today’s outline - january 20, 2016csrri.iit.edu/~segre/phys406/16s/lecture_03.pdftoday’s outline...

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








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









Wba Wbb

)(αβ

)= E 1

(αβ

)

Wij =⟨ψ0i

∣∣H ′∣∣ψ0j

⟩H0ψ0

i = E 0ψ0i ,

⟨ψ0i

∣∣ψ0j

⟩= δij

ψ0 =n∑

i=1

αiψ0i









Wba Wbb

)(αβ

)= E 1

(αβ

)

Wij =⟨ψ0i

∣∣H ′∣∣ψ0j

⟩

H0ψ0i = E 0ψ0

i ,⟨ψ0i

∣∣ψ0j

⟩= δij

ψ0 =n∑

i=1

αiψ0i






i are n-fold degen-erate

and that ψ0 is a general lin-ear combination



Wba Wbb

)(αβ

)= E 1

(αβ

)

Wij =⟨ψ0i

∣∣H ′∣∣ψ0j

⟩

H0ψ0i = E 0ψ0

i ,⟨ψ0i

∣∣ψ0j

⟩= δij

ψ0 =n∑

i=1

αiψ0i






i are n-fold degen-erate

and that ψ0 is a general lin-ear combination



Wba Wbb

)(αβ

)= E 1

(αβ

)

Wij =⟨ψ0i

∣∣H ′∣∣ψ0j

⟩H0ψ0

i = E 0ψ0i ,

⟨ψ0i

∣∣ψ0j

⟩= δij

ψ0 =n∑

i=1

αiψ0i









Wba Wbb

)(αβ

)= E 1

(αβ

)

Wij =⟨ψ0i

∣∣H ′∣∣ψ0j

⟩H0ψ0

i = E 0ψ0i ,

⟨ψ0i

∣∣ψ0j

⟩= δij

ψ0 =n∑

i=1

αiψ0i


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 + · · ·


Problem 6.10


ψ0 =n∑

j=1

αjψ0j .



H0ψ0 =n∑

j=1

αjH0ψ0

j

= E 0n∑

j=1

αjψ0j = E 0ψ0


ψ = ψ0 + ψ1 + · · · E = E 0 + E 1 + · · ·


Problem 6.10


ψ0 =n∑

j=1

αjψ0j .



H0ψ0 =n∑

j=1

αjH0ψ0

j = E 0n∑

j=1

αjψ0j

= E 0ψ0


ψ = ψ0 + ψ1 + · · · E = E 0 + E 1 + · · ·


Problem 6.10


ψ0 =n∑

j=1

αjψ0j .



H0ψ0 =n∑

j=1

αjH0ψ0

j = E 0n∑

j=1

αjψ0j = E 0ψ0


ψ = ψ0 + ψ1 + · · · E = E 0 + E 1 + · · ·


Problem 6.10


ψ0 =n∑

j=1

αjψ0j .



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 + · · ·


Problem 6.10


ψ0 =n∑

j=1

αjψ0j .



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 + · · ·


Problem 6.10


ψ0 =n∑

j=1

αjψ0j .



H0ψ0 =n∑

j=1

αjH0ψ0

j = E 0n∑

j=1

αjψ0j = E 0ψ0


ψ = ψ0 + ψ1 + · · ·

E = E 0 + E 1 + · · ·


Problem 6.10


ψ0 =n∑

j=1

αjψ0j .



H0ψ0 =n∑

j=1

αjH0ψ0

j = E 0n∑

j=1

αjψ0j = E 0ψ0


ψ = ψ0 + ψ1 + · · · E = E 0 + E 1 + · · ·


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




Hψ = Eψ

−→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0


∣∣��⟨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


j





Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0


∣∣��⟨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


j





Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0


∣∣

��⟨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


j





Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0


∣∣⟨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


j





Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0


∣∣��⟨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


j





Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0


∣∣��⟨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


j





Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0


∣∣��⟨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


j





Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0


∣∣��⟨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


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




Hψ = Eψ −→ H0ψ1 + H ′ψ0 = E 0ψ1 + E 1ψ0


∣∣��⟨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


j




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



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


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


Example 6.2


V 0(x , y , z) =

{0, 0 < x , y , z < a

∞, otherwise



(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)


ψa ≡ ψ112, ψb ≡ ψ121, ψc ≡ ψ211, E 01 ≡ 3

π2~2

ma2


Example 6.2


V 0(x , y , z) =

{0, 0 < x , y , z < a

∞, otherwise



(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)


ψa ≡ ψ112, ψb ≡ ψ121, ψc ≡ ψ211, E 01 ≡ 3

π2~2

ma2


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


Example 6.2


H ′ =

{V0, 0 < x < a/2 and 0 < y < a/2

0, otherwise


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


Example 6.2


H ′ =

{V0, 0 < x < a/2 and 0 < y < a/2

0, otherwise


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


Example 6.2


H ′ =

{V0, 0 < x < a/2 and 0 < y < a/2

0, otherwise


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


Example 6.2


H ′ =

{V0, 0 < x < a/2 and 0 < y < a/2

0, otherwise


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


Example 6.2


H ′ =

{V0, 0 < x < a/2 and 0 < y < a/2

0, otherwise


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


Example 6.2


H ′ =

{V0, 0 < x < a/2 and 0 < y < a/2

0, otherwise


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


Example 6.2


H ′ =

{V0, 0 < x < a/2 and 0 < y < a/2

0, otherwise


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


Example 6.2


H ′ =

{V0, 0 < x < a/2 and 0 < y < a/2

0, otherwise


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


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


Example 6.2


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


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


Example 6.2


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


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


Example 6.2


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


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


Example 6.2


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


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


Example 6.2


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


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


Example 6.2


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


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


Example 6.2


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


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


Example 6.2


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


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


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


Example 6.2


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


∫ π/2


=

(4V0

π2

){∫ π/2


}2=

(4V0

π2

){2

[sin3(θ)

3

∣∣∣∣π/20

}2

=16V0

9π2=

V0

4· 64

9π2=

V0

4κ κ ≡

(8

3π

)2≈ 0.7205


Example 6.2


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


∫ π/2


=

(4V0

π2

){∫ π/2


}2=

(4V0

π2

){2

[sin3(θ)

3

∣∣∣∣π/20

}2

=16V0

9π2=

V0

4· 64

9π2=

V0

4κ κ ≡

(8

3π

)2≈ 0.7205


Example 6.2


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


∫ π/2


=

(4V0

π2

){∫ π/2


}2=

(4V0

π2

){2

[sin3(θ)

3

∣∣∣∣π/20

}2

=16V0

9π2=

V0

4· 64

9π2=

V0

4κ κ ≡

(8

3π

)2≈ 0.7205


Example 6.2


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


∫ π/2


=

(4V0

π2

){∫ π/2


}2=

(4V0

π2

){2

[sin3(θ)

3

∣∣∣∣π/20

}2

=16V0

9π2=

V0

4· 64

9π2=

V0

4κ κ ≡

(8

3π

)2≈ 0.7205


Example 6.2


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


∫ π/2


=

(4V0

π2

){∫ π/2


}2=

(4V0

π2

){2

[sin3(θ)

3

∣∣∣∣π/20

}2

=16V0

9π2=

V0

4· 64

9π2=

V0

4κ κ ≡

(8

3π

)2≈ 0.7205


Example 6.2


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


∫ π/2


=

(4V0

π2

){∫ π/2


}2

=

(4V0

π2

){2

[sin3(θ)

3

∣∣∣∣π/20

}2

=16V0

9π2=

V0

4· 64

9π2=

V0

4κ κ ≡

(8

3π

)2≈ 0.7205


Example 6.2


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


∫ π/2


=

(4V0

π2

){∫ π/2


}2=

(4V0

π2

){2

[sin3(θ)

3

∣∣∣∣π/20

}2

=16V0

9π2=

V0

4· 64

9π2=

V0

4κ κ ≡

(8

3π

)2≈ 0.7205


Example 6.2


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


∫ π/2


=

(4V0

π2

){∫ π/2


}2=

(4V0

π2

){2

[sin3(θ)

3

∣∣∣∣π/20

}2

=16V0

9π2

=V0

4· 64

9π2=

V0

4κ κ ≡

(8

3π

)2≈ 0.7205


Example 6.2


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


∫ π/2


=

(4V0

π2

){∫ π/2


}2=

(4V0

π2

){2

[sin3(θ)

3

∣∣∣∣π/20

}2

=16V0

9π2=

V0

4· 64

9π2

=V0

4κ κ ≡

(8

3π

)2≈ 0.7205


Example 6.2


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


∫ π/2


=

(4V0

π2

){∫ π/2


}2=

(4V0

π2

){2

[sin3(θ)

3

∣∣∣∣π/20

}2

=16V0

9π2=

V0

4· 64

9π2=

V0

4κ κ ≡

(8

3π

)2≈ 0.7205


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


Example 6.2



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


Example 6.2



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


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)


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






me → µ =memp

me + mp

≈ 1836

1837me ≈ me

α =e2

4πε0~c' 1

137.04


Bohr energy: α2mc2


Lamb shift: α5mc2







me → µ =memp

me + mp≈ 1836

1837me ≈ me

α =e2

4πε0~c' 1

137.04


Bohr energy: α2mc2


Lamb shift: α5mc2



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







T =1

2mv2

=p2

2m= − ~2

2m∇2

T = γmc2 −mc2

=mc2√

1− (v/c)2−mc2

p = γmv =mv√

1− (v/c)2


γ2

= γ2(m2v2c2 + m2c4

[1− (v/c)2

])= γ2m2c4

=(T + mc2

)2 −→ T =√p2c2 + m2c4 −mc2







T =1

2mv2 =

p2

2m

= − ~2

2m∇2

T = γmc2 −mc2

=mc2√

1− (v/c)2−mc2

p = γmv =mv√

1− (v/c)2


γ2

= γ2(m2v2c2 + m2c4

[1− (v/c)2

])= γ2m2c4

=(T + mc2

)2 −→ T =√p2c2 + m2c4 −mc2
