of the harmonic oscillator. What is the probability that ameasurement of p on a particle in this state would yield a value outsidethe classical range?
The ground state of the harmonic oscillator is
Ψ0(x , t) =(mωπ~
)14e−mωx
2/2~e−iωt/2
Φ(p, t) =1√2π~
(mωπ~
)14e−iωt/2
∫ ∞−∞
e−ipx/~e−mωx2/2~ dx
=1√2π~
(mωπ~
)14e−iωt/2
√2π~mω
e−p2/2mω~
=
(1
πmω~
)14
e−p2/2mω~e−iωt/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 3 / 19
Problem 3.11
Find the momentum space wave function, Φ(p, t), for a particle in theground state of the harmonic oscillator. What is the probability that ameasurement of p on a particle in this state would yield a value outsidethe classical range?
The ground state of the harmonic oscillator is
Ψ0(x , t) =(mωπ~
)14e−mωx
2/2~e−iωt/2
Φ(p, t) =1√2π~
(mωπ~
)14e−iωt/2
∫ ∞−∞
e−ipx/~e−mωx2/2~ dx
=1√2π~
(mωπ~
)14e−iωt/2
√2π~mω
e−p2/2mω~
=
(1
πmω~
)14
e−p2/2mω~e−iωt/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 3 / 19
Problem 3.11
Find the momentum space wave function, Φ(p, t), for a particle in theground state of the harmonic oscillator. What is the probability that ameasurement of p on a particle in this state would yield a value outsidethe classical range?
The ground state of the harmonic oscillator is
Ψ0(x , t) =(mωπ~
)14e−mωx
2/2~e−iωt/2
Φ(p, t) =1√2π~
(mωπ~
)14e−iωt/2
∫ ∞−∞
e−ipx/~e−mωx2/2~ dx
=1√2π~
(mωπ~
)14e−iωt/2
√2π~mω
e−p2/2mω~
=
(1
πmω~
)14
e−p2/2mω~e−iωt/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 3 / 19
Problem 3.11
Find the momentum space wave function, Φ(p, t), for a particle in theground state of the harmonic oscillator. What is the probability that ameasurement of p on a particle in this state would yield a value outsidethe classical range?
The ground state of the harmonic oscillator is
Ψ0(x , t) =(mωπ~
)14e−mωx
2/2~e−iωt/2
Φ(p, t) =1√2π~
(mωπ~
)14e−iωt/2
∫ ∞−∞
e−ipx/~e−mωx2/2~ dx
=1√2π~
(mωπ~
)14e−iωt/2
√2π~mω
e−p2/2mω~
=
(1
πmω~
)14
e−p2/2mω~e−iωt/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 3 / 19
Problem 3.11
Find the momentum space wave function, Φ(p, t), for a particle in theground state of the harmonic oscillator. What is the probability that ameasurement of p on a particle in this state would yield a value outsidethe classical range?
The ground state of the harmonic oscillator is
Ψ0(x , t) =(mωπ~
)14e−mωx
2/2~e−iωt/2
Φ(p, t) =1√2π~
(mωπ~
)14e−iωt/2
∫ ∞−∞
e−ipx/~e−mωx2/2~ dx
=1√2π~
(mωπ~
)14e−iωt/2
√2π~mω
e−p2/2mω~
=
(1
πmω~
)14
e−p2/2mω~e−iωt/2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 3 / 19
Problem 3.11 (cont.)
The maximum classical momentum
p2
2m= E = 1
2~ω → p =√mω~
the probablity of finding the system outside the maximum classical limits
P =
∫ −√mω~−∞
|Φ|2 dp +
∫ ∞√mω~|Φ|2 dp = 2
∫ ∞√mω~|Φ|2 dp
= 1− 2
∫ √mω~0
|Φ|2 dp = 1− 2√πmω~
∫ √mω~0
e−p2/mω~ dp
substituting z ≡ p/√mω~ and dp =
√mω~ dz
P = 1− 2√π
∫ 1
0e−z
2dx = 1− erf(1) = 1− 0.843 = 0.16
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 4 / 19
Problem 3.11 (cont.)
The maximum classical momentum p2
2m= E = 1
2~ω
→ p =√mω~
the probablity of finding the system outside the maximum classical limits
P =
∫ −√mω~−∞
|Φ|2 dp +
∫ ∞√mω~|Φ|2 dp = 2
∫ ∞√mω~|Φ|2 dp
= 1− 2
∫ √mω~0
|Φ|2 dp = 1− 2√πmω~
∫ √mω~0
e−p2/mω~ dp
substituting z ≡ p/√mω~ and dp =
√mω~ dz
P = 1− 2√π
∫ 1
0e−z
2dx = 1− erf(1) = 1− 0.843 = 0.16
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 4 / 19
Problem 3.11 (cont.)
The maximum classical momentum p2
2m= E = 1
2~ω → p =√mω~
the probablity of finding the system outside the maximum classical limits
P =
∫ −√mω~−∞
|Φ|2 dp +
∫ ∞√mω~|Φ|2 dp = 2
∫ ∞√mω~|Φ|2 dp
= 1− 2
∫ √mω~0
|Φ|2 dp = 1− 2√πmω~
∫ √mω~0
e−p2/mω~ dp
substituting z ≡ p/√mω~ and dp =
√mω~ dz
P = 1− 2√π
∫ 1
0e−z
2dx = 1− erf(1) = 1− 0.843 = 0.16
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 4 / 19
Problem 3.11 (cont.)
The maximum classical momentum p2
2m= E = 1
2~ω → p =√mω~
the probablity of finding the system outside the maximum classical limits
P =
∫ −√mω~−∞
|Φ|2 dp +
∫ ∞√mω~|Φ|2 dp = 2
∫ ∞√mω~|Φ|2 dp
= 1− 2
∫ √mω~0
|Φ|2 dp = 1− 2√πmω~
∫ √mω~0
e−p2/mω~ dp
substituting z ≡ p/√mω~ and dp =
√mω~ dz
P = 1− 2√π
∫ 1
0e−z
2dx = 1− erf(1) = 1− 0.843 = 0.16
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 4 / 19
Problem 3.11 (cont.)
The maximum classical momentum p2
2m= E = 1
2~ω → p =√mω~
the probablity of finding the system outside the maximum classical limits
P =
∫ −√mω~−∞
|Φ|2 dp +
∫ ∞√mω~|Φ|2 dp
= 2
∫ ∞√mω~|Φ|2 dp
= 1− 2
∫ √mω~0
|Φ|2 dp = 1− 2√πmω~
∫ √mω~0
e−p2/mω~ dp
substituting z ≡ p/√mω~ and dp =
√mω~ dz
P = 1− 2√π
∫ 1
0e−z
2dx = 1− erf(1) = 1− 0.843 = 0.16
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 4 / 19
Problem 3.11 (cont.)
The maximum classical momentum p2
2m= E = 1
2~ω → p =√mω~
the probablity of finding the system outside the maximum classical limits
P =
∫ −√mω~−∞
|Φ|2 dp +
∫ ∞√mω~|Φ|2 dp = 2
∫ ∞√mω~|Φ|2 dp
= 1− 2
∫ √mω~0
|Φ|2 dp = 1− 2√πmω~
∫ √mω~0
e−p2/mω~ dp
substituting z ≡ p/√mω~ and dp =
√mω~ dz
P = 1− 2√π
∫ 1
0e−z
2dx = 1− erf(1) = 1− 0.843 = 0.16
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 4 / 19
Problem 3.11 (cont.)
The maximum classical momentum p2
2m= E = 1
2~ω → p =√mω~
the probablity of finding the system outside the maximum classical limits
P =
∫ −√mω~−∞
|Φ|2 dp +
∫ ∞√mω~|Φ|2 dp = 2
∫ ∞√mω~|Φ|2 dp
= 1− 2
∫ √mω~0
|Φ|2 dp
= 1− 2√πmω~
∫ √mω~0
e−p2/mω~ dp
substituting z ≡ p/√mω~ and dp =
√mω~ dz
P = 1− 2√π
∫ 1
0e−z
2dx = 1− erf(1) = 1− 0.843 = 0.16
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 4 / 19
Problem 3.11 (cont.)
The maximum classical momentum p2
2m= E = 1
2~ω → p =√mω~
the probablity of finding the system outside the maximum classical limits
P =
∫ −√mω~−∞
|Φ|2 dp +
∫ ∞√mω~|Φ|2 dp = 2
∫ ∞√mω~|Φ|2 dp
= 1− 2
∫ √mω~0
|Φ|2 dp = 1− 2√πmω~
∫ √mω~0
e−p2/mω~ dp
substituting z ≡ p/√mω~ and dp =
√mω~ dz
P = 1− 2√π
∫ 1
0e−z
2dx = 1− erf(1) = 1− 0.843 = 0.16
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 4 / 19
Problem 3.11 (cont.)
The maximum classical momentum p2
2m= E = 1
2~ω → p =√mω~
the probablity of finding the system outside the maximum classical limits
P =
∫ −√mω~−∞
|Φ|2 dp +
∫ ∞√mω~|Φ|2 dp = 2
∫ ∞√mω~|Φ|2 dp
= 1− 2
∫ √mω~0
|Φ|2 dp = 1− 2√πmω~
∫ √mω~0
e−p2/mω~ dp
substituting z ≡ p/√mω~ and dp =
√mω~ dz
P = 1− 2√π
∫ 1
0e−z
2dx = 1− erf(1) = 1− 0.843 = 0.16
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 4 / 19
Problem 3.11 (cont.)
The maximum classical momentum p2
2m= E = 1
2~ω → p =√mω~
the probablity of finding the system outside the maximum classical limits
P =
∫ −√mω~−∞
|Φ|2 dp +
∫ ∞√mω~|Φ|2 dp = 2
∫ ∞√mω~|Φ|2 dp
= 1− 2
∫ √mω~0
|Φ|2 dp = 1− 2√πmω~
∫ √mω~0
e−p2/mω~ dp
substituting z ≡ p/√mω~ and dp =
√mω~ dz
P = 1− 2√π
∫ 1
0e−z
2dx
= 1− erf(1) = 1− 0.843 = 0.16
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 4 / 19
Problem 3.11 (cont.)
The maximum classical momentum p2
2m= E = 1
2~ω → p =√mω~
the probablity of finding the system outside the maximum classical limits
P =
∫ −√mω~−∞
|Φ|2 dp +
∫ ∞√mω~|Φ|2 dp = 2
∫ ∞√mω~|Φ|2 dp
= 1− 2
∫ √mω~0
|Φ|2 dp = 1− 2√πmω~
∫ √mω~0
e−p2/mω~ dp
substituting z ≡ p/√mω~ and dp =
√mω~ dz
P = 1− 2√π
∫ 1
0e−z
2dx = 1− erf(1)
= 1− 0.843 = 0.16
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 4 / 19
Problem 3.11 (cont.)
The maximum classical momentum p2
2m= E = 1
2~ω → p =√mω~
the probablity of finding the system outside the maximum classical limits
P =
∫ −√mω~−∞
|Φ|2 dp +
∫ ∞√mω~|Φ|2 dp = 2
∫ ∞√mω~|Φ|2 dp
= 1− 2
∫ √mω~0
|Φ|2 dp = 1− 2√πmω~
∫ √mω~0
e−p2/mω~ dp
substituting z ≡ p/√mω~ and dp =
√mω~ dz
P = 1− 2√π
∫ 1
0e−z
2dx = 1− erf(1) = 1− 0.843 = 0.16
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 4 / 19
Problem 3.27
An operator A, representing observable A has two normalized eigenstatesψ1 and ψ2, with eigenvalues a1 and a2, respectively. Operator B,representing observable B has two normalized eigenstates φ1 and φ2, witheigenvalues b1 and b2. The eigenstates are related by
ψ1 = 15(3φ1 + 4φ2) ψ2 = 1
5(4φ1 − 3φ2)
(a) Observable A is measured, and the value a1 is obtained. What is thestate of the system immediately after this measurement?
(b) If B is now measured, what are the possible results, and what are theirprobabilities?
(c) Right after the measurement of B, A is measured again. What is theprobability of getting a1?
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 5 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16
25b2〈φ2|φ2〉 = 925b1 + 16
25b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16
25b2〈φ2|φ2〉 = 925b1 + 16
25b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16
25b2〈φ2|φ2〉 = 925b1 + 16
25b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16
25b2〈φ2|φ2〉 = 925b1 + 16
25b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16
25b2〈φ2|φ2〉 = 925b1 + 16
25b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉
= 925b1〈φ1|φ1〉+ 16
25b2〈φ2|φ2〉 = 925b1 + 16
25b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉 = 925b1〈φ1|φ1〉
+ 1625b2〈φ2|φ2〉 = 9
25b1 + 1625b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16
25b2〈φ2|φ2〉
= 925b1 + 16
25b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16
25b2〈φ2|φ2〉 = 925b1 + 16
25b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16
25b2〈φ2|φ2〉 = 925b1 + 16
25b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16
25b2〈φ2|φ2〉 = 925b1 + 16
25b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16
25b2〈φ2|φ2〉 = 925b1 + 16
25b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16
25b2〈φ2|φ2〉 = 925b1 + 16
25b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16
25b2〈φ2|φ2〉 = 925b1 + 16
25b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 3.27
ψ1 = 15(3φ1 + 4φ2)
ψ2 = 15(4φ1 − 3φ2)
φ1 = 15(3ψ1 + 4ψ2)
φ2 = 15(4ψ1 − 3ψ2)
(a) Since the expectation value measured is a1, the system must be in thecorresponding eigenstate ψ1
(b) The probabilities can be obtained by taking the expectation value of Bwith the state ψ1
〈φ1|Bψ1〉 = 925b1〈φ1|φ1〉+ 16
25b2〈φ2|φ2〉 = 925b1 + 16
25b2
(c) We have to consider two cases just after B is measured
the system is in state φ1 with prob-ablity 9
25 and then we measure
the system is in state φ2 with prob-ablity 16
25 and then we measure
Pa1(φ1) = 925
Pa1(φ2) = 1625
Pa1(total) = 925 ·
925 + 16
25 ·1625
= 337625 = 0.5392
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 6 / 19
Problem 4.13
(a) Find 〈r〉 and 〈r2〉 for an electron in the ground state of hydrogen.Express your answer in terms of the Bohr radius.
(b) Find 〈x〉 and 〈x2〉 for an electron in the ground state of hydrogen.
(c) Find 〈x2〉 in the state n = 2, l = 1,m = 1.
(a) The ground state of the hydrogen atom is
ψ =1√πa3
e−r/a
〈rn〉 =1
πa3
∫rne−2r/ar2 sin θ dr dθ dφ =
4π
πa3
∫ ∞0
rn+2e−2r/a dr
〈r〉 =4
a3
∫ ∞0r3e−2r/a dr =
4
a3
(a2
)3
∫ ∞0r2e−2r/a dr
=4
a3
(a2
)23 · 2
∫ ∞0re−2r/a dr =
4
a3
(a2
)33!
∫ ∞0e−2r/a dr =
4
a33!(a
2
)4
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 7 / 19
Problem 4.13
(a) Find 〈r〉 and 〈r2〉 for an electron in the ground state of hydrogen.Express your answer in terms of the Bohr radius.
(b) Find 〈x〉 and 〈x2〉 for an electron in the ground state of hydrogen.
(c) Find 〈x2〉 in the state n = 2, l = 1,m = 1.
(a) The ground state of the hydrogen atom is
ψ =1√πa3
e−r/a
〈rn〉 =1
πa3
∫rne−2r/ar2 sin θ dr dθ dφ =
4π
πa3
∫ ∞0
rn+2e−2r/a dr
〈r〉 =4
a3
∫ ∞0r3e−2r/a dr =
4
a3
(a2
)3
∫ ∞0r2e−2r/a dr
=4
a3
(a2
)23 · 2
∫ ∞0re−2r/a dr =
4
a3
(a2
)33!
∫ ∞0e−2r/a dr =
4
a33!(a
2
)4
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 7 / 19
Problem 4.13
(a) Find 〈r〉 and 〈r2〉 for an electron in the ground state of hydrogen.Express your answer in terms of the Bohr radius.
(b) Find 〈x〉 and 〈x2〉 for an electron in the ground state of hydrogen.
(c) Find 〈x2〉 in the state n = 2, l = 1,m = 1.
(a) The ground state of the hydrogen atom is
ψ =1√πa3
e−r/a
〈rn〉 =1
πa3
∫rne−2r/ar2 sin θ dr dθ dφ =
4π
πa3
∫ ∞0
rn+2e−2r/a dr
〈r〉 =4
a3
∫ ∞0r3e−2r/a dr =
4
a3
(a2
)3
∫ ∞0r2e−2r/a dr
=4
a3
(a2
)23 · 2
∫ ∞0re−2r/a dr =
4
a3
(a2
)33!
∫ ∞0e−2r/a dr =
4
a33!(a
2
)4
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 7 / 19
Problem 4.13
(a) Find 〈r〉 and 〈r2〉 for an electron in the ground state of hydrogen.Express your answer in terms of the Bohr radius.
(b) Find 〈x〉 and 〈x2〉 for an electron in the ground state of hydrogen.
(c) Find 〈x2〉 in the state n = 2, l = 1,m = 1.
(a) The ground state of the hydrogen atom is
ψ =1√πa3
e−r/a
〈rn〉 =1
πa3
∫rne−2r/ar2 sin θ dr dθ dφ
=4π
πa3
∫ ∞0
rn+2e−2r/a dr
〈r〉 =4
a3
∫ ∞0r3e−2r/a dr =
4
a3
(a2
)3
∫ ∞0r2e−2r/a dr
=4
a3
(a2
)23 · 2
∫ ∞0re−2r/a dr =
4
a3
(a2
)33!
∫ ∞0e−2r/a dr =
4
a33!(a
2
)4
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 7 / 19
Problem 4.13
(a) Find 〈r〉 and 〈r2〉 for an electron in the ground state of hydrogen.Express your answer in terms of the Bohr radius.
(b) Find 〈x〉 and 〈x2〉 for an electron in the ground state of hydrogen.
(c) Find 〈x2〉 in the state n = 2, l = 1,m = 1.
(a) The ground state of the hydrogen atom is
ψ =1√πa3
e−r/a
〈rn〉 =1
πa3
∫rne−2r/ar2 sin θ dr dθ dφ =
4π
πa3
∫ ∞0
rn+2e−2r/a dr
〈r〉 =4
a3
∫ ∞0r3e−2r/a dr =
4
a3
(a2
)3
∫ ∞0r2e−2r/a dr
=4
a3
(a2
)23 · 2
∫ ∞0re−2r/a dr =
4
a3
(a2
)33!
∫ ∞0e−2r/a dr =
4
a33!(a
2
)4
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 7 / 19
Problem 4.13
(a) Find 〈r〉 and 〈r2〉 for an electron in the ground state of hydrogen.Express your answer in terms of the Bohr radius.
(b) Find 〈x〉 and 〈x2〉 for an electron in the ground state of hydrogen.
(c) Find 〈x2〉 in the state n = 2, l = 1,m = 1.
(a) The ground state of the hydrogen atom is
ψ =1√πa3
e−r/a
〈rn〉 =1
πa3
∫rne−2r/ar2 sin θ dr dθ dφ =
4π
πa3
∫ ∞0
rn+2e−2r/a dr
〈r〉 =4
a3
∫ ∞0r3e−2r/a dr
=4
a3
(a2
)3
∫ ∞0r2e−2r/a dr
=4
a3
(a2
)23 · 2
∫ ∞0re−2r/a dr =
4
a3
(a2
)33!
∫ ∞0e−2r/a dr =
4
a33!(a
2
)4
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 7 / 19
Problem 4.13
(a) Find 〈r〉 and 〈r2〉 for an electron in the ground state of hydrogen.Express your answer in terms of the Bohr radius.
(b) Find 〈x〉 and 〈x2〉 for an electron in the ground state of hydrogen.
(c) Find 〈x2〉 in the state n = 2, l = 1,m = 1.
(a) The ground state of the hydrogen atom is
ψ =1√πa3
e−r/a
〈rn〉 =1
πa3
∫rne−2r/ar2 sin θ dr dθ dφ =
4π
πa3
∫ ∞0
rn+2e−2r/a dr
〈r〉 =4
a3
∫ ∞0r3e−2r/a dr =
4
a3
(a2
)3
∫ ∞0r2e−2r/a dr
=4
a3
(a2
)23 · 2
∫ ∞0re−2r/a dr =
4
a3
(a2
)33!
∫ ∞0e−2r/a dr =
4
a33!(a
2
)4
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 7 / 19
Problem 4.13
(a) Find 〈r〉 and 〈r2〉 for an electron in the ground state of hydrogen.Express your answer in terms of the Bohr radius.
(b) Find 〈x〉 and 〈x2〉 for an electron in the ground state of hydrogen.
(c) Find 〈x2〉 in the state n = 2, l = 1,m = 1.
(a) The ground state of the hydrogen atom is
ψ =1√πa3
e−r/a
〈rn〉 =1
πa3
∫rne−2r/ar2 sin θ dr dθ dφ =
4π
πa3
∫ ∞0
rn+2e−2r/a dr
〈r〉 =4
a3
∫ ∞0r3e−2r/a dr =
4
a3
(a2
)3
∫ ∞0r2e−2r/a dr
=4
a3
(a2
)23 · 2
∫ ∞0re−2r/a dr
=4
a3
(a2
)33!
∫ ∞0e−2r/a dr =
4
a33!(a
2
)4
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 7 / 19
Problem 4.13
(a) Find 〈r〉 and 〈r2〉 for an electron in the ground state of hydrogen.Express your answer in terms of the Bohr radius.
(b) Find 〈x〉 and 〈x2〉 for an electron in the ground state of hydrogen.
(c) Find 〈x2〉 in the state n = 2, l = 1,m = 1.
(a) The ground state of the hydrogen atom is
ψ =1√πa3
e−r/a
〈rn〉 =1
πa3
∫rne−2r/ar2 sin θ dr dθ dφ =
4π
πa3
∫ ∞0
rn+2e−2r/a dr
〈r〉 =4
a3
∫ ∞0r3e−2r/a dr =
4
a3
(a2
)3
∫ ∞0r2e−2r/a dr
=4
a3
(a2
)23 · 2
∫ ∞0re−2r/a dr =
4
a3
(a2
)33!
∫ ∞0e−2r/a dr
=4
a33!(a
2
)4
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 7 / 19
Problem 4.13
(a) Find 〈r〉 and 〈r2〉 for an electron in the ground state of hydrogen.Express your answer in terms of the Bohr radius.
(b) Find 〈x〉 and 〈x2〉 for an electron in the ground state of hydrogen.
(c) Find 〈x2〉 in the state n = 2, l = 1,m = 1.
(a) The ground state of the hydrogen atom is
ψ =1√πa3
e−r/a
〈rn〉 =1
πa3
∫rne−2r/ar2 sin θ dr dθ dφ =
4π
πa3
∫ ∞0
rn+2e−2r/a dr
〈r〉 =4
a3
∫ ∞0r3e−2r/a dr =
4
a3
(a2
)3
∫ ∞0r2e−2r/a dr
=4
a3
(a2
)23 · 2
∫ ∞0re−2r/a dr =
4
a3
(a2
)33!
∫ ∞0e−2r/a dr =
4
a33!(a
2
)4C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 7 / 19
Problem 4.13 (cont.)
〈r〉 =4
a33!(a
2
)4=
3
2a
⟨r2⟩
=4
a3
∫ ∞0
r4e−2r/a dr =4
a34!(a
2
)5= 3a2
(b) In terms of r we can write x = r sin θ cosφ so the integral over φbecomes ∫ 2π
0cosφ dφ = sinφ
∣∣∣2π0
= 0 −→ 〈x〉 = 0
Since r2 = x2 + y2 + z2, and by symmetry, all three directions are identicalfor this spherical state ⟨
x2⟩
=1
3
⟨r2⟩
= a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 8 / 19
Problem 4.13 (cont.)
〈r〉 =4
a33!(a
2
)4=
3
2a⟨
r2⟩
=4
a3
∫ ∞0
r4e−2r/a dr
=4
a34!(a
2
)5= 3a2
(b) In terms of r we can write x = r sin θ cosφ so the integral over φbecomes ∫ 2π
0cosφ dφ = sinφ
∣∣∣2π0
= 0 −→ 〈x〉 = 0
Since r2 = x2 + y2 + z2, and by symmetry, all three directions are identicalfor this spherical state ⟨
x2⟩
=1
3
⟨r2⟩
= a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 8 / 19
Problem 4.13 (cont.)
〈r〉 =4
a33!(a
2
)4=
3
2a⟨
r2⟩
=4
a3
∫ ∞0
r4e−2r/a dr =4
a34!(a
2
)5
= 3a2
(b) In terms of r we can write x = r sin θ cosφ so the integral over φbecomes ∫ 2π
0cosφ dφ = sinφ
∣∣∣2π0
= 0 −→ 〈x〉 = 0
Since r2 = x2 + y2 + z2, and by symmetry, all three directions are identicalfor this spherical state ⟨
x2⟩
=1
3
⟨r2⟩
= a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 8 / 19
Problem 4.13 (cont.)
〈r〉 =4
a33!(a
2
)4=
3
2a⟨
r2⟩
=4
a3
∫ ∞0
r4e−2r/a dr =4
a34!(a
2
)5= 3a2
(b) In terms of r we can write x = r sin θ cosφ so the integral over φbecomes ∫ 2π
0cosφ dφ = sinφ
∣∣∣2π0
= 0 −→ 〈x〉 = 0
Since r2 = x2 + y2 + z2, and by symmetry, all three directions are identicalfor this spherical state ⟨
x2⟩
=1
3
⟨r2⟩
= a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 8 / 19
Problem 4.13 (cont.)
〈r〉 =4
a33!(a
2
)4=
3
2a⟨
r2⟩
=4
a3
∫ ∞0
r4e−2r/a dr =4
a34!(a
2
)5= 3a2
(b) In terms of r we can write x = r sin θ cosφ so the integral over φbecomes
∫ 2π
0cosφ dφ = sinφ
∣∣∣2π0
= 0 −→ 〈x〉 = 0
Since r2 = x2 + y2 + z2, and by symmetry, all three directions are identicalfor this spherical state ⟨
x2⟩
=1
3
⟨r2⟩
= a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 8 / 19
Problem 4.13 (cont.)
〈r〉 =4
a33!(a
2
)4=
3
2a⟨
r2⟩
=4
a3
∫ ∞0
r4e−2r/a dr =4
a34!(a
2
)5= 3a2
(b) In terms of r we can write x = r sin θ cosφ so the integral over φbecomes ∫ 2π
0cosφ dφ
= sinφ∣∣∣2π0
= 0 −→ 〈x〉 = 0
Since r2 = x2 + y2 + z2, and by symmetry, all three directions are identicalfor this spherical state ⟨
x2⟩
=1
3
⟨r2⟩
= a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 8 / 19
Problem 4.13 (cont.)
〈r〉 =4
a33!(a
2
)4=
3
2a⟨
r2⟩
=4
a3
∫ ∞0
r4e−2r/a dr =4
a34!(a
2
)5= 3a2
(b) In terms of r we can write x = r sin θ cosφ so the integral over φbecomes ∫ 2π
0cosφ dφ = sinφ
∣∣∣2π0
= 0 −→ 〈x〉 = 0
Since r2 = x2 + y2 + z2, and by symmetry, all three directions are identicalfor this spherical state ⟨
x2⟩
=1
3
⟨r2⟩
= a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 8 / 19
Problem 4.13 (cont.)
〈r〉 =4
a33!(a
2
)4=
3
2a⟨
r2⟩
=4
a3
∫ ∞0
r4e−2r/a dr =4
a34!(a
2
)5= 3a2
(b) In terms of r we can write x = r sin θ cosφ so the integral over φbecomes ∫ 2π
0cosφ dφ = sinφ
∣∣∣2π0
= 0 −→ 〈x〉 = 0
Since r2 = x2 + y2 + z2, and by symmetry, all three directions are identicalfor this spherical state ⟨
x2⟩
=1
3
⟨r2⟩
= a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 8 / 19
Problem 4.13 (cont.)
〈r〉 =4
a33!(a
2
)4=
3
2a⟨
r2⟩
=4
a3
∫ ∞0
r4e−2r/a dr =4
a34!(a
2
)5= 3a2
(b) In terms of r we can write x = r sin θ cosφ so the integral over φbecomes ∫ 2π
0cosφ dφ = sinφ
∣∣∣2π0
= 0 −→ 〈x〉 = 0
Since r2 = x2 + y2 + z2, and by symmetry, all three directions are identicalfor this spherical state
⟨x2⟩
=1
3
⟨r2⟩
= a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 8 / 19
Problem 4.13 (cont.)
〈r〉 =4
a33!(a
2
)4=
3
2a⟨
r2⟩
=4
a3
∫ ∞0
r4e−2r/a dr =4
a34!(a
2
)5= 3a2
(b) In terms of r we can write x = r sin θ cosφ so the integral over φbecomes ∫ 2π
0cosφ dφ = sinφ
∣∣∣2π0
= 0 −→ 〈x〉 = 0
Since r2 = x2 + y2 + z2, and by symmetry, all three directions are identicalfor this spherical state ⟨
x2⟩
=1
3
⟨r2⟩
= a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 8 / 19
Problem 4.13 (cont.)
(c) The wavefunction for n = 2, l = 1,m = 1 is
ψ211 = R21Y11 = − 1√
πa
1
8a2re−r/2a sin θe iφ
⟨x2⟩
=1
πa
1
(8a2)2
∫(r2e−r/a sin2 θ)(r2 sin2 θ cos2 φ)r2 sin θ dr dθ dφ
=1
64πa5
∫ ∞0r6e−r/a dr
∫ π
0sin5 θ dθ
∫ 2π
0cos2 φ dφ
=1
64πa5(6!a7
) (2
2 · 41 · 3 · 5
) (1
2· 2π
)= 12a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 9 / 19
Problem 4.13 (cont.)
(c) The wavefunction for n = 2, l = 1,m = 1 is
ψ211 = R21Y11 = − 1√
πa
1
8a2re−r/2a sin θe iφ
⟨x2⟩
=1
πa
1
(8a2)2
∫(r2e−r/a sin2 θ)(r2 sin2 θ cos2 φ)r2 sin θ dr dθ dφ
=1
64πa5
∫ ∞0r6e−r/a dr
∫ π
0sin5 θ dθ
∫ 2π
0cos2 φ dφ
=1
64πa5(6!a7
) (2
2 · 41 · 3 · 5
) (1
2· 2π
)= 12a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 9 / 19
Problem 4.13 (cont.)
(c) The wavefunction for n = 2, l = 1,m = 1 is
ψ211 = R21Y11 = − 1√
πa
1
8a2re−r/2a sin θe iφ
⟨x2⟩
=1
πa
1
(8a2)2
∫(r2e−r/a sin2 θ)(r2 sin2 θ cos2 φ)r2 sin θ dr dθ dφ
=1
64πa5
∫ ∞0r6e−r/a dr
∫ π
0sin5 θ dθ
∫ 2π
0cos2 φ dφ
=1
64πa5(6!a7
) (2
2 · 41 · 3 · 5
) (1
2· 2π
)= 12a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 9 / 19
Problem 4.13 (cont.)
(c) The wavefunction for n = 2, l = 1,m = 1 is
ψ211 = R21Y11 = − 1√
πa
1
8a2re−r/2a sin θe iφ
⟨x2⟩
=1
πa
1
(8a2)2
∫(r2e−r/a sin2 θ)(r2 sin2 θ cos2 φ)r2 sin θ dr dθ dφ
=1
64πa5
∫ ∞0r6e−r/a dr
∫ π
0sin5 θ dθ
∫ 2π
0cos2 φ dφ
=1
64πa5(6!a7
) (2
2 · 41 · 3 · 5
) (1
2· 2π
)= 12a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 9 / 19
Problem 4.13 (cont.)
(c) The wavefunction for n = 2, l = 1,m = 1 is
ψ211 = R21Y11 = − 1√
πa
1
8a2re−r/2a sin θe iφ
⟨x2⟩
=1
πa
1
(8a2)2
∫(r2e−r/a sin2 θ)(r2 sin2 θ cos2 φ)r2 sin θ dr dθ dφ
=1
64πa5
∫ ∞0r6e−r/a dr
∫ π
0sin5 θ dθ
∫ 2π
0cos2 φ dφ
=1
64πa5(6!a7
) (2
2 · 41 · 3 · 5
) (1
2· 2π
)= 12a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 9 / 19
Problem 4.13 (cont.)
(c) The wavefunction for n = 2, l = 1,m = 1 is
ψ211 = R21Y11 = − 1√
πa
1
8a2re−r/2a sin θe iφ
⟨x2⟩
=1
πa
1
(8a2)2
∫(r2e−r/a sin2 θ)(r2 sin2 θ cos2 φ)r2 sin θ dr dθ dφ
=1
64πa5
∫ ∞0r6e−r/a dr
∫ π
0sin5 θ dθ
∫ 2π
0cos2 φ dφ
=1
64πa5(6!a7
)
(2
2 · 41 · 3 · 5
) (1
2· 2π
)= 12a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 9 / 19
Problem 4.13 (cont.)
(c) The wavefunction for n = 2, l = 1,m = 1 is
ψ211 = R21Y11 = − 1√
πa
1
8a2re−r/2a sin θe iφ
⟨x2⟩
=1
πa
1
(8a2)2
∫(r2e−r/a sin2 θ)(r2 sin2 θ cos2 φ)r2 sin θ dr dθ dφ
=1
64πa5
∫ ∞0r6e−r/a dr
∫ π
0sin5 θ dθ
∫ 2π
0cos2 φ dφ
=1
64πa5(6!a7
) (2
2 · 41 · 3 · 5
)
(1
2· 2π
)= 12a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 9 / 19
Problem 4.13 (cont.)
(c) The wavefunction for n = 2, l = 1,m = 1 is
ψ211 = R21Y11 = − 1√
πa
1
8a2re−r/2a sin θe iφ
⟨x2⟩
=1
πa
1
(8a2)2
∫(r2e−r/a sin2 θ)(r2 sin2 θ cos2 φ)r2 sin θ dr dθ dφ
=1
64πa5
∫ ∞0r6e−r/a dr
∫ π
0sin5 θ dθ
∫ 2π
0cos2 φ dφ
=1
64πa5(6!a7
) (2
2 · 41 · 3 · 5
) (1
2· 2π
)
= 12a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 9 / 19
Problem 4.13 (cont.)
(c) The wavefunction for n = 2, l = 1,m = 1 is
ψ211 = R21Y11 = − 1√
πa
1
8a2re−r/2a sin θe iφ
⟨x2⟩
=1
πa
1
(8a2)2
∫(r2e−r/a sin2 θ)(r2 sin2 θ cos2 φ)r2 sin θ dr dθ dφ
=1
64πa5
∫ ∞0r6e−r/a dr
∫ π
0sin5 θ dθ
∫ 2π
0cos2 φ dφ
=1
64πa5(6!a7
) (2
2 · 41 · 3 · 5
) (1
2· 2π
)= 12a2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 9 / 19
Problem 4.14
What is the most probable value of r , in the ground state of hydrogen?
ψ =1√πa3
e−r/a
P = |ψ|24πr2dr =4
a3e−2r/ar2dr = p(r)dr
dp
dr=
4
a3
[2re−2r/a + r2
(−2
ae−2r/a
)]=
8r
a3e−2r/a
(1− r
a
)dp
dr= 0 −→ 1− r
a= 0 −→ r = a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 10 / 19
Problem 4.14
What is the most probable value of r , in the ground state of hydrogen?
ψ =1√πa3
e−r/a
P = |ψ|24πr2dr =4
a3e−2r/ar2dr = p(r)dr
dp
dr=
4
a3
[2re−2r/a + r2
(−2
ae−2r/a
)]=
8r
a3e−2r/a
(1− r
a
)dp
dr= 0 −→ 1− r
a= 0 −→ r = a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 10 / 19
Problem 4.14
What is the most probable value of r , in the ground state of hydrogen?
ψ =1√πa3
e−r/a
P = |ψ|24πr2dr
=4
a3e−2r/ar2dr = p(r)dr
dp
dr=
4
a3
[2re−2r/a + r2
(−2
ae−2r/a
)]=
8r
a3e−2r/a
(1− r
a
)dp
dr= 0 −→ 1− r
a= 0 −→ r = a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 10 / 19
Problem 4.14
What is the most probable value of r , in the ground state of hydrogen?
ψ =1√πa3
e−r/a
P = |ψ|24πr2dr =4
a3e−2r/ar2dr
= p(r)dr
dp
dr=
4
a3
[2re−2r/a + r2
(−2
ae−2r/a
)]=
8r
a3e−2r/a
(1− r
a
)dp
dr= 0 −→ 1− r
a= 0 −→ r = a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 10 / 19
Problem 4.14
What is the most probable value of r , in the ground state of hydrogen?
ψ =1√πa3
e−r/a
P = |ψ|24πr2dr =4
a3e−2r/ar2dr = p(r)dr
dp
dr=
4
a3
[2re−2r/a + r2
(−2
ae−2r/a
)]=
8r
a3e−2r/a
(1− r
a
)dp
dr= 0 −→ 1− r
a= 0 −→ r = a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 10 / 19
Problem 4.14
What is the most probable value of r , in the ground state of hydrogen?
ψ =1√πa3
e−r/a
P = |ψ|24πr2dr =4
a3e−2r/ar2dr = p(r)dr
dp
dr=
4
a3
[2re−2r/a + r2
(−2
ae−2r/a
)]
=8r
a3e−2r/a
(1− r
a
)dp
dr= 0 −→ 1− r
a= 0 −→ r = a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 10 / 19
Problem 4.14
What is the most probable value of r , in the ground state of hydrogen?
ψ =1√πa3
e−r/a
P = |ψ|24πr2dr =4
a3e−2r/ar2dr = p(r)dr
dp
dr=
4
a3
[2re−2r/a + r2
(−2
ae−2r/a
)]=
8r
a3e−2r/a
(1− r
a
)
dp
dr= 0 −→ 1− r
a= 0 −→ r = a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 10 / 19
Problem 4.14
What is the most probable value of r , in the ground state of hydrogen?
ψ =1√πa3
e−r/a
P = |ψ|24πr2dr =4
a3e−2r/ar2dr = p(r)dr
dp
dr=
4
a3
[2re−2r/a + r2
(−2
ae−2r/a
)]=
8r
a3e−2r/a
(1− r
a
)dp
dr= 0
−→ 1− r
a= 0 −→ r = a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 10 / 19
Problem 4.14
What is the most probable value of r , in the ground state of hydrogen?
ψ =1√πa3
e−r/a
P = |ψ|24πr2dr =4
a3e−2r/ar2dr = p(r)dr
dp
dr=
4
a3
[2re−2r/a + r2
(−2
ae−2r/a
)]=
8r
a3e−2r/a
(1− r
a
)dp
dr= 0 −→ 1− r
a= 0
−→ r = a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 10 / 19
Problem 4.14
What is the most probable value of r , in the ground state of hydrogen?
ψ =1√πa3
e−r/a
P = |ψ|24πr2dr =4
a3e−2r/ar2dr = p(r)dr
dp
dr=
4
a3
[2re−2r/a + r2
(−2
ae−2r/a
)]=
8r
a3e−2r/a
(1− r
a
)dp
dr= 0 −→ 1− r
a= 0 −→ r = a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 10 / 19
Problem 4.16
A hydrogenic atoms consists of a single electron orbiting a nucleus with Zprotons. Determine the Bohr energies En(Z ), E1(Z ), the Bohr radius, andthe Rydberg constant. Where in the electromagnetic spectrum would theLyman series fall for Z = 2 and Z = 3?
E1 = −
[m2
2~2
(e2
4πε0
)2]−→ E1(Z ) = −
[m2
2~2
(Ze2
4πε0
)2]
= Z 2E1
En =E1
n2−→ En(Z ) = Z 2En
a ≡ 4πε0~2
me2−→ a(Z ) =
4πε0~2
mZe2=
a
Z
R ≡ m
4πc~3
(e2
4πε0
)2
−→ R(Z ) =m
4πc~3
(Ze2
4πε0
)2
= Z 2R
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 11 / 19
Problem 4.16
A hydrogenic atoms consists of a single electron orbiting a nucleus with Zprotons. Determine the Bohr energies En(Z ), E1(Z ), the Bohr radius, andthe Rydberg constant. Where in the electromagnetic spectrum would theLyman series fall for Z = 2 and Z = 3?
E1 = −
[m2
2~2
(e2
4πε0
)2]
−→ E1(Z ) = −
[m2
2~2
(Ze2
4πε0
)2]
= Z 2E1
En =E1
n2−→ En(Z ) = Z 2En
a ≡ 4πε0~2
me2−→ a(Z ) =
4πε0~2
mZe2=
a
Z
R ≡ m
4πc~3
(e2
4πε0
)2
−→ R(Z ) =m
4πc~3
(Ze2
4πε0
)2
= Z 2R
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 11 / 19
Problem 4.16
A hydrogenic atoms consists of a single electron orbiting a nucleus with Zprotons. Determine the Bohr energies En(Z ), E1(Z ), the Bohr radius, andthe Rydberg constant. Where in the electromagnetic spectrum would theLyman series fall for Z = 2 and Z = 3?
E1 = −
[m2
2~2
(e2
4πε0
)2]−→ E1(Z ) = −
[m2
2~2
(Ze2
4πε0
)2]
= Z 2E1
En =E1
n2−→ En(Z ) = Z 2En
a ≡ 4πε0~2
me2−→ a(Z ) =
4πε0~2
mZe2=
a
Z
R ≡ m
4πc~3
(e2
4πε0
)2
−→ R(Z ) =m
4πc~3
(Ze2
4πε0
)2
= Z 2R
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 11 / 19
Problem 4.16
A hydrogenic atoms consists of a single electron orbiting a nucleus with Zprotons. Determine the Bohr energies En(Z ), E1(Z ), the Bohr radius, andthe Rydberg constant. Where in the electromagnetic spectrum would theLyman series fall for Z = 2 and Z = 3?
E1 = −
[m2
2~2
(e2
4πε0
)2]−→ E1(Z ) = −
[m2
2~2
(Ze2
4πε0
)2]
= Z 2E1
En =E1
n2−→ En(Z ) = Z 2En
a ≡ 4πε0~2
me2−→ a(Z ) =
4πε0~2
mZe2=
a
Z
R ≡ m
4πc~3
(e2
4πε0
)2
−→ R(Z ) =m
4πc~3
(Ze2
4πε0
)2
= Z 2R
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 11 / 19
Problem 4.16
A hydrogenic atoms consists of a single electron orbiting a nucleus with Zprotons. Determine the Bohr energies En(Z ), E1(Z ), the Bohr radius, andthe Rydberg constant. Where in the electromagnetic spectrum would theLyman series fall for Z = 2 and Z = 3?
E1 = −
[m2
2~2
(e2
4πε0
)2]−→ E1(Z ) = −
[m2
2~2
(Ze2
4πε0
)2]
= Z 2E1
En =E1
n2
−→ En(Z ) = Z 2En
a ≡ 4πε0~2
me2−→ a(Z ) =
4πε0~2
mZe2=
a
Z
R ≡ m
4πc~3
(e2
4πε0
)2
−→ R(Z ) =m
4πc~3
(Ze2
4πε0
)2
= Z 2R
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 11 / 19
Problem 4.16
A hydrogenic atoms consists of a single electron orbiting a nucleus with Zprotons. Determine the Bohr energies En(Z ), E1(Z ), the Bohr radius, andthe Rydberg constant. Where in the electromagnetic spectrum would theLyman series fall for Z = 2 and Z = 3?
E1 = −
[m2
2~2
(e2
4πε0
)2]−→ E1(Z ) = −
[m2
2~2
(Ze2
4πε0
)2]
= Z 2E1
En =E1
n2−→ En(Z ) = Z 2En
a ≡ 4πε0~2
me2−→ a(Z ) =
4πε0~2
mZe2=
a
Z
R ≡ m
4πc~3
(e2
4πε0
)2
−→ R(Z ) =m
4πc~3
(Ze2
4πε0
)2
= Z 2R
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 11 / 19
Problem 4.16
A hydrogenic atoms consists of a single electron orbiting a nucleus with Zprotons. Determine the Bohr energies En(Z ), E1(Z ), the Bohr radius, andthe Rydberg constant. Where in the electromagnetic spectrum would theLyman series fall for Z = 2 and Z = 3?
E1 = −
[m2
2~2
(e2
4πε0
)2]−→ E1(Z ) = −
[m2
2~2
(Ze2
4πε0
)2]
= Z 2E1
En =E1
n2−→ En(Z ) = Z 2En
a ≡ 4πε0~2
me2
−→ a(Z ) =4πε0~2
mZe2=
a
Z
R ≡ m
4πc~3
(e2
4πε0
)2
−→ R(Z ) =m
4πc~3
(Ze2
4πε0
)2
= Z 2R
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 11 / 19
Problem 4.16
A hydrogenic atoms consists of a single electron orbiting a nucleus with Zprotons. Determine the Bohr energies En(Z ), E1(Z ), the Bohr radius, andthe Rydberg constant. Where in the electromagnetic spectrum would theLyman series fall for Z = 2 and Z = 3?
E1 = −
[m2
2~2
(e2
4πε0
)2]−→ E1(Z ) = −
[m2
2~2
(Ze2
4πε0
)2]
= Z 2E1
En =E1
n2−→ En(Z ) = Z 2En
a ≡ 4πε0~2
me2−→ a(Z ) =
4πε0~2
mZe2
=a
Z
R ≡ m
4πc~3
(e2
4πε0
)2
−→ R(Z ) =m
4πc~3
(Ze2
4πε0
)2
= Z 2R
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 11 / 19
Problem 4.16
A hydrogenic atoms consists of a single electron orbiting a nucleus with Zprotons. Determine the Bohr energies En(Z ), E1(Z ), the Bohr radius, andthe Rydberg constant. Where in the electromagnetic spectrum would theLyman series fall for Z = 2 and Z = 3?
E1 = −
[m2
2~2
(e2
4πε0
)2]−→ E1(Z ) = −
[m2
2~2
(Ze2
4πε0
)2]
= Z 2E1
En =E1
n2−→ En(Z ) = Z 2En
a ≡ 4πε0~2
me2−→ a(Z ) =
4πε0~2
mZe2=
a
Z
R ≡ m
4πc~3
(e2
4πε0
)2
−→ R(Z ) =m
4πc~3
(Ze2
4πε0
)2
= Z 2R
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 11 / 19
Problem 4.16
A hydrogenic atoms consists of a single electron orbiting a nucleus with Zprotons. Determine the Bohr energies En(Z ), E1(Z ), the Bohr radius, andthe Rydberg constant. Where in the electromagnetic spectrum would theLyman series fall for Z = 2 and Z = 3?
E1 = −
[m2
2~2
(e2
4πε0
)2]−→ E1(Z ) = −
[m2
2~2
(Ze2
4πε0
)2]
= Z 2E1
En =E1
n2−→ En(Z ) = Z 2En
a ≡ 4πε0~2
me2−→ a(Z ) =
4πε0~2
mZe2=
a
Z
R ≡ m
4πc~3
(e2
4πε0
)2
−→ R(Z ) =m
4πc~3
(Ze2
4πε0
)2
= Z 2R
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 11 / 19
Problem 4.16
A hydrogenic atoms consists of a single electron orbiting a nucleus with Zprotons. Determine the Bohr energies En(Z ), E1(Z ), the Bohr radius, andthe Rydberg constant. Where in the electromagnetic spectrum would theLyman series fall for Z = 2 and Z = 3?
E1 = −
[m2
2~2
(e2
4πε0
)2]−→ E1(Z ) = −
[m2
2~2
(Ze2
4πε0
)2]
= Z 2E1
En =E1
n2−→ En(Z ) = Z 2En
a ≡ 4πε0~2
me2−→ a(Z ) =
4πε0~2
mZe2=
a
Z
R ≡ m
4πc~3
(e2
4πε0
)2
−→ R(Z ) =m
4πc~3
(Ze2
4πε0
)2
= Z 2R
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 11 / 19
Problem 4.16
A hydrogenic atoms consists of a single electron orbiting a nucleus with Zprotons. Determine the Bohr energies En(Z ), E1(Z ), the Bohr radius, andthe Rydberg constant. Where in the electromagnetic spectrum would theLyman series fall for Z = 2 and Z = 3?
E1 = −
[m2
2~2
(e2
4πε0
)2]−→ E1(Z ) = −
[m2
2~2
(Ze2
4πε0
)2]
= Z 2E1
En =E1
n2−→ En(Z ) = Z 2En
a ≡ 4πε0~2
me2−→ a(Z ) =
4πε0~2
mZe2=
a
Z
R ≡ m
4πc~3
(e2
4πε0
)2
−→ R(Z ) =m
4πc~3
(Ze2
4πε0
)2
= Z 2R
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 11 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)
= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)
= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.16 (cont.)
The Lyman energies are those for transitions n = 2→ 1 and n =∞→ 1which depend on Z as follows
∆E2→1 = Z 2E1
(1
4− 1
)= Z 2 · 10.2 eV
∆E∞→1 = Z 2E1
(1
∞− 1
)= Z 2 · 13.6 eV
for Z = 2
∆E2→1 = 40.8 eV
λ2→1 = 30.3 nm
∆E∞→1 = 54.4 eV
λ∞→1 = 22.8 nm
for Z = 3
∆E2→1 = 91.8 eV
λ2→1 = 13.5 nm
∆E∞→1 = 122.4 eV
λ∞→1 = 10.1 nm
all these are in the ultraviolet regime
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 12 / 19
Problem 4.55
The electron in a hydrogen atom occupies the combined spin and positionstate
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)and let ~J ≡ ~L + ~S
(a) If you measured L2, what values might you get, and what is theprobability of each?
Repeat for (b) Lz , (c) S2, (d) Sz , (e) J2, (f) Jz .
(g) If you measured the position of the particle, what is the probabilitydensity for finding it at r , θ, φ?
(h) If you measured both the Z component of the spin and the distancefrom the origin, what is the probability density for finding the particle withspin up and at a radius r?
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 13 / 19
Problem 4.55 – solution
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)(a) L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
(b) Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
(c) S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
(d) Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 14 / 19
Problem 4.55 – solution
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)(a) L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1)
= ~2[1(2)] = 2~2
(b) Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
(c) S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
(d) Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 14 / 19
Problem 4.55 – solution
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)(a) L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
(b) Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
(c) S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
(d) Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 14 / 19
Problem 4.55 – solution
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)(a) L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
(b) Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
(c) S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
(d) Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 14 / 19
Problem 4.55 – solution
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)(a) L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
(b) Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
(c) S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
(d) Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 14 / 19
Problem 4.55 – solution
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)(a) L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
(b) Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
(c) S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
(d) Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 14 / 19
Problem 4.55 – solution
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)(a) L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
(b) Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
(c) S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
(d) Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 14 / 19
Problem 4.55 – solution
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)(a) L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
(b) Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
(c) S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1)
= ~2[ 12( 32)] = 3
4~2
(d) Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 14 / 19
Problem 4.55 – solution
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)(a) L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
(b) Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
(c) S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
(d) Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 14 / 19
Problem 4.55 – solution
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)(a) L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
(b) Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
(c) S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
(d) Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 14 / 19
Problem 4.55 – solution
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)(a) L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
(b) Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
(c) S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
(d) Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 14 / 19
Problem 4.55 – solution
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)(a) L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
(b) Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
(c) S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
(d) Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 14 / 19
Problem 4.55 – solution
(e) Rewriting the angular and spin portion of the wave function with thefirst bra-ket corresponding to the angular momentum
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−1
2〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉+
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
(f) both states have mj = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 15 / 19
Problem 4.55 – solution
(e) Rewriting the angular and spin portion of the wave function with thefirst bra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)
√13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−1
2〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉+
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
(f) both states have mj = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 15 / 19
Problem 4.55 – solution
(e) Rewriting the angular and spin portion of the wave function with thefirst bra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−1
2〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉+
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
(f) both states have mj = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 15 / 19
Problem 4.55 – solution
(e) Rewriting the angular and spin portion of the wave function with thefirst bra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−1
2〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable
√13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉+
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
(f) both states have mj = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 15 / 19
Problem 4.55 – solution
(e) Rewriting the angular and spin portion of the wave function with thefirst bra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−1
2〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable
√13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉+
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
(f) both states have mj = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 15 / 19
Problem 4.55 – solution
(e) Rewriting the angular and spin portion of the wave function with thefirst bra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−1
2〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉+
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
(f) both states have mj = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 15 / 19
Problem 4.55 – solution
(e) Rewriting the angular and spin portion of the wave function with thefirst bra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−1
2〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉+
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
(f) both states have mj = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 15 / 19
Problem 4.55 – solution
(e) Rewriting the angular and spin portion of the wave function with thefirst bra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−1
2〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉+
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
(f) both states have mj = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 15 / 19
Problem 4.55 – solution
(e) Rewriting the angular and spin portion of the wave function with thefirst bra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−1
2〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉+
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2
; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
(f) both states have mj = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 15 / 19
Problem 4.55 – solution
(e) Rewriting the angular and spin portion of the wave function with thefirst bra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−1
2〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉+
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
(f) both states have mj = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 15 / 19
Problem 4.55 – solution
(e) Rewriting the angular and spin portion of the wave function with thefirst bra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−1
2〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉+
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
(f) both states have mj = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 15 / 19
Problem 4.55 – solution
(g) The probability density is simply |ψ|2
|ψ|2 = |R21|2{
1
3|Y 0
1 |2(χ†+χ+)
+
√2
3
[Y 0∗1 Y 1
1 (���χ†+χ−) + Y 1∗
1 Y 01 (���χ†−χ+)
]+
2
3|Y 1
1 |2(χ†−χ−)
}=
1
3|R21|2
(|Y 0
1 |2 + 2|Y 11 |2)
=1
3· 1
24· 1
a3· r
2
a2e−r/a
[3
4πcos2 θ + 2
3
8πsin2 θ
]=
1
3 · 24a5r2e−r/a · 3
4π(cos2 θ + sin2 θ) =
1
96πa5r2e−r/a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 16 / 19
Problem 4.55 – solution
(g) The probability density is simply |ψ|2
|ψ|2 = |R21|2{
1
3|Y 0
1 |2(χ†+χ+)
+
√2
3
[Y 0∗1 Y 1
1 (���χ†+χ−) + Y 1∗
1 Y 01 (���χ†−χ+)
]+
2
3|Y 1
1 |2(χ†−χ−)
}
=1
3|R21|2
(|Y 0
1 |2 + 2|Y 11 |2)
=1
3· 1
24· 1
a3· r
2
a2e−r/a
[3
4πcos2 θ + 2
3
8πsin2 θ
]=
1
3 · 24a5r2e−r/a · 3
4π(cos2 θ + sin2 θ) =
1
96πa5r2e−r/a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 16 / 19
Problem 4.55 – solution
(g) The probability density is simply |ψ|2
|ψ|2 = |R21|2{
1
3|Y 0
1 |2(χ†+χ+)
+
√2
3
[Y 0∗1 Y 1
1 (���χ†+χ−) + Y 1∗
1 Y 01 (���χ†−χ+)
]+
2
3|Y 1
1 |2(χ†−χ−)
}
=1
3|R21|2
(|Y 0
1 |2 + 2|Y 11 |2)
=1
3· 1
24· 1
a3· r
2
a2e−r/a
[3
4πcos2 θ + 2
3
8πsin2 θ
]=
1
3 · 24a5r2e−r/a · 3
4π(cos2 θ + sin2 θ) =
1
96πa5r2e−r/a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 16 / 19
Problem 4.55 – solution
(g) The probability density is simply |ψ|2
|ψ|2 = |R21|2{
1
3|Y 0
1 |2(χ†+χ+)
+
√2
3
[Y 0∗1 Y 1
1 (χ†+χ−) + Y 1∗1 Y 0
1 (χ†−χ+)]
+2
3|Y 1
1 |2(χ†−χ−)
}
=1
3|R21|2
(|Y 0
1 |2 + 2|Y 11 |2)
=1
3· 1
24· 1
a3· r
2
a2e−r/a
[3
4πcos2 θ + 2
3
8πsin2 θ
]=
1
3 · 24a5r2e−r/a · 3
4π(cos2 θ + sin2 θ) =
1
96πa5r2e−r/a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 16 / 19
Problem 4.55 – solution
(g) The probability density is simply |ψ|2
|ψ|2 = |R21|2{
1
3|Y 0
1 |2(χ†+χ+)
+
√2
3
[Y 0∗1 Y 1
1 (χ†+χ−) + Y 1∗1 Y 0
1 (χ†−χ+)]
+2
3|Y 1
1 |2(χ†−χ−)
}
=1
3|R21|2
(|Y 0
1 |2 + 2|Y 11 |2)
=1
3· 1
24· 1
a3· r
2
a2e−r/a
[3
4πcos2 θ + 2
3
8πsin2 θ
]=
1
3 · 24a5r2e−r/a · 3
4π(cos2 θ + sin2 θ) =
1
96πa5r2e−r/a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 16 / 19
Problem 4.55 – solution
(g) The probability density is simply |ψ|2
|ψ|2 = |R21|2{
1
3|Y 0
1 |2(χ†+χ+)
+
√2
3
[Y 0∗1 Y 1
1 (���χ†+χ−) + Y 1∗
1 Y 01 (���χ†−χ+)
]+
2
3|Y 1
1 |2(χ†−χ−)
}
=1
3|R21|2
(|Y 0
1 |2 + 2|Y 11 |2)
=1
3· 1
24· 1
a3· r
2
a2e−r/a
[3
4πcos2 θ + 2
3
8πsin2 θ
]=
1
3 · 24a5r2e−r/a · 3
4π(cos2 θ + sin2 θ) =
1
96πa5r2e−r/a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 16 / 19
Problem 4.55 – solution
(g) The probability density is simply |ψ|2
|ψ|2 = |R21|2{
1
3|Y 0
1 |2(χ†+χ+)
+
√2
3
[Y 0∗1 Y 1
1 (���χ†+χ−) + Y 1∗
1 Y 01 (���χ†−χ+)
]+
2
3|Y 1
1 |2(χ†−χ−)
}=
1
3|R21|2
(|Y 0
1 |2 + 2|Y 11 |2)
=1
3· 1
24· 1
a3· r
2
a2e−r/a
[3
4πcos2 θ + 2
3
8πsin2 θ
]=
1
3 · 24a5r2e−r/a · 3
4π(cos2 θ + sin2 θ) =
1
96πa5r2e−r/a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 16 / 19
Problem 4.55 – solution
(g) The probability density is simply |ψ|2
|ψ|2 = |R21|2{
1
3|Y 0
1 |2(χ†+χ+)
+
√2
3
[Y 0∗1 Y 1
1 (���χ†+χ−) + Y 1∗
1 Y 01 (���χ†−χ+)
]+
2
3|Y 1
1 |2(χ†−χ−)
}=
1
3|R21|2
(|Y 0
1 |2 + 2|Y 11 |2)
=1
3· 1
24· 1
a3· r
2
a2e−r/a
[3
4πcos2 θ + 2
3
8πsin2 θ
]
=1
3 · 24a5r2e−r/a · 3
4π(cos2 θ + sin2 θ) =
1
96πa5r2e−r/a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 16 / 19
Problem 4.55 – solution
(g) The probability density is simply |ψ|2
|ψ|2 = |R21|2{
1
3|Y 0
1 |2(χ†+χ+)
+
√2
3
[Y 0∗1 Y 1
1 (���χ†+χ−) + Y 1∗
1 Y 01 (���χ†−χ+)
]+
2
3|Y 1
1 |2(χ†−χ−)
}=
1
3|R21|2
(|Y 0
1 |2 + 2|Y 11 |2)
=1
3· 1
24· 1
a3· r
2
a2e−r/a
[3
4πcos2 θ + 2
3
8πsin2 θ
]=
1
3 · 24a5r2e−r/a · 3
4π(cos2 θ + sin2 θ)
=1
96πa5r2e−r/a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 16 / 19
Problem 4.55 – solution
(g) The probability density is simply |ψ|2
|ψ|2 = |R21|2{
1
3|Y 0
1 |2(χ†+χ+)
+
√2
3
[Y 0∗1 Y 1
1 (���χ†+χ−) + Y 1∗
1 Y 01 (���χ†−χ+)
]+
2
3|Y 1
1 |2(χ†−χ−)
}=
1
3|R21|2
(|Y 0
1 |2 + 2|Y 11 |2)
=1
3· 1
24· 1
a3· r
2
a2e−r/a
[3
4πcos2 θ + 2
3
8πsin2 θ
]=
1
3 · 24a5r2e−r/a · 3
4π(cos2 θ + sin2 θ) =
1
96πa5r2e−r/a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 16 / 19
Problem 4.55 – solution
(h) For a spin up particle only, the probability density becomes
1
3|R21|2
∫|Y 0
1 |2 sin2 θ dθdφ =1
3|R21|2 =
1
72a5r2e−r/a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 17 / 19
Problem 4.55 – solution
(h) For a spin up particle only, the probability density becomes
1
3|R21|2
∫|Y 0
1 |2 sin2 θ dθdφ
=1
3|R21|2 =
1
72a5r2e−r/a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 17 / 19
Problem 4.55 – solution
(h) For a spin up particle only, the probability density becomes
1
3|R21|2
∫|Y 0
1 |2 sin2 θ dθdφ =1
3|R21|2
=1
72a5r2e−r/a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 17 / 19
Problem 4.55 – solution
(h) For a spin up particle only, the probability density becomes
1
3|R21|2
∫|Y 0
1 |2 sin2 θ dθdφ =1
3|R21|2 =
1
72a5r2e−r/a
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 17 / 19
Stern-Gerlach problem
A beam of atoms with spin quantum number 12 and zero orbital angular
momentum passes through a Stern-Gerlach magnet whose magnetic fieldis along a direction D at an angle θ to the z axis. The emerging beamwith spins along D is passed through a second Stern-Gerlach magnet withits field along the z axis. What is the ratio of the spin “up” and spin“down” atoms after the second magnet?
First compute the spin operatoralong D
The corresponding Pauli spin ma-trix is thus
Solving the eigenvalue equation
SD = cos θSz + sin θSx
σD =
(cos θ sin θsin θ − cos θ
)0 = det
∣∣∣∣ cos θ − λ sin θsin θ − cos θ − λ
∣∣∣∣0 = −(cos2 θ − λ2)− sin2 θ
→ λ2 = 1 → λ = ±1
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 18 / 19
Stern-Gerlach problem
A beam of atoms with spin quantum number 12 and zero orbital angular
momentum passes through a Stern-Gerlach magnet whose magnetic fieldis along a direction D at an angle θ to the z axis. The emerging beamwith spins along D is passed through a second Stern-Gerlach magnet withits field along the z axis. What is the ratio of the spin “up” and spin“down” atoms after the second magnet?
First compute the spin operatoralong D
The corresponding Pauli spin ma-trix is thus
Solving the eigenvalue equation
SD = cos θSz + sin θSx
σD =
(cos θ sin θsin θ − cos θ
)0 = det
∣∣∣∣ cos θ − λ sin θsin θ − cos θ − λ
∣∣∣∣0 = −(cos2 θ − λ2)− sin2 θ
→ λ2 = 1 → λ = ±1
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 18 / 19
Stern-Gerlach problem
A beam of atoms with spin quantum number 12 and zero orbital angular
momentum passes through a Stern-Gerlach magnet whose magnetic fieldis along a direction D at an angle θ to the z axis. The emerging beamwith spins along D is passed through a second Stern-Gerlach magnet withits field along the z axis. What is the ratio of the spin “up” and spin“down” atoms after the second magnet?
First compute the spin operatoralong D
The corresponding Pauli spin ma-trix is thus
Solving the eigenvalue equation
SD = cos θSz + sin θSx
σD =
(cos θ sin θsin θ − cos θ
)0 = det
∣∣∣∣ cos θ − λ sin θsin θ − cos θ − λ
∣∣∣∣0 = −(cos2 θ − λ2)− sin2 θ
→ λ2 = 1 → λ = ±1
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 18 / 19
Stern-Gerlach problem
A beam of atoms with spin quantum number 12 and zero orbital angular
momentum passes through a Stern-Gerlach magnet whose magnetic fieldis along a direction D at an angle θ to the z axis. The emerging beamwith spins along D is passed through a second Stern-Gerlach magnet withits field along the z axis. What is the ratio of the spin “up” and spin“down” atoms after the second magnet?
First compute the spin operatoralong D
The corresponding Pauli spin ma-trix is thus
Solving the eigenvalue equation
SD = cos θSz + sin θSx
σD =
(cos θ sin θsin θ − cos θ
)0 = det
∣∣∣∣ cos θ − λ sin θsin θ − cos θ − λ
∣∣∣∣0 = −(cos2 θ − λ2)− sin2 θ
→ λ2 = 1 → λ = ±1
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 18 / 19
Stern-Gerlach problem
A beam of atoms with spin quantum number 12 and zero orbital angular
momentum passes through a Stern-Gerlach magnet whose magnetic fieldis along a direction D at an angle θ to the z axis. The emerging beamwith spins along D is passed through a second Stern-Gerlach magnet withits field along the z axis. What is the ratio of the spin “up” and spin“down” atoms after the second magnet?
First compute the spin operatoralong D
The corresponding Pauli spin ma-trix is thus
Solving the eigenvalue equation
SD = cos θSz + sin θSx
σD =
(cos θ sin θsin θ − cos θ
)
0 = det
∣∣∣∣ cos θ − λ sin θsin θ − cos θ − λ
∣∣∣∣0 = −(cos2 θ − λ2)− sin2 θ
→ λ2 = 1 → λ = ±1
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 18 / 19
Stern-Gerlach problem
A beam of atoms with spin quantum number 12 and zero orbital angular
momentum passes through a Stern-Gerlach magnet whose magnetic fieldis along a direction D at an angle θ to the z axis. The emerging beamwith spins along D is passed through a second Stern-Gerlach magnet withits field along the z axis. What is the ratio of the spin “up” and spin“down” atoms after the second magnet?
First compute the spin operatoralong D
The corresponding Pauli spin ma-trix is thus
Solving the eigenvalue equation
SD = cos θSz + sin θSx
σD =
(cos θ sin θsin θ − cos θ
)
0 = det
∣∣∣∣ cos θ − λ sin θsin θ − cos θ − λ
∣∣∣∣0 = −(cos2 θ − λ2)− sin2 θ
→ λ2 = 1 → λ = ±1
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 18 / 19
Stern-Gerlach problem
A beam of atoms with spin quantum number 12 and zero orbital angular
momentum passes through a Stern-Gerlach magnet whose magnetic fieldis along a direction D at an angle θ to the z axis. The emerging beamwith spins along D is passed through a second Stern-Gerlach magnet withits field along the z axis. What is the ratio of the spin “up” and spin“down” atoms after the second magnet?
First compute the spin operatoralong D
The corresponding Pauli spin ma-trix is thus
Solving the eigenvalue equation
SD = cos θSz + sin θSx
σD =
(cos θ sin θsin θ − cos θ
)0 = det
∣∣∣∣ cos θ − λ sin θsin θ − cos θ − λ
∣∣∣∣
0 = −(cos2 θ − λ2)− sin2 θ
→ λ2 = 1 → λ = ±1
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 18 / 19
Stern-Gerlach problem
A beam of atoms with spin quantum number 12 and zero orbital angular
momentum passes through a Stern-Gerlach magnet whose magnetic fieldis along a direction D at an angle θ to the z axis. The emerging beamwith spins along D is passed through a second Stern-Gerlach magnet withits field along the z axis. What is the ratio of the spin “up” and spin“down” atoms after the second magnet?
First compute the spin operatoralong D
The corresponding Pauli spin ma-trix is thus
Solving the eigenvalue equation
SD = cos θSz + sin θSx
σD =
(cos θ sin θsin θ − cos θ
)0 = det
∣∣∣∣ cos θ − λ sin θsin θ − cos θ − λ
∣∣∣∣0 = −(cos2 θ − λ2)− sin2 θ
→ λ2 = 1 → λ = ±1
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 18 / 19
Stern-Gerlach problem
A beam of atoms with spin quantum number 12 and zero orbital angular
momentum passes through a Stern-Gerlach magnet whose magnetic fieldis along a direction D at an angle θ to the z axis. The emerging beamwith spins along D is passed through a second Stern-Gerlach magnet withits field along the z axis. What is the ratio of the spin “up” and spin“down” atoms after the second magnet?
First compute the spin operatoralong D
The corresponding Pauli spin ma-trix is thus
Solving the eigenvalue equation
SD = cos θSz + sin θSx
σD =
(cos θ sin θsin θ − cos θ
)0 = det
∣∣∣∣ cos θ − λ sin θsin θ − cos θ − λ
∣∣∣∣0 = −(cos2 θ − λ2)− sin2 θ → λ2 = 1
→ λ = ±1
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 18 / 19
Stern-Gerlach problem
A beam of atoms with spin quantum number 12 and zero orbital angular
momentum passes through a Stern-Gerlach magnet whose magnetic fieldis along a direction D at an angle θ to the z axis. The emerging beamwith spins along D is passed through a second Stern-Gerlach magnet withits field along the z axis. What is the ratio of the spin “up” and spin“down” atoms after the second magnet?
First compute the spin operatoralong D
The corresponding Pauli spin ma-trix is thus
Solving the eigenvalue equation
SD = cos θSz + sin θSx
σD =
(cos θ sin θsin θ − cos θ
)0 = det
∣∣∣∣ cos θ − λ sin θsin θ − cos θ − λ
∣∣∣∣0 = −(cos2 θ − λ2)− sin2 θ → λ2 = 1 → λ = ±1
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 18 / 19
Stern-Gerlach problem (cont.)
Let’s calculate the eigenspinors of the σD operator in terms of the spinorsin the z direction
(cos θ sin θsin θ − cos θ
)(αβ
)= λ
(αβ
)α cos θ + β sin θ = λα α sin θ − β cos θ = λβ
α
β=
sin θ
λ− cos θ
=λ+ cos θ
sin θ
Choosing the eigenvalue λ = +1 (since the first magnet produces a beampolarized in the D direction), the ratio of spin up to spin down after thesecond magnet is simply(
α
β
)2
=
(sin θ
1− cos θ
)2
=
(2 sin θ
2 cos θ21− cos2 θ2 + sin2 θ
2
)2
=cos2 θ2sin2 θ
2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 19 / 19
Stern-Gerlach problem (cont.)
Let’s calculate the eigenspinors of the σD operator in terms of the spinorsin the z direction(
cos θ sin θsin θ − cos θ
)(αβ
)= λ
(αβ
)
α cos θ + β sin θ = λα α sin θ − β cos θ = λβ
α
β=
sin θ
λ− cos θ
=λ+ cos θ
sin θ
Choosing the eigenvalue λ = +1 (since the first magnet produces a beampolarized in the D direction), the ratio of spin up to spin down after thesecond magnet is simply(
α
β
)2
=
(sin θ
1− cos θ
)2
=
(2 sin θ
2 cos θ21− cos2 θ2 + sin2 θ
2
)2
=cos2 θ2sin2 θ
2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 19 / 19
Stern-Gerlach problem (cont.)
Let’s calculate the eigenspinors of the σD operator in terms of the spinorsin the z direction(
cos θ sin θsin θ − cos θ
)(αβ
)= λ
(αβ
)α cos θ + β sin θ = λα
α sin θ − β cos θ = λβ
α
β=
sin θ
λ− cos θ
=λ+ cos θ
sin θ
Choosing the eigenvalue λ = +1 (since the first magnet produces a beampolarized in the D direction), the ratio of spin up to spin down after thesecond magnet is simply(
α
β
)2
=
(sin θ
1− cos θ
)2
=
(2 sin θ
2 cos θ21− cos2 θ2 + sin2 θ
2
)2
=cos2 θ2sin2 θ
2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 19 / 19
Stern-Gerlach problem (cont.)
Let’s calculate the eigenspinors of the σD operator in terms of the spinorsin the z direction(
cos θ sin θsin θ − cos θ
)(αβ
)= λ
(αβ
)α cos θ + β sin θ = λα α sin θ − β cos θ = λβ
α
β=
sin θ
λ− cos θ
=λ+ cos θ
sin θ
Choosing the eigenvalue λ = +1 (since the first magnet produces a beampolarized in the D direction), the ratio of spin up to spin down after thesecond magnet is simply(
α
β
)2
=
(sin θ
1− cos θ
)2
=
(2 sin θ
2 cos θ21− cos2 θ2 + sin2 θ
2
)2
=cos2 θ2sin2 θ
2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 19 / 19
Stern-Gerlach problem (cont.)
Let’s calculate the eigenspinors of the σD operator in terms of the spinorsin the z direction(
cos θ sin θsin θ − cos θ
)(αβ
)= λ
(αβ
)α cos θ + β sin θ = λα α sin θ − β cos θ = λβ
α
β=
sin θ
λ− cos θ
=λ+ cos θ
sin θ
Choosing the eigenvalue λ = +1 (since the first magnet produces a beampolarized in the D direction), the ratio of spin up to spin down after thesecond magnet is simply(
α
β
)2
=
(sin θ
1− cos θ
)2
=
(2 sin θ
2 cos θ21− cos2 θ2 + sin2 θ
2
)2
=cos2 θ2sin2 θ
2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 19 / 19
Stern-Gerlach problem (cont.)
Let’s calculate the eigenspinors of the σD operator in terms of the spinorsin the z direction(
cos θ sin θsin θ − cos θ
)(αβ
)= λ
(αβ
)α cos θ + β sin θ = λα α sin θ − β cos θ = λβ
α
β=
sin θ
λ− cos θ=λ+ cos θ
sin θ
Choosing the eigenvalue λ = +1 (since the first magnet produces a beampolarized in the D direction), the ratio of spin up to spin down after thesecond magnet is simply(
α
β
)2
=
(sin θ
1− cos θ
)2
=
(2 sin θ
2 cos θ21− cos2 θ2 + sin2 θ
2
)2
=cos2 θ2sin2 θ
2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 19 / 19
Stern-Gerlach problem (cont.)
Let’s calculate the eigenspinors of the σD operator in terms of the spinorsin the z direction(
cos θ sin θsin θ − cos θ
)(αβ
)= λ
(αβ
)α cos θ + β sin θ = λα α sin θ − β cos θ = λβ
α
β=
sin θ
λ− cos θ=λ+ cos θ
sin θ
Choosing the eigenvalue λ = +1 (since the first magnet produces a beampolarized in the D direction), the ratio of spin up to spin down after thesecond magnet is simply
(α
β
)2
=
(sin θ
1− cos θ
)2
=
(2 sin θ
2 cos θ21− cos2 θ2 + sin2 θ
2
)2
=cos2 θ2sin2 θ
2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 19 / 19
Stern-Gerlach problem (cont.)
Let’s calculate the eigenspinors of the σD operator in terms of the spinorsin the z direction(
cos θ sin θsin θ − cos θ
)(αβ
)= λ
(αβ
)α cos θ + β sin θ = λα α sin θ − β cos θ = λβ
α
β=
sin θ
λ− cos θ=λ+ cos θ
sin θ
Choosing the eigenvalue λ = +1 (since the first magnet produces a beampolarized in the D direction), the ratio of spin up to spin down after thesecond magnet is simply(
α
β
)2
=
(sin θ
1− cos θ
)2
=
(2 sin θ
2 cos θ21− cos2 θ2 + sin2 θ
2
)2
=cos2 θ2sin2 θ
2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 19 / 19
Stern-Gerlach problem (cont.)
Let’s calculate the eigenspinors of the σD operator in terms of the spinorsin the z direction(
cos θ sin θsin θ − cos θ
)(αβ
)= λ
(αβ
)α cos θ + β sin θ = λα α sin θ − β cos θ = λβ
α
β=
sin θ
λ− cos θ=λ+ cos θ
sin θ
Choosing the eigenvalue λ = +1 (since the first magnet produces a beampolarized in the D direction), the ratio of spin up to spin down after thesecond magnet is simply(
α
β
)2
=
(sin θ
1− cos θ
)2
=
(2 sin θ
2 cos θ21− cos2 θ2 + sin2 θ
2
)2
=cos2 θ2sin2 θ
2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 19 / 19
Stern-Gerlach problem (cont.)
Let’s calculate the eigenspinors of the σD operator in terms of the spinorsin the z direction(
cos θ sin θsin θ − cos θ
)(αβ
)= λ
(αβ
)α cos θ + β sin θ = λα α sin θ − β cos θ = λβ
α
β=
sin θ
λ− cos θ=λ+ cos θ
sin θ
Choosing the eigenvalue λ = +1 (since the first magnet produces a beampolarized in the D direction), the ratio of spin up to spin down after thesecond magnet is simply(
α
β
)2
=
(sin θ
1− cos θ
)2
=
(2 sin θ
2 cos θ21− cos2 θ2 + sin2 θ
2
)2
=cos2 θ2sin2 θ
2
C. Segre (IIT) PHYS 405 - Fall 2015 November 11, 2015 19 / 19
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