Upload
others
View
0
Download
0
Embed Size (px)
Citation preview
Today’s Outline - January 20, 2016
• Multiple degenerate states
• Fine structure
• Relativistic correction
• Spin-orbit coupling
Homework Assignment #02:Chapter 5:27,30; Chapter 6:1,4,6,29due Wednesday, January 27, 2016
Homework Assignment #03:Chapter 6:9,17,21,24,25,32due Wednesday, February 3, 2016
Office hours:Monday & Wednesday, 13:50–15:05
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 1 / 14
Today’s Outline - January 20, 2016
• Multiple degenerate states
• Fine structure
• Relativistic correction
• Spin-orbit coupling
Homework Assignment #02:Chapter 5:27,30; Chapter 6:1,4,6,29due Wednesday, January 27, 2016
Homework Assignment #03:Chapter 6:9,17,21,24,25,32due Wednesday, February 3, 2016
Office hours:Monday & Wednesday, 13:50–15:05
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 1 / 14
Today’s Outline - January 20, 2016
• Multiple degenerate states
• Fine structure
• Relativistic correction
• Spin-orbit coupling
Homework Assignment #02:Chapter 5:27,30; Chapter 6:1,4,6,29due Wednesday, January 27, 2016
Homework Assignment #03:Chapter 6:9,17,21,24,25,32due Wednesday, February 3, 2016
Office hours:Monday & Wednesday, 13:50–15:05
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 1 / 14
Today’s Outline - January 20, 2016
• Multiple degenerate states
• Fine structure
• Relativistic correction
• Spin-orbit coupling
Homework Assignment #02:Chapter 5:27,30; Chapter 6:1,4,6,29due Wednesday, January 27, 2016
Homework Assignment #03:Chapter 6:9,17,21,24,25,32due Wednesday, February 3, 2016
Office hours:Monday & Wednesday, 13:50–15:05
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 1 / 14
Today’s Outline - January 20, 2016
• Multiple degenerate states
• Fine structure
• Relativistic correction
• Spin-orbit coupling
Homework Assignment #02:Chapter 5:27,30; Chapter 6:1,4,6,29due Wednesday, January 27, 2016
Homework Assignment #03:Chapter 6:9,17,21,24,25,32due Wednesday, February 3, 2016
Office hours:Monday & Wednesday, 13:50–15:05
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 1 / 14
Today’s Outline - January 20, 2016
• Multiple degenerate states
• Fine structure
• Relativistic correction
• Spin-orbit coupling
Homework Assignment #02:Chapter 5:27,30; Chapter 6:1,4,6,29due Wednesday, January 27, 2016
Homework Assignment #03:Chapter 6:9,17,21,24,25,32due Wednesday, February 3, 2016
Office hours:Monday & Wednesday, 13:50–15:05
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 1 / 14
Today’s Outline - January 20, 2016
• Multiple degenerate states
• Fine structure
• Relativistic correction
• Spin-orbit coupling
Homework Assignment #02:Chapter 5:27,30; Chapter 6:1,4,6,29due Wednesday, January 27, 2016
Homework Assignment #03:Chapter 6:9,17,21,24,25,32due Wednesday, February 3, 2016
Office hours:Monday & Wednesday, 13:50–15:05
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 1 / 14
Today’s Outline - January 20, 2016
• Multiple degenerate states
• Fine structure
• Relativistic correction
• Spin-orbit coupling
Homework Assignment #02:Chapter 5:27,30; Chapter 6:1,4,6,29due Wednesday, January 27, 2016
Homework Assignment #03:Chapter 6:9,17,21,24,25,32due Wednesday, February 3, 2016
Office hours:Monday & Wednesday, 13:50–15:05
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 1 / 14
Higher order degeneracy
When there are more than two de-generate states, a matrix formula-tion is more useful
The E 1 are the eigenvalues of theW -matrix and the “good” linearcombinations of the unperturbedstates are the eigenvectors of Wij
This can be proven as follows, as-sume that the set of unperturbedeigenfunctions ψ0
i are n-fold degen-erate and that ψ0 is a general lin-ear combination
αWaa + βWab = αE 1
αWba + βWbb = βE 1(Waa Wab
Wba Wbb
)(αβ
)= E 1
(αβ
)
Wij =⟨ψ0i
∣∣H ′∣∣ψ0j
⟩H0ψ0
i = E 0ψ0i ,
⟨ψ0i
∣∣ψ0j
⟩= δij
ψ0 =n∑
i=1
αiψ0i
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 2 / 14
Higher order degeneracy
When there are more than two de-generate states, a matrix formula-tion is more useful
The E 1 are the eigenvalues of theW -matrix and the “good” linearcombinations of the unperturbedstates are the eigenvectors of Wij
This can be proven as follows, as-sume that the set of unperturbedeigenfunctions ψ0
i are n-fold degen-erate and that ψ0 is a general lin-ear combination
αWaa + βWab = αE 1
αWba + βWbb = βE 1
(Waa Wab
Wba Wbb
)(αβ
)= E 1
(αβ
)
Wij =⟨ψ0i
∣∣H ′∣∣ψ0j
⟩H0ψ0
i = E 0ψ0i ,
⟨ψ0i
∣∣ψ0j
⟩= δij
ψ0 =n∑
i=1
αiψ0i
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 2 / 14
Higher order degeneracy
When there are more than two de-generate states, a matrix formula-tion is more useful
The E 1 are the eigenvalues of theW -matrix and the “good” linearcombinations of the unperturbedstates are the eigenvectors of Wij
This can be proven as follows, as-sume that the set of unperturbedeigenfunctions ψ0
i are n-fold degen-erate and that ψ0 is a general lin-ear combination
αWaa + βWab = αE 1
αWba + βWbb = βE 1(Waa Wab
Wba Wbb
)(αβ
)= E 1
(αβ
)
Wij =⟨ψ0i
∣∣H ′∣∣ψ0j
⟩H0ψ0
i = E 0ψ0i ,
⟨ψ0i
∣∣ψ0j
⟩= δij
ψ0 =n∑
i=1
αiψ0i
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 2 / 14
Higher order degeneracy
When there are more than two de-generate states, a matrix formula-tion is more useful
The E 1 are the eigenvalues of theW -matrix and the “good” linearcombinations of the unperturbedstates are the eigenvectors of Wij
This can be proven as follows, as-sume that the set of unperturbedeigenfunctions ψ0
i are n-fold degen-erate and that ψ0 is a general lin-ear combination
αWaa + βWab = αE 1
αWba + βWbb = βE 1(Waa Wab
Wba Wbb
)(αβ
)= E 1
(αβ
)
Wij =⟨ψ0i
∣∣H ′∣∣ψ0j
⟩H0ψ0
i = E 0ψ0i ,
⟨ψ0i
∣∣ψ0j
⟩= δij
ψ0 =n∑
i=1
αiψ0i
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 2 / 14
Higher order degeneracy
When there are more than two de-generate states, a matrix formula-tion is more useful
The E 1 are the eigenvalues of theW -matrix and the “good” linearcombinations of the unperturbedstates are the eigenvectors of Wij
This can be proven as follows, as-sume that the set of unperturbedeigenfunctions ψ0
i are n-fold degen-erate and that ψ0 is a general lin-ear combination
αWaa + βWab = αE 1
αWba + βWbb = βE 1(Waa Wab
Wba Wbb
)(αβ
)= E 1
(αβ
)
Wij =⟨ψ0i
∣∣H ′∣∣ψ0j
⟩
H0ψ0i = E 0ψ0
i ,⟨ψ0i
∣∣ψ0j
⟩= δij
ψ0 =n∑
i=1
αiψ0i
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 2 / 14
Higher order degeneracy
When there are more than two de-generate states, a matrix formula-tion is more useful
The E 1 are the eigenvalues of theW -matrix and the “good” linearcombinations of the unperturbedstates are the eigenvectors of Wij
This can be proven as follows, as-sume that the set of unperturbedeigenfunctions ψ0
i are n-fold degen-erate
and that ψ0 is a general lin-ear combination
αWaa + βWab = αE 1
αWba + βWbb = βE 1(Waa Wab
Wba Wbb
)(αβ
)= E 1
(αβ
)
Wij =⟨ψ0i
∣∣H ′∣∣ψ0j
⟩
H0ψ0i = E 0ψ0
i ,⟨ψ0i
∣∣ψ0j
⟩= δij
ψ0 =n∑
i=1
αiψ0i
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 2 / 14
Higher order degeneracy
When there are more than two de-generate states, a matrix formula-tion is more useful
The E 1 are the eigenvalues of theW -matrix and the “good” linearcombinations of the unperturbedstates are the eigenvectors of Wij
This can be proven as follows, as-sume that the set of unperturbedeigenfunctions ψ0
i are n-fold degen-erate
and that ψ0 is a general lin-ear combination
αWaa + βWab = αE 1
αWba + βWbb = βE 1(Waa Wab
Wba Wbb
)(αβ
)= E 1
(αβ
)
Wij =⟨ψ0i
∣∣H ′∣∣ψ0j
⟩H0ψ0
i = E 0ψ0i ,
⟨ψ0i
∣∣ψ0j
⟩= δij
ψ0 =n∑
i=1
αiψ0i
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 2 / 14
Higher order degeneracy
When there are more than two de-generate states, a matrix formula-tion is more useful
The E 1 are the eigenvalues of theW -matrix and the “good” linearcombinations of the unperturbedstates are the eigenvectors of Wij
This can be proven as follows, as-sume that the set of unperturbedeigenfunctions ψ0
i are n-fold degen-erate and that ψ0 is a general lin-ear combination
αWaa + βWab = αE 1
αWba + βWbb = βE 1(Waa Wab
Wba Wbb
)(αβ
)= E 1
(αβ
)
Wij =⟨ψ0i
∣∣H ′∣∣ψ0j
⟩H0ψ0
i = E 0ψ0i ,
⟨ψ0i
∣∣ψ0j
⟩= δij
ψ0 =n∑
i=1
αiψ0i
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 2 / 14
Higher order degeneracy
When there are more than two de-generate states, a matrix formula-tion is more useful
The E 1 are the eigenvalues of theW -matrix and the “good” linearcombinations of the unperturbedstates are the eigenvectors of Wij
This can be proven as follows, as-sume that the set of unperturbedeigenfunctions ψ0
i are n-fold degen-erate and that ψ0 is a general lin-ear combination
αWaa + βWab = αE 1
αWba + βWbb = βE 1(Waa Wab
Wba Wbb
)(αβ
)= E 1
(αβ
)
Wij =⟨ψ0i
∣∣H ′∣∣ψ0j
⟩H0ψ0
i = E 0ψ0i ,
⟨ψ0i
∣∣ψ0j
⟩= δij
ψ0 =n∑
i=1
αiψ0i
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 2 / 14
Problem 6.10
In the text it is asserted that the first-order corrections to an n-folddegenerate energy are the eigenvalues of the W matrix, and that this isthe “natural” generalization fo the n = 2 case. Prove this by starting with
ψ0 =n∑
j=1
αjψ0j .
First we note that since each of the n degenerate solutions, ψ0j of H0 has
eigenvalue E 0 and therefore
H0ψ0 =n∑
j=1
αjH0ψ0
j = E 0n∑
j=1
αjψ0j = E 0ψ0
we want to solve for the corrections in the presence of a perturbingHamiltonian, H ′, by expanding the presumed eigenfunction and eigenvalues
ψ = ψ0 + ψ1 + · · · E = E 0 + E 1 + · · ·
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 3 / 14
Problem 6.10
In the text it is asserted that the first-order corrections to an n-folddegenerate energy are the eigenvalues of the W matrix, and that this isthe “natural” generalization fo the n = 2 case. Prove this by starting with
ψ0 =n∑
j=1
αjψ0j .
First we note that since each of the n degenerate solutions, ψ0j of H0 has
eigenvalue E 0 and therefore
H0ψ0 =n∑
j=1
αjH0ψ0
j = E 0n∑
j=1
αjψ0j = E 0ψ0
we want to solve for the corrections in the presence of a perturbingHamiltonian, H ′, by expanding the presumed eigenfunction and eigenvalues
ψ = ψ0 + ψ1 + · · · E = E 0 + E 1 + · · ·
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 3 / 14
Problem 6.10
In the text it is asserted that the first-order corrections to an n-folddegenerate energy are the eigenvalues of the W matrix, and that this isthe “natural” generalization fo the n = 2 case. Prove this by starting with
ψ0 =n∑
j=1
αjψ0j .
First we note that since each of the n degenerate solutions, ψ0j of H0 has
eigenvalue E 0 and therefore
H0ψ0 =n∑
j=1
αjH0ψ0
j
= E 0n∑
j=1
αjψ0j = E 0ψ0
we want to solve for the corrections in the presence of a perturbingHamiltonian, H ′, by expanding the presumed eigenfunction and eigenvalues
ψ = ψ0 + ψ1 + · · · E = E 0 + E 1 + · · ·
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 3 / 14
Problem 6.10
In the text it is asserted that the first-order corrections to an n-folddegenerate energy are the eigenvalues of the W matrix, and that this isthe “natural” generalization fo the n = 2 case. Prove this by starting with
ψ0 =n∑
j=1
αjψ0j .
First we note that since each of the n degenerate solutions, ψ0j of H0 has
eigenvalue E 0 and therefore
H0ψ0 =n∑
j=1
αjH0ψ0
j = E 0n∑
j=1
αjψ0j
= E 0ψ0
we want to solve for the corrections in the presence of a perturbingHamiltonian, H ′, by expanding the presumed eigenfunction and eigenvalues
ψ = ψ0 + ψ1 + · · · E = E 0 + E 1 + · · ·
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 3 / 14
Problem 6.10
In the text it is asserted that the first-order corrections to an n-folddegenerate energy are the eigenvalues of the W matrix, and that this isthe “natural” generalization fo the n = 2 case. Prove this by starting with
ψ0 =n∑
j=1
αjψ0j .
First we note that since each of the n degenerate solutions, ψ0j of H0 has
eigenvalue E 0 and therefore
H0ψ0 =n∑
j=1
αjH0ψ0
j = E 0n∑
j=1
αjψ0j = E 0ψ0
we want to solve for the corrections in the presence of a perturbingHamiltonian, H ′, by expanding the presumed eigenfunction and eigenvalues
ψ = ψ0 + ψ1 + · · · E = E 0 + E 1 + · · ·
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 3 / 14
Problem 6.10
In the text it is asserted that the first-order corrections to an n-folddegenerate energy are the eigenvalues of the W matrix, and that this isthe “natural” generalization fo the n = 2 case. Prove this by starting with
ψ0 =n∑
j=1
αjψ0j .
First we note that since each of the n degenerate solutions, ψ0j of H0 has
eigenvalue E 0 and therefore
H0ψ0 =n∑
j=1
αjH0ψ0
j = E 0n∑
j=1
αjψ0j = E 0ψ0
we want to solve for the corrections in the presence of a perturbingHamiltonian, H ′, by expanding the presumed eigenfunction
and eigenvalues
ψ = ψ0 + ψ1 + · · · E = E 0 + E 1 + · · ·
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 3 / 14
Problem 6.10
In the text it is asserted that the first-order corrections to an n-folddegenerate energy are the eigenvalues of the W matrix, and that this isthe “natural” generalization fo the n = 2 case. Prove this by starting with
ψ0 =n∑
j=1
αjψ0j .
First we note that since each of the n degenerate solutions, ψ0j of H0 has
eigenvalue E 0 and therefore
H0ψ0 =n∑
j=1
αjH0ψ0
j = E 0n∑
j=1
αjψ0j = E 0ψ0
we want to solve for the corrections in the presence of a perturbingHamiltonian, H ′, by expanding the presumed eigenfunction
and eigenvalues
ψ = ψ0 + ψ1 + · · ·
E = E 0 + E 1 + · · ·
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 3 / 14
Problem 6.10
In the text it is asserted that the first-order corrections to an n-folddegenerate energy are the eigenvalues of the W matrix, and that this isthe “natural” generalization fo the n = 2 case. Prove this by starting with
ψ0 =n∑
j=1
αjψ0j .
First we note that since each of the n degenerate solutions, ψ0j of H0 has
eigenvalue E 0 and therefore
H0ψ0 =n∑
j=1
αjH0ψ0
j = E 0n∑
j=1
αjψ0j = E 0ψ0
we want to solve for the corrections in the presence of a perturbingHamiltonian, H ′, by expanding the presumed eigenfunction and eigenvalues
ψ = ψ0 + ψ1 + · · ·
E = E 0 + E 1 + · · ·
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 3 / 14
Problem 6.10
In the text it is asserted that the first-order corrections to an n-folddegenerate energy are the eigenvalues of the W matrix, and that this isthe “natural” generalization fo the n = 2 case. Prove this by starting with
ψ0 =n∑
j=1
αjψ0j .
First we note that since each of the n degenerate solutions, ψ0j of H0 has
eigenvalue E 0 and therefore
H0ψ0 =n∑
j=1
αjH0ψ0
j = E 0n∑
j=1
αjψ0j = E 0ψ0
we want to solve for the corrections in the presence of a perturbingHamiltonian, H ′, by expanding the presumed eigenfunction and eigenvalues
ψ = ψ0 + ψ1 + · · · E = E 0 + E 1 + · · ·
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 3 / 14
Problem 6.10 (cont.)
Gather all the first order combinations of the eigenvalue equation
Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0
take the inner product with⟨ψ0i
∣∣������⟨ψ0i |H0ψ1
⟩+⟨ψ0i |H ′ψ0
⟩= ������
E 0⟨ψ0i |ψ1
⟩+ E 1
⟨ψ0i |ψ0
⟩n∑
j=1
αj
⟨ψ0i |H ′ψ0
j
⟩= E 1
n∑j=1
αj
⟨ψ0i |ψ0
j
⟩= E 1αi
n∑j=1
αjWij = E 1αi , Wij ≡⟨ψ0i |H ′|ψ0
j
⟩this defines n equations in n unknowns, the αi ’s, which together give annth order equation in E 1 whose n solutions are the elements of thediagonalized W -matrix
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 4 / 14
Problem 6.10 (cont.)
Gather all the first order combinations of the eigenvalue equation
Hψ = Eψ
−→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0
take the inner product with⟨ψ0i
∣∣������⟨ψ0i |H0ψ1
⟩+⟨ψ0i |H ′ψ0
⟩= ������
E 0⟨ψ0i |ψ1
⟩+ E 1
⟨ψ0i |ψ0
⟩n∑
j=1
αj
⟨ψ0i |H ′ψ0
j
⟩= E 1
n∑j=1
αj
⟨ψ0i |ψ0
j
⟩= E 1αi
n∑j=1
αjWij = E 1αi , Wij ≡⟨ψ0i |H ′|ψ0
j
⟩this defines n equations in n unknowns, the αi ’s, which together give annth order equation in E 1 whose n solutions are the elements of thediagonalized W -matrix
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 4 / 14
Problem 6.10 (cont.)
Gather all the first order combinations of the eigenvalue equation
Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0
take the inner product with⟨ψ0i
∣∣������⟨ψ0i |H0ψ1
⟩+⟨ψ0i |H ′ψ0
⟩= ������
E 0⟨ψ0i |ψ1
⟩+ E 1
⟨ψ0i |ψ0
⟩n∑
j=1
αj
⟨ψ0i |H ′ψ0
j
⟩= E 1
n∑j=1
αj
⟨ψ0i |ψ0
j
⟩= E 1αi
n∑j=1
αjWij = E 1αi , Wij ≡⟨ψ0i |H ′|ψ0
j
⟩this defines n equations in n unknowns, the αi ’s, which together give annth order equation in E 1 whose n solutions are the elements of thediagonalized W -matrix
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 4 / 14
Problem 6.10 (cont.)
Gather all the first order combinations of the eigenvalue equation
Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0
take the inner product with⟨ψ0i
∣∣
������⟨ψ0i |H0ψ1
⟩+⟨ψ0i |H ′ψ0
⟩= ������
E 0⟨ψ0i |ψ1
⟩+ E 1
⟨ψ0i |ψ0
⟩n∑
j=1
αj
⟨ψ0i |H ′ψ0
j
⟩= E 1
n∑j=1
αj
⟨ψ0i |ψ0
j
⟩= E 1αi
n∑j=1
αjWij = E 1αi , Wij ≡⟨ψ0i |H ′|ψ0
j
⟩this defines n equations in n unknowns, the αi ’s, which together give annth order equation in E 1 whose n solutions are the elements of thediagonalized W -matrix
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 4 / 14
Problem 6.10 (cont.)
Gather all the first order combinations of the eigenvalue equation
Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0
take the inner product with⟨ψ0i
∣∣⟨ψ0i |H0ψ1
⟩+⟨ψ0i |H ′ψ0
⟩= E 0
⟨ψ0i |ψ1
⟩+ E 1
⟨ψ0i |ψ0
⟩
n∑j=1
αj
⟨ψ0i |H ′ψ0
j
⟩= E 1
n∑j=1
αj
⟨ψ0i |ψ0
j
⟩= E 1αi
n∑j=1
αjWij = E 1αi , Wij ≡⟨ψ0i |H ′|ψ0
j
⟩this defines n equations in n unknowns, the αi ’s, which together give annth order equation in E 1 whose n solutions are the elements of thediagonalized W -matrix
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 4 / 14
Problem 6.10 (cont.)
Gather all the first order combinations of the eigenvalue equation
Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0
take the inner product with⟨ψ0i
∣∣������⟨ψ0i |H0ψ1
⟩+⟨ψ0i |H ′ψ0
⟩= ������
E 0⟨ψ0i |ψ1
⟩+ E 1
⟨ψ0i |ψ0
⟩
n∑j=1
αj
⟨ψ0i |H ′ψ0
j
⟩= E 1
n∑j=1
αj
⟨ψ0i |ψ0
j
⟩= E 1αi
n∑j=1
αjWij = E 1αi , Wij ≡⟨ψ0i |H ′|ψ0
j
⟩this defines n equations in n unknowns, the αi ’s, which together give annth order equation in E 1 whose n solutions are the elements of thediagonalized W -matrix
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 4 / 14
Problem 6.10 (cont.)
Gather all the first order combinations of the eigenvalue equation
Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0
take the inner product with⟨ψ0i
∣∣������⟨ψ0i |H0ψ1
⟩+⟨ψ0i |H ′ψ0
⟩= ������
E 0⟨ψ0i |ψ1
⟩+ E 1
⟨ψ0i |ψ0
⟩n∑
j=1
αj
⟨ψ0i |H ′ψ0
j
⟩= E 1
n∑j=1
αj
⟨ψ0i |ψ0
j
⟩
= E 1αi
n∑j=1
αjWij = E 1αi , Wij ≡⟨ψ0i |H ′|ψ0
j
⟩this defines n equations in n unknowns, the αi ’s, which together give annth order equation in E 1 whose n solutions are the elements of thediagonalized W -matrix
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 4 / 14
Problem 6.10 (cont.)
Gather all the first order combinations of the eigenvalue equation
Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0
take the inner product with⟨ψ0i
∣∣������⟨ψ0i |H0ψ1
⟩+⟨ψ0i |H ′ψ0
⟩= ������
E 0⟨ψ0i |ψ1
⟩+ E 1
⟨ψ0i |ψ0
⟩n∑
j=1
αj
⟨ψ0i |H ′ψ0
j
⟩= E 1
n∑j=1
αj
⟨ψ0i |ψ0
j
⟩= E 1αi
n∑j=1
αjWij = E 1αi , Wij ≡⟨ψ0i |H ′|ψ0
j
⟩this defines n equations in n unknowns, the αi ’s, which together give annth order equation in E 1 whose n solutions are the elements of thediagonalized W -matrix
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 4 / 14
Problem 6.10 (cont.)
Gather all the first order combinations of the eigenvalue equation
Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0
take the inner product with⟨ψ0i
∣∣������⟨ψ0i |H0ψ1
⟩+⟨ψ0i |H ′ψ0
⟩= ������
E 0⟨ψ0i |ψ1
⟩+ E 1
⟨ψ0i |ψ0
⟩n∑
j=1
αj
⟨ψ0i |H ′ψ0
j
⟩= E 1
n∑j=1
αj
⟨ψ0i |ψ0
j
⟩= E 1αi
n∑j=1
αjWij = E 1αi , Wij ≡⟨ψ0i |H ′|ψ0
j
⟩
this defines n equations in n unknowns, the αi ’s, which together give annth order equation in E 1 whose n solutions are the elements of thediagonalized W -matrix
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 4 / 14
Problem 6.10 (cont.)
Gather all the first order combinations of the eigenvalue equation
Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0
take the inner product with⟨ψ0i
∣∣������⟨ψ0i |H0ψ1
⟩+⟨ψ0i |H ′ψ0
⟩= ������
E 0⟨ψ0i |ψ1
⟩+ E 1
⟨ψ0i |ψ0
⟩n∑
j=1
αj
⟨ψ0i |H ′ψ0
j
⟩= E 1
n∑j=1
αj
⟨ψ0i |ψ0
j
⟩= E 1αi
n∑j=1
αjWij = E 1αi , Wij ≡⟨ψ0i |H ′|ψ0
j
⟩this defines n equations in n unknowns, the αi ’s, which together give annth order equation in E 1 whose n solutions are the elements of thediagonalized W -matrix
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 4 / 14
Problem 6.10 (cont.)
To see this more clearly, rewrite the defining equations in matrix form
W11 W12 W13 · · · W1n
W21 W22 W23 · · · W2n
W31 W32 W33 · · · W3n...
......
. . ....
Wn1 Wn2 Wn3 · · · Wnn
α1
α2
α3...αn
= E 1
α1
α2
α3...αn
which is solved by a determinant
det
∣∣∣∣∣∣∣∣∣∣∣
W11 − E 1 W12 W13 · · · W1n
W21 W22 − E 1 W23 · · · W2n
W31 W32 W33 − E 1 · · · W3n...
......
. . ....
Wn1 Wn2 Wn3 · · · Wnn − E 1
∣∣∣∣∣∣∣∣∣∣∣= 0
giving an nth order equation in E 1
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 5 / 14
Problem 6.10 (cont.)
To see this more clearly, rewrite the defining equations in matrix formW11 W12 W13 · · · W1n
W21 W22 W23 · · · W2n
W31 W32 W33 · · · W3n...
......
. . ....
Wn1 Wn2 Wn3 · · · Wnn
α1
α2
α3...αn
= E 1
α1
α2
α3...αn
which is solved by a determinant
det
∣∣∣∣∣∣∣∣∣∣∣
W11 − E 1 W12 W13 · · · W1n
W21 W22 − E 1 W23 · · · W2n
W31 W32 W33 − E 1 · · · W3n...
......
. . ....
Wn1 Wn2 Wn3 · · · Wnn − E 1
∣∣∣∣∣∣∣∣∣∣∣= 0
giving an nth order equation in E 1
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 5 / 14
Problem 6.10 (cont.)
To see this more clearly, rewrite the defining equations in matrix formW11 W12 W13 · · · W1n
W21 W22 W23 · · · W2n
W31 W32 W33 · · · W3n...
......
. . ....
Wn1 Wn2 Wn3 · · · Wnn
α1
α2
α3...αn
= E 1
α1
α2
α3...αn
which is solved by a determinant
det
∣∣∣∣∣∣∣∣∣∣∣
W11 − E 1 W12 W13 · · · W1n
W21 W22 − E 1 W23 · · · W2n
W31 W32 W33 − E 1 · · · W3n...
......
. . ....
Wn1 Wn2 Wn3 · · · Wnn − E 1
∣∣∣∣∣∣∣∣∣∣∣= 0
giving an nth order equation in E 1
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 5 / 14
Problem 6.10 (cont.)
To see this more clearly, rewrite the defining equations in matrix formW11 W12 W13 · · · W1n
W21 W22 W23 · · · W2n
W31 W32 W33 · · · W3n...
......
. . ....
Wn1 Wn2 Wn3 · · · Wnn
α1
α2
α3...αn
= E 1
α1
α2
α3...αn
which is solved by a determinant
det
∣∣∣∣∣∣∣∣∣∣∣
W11 − E 1 W12 W13 · · · W1n
W21 W22 − E 1 W23 · · · W2n
W31 W32 W33 − E 1 · · · W3n...
......
. . ....
Wn1 Wn2 Wn3 · · · Wnn − E 1
∣∣∣∣∣∣∣∣∣∣∣= 0
giving an nth order equation in E 1
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 5 / 14
Problem 6.10 (cont.)
To see this more clearly, rewrite the defining equations in matrix formW11 W12 W13 · · · W1n
W21 W22 W23 · · · W2n
W31 W32 W33 · · · W3n...
......
. . ....
Wn1 Wn2 Wn3 · · · Wnn
α1
α2
α3...αn
= E 1
α1
α2
α3...αn
which is solved by a determinant
det
∣∣∣∣∣∣∣∣∣∣∣
W11 − E 1 W12 W13 · · · W1n
W21 W22 − E 1 W23 · · · W2n
W31 W32 W33 − E 1 · · · W3n...
......
. . ....
Wn1 Wn2 Wn3 · · · Wnn − E 1
∣∣∣∣∣∣∣∣∣∣∣= 0
giving an nth order equation in E 1
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 5 / 14
Example 6.2
Consider the three-dimensional infinite cubical well
V 0(x , y , z) =
{0, 0 < x , y , z < a
∞, otherwise
The stationary states are
ψ0nxnynz (x , y , z) =
(2
a
)3/2
sin(nxπ
ax)
sin(nyπ
ay)
sin(nzπ
az)
where nx , ny , and nz are positive integers. The corresponding allowedenergies are
E 0nxnynz =
π2~2
2ma2(n2x + n2y + n2z)
The first excited state is triply degenerate
ψa ≡ ψ112, ψb ≡ ψ121, ψc ≡ ψ211, E 01 ≡ 3
π2~2
ma2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 6 / 14
Example 6.2
Consider the three-dimensional infinite cubical well
V 0(x , y , z) =
{0, 0 < x , y , z < a
∞, otherwise
The stationary states are
ψ0nxnynz (x , y , z) =
(2
a
)3/2
sin(nxπ
ax)
sin(nyπ
ay)
sin(nzπ
az)
where nx , ny , and nz are positive integers.
The corresponding allowedenergies are
E 0nxnynz =
π2~2
2ma2(n2x + n2y + n2z)
The first excited state is triply degenerate
ψa ≡ ψ112, ψb ≡ ψ121, ψc ≡ ψ211, E 01 ≡ 3
π2~2
ma2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 6 / 14
Example 6.2
Consider the three-dimensional infinite cubical well
V 0(x , y , z) =
{0, 0 < x , y , z < a
∞, otherwise
The stationary states are
ψ0nxnynz (x , y , z) =
(2
a
)3/2
sin(nxπ
ax)
sin(nyπ
ay)
sin(nzπ
az)
where nx , ny , and nz are positive integers. The corresponding allowedenergies are
E 0nxnynz =
π2~2
2ma2(n2x + n2y + n2z)
The first excited state is triply degenerate
ψa ≡ ψ112, ψb ≡ ψ121, ψc ≡ ψ211, E 01 ≡ 3
π2~2
ma2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 6 / 14
Example 6.2
Consider the three-dimensional infinite cubical well
V 0(x , y , z) =
{0, 0 < x , y , z < a
∞, otherwise
The stationary states are
ψ0nxnynz (x , y , z) =
(2
a
)3/2
sin(nxπ
ax)
sin(nyπ
ay)
sin(nzπ
az)
where nx , ny , and nz are positive integers. The corresponding allowedenergies are
E 0nxnynz =
π2~2
2ma2(n2x + n2y + n2z)
The first excited state is triply degenerate
ψa ≡ ψ112, ψb ≡ ψ121, ψc ≡ ψ211, E 01 ≡ 3
π2~2
ma2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 6 / 14
Example 6.2
Suppose we apply a perturbation
H ′ =
{V0, 0 < x < a/2 and 0 < y < a/2
0, otherwise
We need to calculate the matrix elements of the the 3× 3, W -matrix
Waa =⟨ψ112|H ′|ψ112
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(πay)dy
∫ a
0sin2
(2π
az
)dz
=
(2
a
)3
V0
{[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
}2[z
2− a
8πsin
(4π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
)2 (a2
)=
V0
4
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 7 / 14
Example 6.2
Suppose we apply a perturbation
H ′ =
{V0, 0 < x < a/2 and 0 < y < a/2
0, otherwise
We need to calculate the matrix elements of the the 3× 3, W -matrix
Waa =⟨ψ112|H ′|ψ112
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(πay)dy
∫ a
0sin2
(2π
az
)dz
=
(2
a
)3
V0
{[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
}2[z
2− a
8πsin
(4π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
)2 (a2
)=
V0
4
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 7 / 14
Example 6.2
Suppose we apply a perturbation
H ′ =
{V0, 0 < x < a/2 and 0 < y < a/2
0, otherwise
We need to calculate the matrix elements of the the 3× 3, W -matrix
Waa =⟨ψ112|H ′|ψ112
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(πay)dy
∫ a
0sin2
(2π
az
)dz
=
(2
a
)3
V0
{[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
}2[z
2− a
8πsin
(4π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
)2 (a2
)=
V0
4
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 7 / 14
Example 6.2
Suppose we apply a perturbation
H ′ =
{V0, 0 < x < a/2 and 0 < y < a/2
0, otherwise
We need to calculate the matrix elements of the the 3× 3, W -matrix
Waa =⟨ψ112|H ′|ψ112
⟩
=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(πay)dy
∫ a
0sin2
(2π
az
)dz
=
(2
a
)3
V0
{[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
}2[z
2− a
8πsin
(4π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
)2 (a2
)=
V0
4
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 7 / 14
Example 6.2
Suppose we apply a perturbation
H ′ =
{V0, 0 < x < a/2 and 0 < y < a/2
0, otherwise
We need to calculate the matrix elements of the the 3× 3, W -matrix
Waa =⟨ψ112|H ′|ψ112
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(πay)dy
∫ a
0sin2
(2π
az
)dz
=
(2
a
)3
V0
{[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
}2[z
2− a
8πsin
(4π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
)2 (a2
)=
V0
4
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 7 / 14
Example 6.2
Suppose we apply a perturbation
H ′ =
{V0, 0 < x < a/2 and 0 < y < a/2
0, otherwise
We need to calculate the matrix elements of the the 3× 3, W -matrix
Waa =⟨ψ112|H ′|ψ112
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(πay)dy
∫ a
0sin2
(2π
az
)dz
=
(2
a
)3
V0
{[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
}2[z
2− a
8πsin
(4π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
)2 (a2
)=
V0
4
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 7 / 14
Example 6.2
Suppose we apply a perturbation
H ′ =
{V0, 0 < x < a/2 and 0 < y < a/2
0, otherwise
We need to calculate the matrix elements of the the 3× 3, W -matrix
Waa =⟨ψ112|H ′|ψ112
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(πay)dy
∫ a
0sin2
(2π
az
)dz
=
(2
a
)3
V0
{
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
}2[z
2− a
8πsin
(4π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
)2 (a2
)=
V0
4
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 7 / 14
Example 6.2
Suppose we apply a perturbation
H ′ =
{V0, 0 < x < a/2 and 0 < y < a/2
0, otherwise
We need to calculate the matrix elements of the the 3× 3, W -matrix
Waa =⟨ψ112|H ′|ψ112
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(πay)dy
∫ a
0sin2
(2π
az
)dz
=
(2
a
)3
V0
{[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
}2
[z
2− a
8πsin
(4π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
)2 (a2
)=
V0
4
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 7 / 14
Example 6.2
Suppose we apply a perturbation
H ′ =
{V0, 0 < x < a/2 and 0 < y < a/2
0, otherwise
We need to calculate the matrix elements of the the 3× 3, W -matrix
Waa =⟨ψ112|H ′|ψ112
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(πay)dy
∫ a
0sin2
(2π
az
)dz
=
(2
a
)3
V0
{[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
}2[z
2− a
8πsin
(4π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
)2 (a2
)=
V0
4
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 7 / 14
Example 6.2
Suppose we apply a perturbation
H ′ =
{V0, 0 < x < a/2 and 0 < y < a/2
0, otherwise
We need to calculate the matrix elements of the the 3× 3, W -matrix
Waa =⟨ψ112|H ′|ψ112
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(πay)dy
∫ a
0sin2
(2π
az
)dz
=
(2
a
)3
V0
{[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
}2[z
2− a
8πsin
(4π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
)2 (a2
)=
V0
4
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 7 / 14
Example 6.2
Suppose we apply a perturbation
H ′ =
{V0, 0 < x < a/2 and 0 < y < a/2
0, otherwise
We need to calculate the matrix elements of the the 3× 3, W -matrix
Waa =⟨ψ112|H ′|ψ112
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(πay)dy
∫ a
0sin2
(2π
az
)dz
=
(2
a
)3
V0
{[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
}2[z
2− a
8πsin
(4π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
)2
(a2
)=
V0
4
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 7 / 14
Example 6.2
Suppose we apply a perturbation
H ′ =
{V0, 0 < x < a/2 and 0 < y < a/2
0, otherwise
We need to calculate the matrix elements of the the 3× 3, W -matrix
Waa =⟨ψ112|H ′|ψ112
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(πay)dy
∫ a
0sin2
(2π
az
)dz
=
(2
a
)3
V0
{[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
}2[z
2− a
8πsin
(4π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
)2 (a2
)
=V0
4
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 7 / 14
Example 6.2
Suppose we apply a perturbation
H ′ =
{V0, 0 < x < a/2 and 0 < y < a/2
0, otherwise
We need to calculate the matrix elements of the the 3× 3, W -matrix
Waa =⟨ψ112|H ′|ψ112
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(πay)dy
∫ a
0sin2
(2π
az
)dz
=
(2
a
)3
V0
{[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
}2[z
2− a
8πsin
(4π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
)2 (a2
)=
V0
4
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 7 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩
=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
) (a2
)=
V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
) (a2
)=
V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
) (a2
)=
V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
) (a2
)=
V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
) (a2
)=
V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
) (a2
)=
V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
) (a2
)=
V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
)
(a4
) (a2
)=
V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
)
(a2
)=
V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
) (a2
)
=V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
) (a2
)=
V0
4
= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
) (a2
)=
V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
) (a2
)=
V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
) (a2
)=
V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
) (a2
)=
V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
∫ a
0sin
(2π
az
)sin(πaz)dz
]
= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The next diagonal element is slightly different
Wbb =⟨ψ121|H ′|ψ121
⟩=
(2
a
)3
V0
∫ a/2
0sin2
(πax)dx
∫ a/2
0sin2
(2π
ay
)dy
∫ a
0sin2
(πaz)dz
=
(2
a
)3
V0
[x
2− a
4πsin
(2π
ax
)∣∣∣∣a/20
[y
2− a
8πsin
(4π
ay
)∣∣∣∣a/20
×[z
2− a
4πsin
(2π
az
)∣∣∣∣a0
=
(2
a
)3
V0
(a4
) (a4
) (a2
)=
V0
4= Wcc
now compute the off-diagonal elements of the matrix
Wab =⟨ψ112|H ′|ψ121
⟩=
(2
a
)3V0
[∫ a/2
0sin2
(πax)dx
×∫ a/2
0sin(πay)
sin
(2π
ay
)dy
�������������∫ a
0sin
(2π
az
)sin(πaz)dz
]= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 8 / 14
Example 6.2
The same integrals hold for Wac , thus Wab = Wac = 0, and all thatremains is Wbc
Wbc =⟨ψ121|H ′|ψ211
⟩=
(2
a
)3V0
[∫ a/2
0sin(πax)
sin
(2π
ax
)dx
×∫ a/2
0sin
(2π
ay
)sin(πay)dy
∫ a
0sin2
(πaz)dz
]
=
(2
a
)3V0
(a2
)( aπ
)
2
∫ π/2
0sin(θ) sin(2θ) dθ
∫ π/2
0sin(2φ) sin(φ) dφ
=
(4V0
π2
){∫ π/2
02 sin2(θ) cos(θ) dθ
}2=
(4V0
π2
){2
[sin3(θ)
3
∣∣∣∣π/20
}2
=16V0
9π2=
V0
4· 64
9π2=
V0
4κ κ ≡
(8
3π
)2≈ 0.7205
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 9 / 14
Example 6.2
The same integrals hold for Wac , thus Wab = Wac = 0, and all thatremains is Wbc
Wbc =⟨ψ121|H ′|ψ211
⟩
=
(2
a
)3V0
[∫ a/2
0sin(πax)
sin
(2π
ax
)dx
×∫ a/2
0sin
(2π
ay
)sin(πay)dy
∫ a
0sin2
(πaz)dz
]
=
(2
a
)3V0
(a2
)( aπ
)
2
∫ π/2
0sin(θ) sin(2θ) dθ
∫ π/2
0sin(2φ) sin(φ) dφ
=
(4V0
π2
){∫ π/2
02 sin2(θ) cos(θ) dθ
}2=
(4V0
π2
){2
[sin3(θ)
3
∣∣∣∣π/20
}2
=16V0
9π2=
V0
4· 64
9π2=
V0
4κ κ ≡
(8
3π
)2≈ 0.7205
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 9 / 14
Example 6.2
The same integrals hold for Wac , thus Wab = Wac = 0, and all thatremains is Wbc
Wbc =⟨ψ121|H ′|ψ211
⟩=
(2
a
)3V0
[∫ a/2
0sin(πax)
sin
(2π
ax
)dx
×∫ a/2
0sin
(2π
ay
)sin(πay)dy
∫ a
0sin2
(πaz)dz
]
=
(2
a
)3V0
(a2
)( aπ
)
2
∫ π/2
0sin(θ) sin(2θ) dθ
∫ π/2
0sin(2φ) sin(φ) dφ
=
(4V0
π2
){∫ π/2
02 sin2(θ) cos(θ) dθ
}2=
(4V0
π2
){2
[sin3(θ)
3
∣∣∣∣π/20
}2
=16V0
9π2=
V0
4· 64
9π2=
V0
4κ κ ≡
(8
3π
)2≈ 0.7205
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 9 / 14
Example 6.2
The same integrals hold for Wac , thus Wab = Wac = 0, and all thatremains is Wbc
Wbc =⟨ψ121|H ′|ψ211
⟩=
(2
a
)3V0
[∫ a/2
0sin(πax)
sin
(2π
ax
)dx
×∫ a/2
0sin
(2π
ay
)sin(πay)dy
∫ a
0sin2
(πaz)dz
]
=
(2
a
)3V0
(a2
)( aπ
)
2
∫ π/2
0sin(θ) sin(2θ) dθ
∫ π/2
0sin(2φ) sin(φ) dφ
=
(4V0
π2
){∫ π/2
02 sin2(θ) cos(θ) dθ
}2=
(4V0
π2
){2
[sin3(θ)
3
∣∣∣∣π/20
}2
=16V0
9π2=
V0
4· 64
9π2=
V0
4κ κ ≡
(8
3π
)2≈ 0.7205
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 9 / 14
Example 6.2
The same integrals hold for Wac , thus Wab = Wac = 0, and all thatremains is Wbc
Wbc =⟨ψ121|H ′|ψ211
⟩=
(2
a
)3V0
[∫ a/2
0sin(πax)
sin
(2π
ax
)dx
×∫ a/2
0sin
(2π
ay
)sin(πay)dy
∫ a
0sin2
(πaz)dz
]
=
(2
a
)3V0
(a2
)
( aπ
)
2
∫ π/2
0sin(θ) sin(2θ) dθ
∫ π/2
0sin(2φ) sin(φ) dφ
=
(4V0
π2
){∫ π/2
02 sin2(θ) cos(θ) dθ
}2=
(4V0
π2
){2
[sin3(θ)
3
∣∣∣∣π/20
}2
=16V0
9π2=
V0
4· 64
9π2=
V0
4κ κ ≡
(8
3π
)2≈ 0.7205
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 9 / 14
Example 6.2
The same integrals hold for Wac , thus Wab = Wac = 0, and all thatremains is Wbc
Wbc =⟨ψ121|H ′|ψ211
⟩=
(2
a
)3V0
[∫ a/2
0sin(πax)
sin
(2π
ax
)dx
×∫ a/2
0sin
(2π
ay
)sin(πay)dy
∫ a
0sin2
(πaz)dz
]
=
(2
a
)3V0
(a2
)
( aπ
)2∫ π/2
0sin(θ) sin(2θ) dθ
∫ π/2
0sin(2φ) sin(φ) dφ
=
(4V0
π2
){∫ π/2
02 sin2(θ) cos(θ) dθ
}2=
(4V0
π2
){2
[sin3(θ)
3
∣∣∣∣π/20
}2
=16V0
9π2=
V0
4· 64
9π2=
V0
4κ κ ≡
(8
3π
)2≈ 0.7205
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 9 / 14
Example 6.2
The same integrals hold for Wac , thus Wab = Wac = 0, and all thatremains is Wbc
Wbc =⟨ψ121|H ′|ψ211
⟩=
(2
a
)3V0
[∫ a/2
0sin(πax)
sin
(2π
ax
)dx
×∫ a/2
0sin
(2π
ay
)sin(πay)dy
∫ a
0sin2
(πaz)dz
]
=
(2
a
)3V0
(a2
)( aπ
)2∫ π/2
0sin(θ) sin(2θ) dθ
∫ π/2
0sin(2φ) sin(φ) dφ
=
(4V0
π2
){∫ π/2
02 sin2(θ) cos(θ) dθ
}2=
(4V0
π2
){2
[sin3(θ)
3
∣∣∣∣π/20
}2
=16V0
9π2=
V0
4· 64
9π2=
V0
4κ κ ≡
(8
3π
)2≈ 0.7205
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 9 / 14
Example 6.2
The same integrals hold for Wac , thus Wab = Wac = 0, and all thatremains is Wbc
Wbc =⟨ψ121|H ′|ψ211
⟩=
(2
a
)3V0
[∫ a/2
0sin(πax)
sin
(2π
ax
)dx
×∫ a/2
0sin
(2π
ay
)sin(πay)dy
∫ a
0sin2
(πaz)dz
]
=
(2
a
)3V0
(a2
)( aπ
)2∫ π/2
0sin(θ) sin(2θ) dθ
∫ π/2
0sin(2φ) sin(φ) dφ
=
(4V0
π2
){∫ π/2
02 sin2(θ) cos(θ) dθ
}2
=
(4V0
π2
){2
[sin3(θ)
3
∣∣∣∣π/20
}2
=16V0
9π2=
V0
4· 64
9π2=
V0
4κ κ ≡
(8
3π
)2≈ 0.7205
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 9 / 14
Example 6.2
The same integrals hold for Wac , thus Wab = Wac = 0, and all thatremains is Wbc
Wbc =⟨ψ121|H ′|ψ211
⟩=
(2
a
)3V0
[∫ a/2
0sin(πax)
sin
(2π
ax
)dx
×∫ a/2
0sin
(2π
ay
)sin(πay)dy
∫ a
0sin2
(πaz)dz
]
=
(2
a
)3V0
(a2
)( aπ
)2∫ π/2
0sin(θ) sin(2θ) dθ
∫ π/2
0sin(2φ) sin(φ) dφ
=
(4V0
π2
){∫ π/2
02 sin2(θ) cos(θ) dθ
}2=
(4V0
π2
){2
[sin3(θ)
3
∣∣∣∣π/20
}2
=16V0
9π2=
V0
4· 64
9π2=
V0
4κ κ ≡
(8
3π
)2≈ 0.7205
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 9 / 14
Example 6.2
The same integrals hold for Wac , thus Wab = Wac = 0, and all thatremains is Wbc
Wbc =⟨ψ121|H ′|ψ211
⟩=
(2
a
)3V0
[∫ a/2
0sin(πax)
sin
(2π
ax
)dx
×∫ a/2
0sin
(2π
ay
)sin(πay)dy
∫ a
0sin2
(πaz)dz
]
=
(2
a
)3V0
(a2
)( aπ
)2∫ π/2
0sin(θ) sin(2θ) dθ
∫ π/2
0sin(2φ) sin(φ) dφ
=
(4V0
π2
){∫ π/2
02 sin2(θ) cos(θ) dθ
}2=
(4V0
π2
){2
[sin3(θ)
3
∣∣∣∣π/20
}2
=16V0
9π2
=V0
4· 64
9π2=
V0
4κ κ ≡
(8
3π
)2≈ 0.7205
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 9 / 14
Example 6.2
The same integrals hold for Wac , thus Wab = Wac = 0, and all thatremains is Wbc
Wbc =⟨ψ121|H ′|ψ211
⟩=
(2
a
)3V0
[∫ a/2
0sin(πax)
sin
(2π
ax
)dx
×∫ a/2
0sin
(2π
ay
)sin(πay)dy
∫ a
0sin2
(πaz)dz
]
=
(2
a
)3V0
(a2
)( aπ
)2∫ π/2
0sin(θ) sin(2θ) dθ
∫ π/2
0sin(2φ) sin(φ) dφ
=
(4V0
π2
){∫ π/2
02 sin2(θ) cos(θ) dθ
}2=
(4V0
π2
){2
[sin3(θ)
3
∣∣∣∣π/20
}2
=16V0
9π2=
V0
4· 64
9π2
=V0
4κ κ ≡
(8
3π
)2≈ 0.7205
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 9 / 14
Example 6.2
The same integrals hold for Wac , thus Wab = Wac = 0, and all thatremains is Wbc
Wbc =⟨ψ121|H ′|ψ211
⟩=
(2
a
)3V0
[∫ a/2
0sin(πax)
sin
(2π
ax
)dx
×∫ a/2
0sin
(2π
ay
)sin(πay)dy
∫ a
0sin2
(πaz)dz
]
=
(2
a
)3V0
(a2
)( aπ
)2∫ π/2
0sin(θ) sin(2θ) dθ
∫ π/2
0sin(2φ) sin(φ) dφ
=
(4V0
π2
){∫ π/2
02 sin2(θ) cos(θ) dθ
}2=
(4V0
π2
){2
[sin3(θ)
3
∣∣∣∣π/20
}2
=16V0
9π2=
V0
4· 64
9π2=
V0
4κ κ ≡
(8
3π
)2≈ 0.7205
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 9 / 14
Example 6.2
The W -matrix thus becomes
the matrix can be diagonalizedin the usual way
the solutions are:
w3 = 1 + κ ≈ 1.7205,
w1 = 1,
w2 = 1− κ ≈ 0.2795
E1(λ) =
E 01 + λ(1 + κ)V0/4
E 01 + λV0/4
E 01 + λ(1− κ)V0/4
W =V0
4
1 0 00 1 κ0 κ 1
0 = det
∣∣∣∣∣∣1− w 0 0
0 1− w κ0 κ 1− w
∣∣∣∣∣∣= (1− w)[(1− w)2 − κ2]
= (1− w)(1− w − κ)(1− w + κ)
λ1
E1
κ
V
4
0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 10 / 14
Example 6.2
The W -matrix thus becomes
the matrix can be diagonalizedin the usual way
the solutions are:
w3 = 1 + κ ≈ 1.7205,
w1 = 1,
w2 = 1− κ ≈ 0.2795
E1(λ) =
E 01 + λ(1 + κ)V0/4
E 01 + λV0/4
E 01 + λ(1− κ)V0/4
W =V0
4
1 0 00 1 κ0 κ 1
0 = det
∣∣∣∣∣∣1− w 0 0
0 1− w κ0 κ 1− w
∣∣∣∣∣∣= (1− w)[(1− w)2 − κ2]
= (1− w)(1− w − κ)(1− w + κ)
λ1
E1
κ
V
4
0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 10 / 14
Example 6.2
The W -matrix thus becomes
the matrix can be diagonalizedin the usual way
the solutions are:
w3 = 1 + κ ≈ 1.7205,
w1 = 1,
w2 = 1− κ ≈ 0.2795
E1(λ) =
E 01 + λ(1 + κ)V0/4
E 01 + λV0/4
E 01 + λ(1− κ)V0/4
W =V0
4
1 0 00 1 κ0 κ 1
0 = det
∣∣∣∣∣∣1− w 0 0
0 1− w κ0 κ 1− w
∣∣∣∣∣∣= (1− w)[(1− w)2 − κ2]
= (1− w)(1− w − κ)(1− w + κ)
λ1
E1
κ
V
4
0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 10 / 14
Example 6.2
The W -matrix thus becomes
the matrix can be diagonalizedin the usual way
the solutions are:
w3 = 1 + κ ≈ 1.7205,
w1 = 1,
w2 = 1− κ ≈ 0.2795
E1(λ) =
E 01 + λ(1 + κ)V0/4
E 01 + λV0/4
E 01 + λ(1− κ)V0/4
W =V0
4
1 0 00 1 κ0 κ 1
0 = det
∣∣∣∣∣∣1− w 0 0
0 1− w κ0 κ 1− w
∣∣∣∣∣∣
= (1− w)[(1− w)2 − κ2]
= (1− w)(1− w − κ)(1− w + κ)
λ1
E1
κ
V
4
0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 10 / 14
Example 6.2
The W -matrix thus becomes
the matrix can be diagonalizedin the usual way
the solutions are:
w3 = 1 + κ ≈ 1.7205,
w1 = 1,
w2 = 1− κ ≈ 0.2795
E1(λ) =
E 01 + λ(1 + κ)V0/4
E 01 + λV0/4
E 01 + λ(1− κ)V0/4
W =V0
4
1 0 00 1 κ0 κ 1
0 = det
∣∣∣∣∣∣1− w 0 0
0 1− w κ0 κ 1− w
∣∣∣∣∣∣= (1− w)[(1− w)2 − κ2]
= (1− w)(1− w − κ)(1− w + κ)
λ1
E1
κ
V
4
0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 10 / 14
Example 6.2
The W -matrix thus becomes
the matrix can be diagonalizedin the usual way
the solutions are:
w3 = 1 + κ ≈ 1.7205,
w1 = 1,
w2 = 1− κ ≈ 0.2795
E1(λ) =
E 01 + λ(1 + κ)V0/4
E 01 + λV0/4
E 01 + λ(1− κ)V0/4
W =V0
4
1 0 00 1 κ0 κ 1
0 = det
∣∣∣∣∣∣1− w 0 0
0 1− w κ0 κ 1− w
∣∣∣∣∣∣= (1− w)[(1− w)2 − κ2]
= (1− w)(1− w − κ)(1− w + κ)
λ1
E1
κ
V
4
0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 10 / 14
Example 6.2
The W -matrix thus becomes
the matrix can be diagonalizedin the usual way
the solutions are:
w3 = 1 + κ ≈ 1.7205,
w1 = 1,
w2 = 1− κ ≈ 0.2795
E1(λ) =
E 01 + λ(1 + κ)V0/4
E 01 + λV0/4
E 01 + λ(1− κ)V0/4
W =V0
4
1 0 00 1 κ0 κ 1
0 = det
∣∣∣∣∣∣1− w 0 0
0 1− w κ0 κ 1− w
∣∣∣∣∣∣= (1− w)[(1− w)2 − κ2]
= (1− w)(1− w − κ)(1− w + κ)
λ1
E1
κ
V
4
0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 10 / 14
Example 6.2
The W -matrix thus becomes
the matrix can be diagonalizedin the usual way
the solutions are:
w3 = 1 + κ ≈ 1.7205,
w1 = 1,
w2 = 1− κ ≈ 0.2795
E1(λ) =
E 01 + λ(1 + κ)V0/4
E 01 + λV0/4
E 01 + λ(1− κ)V0/4
W =V0
4
1 0 00 1 κ0 κ 1
0 = det
∣∣∣∣∣∣1− w 0 0
0 1− w κ0 κ 1− w
∣∣∣∣∣∣= (1− w)[(1− w)2 − κ2]
= (1− w)(1− w − κ)(1− w + κ)
λ1
E1
κ
V
4
0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 10 / 14
Example 6.2
The W -matrix thus becomes
the matrix can be diagonalizedin the usual way
the solutions are:
w3 = 1 + κ ≈ 1.7205,
w1 = 1,
w2 = 1− κ ≈ 0.2795
E1(λ) =
E 01 + λ(1 + κ)V0/4
E 01 + λV0/4
E 01 + λ(1− κ)V0/4
W =V0
4
1 0 00 1 κ0 κ 1
0 = det
∣∣∣∣∣∣1− w 0 0
0 1− w κ0 κ 1− w
∣∣∣∣∣∣= (1− w)[(1− w)2 − κ2]
= (1− w)(1− w − κ)(1− w + κ)
λ1
E1
κ
V
4
0
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 10 / 14
Example 6.2
Now we can determine the“good” linear combinations
ψ0 = αψ112 + βψ121 + γψ211
expanding into three equations
when w = 1 the only solution isα = 1, β = γ = 0
the other two eigenvalues give asolution where α = 0 and β =±γ
the three “good” states arethus
what operator would fullfill thetheorem?
1 0 00 1 κ0 κ 1
αβγ
= w
αβγ
wα = α
wβ = β + κγ
wγ = κβ + γ
ψ0 =
1√2
(ψ121 + ψ211)
ψ112
1√2
(ψ121 − ψ211)
PxyQz f (x , y , z) = f (y , x , a− z)
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 11 / 14
Example 6.2
Now we can determine the“good” linear combinations
ψ0 = αψ112 + βψ121 + γψ211
expanding into three equations
when w = 1 the only solution isα = 1, β = γ = 0
the other two eigenvalues give asolution where α = 0 and β =±γ
the three “good” states arethus
what operator would fullfill thetheorem?
1 0 00 1 κ0 κ 1
αβγ
= w
αβγ
wα = α
wβ = β + κγ
wγ = κβ + γ
ψ0 =
1√2
(ψ121 + ψ211)
ψ112
1√2
(ψ121 − ψ211)
PxyQz f (x , y , z) = f (y , x , a− z)
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 11 / 14
Example 6.2
Now we can determine the“good” linear combinations
ψ0 = αψ112 + βψ121 + γψ211
expanding into three equations
when w = 1 the only solution isα = 1, β = γ = 0
the other two eigenvalues give asolution where α = 0 and β =±γ
the three “good” states arethus
what operator would fullfill thetheorem?
1 0 00 1 κ0 κ 1
αβγ
= w
αβγ
wα = α
wβ = β + κγ
wγ = κβ + γ
ψ0 =
1√2
(ψ121 + ψ211)
ψ112
1√2
(ψ121 − ψ211)
PxyQz f (x , y , z) = f (y , x , a− z)
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 11 / 14
Example 6.2
Now we can determine the“good” linear combinations
ψ0 = αψ112 + βψ121 + γψ211
expanding into three equations
when w = 1 the only solution isα = 1, β = γ = 0
the other two eigenvalues give asolution where α = 0 and β =±γ
the three “good” states arethus
what operator would fullfill thetheorem?
1 0 00 1 κ0 κ 1
αβγ
= w
αβγ
wα = α
wβ = β + κγ
wγ = κβ + γ
ψ0 =
1√2
(ψ121 + ψ211)
ψ112
1√2
(ψ121 − ψ211)
PxyQz f (x , y , z) = f (y , x , a− z)
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 11 / 14
Example 6.2
Now we can determine the“good” linear combinations
ψ0 = αψ112 + βψ121 + γψ211
expanding into three equations
when w = 1 the only solution isα = 1, β = γ = 0
the other two eigenvalues give asolution where α = 0 and β =±γ
the three “good” states arethus
what operator would fullfill thetheorem?
1 0 00 1 κ0 κ 1
αβγ
= w
αβγ
wα = α
wβ = β + κγ
wγ = κβ + γ
ψ0 =
1√2
(ψ121 + ψ211)
ψ112
1√2
(ψ121 − ψ211)
PxyQz f (x , y , z) = f (y , x , a− z)
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 11 / 14
Example 6.2
Now we can determine the“good” linear combinations
ψ0 = αψ112 + βψ121 + γψ211
expanding into three equations
when w = 1 the only solution isα = 1, β = γ = 0
the other two eigenvalues give asolution where α = 0 and β =±γ
the three “good” states arethus
what operator would fullfill thetheorem?
1 0 00 1 κ0 κ 1
αβγ
= w
αβγ
wα = α
wβ = β + κγ
wγ = κβ + γ
ψ0 =
1√2
(ψ121 + ψ211)
ψ112
1√2
(ψ121 − ψ211)
PxyQz f (x , y , z) = f (y , x , a− z)
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 11 / 14
Example 6.2
Now we can determine the“good” linear combinations
ψ0 = αψ112 + βψ121 + γψ211
expanding into three equations
when w = 1 the only solution isα = 1, β = γ = 0
the other two eigenvalues give asolution where α = 0 and β =±γ
the three “good” states arethus
what operator would fullfill thetheorem?
1 0 00 1 κ0 κ 1
αβγ
= w
αβγ
wα = α
wβ = β + κγ
wγ = κβ + γ
ψ0 =
1√2
(ψ121 + ψ211)
ψ112
1√2
(ψ121 − ψ211)
PxyQz f (x , y , z) = f (y , x , a− z)
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 11 / 14
Example 6.2
Now we can determine the“good” linear combinations
ψ0 = αψ112 + βψ121 + γψ211
expanding into three equations
when w = 1 the only solution isα = 1, β = γ = 0
the other two eigenvalues give asolution where α = 0 and β =±γ
the three “good” states arethus
what operator would fullfill thetheorem?
1 0 00 1 κ0 κ 1
αβγ
= w
αβγ
wα = α
wβ = β + κγ
wγ = κβ + γ
ψ0 =
1√2
(ψ121 + ψ211)
ψ112
1√2
(ψ121 − ψ211)
PxyQz f (x , y , z) = f (y , x , a− z)
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 11 / 14
Example 6.2
Now we can determine the“good” linear combinations
ψ0 = αψ112 + βψ121 + γψ211
expanding into three equations
when w = 1 the only solution isα = 1, β = γ = 0
the other two eigenvalues give asolution where α = 0 and β =±γ
the three “good” states arethus
what operator would fullfill thetheorem?
1 0 00 1 κ0 κ 1
αβγ
= w
αβγ
wα = α
wβ = β + κγ
wγ = κβ + γ
ψ0 =
1√2
(ψ121 + ψ211)
ψ112
1√2
(ψ121 − ψ211)
PxyQz f (x , y , z) = f (y , x , a− z)
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 11 / 14
Example 6.2
Now we can determine the“good” linear combinations
ψ0 = αψ112 + βψ121 + γψ211
expanding into three equations
when w = 1 the only solution isα = 1, β = γ = 0
the other two eigenvalues give asolution where α = 0 and β =±γ
the three “good” states arethus
what operator would fullfill thetheorem?
1 0 00 1 κ0 κ 1
αβγ
= w
αβγ
wα = α
wβ = β + κγ
wγ = κβ + γ
ψ0 =
1√2
(ψ121 + ψ211)
ψ112
1√2
(ψ121 − ψ211)
PxyQz f (x , y , z) = f (y , x , a− z)
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 11 / 14
Example 6.2
Now we can determine the“good” linear combinations
ψ0 = αψ112 + βψ121 + γψ211
expanding into three equations
when w = 1 the only solution isα = 1, β = γ = 0
the other two eigenvalues give asolution where α = 0 and β =±γ
the three “good” states arethus
what operator would fullfill thetheorem?
1 0 00 1 κ0 κ 1
αβγ
= w
αβγ
wα = α
wβ = β + κγ
wγ = κβ + γ
ψ0 =
1√2
(ψ121 + ψ211)
ψ112
1√2
(ψ121 − ψ211)
PxyQz f (x , y , z) = f (y , x , a− z)
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 11 / 14
Example 6.2
Now we can determine the“good” linear combinations
ψ0 = αψ112 + βψ121 + γψ211
expanding into three equations
when w = 1 the only solution isα = 1, β = γ = 0
the other two eigenvalues give asolution where α = 0 and β =±γ
the three “good” states arethus
what operator would fullfill thetheorem?
1 0 00 1 κ0 κ 1
αβγ
= w
αβγ
wα = α
wβ = β + κγ
wγ = κβ + γ
ψ0 =
1√2
(ψ121 + ψ211)
ψ112
1√2
(ψ121 − ψ211)
PxyQz f (x , y , z) = f (y , x , a− z)
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 11 / 14
Example 6.2
Now we can determine the“good” linear combinations
ψ0 = αψ112 + βψ121 + γψ211
expanding into three equations
when w = 1 the only solution isα = 1, β = γ = 0
the other two eigenvalues give asolution where α = 0 and β =±γ
the three “good” states arethus
what operator would fullfill thetheorem?
1 0 00 1 κ0 κ 1
αβγ
= w
αβγ
wα = α
wβ = β + κγ
wγ = κβ + γ
ψ0 =
1√2
(ψ121 + ψ211)
ψ112
1√2
(ψ121 − ψ211)
PxyQz f (x , y , z) = f (y , x , a− z)
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 11 / 14
Hydrogen fine structure
The Hamiltonian that can besolved exactly for the hydrogenatom is merely an approxima-tion. The first is the correctionfor the motion of the proton.
The other important correctionscan be scaled according to thefine structure constant:
The largest correction is the finestructure and this consists oftwo terms: the relativistic cor-rection and the spin-orbit cor-rection.
me → µ =memp
me + mp≈ 1836
1837me ≈ me
α =e2
4πε0~c' 1
137.04
Some of the relevant energy scales are:
Bohr energy: α2mc2
Fine structure: α4mc2
Lamb shift: α5mc2
Hyperfine splitting: (m/mp)α4mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 12 / 14
Hydrogen fine structure
The Hamiltonian that can besolved exactly for the hydrogenatom is merely an approxima-tion. The first is the correctionfor the motion of the proton.
The other important correctionscan be scaled according to thefine structure constant:
The largest correction is the finestructure and this consists oftwo terms: the relativistic cor-rection and the spin-orbit cor-rection.
me → µ =memp
me + mp
≈ 1836
1837me ≈ me
α =e2
4πε0~c' 1
137.04
Some of the relevant energy scales are:
Bohr energy: α2mc2
Fine structure: α4mc2
Lamb shift: α5mc2
Hyperfine splitting: (m/mp)α4mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 12 / 14
Hydrogen fine structure
The Hamiltonian that can besolved exactly for the hydrogenatom is merely an approxima-tion. The first is the correctionfor the motion of the proton.
The other important correctionscan be scaled according to thefine structure constant:
The largest correction is the finestructure and this consists oftwo terms: the relativistic cor-rection and the spin-orbit cor-rection.
me → µ =memp
me + mp≈ 1836
1837me ≈ me
α =e2
4πε0~c' 1
137.04
Some of the relevant energy scales are:
Bohr energy: α2mc2
Fine structure: α4mc2
Lamb shift: α5mc2
Hyperfine splitting: (m/mp)α4mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 12 / 14
Hydrogen fine structure
The Hamiltonian that can besolved exactly for the hydrogenatom is merely an approxima-tion. The first is the correctionfor the motion of the proton.
The other important correctionscan be scaled according to thefine structure constant:
The largest correction is the finestructure and this consists oftwo terms: the relativistic cor-rection and the spin-orbit cor-rection.
me → µ =memp
me + mp≈ 1836
1837me ≈ me
α =e2
4πε0~c' 1
137.04
Some of the relevant energy scales are:
Bohr energy: α2mc2
Fine structure: α4mc2
Lamb shift: α5mc2
Hyperfine splitting: (m/mp)α4mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 12 / 14
Hydrogen fine structure
The Hamiltonian that can besolved exactly for the hydrogenatom is merely an approxima-tion. The first is the correctionfor the motion of the proton.
The other important correctionscan be scaled according to thefine structure constant:
The largest correction is the finestructure and this consists oftwo terms: the relativistic cor-rection and the spin-orbit cor-rection.
me → µ =memp
me + mp≈ 1836
1837me ≈ me
α =e2
4πε0~c' 1
137.04
Some of the relevant energy scales are:
Bohr energy: α2mc2
Fine structure: α4mc2
Lamb shift: α5mc2
Hyperfine splitting: (m/mp)α4mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 12 / 14
Hydrogen fine structure
The Hamiltonian that can besolved exactly for the hydrogenatom is merely an approxima-tion. The first is the correctionfor the motion of the proton.
The other important correctionscan be scaled according to thefine structure constant:
The largest correction is the finestructure and this consists oftwo terms: the relativistic cor-rection and the spin-orbit cor-rection.
me → µ =memp
me + mp≈ 1836
1837me ≈ me
α =e2
4πε0~c' 1
137.04
Some of the relevant energy scales are:
Bohr energy: α2mc2
Fine structure: α4mc2
Lamb shift: α5mc2
Hyperfine splitting: (m/mp)α4mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 12 / 14
Hydrogen fine structure
The Hamiltonian that can besolved exactly for the hydrogenatom is merely an approxima-tion. The first is the correctionfor the motion of the proton.
The other important correctionscan be scaled according to thefine structure constant:
The largest correction is the finestructure and this consists oftwo terms: the relativistic cor-rection and the spin-orbit cor-rection.
me → µ =memp
me + mp≈ 1836
1837me ≈ me
α =e2
4πε0~c' 1
137.04
Some of the relevant energy scales are:
Bohr energy: α2mc2
Fine structure: α4mc2
Lamb shift: α5mc2
Hyperfine splitting: (m/mp)α4mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 12 / 14
Hydrogen fine structure
The Hamiltonian that can besolved exactly for the hydrogenatom is merely an approxima-tion. The first is the correctionfor the motion of the proton.
The other important correctionscan be scaled according to thefine structure constant:
The largest correction is the finestructure and this consists oftwo terms: the relativistic cor-rection and the spin-orbit cor-rection.
me → µ =memp
me + mp≈ 1836
1837me ≈ me
α =e2
4πε0~c' 1
137.04
Some of the relevant energy scales are:
Bohr energy: α2mc2
Fine structure: α4mc2
Lamb shift: α5mc2
Hyperfine splitting: (m/mp)α4mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 12 / 14
Hydrogen fine structure
The Hamiltonian that can besolved exactly for the hydrogenatom is merely an approxima-tion. The first is the correctionfor the motion of the proton.
The other important correctionscan be scaled according to thefine structure constant:
The largest correction is the finestructure and this consists oftwo terms: the relativistic cor-rection and the spin-orbit cor-rection.
me → µ =memp
me + mp≈ 1836
1837me ≈ me
α =e2
4πε0~c' 1
137.04
Some of the relevant energy scales are:
Bohr energy: α2mc2
Fine structure: α4mc2
Lamb shift: α5mc2
Hyperfine splitting: (m/mp)α4mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 12 / 14
Hydrogen fine structure
The Hamiltonian that can besolved exactly for the hydrogenatom is merely an approxima-tion. The first is the correctionfor the motion of the proton.
The other important correctionscan be scaled according to thefine structure constant:
The largest correction is the finestructure and this consists oftwo terms: the relativistic cor-rection and the spin-orbit cor-rection.
me → µ =memp
me + mp≈ 1836
1837me ≈ me
α =e2
4πε0~c' 1
137.04
Some of the relevant energy scales are:
Bohr energy: α2mc2
Fine structure: α4mc2
Lamb shift: α5mc2
Hyperfine splitting: (m/mp)α4mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 12 / 14
Hydrogen fine structure
The Hamiltonian that can besolved exactly for the hydrogenatom is merely an approxima-tion. The first is the correctionfor the motion of the proton.
The other important correctionscan be scaled according to thefine structure constant:
The largest correction is the finestructure and this consists oftwo terms: the relativistic cor-rection and the spin-orbit cor-rection.
me → µ =memp
me + mp≈ 1836
1837me ≈ me
α =e2
4πε0~c' 1
137.04
Some of the relevant energy scales are:
Bohr energy: α2mc2
Fine structure: α4mc2
Lamb shift: α5mc2
Hyperfine splitting: (m/mp)α4mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 12 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2
=p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m
= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m
= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv
=mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4
= γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4
= γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])
= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2
−→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
Relativistic correction
The first term in the Hamiltonianrepresents the kinetic energy
if we substitute p = (~/i)∇, thefamiliar operator form emerges
this is not good enough for relativis-tic electrons, instead we must sub-tract the rest energy from the totalenergy
and the relativistic momentum
T =1
2mv2 =
p2
2m= − ~2
2m∇2
T = γmc2 −mc2
=mc2√
1− (v/c)2−mc2
p = γmv =mv√
1− (v/c)2
p2c2 + m2c4 = γ2m2v2c2 + m2c4 = γ2m2v2c2 + γ2m2c41
γ2
= γ2(m2v2c2 + m2c4
[1− (v/c)2
])= γ2m2c4
=(T + mc2
)2 −→ T =√p2c2 + m2c4 −mc2
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 13 / 14
First order relativistic correction
This expression must reduce to theclassical value when p � mc
so ex-panding in powers of (p/mc)
T =√p2c2 + m2c4 −mc2
T = mc2
[√1 +
( p
mc
)2− 1
]= mc2
[
1 +1
2
( p
mc
)2− 1
8
( p
mc
)4· · ·
− 1
]=
p2
2m− p4
8m3c2+ · · ·
The second term is the first orderrelativistic correction
and the correction to the energy isthus
H ′r = − p4
8m3c2
E(1)r =
⟨H ′r
⟩= − 1
8m3c2〈ψ| p4ψ
⟩= − 1
8m3c2⟨p2ψ
∣∣ p2ψ⟩
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 14 / 14
First order relativistic correction
This expression must reduce to theclassical value when p � mc so ex-panding in powers of (p/mc)
T =√p2c2 + m2c4 −mc2
T = mc2
[√1 +
( p
mc
)2− 1
]= mc2
[
1 +1
2
( p
mc
)2− 1
8
( p
mc
)4· · ·
− 1
]=
p2
2m− p4
8m3c2+ · · ·
The second term is the first orderrelativistic correction
and the correction to the energy isthus
H ′r = − p4
8m3c2
E(1)r =
⟨H ′r
⟩= − 1
8m3c2〈ψ| p4ψ
⟩= − 1
8m3c2⟨p2ψ
∣∣ p2ψ⟩
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 14 / 14
First order relativistic correction
This expression must reduce to theclassical value when p � mc so ex-panding in powers of (p/mc)
T =√p2c2 + m2c4 −mc2
T = mc2
[√1 +
( p
mc
)2− 1
]
= mc2[
1 +1
2
( p
mc
)2− 1
8
( p
mc
)4· · ·
− 1
]=
p2
2m− p4
8m3c2+ · · ·
The second term is the first orderrelativistic correction
and the correction to the energy isthus
H ′r = − p4
8m3c2
E(1)r =
⟨H ′r
⟩= − 1
8m3c2〈ψ| p4ψ
⟩= − 1
8m3c2⟨p2ψ
∣∣ p2ψ⟩
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 14 / 14
First order relativistic correction
This expression must reduce to theclassical value when p � mc so ex-panding in powers of (p/mc)
T =√p2c2 + m2c4 −mc2
T = mc2
[√1 +
( p
mc
)2− 1
]= mc2
[
1 +1
2
( p
mc
)2− 1
8
( p
mc
)4· · ·
− 1
]
=p2
2m− p4
8m3c2+ · · ·
The second term is the first orderrelativistic correction
and the correction to the energy isthus
H ′r = − p4
8m3c2
E(1)r =
⟨H ′r
⟩= − 1
8m3c2〈ψ| p4ψ
⟩= − 1
8m3c2⟨p2ψ
∣∣ p2ψ⟩
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 14 / 14
First order relativistic correction
This expression must reduce to theclassical value when p � mc so ex-panding in powers of (p/mc)
T =√p2c2 + m2c4 −mc2
T = mc2
[√1 +
( p
mc
)2− 1
]= mc2
[1
+1
2
( p
mc
)2− 1
8
( p
mc
)4· · ·
− 1
]
=p2
2m− p4
8m3c2+ · · ·
The second term is the first orderrelativistic correction
and the correction to the energy isthus
H ′r = − p4
8m3c2
E(1)r =
⟨H ′r
⟩= − 1
8m3c2〈ψ| p4ψ
⟩= − 1
8m3c2⟨p2ψ
∣∣ p2ψ⟩
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 14 / 14
First order relativistic correction
This expression must reduce to theclassical value when p � mc so ex-panding in powers of (p/mc)
T =√p2c2 + m2c4 −mc2
T = mc2
[√1 +
( p
mc
)2− 1
]= mc2
[1 +
1
2
( p
mc
)2
− 1
8
( p
mc
)4· · ·
− 1
]
=p2
2m− p4
8m3c2+ · · ·
The second term is the first orderrelativistic correction
and the correction to the energy isthus
H ′r = − p4
8m3c2
E(1)r =
⟨H ′r
⟩= − 1
8m3c2〈ψ| p4ψ
⟩= − 1
8m3c2⟨p2ψ
∣∣ p2ψ⟩
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 14 / 14
First order relativistic correction
This expression must reduce to theclassical value when p � mc so ex-panding in powers of (p/mc)
T =√p2c2 + m2c4 −mc2
T = mc2
[√1 +
( p
mc
)2− 1
]= mc2
[1 +
1
2
( p
mc
)2− 1
8
( p
mc
)4· · · − 1
]
=p2
2m− p4
8m3c2+ · · ·
The second term is the first orderrelativistic correction
and the correction to the energy isthus
H ′r = − p4
8m3c2
E(1)r =
⟨H ′r
⟩= − 1
8m3c2〈ψ| p4ψ
⟩= − 1
8m3c2⟨p2ψ
∣∣ p2ψ⟩
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 14 / 14
First order relativistic correction
This expression must reduce to theclassical value when p � mc so ex-panding in powers of (p/mc)
T =√p2c2 + m2c4 −mc2
T = mc2
[√1 +
( p
mc
)2− 1
]= mc2
[1 +
1
2
( p
mc
)2− 1
8
( p
mc
)4· · · − 1
]=
p2
2m− p4
8m3c2+ · · ·
The second term is the first orderrelativistic correction
and the correction to the energy isthus
H ′r = − p4
8m3c2
E(1)r =
⟨H ′r
⟩= − 1
8m3c2〈ψ| p4ψ
⟩= − 1
8m3c2⟨p2ψ
∣∣ p2ψ⟩
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 14 / 14
First order relativistic correction
This expression must reduce to theclassical value when p � mc so ex-panding in powers of (p/mc)
T =√p2c2 + m2c4 −mc2
T = mc2
[√1 +
( p
mc
)2− 1
]= mc2
[1 +
1
2
( p
mc
)2− 1
8
( p
mc
)4· · · − 1
]=
p2
2m− p4
8m3c2+ · · ·
The second term is the first orderrelativistic correction
and the correction to the energy isthus
H ′r = − p4
8m3c2
E(1)r =
⟨H ′r
⟩= − 1
8m3c2〈ψ| p4ψ
⟩= − 1
8m3c2⟨p2ψ
∣∣ p2ψ⟩
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 14 / 14
First order relativistic correction
This expression must reduce to theclassical value when p � mc so ex-panding in powers of (p/mc)
T =√p2c2 + m2c4 −mc2
T = mc2
[√1 +
( p
mc
)2− 1
]= mc2
[1 +
1
2
( p
mc
)2− 1
8
( p
mc
)4· · · − 1
]=
p2
2m− p4
8m3c2+ · · ·
The second term is the first orderrelativistic correction
and the correction to the energy isthus
H ′r = − p4
8m3c2
E(1)r =
⟨H ′r
⟩= − 1
8m3c2〈ψ| p4ψ
⟩= − 1
8m3c2⟨p2ψ
∣∣ p2ψ⟩
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 14 / 14
First order relativistic correction
This expression must reduce to theclassical value when p � mc so ex-panding in powers of (p/mc)
T =√p2c2 + m2c4 −mc2
T = mc2
[√1 +
( p
mc
)2− 1
]= mc2
[1 +
1
2
( p
mc
)2− 1
8
( p
mc
)4· · · − 1
]=
p2
2m− p4
8m3c2+ · · ·
The second term is the first orderrelativistic correction
and the correction to the energy isthus
H ′r = − p4
8m3c2
E(1)r =
⟨H ′r
⟩= − 1
8m3c2〈ψ| p4ψ
⟩= − 1
8m3c2⟨p2ψ
∣∣ p2ψ⟩
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 14 / 14
First order relativistic correction
This expression must reduce to theclassical value when p � mc so ex-panding in powers of (p/mc)
T =√p2c2 + m2c4 −mc2
T = mc2
[√1 +
( p
mc
)2− 1
]= mc2
[1 +
1
2
( p
mc
)2− 1
8
( p
mc
)4· · · − 1
]=
p2
2m− p4
8m3c2+ · · ·
The second term is the first orderrelativistic correction
and the correction to the energy isthus
H ′r = − p4
8m3c2
E(1)r =
⟨H ′r
⟩
= − 1
8m3c2〈ψ| p4ψ
⟩= − 1
8m3c2⟨p2ψ
∣∣ p2ψ⟩
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 14 / 14
First order relativistic correction
This expression must reduce to theclassical value when p � mc so ex-panding in powers of (p/mc)
T =√p2c2 + m2c4 −mc2
T = mc2
[√1 +
( p
mc
)2− 1
]= mc2
[1 +
1
2
( p
mc
)2− 1
8
( p
mc
)4· · · − 1
]=
p2
2m− p4
8m3c2+ · · ·
The second term is the first orderrelativistic correction
and the correction to the energy isthus
H ′r = − p4
8m3c2
E(1)r =
⟨H ′r
⟩= − 1
8m3c2〈ψ| p4ψ
⟩
= − 1
8m3c2⟨p2ψ
∣∣ p2ψ⟩
C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 14 / 14
First order relativistic correction
This expression must reduce to theclassical value when p � mc so ex-panding in powers of (p/mc)
T =√p2c2 + m2c4 −mc2
T = mc2
[√1 +
( p
mc
)2− 1
]= mc2
[1 +
1
2
( p
mc
)2− 1
8
( p
mc
)4· · · − 1
]=
p2
2m− p4
8m3c2+ · · ·
The second term is the first orderrelativistic correction
and the correction to the energy isthus
H ′r = − p4
8m3c2
E(1)r =
⟨H ′r
⟩= − 1
8m3c2〈ψ| p4ψ
⟩= − 1
8m3c2⟨p2ψ
∣∣ p2ψ⟩C. Segre (IIT) PHYS 406 - Spring 2016 January 20, 2016 14 / 14