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Today’s Outline - January 27, 2015

• Spin-orbit coupling

• Zeeman effect

No class on Thursday, January 29, 2015

Homework Assignment #02:Chapter 5: 18,20,21,23,24,26due Tuesday, January 27, 2015

Homework Assignment #03:Chapter 5: 27, 30; Chapter 6: 1, 4, 6, 29due Tuesday, February 3, 2015

Tutoring sessions:Monday, Wednesday, & Friday, 12:45–13:45, 116 LS

C. Segre (IIT) PHYS 406 - Spring 2015 January 27, 2015 1 / 14



• Zeeman effect








• Zeeman effect








• Zeeman effect








• Zeeman effect








• Zeeman effect








• Zeeman effect








• Zeeman effect








• Zeeman effect






Magnetic field at the electron

p

p

B L

e

r

The “normal” view of a hydro-gen atom has the electron rotatingabout the proton (really just havingangular momentum)

In the electron’s frame of referenceit is the opposite

The “rotating” proton has angular momentum and produces a magneticfield at the position of the electron

This magnetic field produces a per-turbative torque on the electron’smagnetic moment from the Biot-Savart law

L = rmv

=2πmr2

τ

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2r=

i

2ε0c2r=

e

2ε0c2τ r

~B =1

4πε0

e

mc2r3~L



L

e

p

B

r





L = rmv

=2πmr2

τ

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2r=

i

2ε0c2r=

e

2ε0c2τ r

~B =1

4πε0

e

mc2r3~L



e

p

B L

r



The “rotating” proton has angular momentum

and produces a magneticfield at the position of the electron


L = rmv

=2πmr2

τ

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2r=

i

2ε0c2r=

e

2ε0c2τ r

~B =1

4πε0

e

mc2r3~L



e

p

B L

r





L = rmv

=2πmr2

τ

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2r=

i

2ε0c2r=

e

2ε0c2τ r

~B =1

4πε0

e

mc2r3~L



e

p

B L

r




This magnetic field produces a per-turbative torque on the electron’smagnetic moment

from the Biot-Savart law

L = rmv

=2πmr2

τ

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2r=

i

2ε0c2r=

e

2ε0c2τ r

~B =1

4πε0

e

mc2r3~L



e

p

B L

r




This magnetic field produces a per-turbative torque on the electron’smagnetic moment

from the Biot-Savart law

L = rmv

=2πmr2

τ

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2r=

i

2ε0c2r=

e

2ε0c2τ r

~B =1

4πε0

e

mc2r3~L



e

p

B L

r





L = rmv

=2πmr2

τ

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2r=

i

2ε0c2r=

e

2ε0c2τ r

~B =1

4πε0

e

mc2r3~L



e

p

B L

r





L = rmv

=2πmr2

τ

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2r

=i

2ε0c2r=

e

2ε0c2τ r

~B =1

4πε0

e

mc2r3~L



e

p

B L

r





L = rmv

=2πmr2

τ

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2r=

i

2ε0c2r

=e

2ε0c2τ r

~B =1

4πε0

e

mc2r3~L



e

p

B L

r





L = rmv

=2πmr2

τ

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2r=

i

2ε0c2r=

e

2ε0c2τ r

~B =1

4πε0

e

mc2r3~L



e

p

B L

r





L = rmv

=2πmr2

τ

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2r=

i

2ε0c2r=

e

2ε0c2τ r

~B =1

4πε0

e

mc2r3~L



e

p

B L

r





L = rmv =2πmr2

τ

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2r=

i

2ε0c2r=

e

2ε0c2τ r

~B =1

4πε0

e

mc2r3~L



e

p

B L

r





L = rmv =2πmr2

τ

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2r=

i

2ε0c2r=

e

2ε0c2τ r

~B =1

4πε0

e

mc2r3~L



e

p

B L

r





L = rmv =2πmr2

τ

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2r=

i

2ε0c2r=

e

2ε0c2τ r

~B =1

4πε0

e

mc2r3~L


Electron magnetic dipole

This magnetic field interacts withthe electron dipole ~µ which can becalculated classically

Consider a ring of charge q andmass m rotating about its axis withperiod τ

the magnetic moment is the currenttimes the area of the loop while itsangular momentum, S is the mo-ment of inertia times the angularvelocity

The so-called, gyromagnetic ratio,γcl is the ratio of these two quanti-ties independent of r and τ

q,m

S µ

r

µ = iA =qπr2

τ

S = Iω =2πmr2

τ

γcl =µ

S=

qπr2

τ· τ

2πmr2=

q

2m







q,m

S µ

r

µ = iA =qπr2

τ

S = Iω =2πmr2

τ

γcl =µ

S=

qπr2

τ· τ

2πmr2=

q

2m







q,m

S µ

r

µ = iA =qπr2

τ

S = Iω =2πmr2

τ

γcl =µ

S=

qπr2

τ· τ

2πmr2=

q

2m





the magnetic moment is the currenttimes the area of the loop

while itsangular momentum, S is the mo-ment of inertia times the angularvelocity


q,m

S µ

r

µ = iA =qπr2

τ

S = Iω =2πmr2

τ

γcl =µ

S=

qπr2

τ· τ

2πmr2=

q

2m








q,m

S µ

r

µ = iA

=qπr2

τ

S = Iω =2πmr2

τ

γcl =µ

S=

qπr2

τ· τ

2πmr2=

q

2m








q,m

S µ

r

µ = iA =qπr2

τ

S = Iω =2πmr2

τ

γcl =µ

S=

qπr2

τ· τ

2πmr2=

q

2m







q,m

S µ

r

µ = iA =qπr2

τ

S = Iω =2πmr2

τ

γcl =µ

S=

qπr2

τ· τ

2πmr2=

q

2m







q,m

S µ

r

µ = iA =qπr2

τ

S = Iω

=2πmr2

τ

γcl =µ

S=

qπr2

τ· τ

2πmr2=

q

2m







q,m

S µ

r

µ = iA =qπr2

τ

S = Iω =2πmr2

τ

γcl =µ

S=

qπr2

τ· τ

2πmr2=

q

2m






The so-called, gyromagnetic ratio,γcl is the ratio of these two quanti-ties

independent of r and τ

q,m

S µ

r

µ = iA =qπr2

τ

S = Iω =2πmr2

τ

γcl =µ

S=

qπr2

τ· τ

2πmr2=

q

2m








q,m

S µ

r

µ = iA =qπr2

τ

S = Iω =2πmr2

τ

γcl =µ

S

=qπr2

τ· τ

2πmr2=

q

2m








q,m

S µ

r

µ = iA =qπr2

τ

S = Iω =2πmr2

τ

γcl =µ

S=

qπr2

τ· τ

2πmr2

=q

2m








q,m

S µ

r

µ = iA =qπr2

τ

S = Iω =2πmr2

τ

γcl =µ

S=

qπr2

τ· τ

2πmr2=

q

2m







q,m

S µ

r

µ = iA =qπr2

τ

S = Iω =2πmr2

τ

γcl =µ

S=

qπr2

τ· τ

2πmr2=

q

2m


Quantum gyromagnetic ratio

For an electron, the magnetic moment is a result of the intrinsic angularmomentum, or “spin”

This is not the classical value butclose to two times bigger, due torelativistic theory

where ge ≈ 2 + α/π = 2.002

The perturbing Hamiltonian is thus

but since the electron is in an ac-celerating frame of reference, theThomas precession correction mustbe applied which leads to replacingge → (ge − 1) and results in thefinal spin-orbit correction Hamilto-nian

~µ = γe~S = −gee

2m~S

H ′ = −~µ · ~B =gee

2m

1

4πε0

e

mc2r3~S · ~L

= ge

(e2

8πε0

)1

m2c2r3~S · ~L

H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L





where ge ≈ 2 + α/π = 2.002



~µ = γe~S

= −gee

2m~S

H ′ = −~µ · ~B =gee

2m

1

4πε0

e

mc2r3~S · ~L

= ge

(e2

8πε0

)1

m2c2r3~S · ~L

H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L





where ge ≈ 2 + α/π = 2.002



~µ = γe~S

= −gee

2m~S

H ′ = −~µ · ~B =gee

2m

1

4πε0

e

mc2r3~S · ~L

= ge

(e2

8πε0

)1

m2c2r3~S · ~L

H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L





where ge ≈ 2 + α/π = 2.002




2m~S

H ′ = −~µ · ~B =gee

2m

1

4πε0

e

mc2r3~S · ~L

= ge

(e2

8πε0

)1

m2c2r3~S · ~L

H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L





where ge ≈ 2 + α/π = 2.002




2m~S

H ′ = −~µ · ~B =gee

2m

1

4πε0

e

mc2r3~S · ~L

= ge

(e2

8πε0

)1

m2c2r3~S · ~L

H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L





where ge ≈ 2 + α/π = 2.002




2m~S

H ′ = −~µ · ~B =gee

2m

1

4πε0

e

mc2r3~S · ~L

= ge

(e2

8πε0

)1

m2c2r3~S · ~L

H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L





where ge ≈ 2 + α/π = 2.002




2m~S

H ′ = −~µ · ~B

=gee

2m

1

4πε0

e

mc2r3~S · ~L

= ge

(e2

8πε0

)1

m2c2r3~S · ~L

H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L





where ge ≈ 2 + α/π = 2.002




2m~S

H ′ = −~µ · ~B =gee

2m

1

4πε0

e

mc2r3~S · ~L

= ge

(e2

8πε0

)1

m2c2r3~S · ~L

H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L





where ge ≈ 2 + α/π = 2.002




2m~S

H ′ = −~µ · ~B =gee

2m

1

4πε0

e

mc2r3~S · ~L

= ge

(e2

8πε0

)1

m2c2r3~S · ~L

H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L





where ge ≈ 2 + α/π = 2.002




2m~S

H ′ = −~µ · ~B =gee

2m

1

4πε0

e

mc2r3~S · ~L

= ge

(e2

8πε0

)1

m2c2r3~S · ~L

H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L





where ge ≈ 2 + α/π = 2.002




2m~S

H ′ = −~µ · ~B =gee

2m

1

4πε0

e

mc2r3~S · ~L

= ge

(e2

8πε0

)1

m2c2r3~S · ~L

H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L


Spin-orbit correction

The spin-orbit interaction does not commute with ~L or ~S and so the spinand orbital angular momentum are no longer separately conserved (ml andms are not “good” quantum numbers).

But H ′so still commutes with L2 and S2 as well as the total angularmomentum ~J = ~L + ~S so these quantities are conserved (and l , s, j , mj

are all “good” quantum numbers!

This can be used to recast the spin-orbit Hamiltonian

J2 = (~L + ~S) · (~L + ~S) = L2 + S2 + 2~L · ~S

~L · ~S =1

2(J2 − L2 − S2)







J2 = (~L + ~S) · (~L + ~S) = L2 + S2 + 2~L · ~S

~L · ~S =1

2(J2 − L2 − S2)







J2 = (~L + ~S) · (~L + ~S) = L2 + S2 + 2~L · ~S

~L · ~S =1

2(J2 − L2 − S2)







J2 = (~L + ~S) · (~L + ~S)

= L2 + S2 + 2~L · ~S

~L · ~S =1

2(J2 − L2 − S2)







J2 = (~L + ~S) · (~L + ~S) = L2 + S2 + 2~L · ~S

~L · ~S =1

2(J2 − L2 − S2)







J2 = (~L + ~S) · (~L + ~S) = L2 + S2 + 2~L · ~S

~L · ~S =1

2(J2 − L2 − S2)


Spin-orbit eigenvalues

Looking at the spin-orbit Hamiltonian again

H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L

The eigenvalues of ~L · ~S are⟨~L · ~S

⟩=

~2

2[j(j + 1)− l(l + 1)− s(s + 1)]

and the expectation value of 1/r3

(the other term in the spin-orbitHamiltonian) is

⟨1

r3

⟩=

1

l(l + 1/2)(l + 1)n3a3

E(1)so =

⟨H ′so⟩

=e2

8πε0· 1

m2c2· ~

2

2·j(j + 1)− l(l + 1)− 3

4

l(l + 12)(l + 1)n3a3




H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L


⟩=

~2

2[j(j + 1)− l(l + 1)− s(s + 1)]



⟨1

r3

⟩=

1

l(l + 1/2)(l + 1)n3a3

E(1)so =

⟨H ′so⟩

=e2

8πε0· 1

m2c2· ~

2

2·j(j + 1)− l(l + 1)− 3

4

l(l + 12)(l + 1)n3a3




H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L


⟩=

~2

2[j(j + 1)− l(l + 1)− s(s + 1)]



⟨1

r3

⟩=

1

l(l + 1/2)(l + 1)n3a3

E(1)so =

⟨H ′so⟩

=e2

8πε0· 1

m2c2· ~

2

2·j(j + 1)− l(l + 1)− 3

4

l(l + 12)(l + 1)n3a3




H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L


⟩=

~2

2[j(j + 1)− l(l + 1)− s(s + 1)]



⟨1

r3

⟩=

1

l(l + 1/2)(l + 1)n3a3

E(1)so =

⟨H ′so⟩

=e2

8πε0· 1

m2c2· ~

2

2·j(j + 1)− l(l + 1)− 3

4

l(l + 12)(l + 1)n3a3




H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L


⟩=

~2

2[j(j + 1)− l(l + 1)− s(s + 1)]



⟨1

r3

⟩=

1

l(l + 1/2)(l + 1)n3a3

E(1)so =

⟨H ′so⟩

=e2

8πε0· 1

m2c2· ~

2

2·j(j + 1)− l(l + 1)− 3

4

l(l + 12)(l + 1)n3a3




H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L


⟩=

~2

2[j(j + 1)− l(l + 1)− s(s + 1)]



⟨1

r3

⟩=

1

l(l + 1/2)(l + 1)n3a3

E(1)so =

⟨H ′so⟩

=e2

8πε0· 1

m2c2· ~

2

2·j(j + 1)− l(l + 1)− 3

4

l(l + 12)(l + 1)n3a3




H ′so ≈(

e2

8πε0

)1

m2c2r3~S · ~L


⟩=

~2

2[j(j + 1)− l(l + 1)− s(s + 1)]



⟨1

r3

⟩=

1

l(l + 1/2)(l + 1)n3a3

E(1)so =

⟨H ′so⟩

=e2

8πε0· 1

m2c2· ~

2

2·j(j + 1)− l(l + 1)− 3

4

l(l + 12)(l + 1)n3a3


Fine structure energy

In terms of En we can express E(1)so as

E(1)so =

E 2n

mc2

{n[j(j + 1)− l(l + 1)− 3

4 ]

l(l + 12)(l + 1)

}

recalling the relativistic correctionE(1)r = − E 2

n

2mc2

[4n

l + 12

− 3

]we get the total fine structure correction of

E(1)fs =

E 2n

2mc2

(3− 4n

j + 12

)the full energy ofor the hydrogen atom, including fine structure is thus

Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]




E(1)so =

E 2n

mc2

{n[j(j + 1)− l(l + 1)− 3

4 ]

l(l + 12)(l + 1)

}


n

2mc2

[4n

l + 12

− 3


E(1)fs =

E 2n

2mc2

(3− 4n

j + 12


Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]




E(1)so =

E 2n

mc2

{n[j(j + 1)− l(l + 1)− 3

4 ]

l(l + 12)(l + 1)

}

recalling the relativistic correction

E(1)r = − E 2

n

2mc2

[4n

l + 12

− 3


E(1)fs =

E 2n

2mc2

(3− 4n

j + 12


Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]




E(1)so =

E 2n

mc2

{n[j(j + 1)− l(l + 1)− 3

4 ]

l(l + 12)(l + 1)

}


n

2mc2

[4n

l + 12

− 3

]

we get the total fine structure correction of

E(1)fs =

E 2n

2mc2

(3− 4n

j + 12


Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]




E(1)so =

E 2n

mc2

{n[j(j + 1)− l(l + 1)− 3

4 ]

l(l + 12)(l + 1)

}


n

2mc2

[4n

l + 12

− 3


E(1)fs =

E 2n

2mc2

(3− 4n

j + 12


Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]




E(1)so =

E 2n

mc2

{n[j(j + 1)− l(l + 1)− 3

4 ]

l(l + 12)(l + 1)

}


n

2mc2

[4n

l + 12

− 3


E(1)fs =

E 2n

2mc2

(3− 4n

j + 12

)

the full energy ofor the hydrogen atom, including fine structure is thus

Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]




E(1)so =

E 2n

mc2

{n[j(j + 1)− l(l + 1)− 3

4 ]

l(l + 12)(l + 1)

}


n

2mc2

[4n

l + 12

− 3


E(1)fs =

E 2n

2mc2

(3− 4n

j + 12


Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]




E(1)so =

E 2n

mc2

{n[j(j + 1)− l(l + 1)− 3

4 ]

l(l + 12)(l + 1)

}


n

2mc2

[4n

l + 12

− 3


E(1)fs =

E 2n

2mc2

(3− 4n

j + 12


Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]C. Segre (IIT) PHYS 406 - Spring 2015 January 27, 2015 7 / 14

Fine structure of hydrogen

j=1/2

j=7/2

j=5/2

j=3/2

j=1/2

j=1/2

j=1/2

j=3/2

j=3/2

j=5/2

l=0 l=1 l=2 l=3

n=1

n=2

n=3

n=4


Sample calculations

Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]

Suppose we have n = 1, l = 0, j = 12

E1 12

= −13.6

12

[1 +

α2

12

(1

12 + 1

2

− 3

4

)]= −13.6

[1 +

α2

4

]For n = 2, l = 0, j = 1

2

E2 12

= −13.6

22

[1 +

α2

22

(2

12 + 1

2

− 3

4

)]= −13.6

4

[1 +

3α2

16

]

E2 32

= −13.6

22

[1 +

α2

22

(2

32 + 1

2

− 3

4

)]= −13.6

4

[1 +

α2

16

]


Sample calculations

Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]


E1 12

= −13.6

12

[1 +

α2

12

(1

12 + 1

2

− 3

4

)]= −13.6

[1 +

α2

4

]For n = 2, l = 0, j = 1

2

E2 12

= −13.6

22

[1 +

α2

22

(2

12 + 1

2

− 3

4

)]= −13.6

4

[1 +

3α2

16

]

E2 32

= −13.6

22

[1 +

α2

22

(2

32 + 1

2

− 3

4

)]= −13.6

4

[1 +

α2

16

]


Sample calculations

Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]


E1 12

= −13.6

12

[1 +

α2

12

(1

12 + 1

2

− 3

4

)]

= −13.6

[1 +

α2

4

]For n = 2, l = 0, j = 1

2

E2 12

= −13.6

22

[1 +

α2

22

(2

12 + 1

2

− 3

4

)]= −13.6

4

[1 +

3α2

16

]

E2 32

= −13.6

22

[1 +

α2

22

(2

32 + 1

2

− 3

4

)]= −13.6

4

[1 +

α2

16

]


Sample calculations

Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]


E1 12

= −13.6

12

[1 +

α2

12

(1

12 + 1

2

− 3

4

)]= −13.6

[1 +

α2

4

]

For n = 2, l = 0, j = 12

E2 12

= −13.6

22

[1 +

α2

22

(2

12 + 1

2

− 3

4

)]= −13.6

4

[1 +

3α2

16

]

E2 32

= −13.6

22

[1 +

α2

22

(2

32 + 1

2

− 3

4

)]= −13.6

4

[1 +

α2

16

]


Sample calculations

Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]


E1 12

= −13.6

12

[1 +

α2

12

(1

12 + 1

2

− 3

4

)]= −13.6

[1 +

α2

4

]For n = 2, l = 0, j = 1

2

E2 12

= −13.6

22

[1 +

α2

22

(2

12 + 1

2

− 3

4

)]= −13.6

4

[1 +

3α2

16

]

E2 32

= −13.6

22

[1 +

α2

22

(2

32 + 1

2

− 3

4

)]= −13.6

4

[1 +

α2

16

]


Sample calculations

Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]


E1 12

= −13.6

12

[1 +

α2

12

(1

12 + 1

2

− 3

4

)]= −13.6

[1 +

α2

4

]For n = 2, l = 0, j = 1

2

E2 12

= −13.6

22

[1 +

α2

22

(2

12 + 1

2

− 3

4

)]

= −13.6

4

[1 +

3α2

16

]

E2 32

= −13.6

22

[1 +

α2

22

(2

32 + 1

2

− 3

4

)]= −13.6

4

[1 +

α2

16

]


Sample calculations

Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]


E1 12

= −13.6

12

[1 +

α2

12

(1

12 + 1

2

− 3

4

)]= −13.6

[1 +

α2

4

]For n = 2, l = 0, j = 1

2

E2 12

= −13.6

22

[1 +

α2

22

(2

12 + 1

2

− 3

4

)]= −13.6

4

[1 +

3α2

16

]

E2 32

= −13.6

22

[1 +

α2

22

(2

32 + 1

2

− 3

4

)]= −13.6

4

[1 +

α2

16

]


Sample calculations

Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]


E1 12

= −13.6

12

[1 +

α2

12

(1

12 + 1

2

− 3

4

)]= −13.6

[1 +

α2

4

]For n = 2, l = 0, j = 1

2

E2 12

= −13.6

22

[1 +

α2

22

(2

12 + 1

2

− 3

4

)]= −13.6

4

[1 +

3α2

16

]

E2 32

= −13.6

22

[1 +

α2

22

(2

32 + 1

2

− 3

4

)]

= −13.6

4

[1 +

α2

16

]


Sample calculations

Enj = −13.6eV

n2

[1 +

α2

n2

(n

j + 12

− 3

4

)]


E1 12

= −13.6

12

[1 +

α2

12

(1

12 + 1

2

− 3

4

)]= −13.6

[1 +

α2

4

]For n = 2, l = 0, j = 1

2

E2 12

= −13.6

22

[1 +

α2

22

(2

12 + 1

2

− 3

4

)]= −13.6

4

[1 +

3α2

16

]

E2 32

= −13.6

22

[1 +

α2

22

(2

32 + 1

2

− 3

4

)]= −13.6

4

[1 +

α2

16

]


Zeeman effect

When an atom is in a uniform mag-netic field ~Bext , the energy levelsare shifted by the Zeeman effect

~µs = − e

m~S ~µl = − e

2m~L

H ′Z = −(~µl + ~µs) · ~Bext

H ′Z =e

2m(~L + 2~S) · ~Bext

The nature of the Zeeman effect is dependent on the relative strengths ofthe external and internal (spin-orbit) magnetic fields

Bext � Bint weak-fieldBext ≈ Bint intermediate-fieldBext � Bint strong-field

depending on the regime, we can use different kinds of perturbation theory


Zeeman effect


~µs = − e

m~S ~µl = − e

2m~L

H ′Z = −(~µl + ~µs) · ~Bext

H ′Z =e

2m(~L + 2~S) · ~Bext





Zeeman effect


~µs = − e

m~S

~µl = − e

2m~L

H ′Z = −(~µl + ~µs) · ~Bext

H ′Z =e

2m(~L + 2~S) · ~Bext





Zeeman effect


~µs = − e

m~S ~µl = − e

2m~L

H ′Z = −(~µl + ~µs) · ~Bext

H ′Z =e

2m(~L + 2~S) · ~Bext





Zeeman effect


~µs = − e

m~S ~µl = − e

2m~L

H ′Z = −(~µl + ~µs) · ~Bext

H ′Z =e

2m(~L + 2~S) · ~Bext





Zeeman effect


~µs = − e

m~S ~µl = − e

2m~L

H ′Z = −(~µl + ~µs) · ~Bext

H ′Z =e

2m(~L + 2~S) · ~Bext





Zeeman effect


~µs = − e

m~S ~µl = − e

2m~L

H ′Z = −(~µl + ~µs) · ~Bext

H ′Z =e

2m(~L + 2~S) · ~Bext


Bext � Bint weak-field

Bext ≈ Bint intermediate-fieldBext � Bint strong-field



Zeeman effect


~µs = − e

m~S ~µl = − e

2m~L

H ′Z = −(~µl + ~µs) · ~Bext

H ′Z =e

2m(~L + 2~S) · ~Bext


Bext � Bint weak-fieldBext ≈ Bint intermediate-field

Bext � Bint strong-field



Zeeman effect


~µs = − e

m~S ~µl = − e

2m~L

H ′Z = −(~µl + ~µs) · ~Bext

H ′Z =e

2m(~L + 2~S) · ~Bext





Zeeman effect


~µs = − e

m~S ~µl = − e

2m~L

H ′Z = −(~µl + ~µs) · ~Bext

H ′Z =e

2m(~L + 2~S) · ~Bext





Weak-field Zeeman effect

When Bext � Bint fine structure dominates and the good quantumnumbers are n, l , j , mj

apply first order perturbation the-ory to get

we can rewrite this using ~J = ~L+~S

this can be evaluated by realizingthat ~J is constant and that the ti-ime average of ~S is

E 1Z =

⟨nljmj

∣∣H ′Z ∣∣ nljmj

⟩=

e

2m~Bext ·

⟨~L + 2~S

⟩=

e

2m~Bext ·

⟨~J + ~S

⟩

~Save =~S · ~JJ2

~J

~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S

~S · ~J =1

2(J2 + S2 − L2) =

~2

2[j(j + 1) + s(s + 1)− l(l + 1)]







E 1Z =

⟨nljmj


⟩=

e

2m~Bext ·

⟨~L + 2~S

⟩=

e

2m~Bext ·

⟨~J + ~S

⟩

~Save =~S · ~JJ2

~J

~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S

~S · ~J =1

2(J2 + S2 − L2) =

~2

2[j(j + 1) + s(s + 1)− l(l + 1)]







E 1Z =

⟨nljmj


⟩

=e

2m~Bext ·

⟨~L + 2~S

⟩=

e

2m~Bext ·

⟨~J + ~S

⟩

~Save =~S · ~JJ2

~J

~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S

~S · ~J =1

2(J2 + S2 − L2) =

~2

2[j(j + 1) + s(s + 1)− l(l + 1)]







E 1Z =

⟨nljmj


⟩=

e

2m~Bext ·

⟨~L + 2~S

⟩

=e

2m~Bext ·

⟨~J + ~S

⟩

~Save =~S · ~JJ2

~J

~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S

~S · ~J =1

2(J2 + S2 − L2) =

~2

2[j(j + 1) + s(s + 1)− l(l + 1)]







E 1Z =

⟨nljmj


⟩=

e

2m~Bext ·

⟨~L + 2~S

⟩

=e

2m~Bext ·

⟨~J + ~S

⟩

~Save =~S · ~JJ2

~J

~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S

~S · ~J =1

2(J2 + S2 − L2) =

~2

2[j(j + 1) + s(s + 1)− l(l + 1)]







E 1Z =

⟨nljmj


⟩=

e

2m~Bext ·

⟨~L + 2~S

⟩=

e

2m~Bext ·

⟨~J + ~S

⟩

~Save =~S · ~JJ2

~J

~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S

~S · ~J =1

2(J2 + S2 − L2) =

~2

2[j(j + 1) + s(s + 1)− l(l + 1)]







E 1Z =

⟨nljmj


⟩=

e

2m~Bext ·

⟨~L + 2~S

⟩=

e

2m~Bext ·

⟨~J + ~S

⟩

~Save =~S · ~JJ2

~J

~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S

~S · ~J =1

2(J2 + S2 − L2) =

~2

2[j(j + 1) + s(s + 1)− l(l + 1)]







E 1Z =

⟨nljmj


⟩=

e

2m~Bext ·

⟨~L + 2~S

⟩=

e

2m~Bext ·

⟨~J + ~S

⟩

~Save =~S · ~JJ2

~J

~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S

~S · ~J =1

2(J2 + S2 − L2) =

~2

2[j(j + 1) + s(s + 1)− l(l + 1)]







E 1Z =

⟨nljmj


⟩=

e

2m~Bext ·

⟨~L + 2~S

⟩=

e

2m~Bext ·

⟨~J + ~S

⟩

~Save =~S · ~JJ2

~J

~L = ~J − ~S

→ L2 = J2 + S2 − 2~J · ~S

~S · ~J =1

2(J2 + S2 − L2) =

~2

2[j(j + 1) + s(s + 1)− l(l + 1)]







E 1Z =

⟨nljmj


⟩=

e

2m~Bext ·

⟨~L + 2~S

⟩=

e

2m~Bext ·

⟨~J + ~S

⟩

~Save =~S · ~JJ2

~J

~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S

~S · ~J =1

2(J2 + S2 − L2) =

~2

2[j(j + 1) + s(s + 1)− l(l + 1)]







E 1Z =

⟨nljmj


⟩=

e

2m~Bext ·

⟨~L + 2~S

⟩=

e

2m~Bext ·

⟨~J + ~S

⟩

~Save =~S · ~JJ2

~J

~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S

~S · ~J =1

2(J2 + S2 − L2)

=~2

2[j(j + 1) + s(s + 1)− l(l + 1)]







E 1Z =

⟨nljmj


⟩=

e

2m~Bext ·

⟨~L + 2~S

⟩=

e

2m~Bext ·

⟨~J + ~S

⟩

~Save =~S · ~JJ2

~J

~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S

~S · ~J =1

2(J2 + S2 − L2) =

~2

2[j(j + 1) + s(s + 1)− l(l + 1)]


Lande g-factor

Thus the expectation value in the Zeeman energy correction becomes

⟨~L + 2~S

⟩=⟨~J + ~S

⟩=

⟨(1 +

~S · ~JJ2

)~J

⟩

=

[1 +

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

]〈~J〉 ≡ gJ〈~J〉

and the full energy correction becomes

E(1)Z = µBgJBextmj µB ≡

e~2m

= 5.788× 10−5eV/T

the total energy includes both the spin-orbit and Zeeman corrections andthe 2j + 1 states then have unique energies


Lande g-factor


⟨~L + 2~S

⟩=⟨~J + ~S

⟩

=

⟨(1 +

~S · ~JJ2

)~J

⟩

=

[1 +

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

]〈~J〉 ≡ gJ〈~J〉



e~2m

= 5.788× 10−5eV/T



Lande g-factor


⟨~L + 2~S

⟩=⟨~J + ~S

⟩=

⟨(1 +

~S · ~JJ2

)~J

⟩

=

[1 +

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

]〈~J〉 ≡ gJ〈~J〉



e~2m

= 5.788× 10−5eV/T



Lande g-factor


⟨~L + 2~S

⟩=⟨~J + ~S

⟩=

⟨(1 +

~S · ~JJ2

)~J

⟩

=

[1 +

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

]〈~J〉

≡ gJ〈~J〉



e~2m

= 5.788× 10−5eV/T



Lande g-factor


⟨~L + 2~S

⟩=⟨~J + ~S

⟩=

⟨(1 +

~S · ~JJ2

)~J

⟩

=

[1 +

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

]〈~J〉 ≡ gJ〈~J〉



e~2m

= 5.788× 10−5eV/T



Lande g-factor


⟨~L + 2~S

⟩=⟨~J + ~S

⟩=

⟨(1 +

~S · ~JJ2

)~J

⟩

=

[1 +

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

]〈~J〉 ≡ gJ〈~J〉



e~2m

= 5.788× 10−5eV/T



Lande g-factor


⟨~L + 2~S

⟩=⟨~J + ~S

⟩=

⟨(1 +

~S · ~JJ2

)~J

⟩

=

[1 +

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

]〈~J〉 ≡ gJ〈~J〉


E(1)Z = µBgJBextmj

µB ≡e~2m

= 5.788× 10−5eV/T



Lande g-factor


⟨~L + 2~S

⟩=⟨~J + ~S

⟩=

⟨(1 +

~S · ~JJ2

)~J

⟩

=

[1 +

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

]〈~J〉 ≡ gJ〈~J〉



e~2m

= 5.788× 10−5eV/T



Lande g-factor


⟨~L + 2~S

⟩=⟨~J + ~S

⟩=

⟨(1 +

~S · ~JJ2

)~J

⟩

=

[1 +

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

]〈~J〉 ≡ gJ〈~J〉



e~2m

= 5.788× 10−5eV/T



Lande g-factor


⟨~L + 2~S

⟩=⟨~J + ~S

⟩=

⟨(1 +

~S · ~JJ2

)~J

⟩

=

[1 +

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

]〈~J〉 ≡ gJ〈~J〉



e~2m

= 5.788× 10−5eV/T



Strong-field Zeeman effect

When Bext � Bint , the spin-orbit coupling must be treated as theperturbation and the solutions must be eigenfunctions of the unperturbedwave functions with good quantum numbers.

If Bext is in the z direction,the Zeeman Hamiltonian is

and the energies (withoutfine structure), are

Applying perturbation theoryto the fine structure Hamilto-nian

H ′Z =e

2mBext(Lz + 2Sz)

Enmlms = −13.6eV

n2+ µBBext(ml + 2ms)

E 1fs =

⟨nlmlms

∣∣(H ′r + H ′so)∣∣ nlmlms

⟩〈~S · ~L〉 =

��〈Sx〉〈Lx〉+��〈Sy 〉〈Ly 〉+ 〈Sz〉〈Lz〉 = ~2mlms

E 1fs =

13.6eV

n3α2

{3

4n−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

]}, l > 0







H ′Z =e

2mBext(Lz + 2Sz)

Enmlms = −13.6eV


E 1fs =

⟨nlmlms


⟩〈~S · ~L〉 =


E 1fs =

13.6eV

n3α2

{3


l(l + 1/2)(l + 1)

]}, l > 0







H ′Z =e

2mBext(Lz + 2Sz)

Enmlms = −13.6eV


E 1fs =

⟨nlmlms


⟩〈~S · ~L〉 =


E 1fs =

13.6eV

n3α2

{3


l(l + 1/2)(l + 1)

]}, l > 0







H ′Z =e

2mBext(Lz + 2Sz)

Enmlms = −13.6eV


E 1fs =

⟨nlmlms


⟩〈~S · ~L〉 =


E 1fs =

13.6eV

n3α2

{3


l(l + 1/2)(l + 1)

]}, l > 0







H ′Z =e

2mBext(Lz + 2Sz)

Enmlms = −13.6eV


E 1fs =

⟨nlmlms


⟩〈~S · ~L〉 =


E 1fs =

13.6eV

n3α2

{3


l(l + 1/2)(l + 1)

]}, l > 0







H ′Z =e

2mBext(Lz + 2Sz)

Enmlms = −13.6eV


E 1fs =

⟨nlmlms


⟩〈~S · ~L〉 =


E 1fs =

13.6eV

n3α2

{3


l(l + 1/2)(l + 1)

]}, l > 0







H ′Z =e

2mBext(Lz + 2Sz)

Enmlms = −13.6eV


E 1fs =

⟨nlmlms


⟩

〈~S · ~L〉 =


E 1fs =

13.6eV

n3α2

{3


l(l + 1/2)(l + 1)

]}, l > 0







H ′Z =e

2mBext(Lz + 2Sz)

Enmlms = −13.6eV


E 1fs =

⟨nlmlms


⟩〈~S · ~L〉 =


E 1fs =

13.6eV

n3α2

{3


l(l + 1/2)(l + 1)

]}, l > 0







H ′Z =e

2mBext(Lz + 2Sz)

Enmlms = −13.6eV


E 1fs =

⟨nlmlms


⟩〈~S · ~L〉 = 〈Sx〉〈Lx〉+ 〈Sy 〉〈Ly 〉+ 〈Sz〉〈Lz〉

= ~2mlms

E 1fs =

13.6eV

n3α2

{3


l(l + 1/2)(l + 1)

]}, l > 0







H ′Z =e

2mBext(Lz + 2Sz)

Enmlms = −13.6eV


E 1fs =

⟨nlmlms


⟩〈~S · ~L〉 = ��〈Sx〉〈Lx〉+��〈Sy 〉〈Ly 〉+ 〈Sz〉〈Lz〉

= ~2mlms

E 1fs =

13.6eV

n3α2

{3


l(l + 1/2)(l + 1)

]}, l > 0







H ′Z =e

2mBext(Lz + 2Sz)

Enmlms = −13.6eV


E 1fs =

⟨nlmlms


⟩〈~S · ~L〉 = ��〈Sx〉〈Lx〉+��〈Sy 〉〈Ly 〉+ 〈Sz〉〈Lz〉 = ~2mlms

E 1fs =

13.6eV

n3α2

{3


l(l + 1/2)(l + 1)

]}, l > 0







H ′Z =e

2mBext(Lz + 2Sz)

Enmlms = −13.6eV


E 1fs =

⟨nlmlms


⟩〈~S · ~L〉 = ��〈Sx〉〈Lx〉+��〈Sy 〉〈Ly 〉+ 〈Sz〉〈Lz〉 = ~2mlms

E 1fs =

13.6eV

n3α2

{3


l(l + 1/2)(l + 1)

]}, l > 0


Intermediate-field Zeeman effect

When H ′Z ≈ H ′fs , we must apply degenerate perturbation theory withH ′ = H ′Z + H ′fs .

Consider the n = 2 case with l = 0, 1 and taking stateswith quantum numbers l , j , and mj (instead of l , ml , s, and ms), we usethe Clebsch-Gordan coefficients to obtain 8 eigenstates

l = 0

{

ψ1 ≡∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩



When H ′Z ≈ H ′fs , we must apply degenerate perturbation theory withH ′ = H ′Z + H ′fs . Consider the n = 2 case with l = 0, 1 and taking stateswith quantum numbers l , j , and mj (instead of l , ml , s, and ms), we usethe Clebsch-Gordan coefficients to obtain 8 eigenstates

l = 0

{

ψ1 ≡∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{

ψ1 ≡∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩

= |00〉∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩

ψ2 ≡∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩

= |00〉∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩

= |11〉∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩

ψ4 ≡∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩

= |1−1〉∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩

ψ5 ≡∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩

=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩

ψ6 ≡∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩

= −√

13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩

ψ7 ≡∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩

=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩

ψ8 ≡∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩

= −√

23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩




l = 0

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12−12

⟩= |00〉

∣∣12−12

⟩

l = 1

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32−32

⟩= |1−1〉

∣∣12−12

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12−12

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12−12

⟩ψ7 ≡

∣∣32−12

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12−12

⟩ψ8 ≡

∣∣12−12

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12−12

⟩C. Segre (IIT) PHYS 406 - Spring 2015 January 27, 2015 14 / 14

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