today’s outline - february 03, 2014csrri.iit.edu/~segre/phys406/14s/lecture_05.pdftoday’s...

Today’s Outline - February 03, 2014

• Relativistic correction

• Spin-orbit coupling

• Zeeman effect

Office Hours: Monday & Wednesday 10:00-11:00 or by Appt.

Homework Assignment #03:Chapter 6: 9, 17, 21, 24, 25, 32due Monday, February 10, 2014

Tutoring sessions:Thursday & Friday, 12:00–13:45, 121 E1

C. Segre (IIT) PHYS 406 - Spring 2014 February 03, 2014 1 / 7

• Zeeman effect

Relativistic correction review

The lowest order relativisticcorrection is

applying first order perturba-tion theory

with the Schrodinger equa-tion

p2ψ = 2m(E − V )ψ

given that⟨1

n2a⟨1

(l + 1/2)n3a2

H ′r = − p4

E 1r =

⟨H ′r

⟩= − 1

8m3c2⟨p2ψ|p2ψ

⟩= − 1

2mc2[e2 − 2E 〈V 〉+ 〈V 2〉]

= − 1

[(En)2 + 2En

4πε0

)2⟨1

E 1r = −(En)2

l + 1/2− 3

p2ψ = 2m(E − V )ψ

given that⟨1

n2a⟨1

(l + 1/2)n3a2

H ′r = − p4

E 1r =

⟨H ′r

⟩= − 1

8m3c2⟨p2ψ|p2ψ

⟩= − 1

2mc2[e2 − 2E 〈V 〉+ 〈V 2〉]

= − 1

[(En)2 + 2En

4πε0

)2⟨1

E 1r = −(En)2

l + 1/2− 3

p2ψ = 2m(E − V )ψ

given that⟨1

n2a⟨1

(l + 1/2)n3a2

H ′r = − p4

E 1r =

⟨H ′r

⟩= − 1

8m3c2⟨p2ψ|p2ψ

⟩= − 1

2mc2[e2 − 2E 〈V 〉+ 〈V 2〉]

= − 1

[(En)2 + 2En

4πε0

)2⟨1

E 1r = −(En)2

l + 1/2− 3

p2ψ = 2m(E − V )ψ

given that⟨1

n2a⟨1

(l + 1/2)n3a2

H ′r = − p4

E 1r =

⟨H ′r

⟩= − 1

8m3c2⟨p2ψ|p2ψ

= − 1

2mc2[e2 − 2E 〈V 〉+ 〈V 2〉]

= − 1

[(En)2 + 2En

4πε0

)2⟨1

E 1r = −(En)2

l + 1/2− 3

p2ψ = 2m(E − V )ψ

given that⟨1

n2a⟨1

(l + 1/2)n3a2

H ′r = − p4

E 1r =

⟨H ′r

⟩= − 1

8m3c2⟨p2ψ|p2ψ

= − 1

2mc2[e2 − 2E 〈V 〉+ 〈V 2〉]

= − 1

[(En)2 + 2En

4πε0

)2⟨1

E 1r = −(En)2

l + 1/2− 3

p2ψ = 2m(E − V )ψ

given that⟨1

n2a⟨1

(l + 1/2)n3a2

H ′r = − p4

E 1r =

⟨H ′r

⟩= − 1

8m3c2⟨p2ψ|p2ψ

⟩= − 1

2mc2[e2 − 2E 〈V 〉+ 〈V 2〉]

= − 1

[(En)2 + 2En

4πε0

)2⟨1

E 1r = −(En)2

l + 1/2− 3

p2ψ = 2m(E − V )ψ

given that⟨1

n2a⟨1

(l + 1/2)n3a2

H ′r = − p4

E 1r =

⟨H ′r

⟩= − 1

8m3c2⟨p2ψ|p2ψ

⟩= − 1

2mc2[e2 − 2E 〈V 〉+ 〈V 2〉]

= − 1

[(En)2 + 2En

4πε0

)2⟨1

E 1r = −(En)2

l + 1/2− 3

p2ψ = 2m(E − V )ψ

given that⟨1

n2a⟨1

(l + 1/2)n3a2

H ′r = − p4

E 1r =

⟨H ′r

⟩= − 1

8m3c2⟨p2ψ|p2ψ

⟩= − 1

2mc2[e2 − 2E 〈V 〉+ 〈V 2〉]

= − 1

[(En)2 + 2En

4πε0

)2⟨1

E 1r = −(En)2

l + 1/2− 3

p2ψ = 2m(E − V )ψ

given that⟨1

n2a⟨1

(l + 1/2)n3a2

H ′r = − p4

E 1r =

⟨H ′r

⟩= − 1

8m3c2⟨p2ψ|p2ψ

⟩= − 1

2mc2[e2 − 2E 〈V 〉+ 〈V 2〉]

= − 1

[(En)2 + 2En

4πε0

)2⟨1

E 1r = −(En)2

l + 1/2− 3

p2ψ = 2m(E − V )ψ

given that⟨1

n2a⟨1

(l + 1/2)n3a2

H ′r = − p4

E 1r =

⟨H ′r

⟩= − 1

8m3c2⟨p2ψ|p2ψ

⟩= − 1

2mc2[e2 − 2E 〈V 〉+ 〈V 2〉]

= − 1

[(En)2 + 2En

4πε0

)2⟨1

E 1r = −(En)2

l + 1/2− 3

]C. Segre (IIT) PHYS 406 - Spring 2014 February 03, 2014 2 / 7

Magnetic field at the electron

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

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2ε0c2r=

2ε0c2τ r

4πε0

mc2r3~L

L = rmv

=2πmr2

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2ε0c2r=

2ε0c2τ r

4πε0

mc2r3~L

The “rotating” proton has angular momentum

and produces a magneticfield at the position of the electron

L = rmv

=2πmr2

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2ε0c2r=

2ε0c2τ r

4πε0

mc2r3~L

L = rmv

=2πmr2

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2ε0c2r=

2ε0c2τ r

4πε0

mc2r3~L

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

from the Biot-Savart law

L = rmv

=2πmr2

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2ε0c2r=

2ε0c2τ r

4πε0

mc2r3~L

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

from the Biot-Savart law

L = rmv

=2πmr2

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2ε0c2r=

2ε0c2τ r

4πε0

mc2r3~L

L = rmv

=2πmr2

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2ε0c2r=

2ε0c2τ r

4πε0

mc2r3~L

L = rmv

=2πmr2

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2ε0c2r=

2ε0c2τ r

4πε0

mc2r3~L

L = rmv

=2πmr2

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2ε0c2r

2ε0c2τ r

4πε0

mc2r3~L

L = rmv

=2πmr2

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2ε0c2r=

2ε0c2τ r

4πε0

mc2r3~L

L = rmv

=2πmr2

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2ε0c2r=

2ε0c2τ r

4πε0

mc2r3~L

L = rmv =2πmr2

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2ε0c2r=

2ε0c2τ r

4πε0

mc2r3~L

L = rmv =2πmr2

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2ε0c2r=

2ε0c2τ r

4πε0

mc2r3~L

L = rmv =2πmr2

since ~B ‖ ~L

H = −~µ · ~B

B =µ0i

2ε0c2r=

2ε0c2τ r

4πε0

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

µ = iA =qπr2

S = Iω =2πmr2

γcl =µ

τ· τ

2πmr2=

µ = iA =qπr2

S = Iω =2πmr2

γcl =µ

τ· τ

2πmr2=

µ = iA =qπr2

S = Iω =2πmr2

γcl =µ

τ· τ

2πmr2=

the magnetic moment is the currenttimes the area of the loop

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

µ = iA =qπr2

S = Iω =2πmr2

γcl =µ

τ· τ

2πmr2=

µ = iA

=qπr2

S = Iω =2πmr2

γcl =µ

τ· τ

2πmr2=

µ = iA =qπr2

S = Iω =2πmr2

γcl =µ

τ· τ

2πmr2=

µ = iA =qπr2

S = Iω =2πmr2

γcl =µ

τ· τ

2πmr2=

µ = iA =qπr2

S = Iω

=2πmr2

γcl =µ

τ· τ

2πmr2=

µ = iA =qπr2

S = Iω =2πmr2

γcl =µ

τ· τ

2πmr2=

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

independent of r and τ

µ = iA =qπr2

S = Iω =2πmr2

γcl =µ

τ· τ

2πmr2=

µ = iA =qπr2

S = Iω =2πmr2

γcl =µ

=qπr2

τ· τ

2πmr2=

µ = iA =qπr2

S = Iω =2πmr2

γcl =µ

τ· τ

2πmr2

µ = iA =qπr2

S = Iω =2πmr2

γcl =µ

τ· τ

2πmr2=

µ = iA =qπr2

S = Iω =2πmr2

γcl =µ

τ· τ

2πmr2=

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

H ′ = −~µ · ~B =gee

4πε0

mc2r3~S · ~L

8πε0

m2c2r3~S · ~L

H ′so ≈

8πε0

m2c2r3~S · ~L

where ge ≈ 2 + α/π = 2.002

~µ = γe~S

= −gee

H ′ = −~µ · ~B =gee

4πε0

mc2r3~S · ~L

8πε0

m2c2r3~S · ~L

H ′so ≈

8πε0

m2c2r3~S · ~L

where ge ≈ 2 + α/π = 2.002

~µ = γe~S

= −gee

H ′ = −~µ · ~B =gee

4πε0

mc2r3~S · ~L

8πε0

m2c2r3~S · ~L

H ′so ≈

8πε0

m2c2r3~S · ~L

where ge ≈ 2 + α/π = 2.002

H ′ = −~µ · ~B =gee

4πε0

mc2r3~S · ~L

8πε0

m2c2r3~S · ~L

H ′so ≈

8πε0

m2c2r3~S · ~L

where ge ≈ 2 + α/π = 2.002

H ′ = −~µ · ~B =gee

4πε0

mc2r3~S · ~L

8πε0

m2c2r3~S · ~L

H ′so ≈

8πε0

m2c2r3~S · ~L

where ge ≈ 2 + α/π = 2.002

H ′ = −~µ · ~B =gee

4πε0

mc2r3~S · ~L

8πε0

m2c2r3~S · ~L

H ′so ≈

8πε0

m2c2r3~S · ~L

where ge ≈ 2 + α/π = 2.002

H ′ = −~µ · ~B

4πε0

mc2r3~S · ~L

8πε0

m2c2r3~S · ~L

H ′so ≈

8πε0

m2c2r3~S · ~L

where ge ≈ 2 + α/π = 2.002

H ′ = −~µ · ~B =gee

4πε0

mc2r3~S · ~L

8πε0

m2c2r3~S · ~L

H ′so ≈

8πε0

m2c2r3~S · ~L

where ge ≈ 2 + α/π = 2.002

H ′ = −~µ · ~B =gee

4πε0

mc2r3~S · ~L

8πε0

m2c2r3~S · ~L

H ′so ≈

8πε0

m2c2r3~S · ~L

where ge ≈ 2 + α/π = 2.002

H ′ = −~µ · ~B =gee

4πε0

mc2r3~S · ~L

8πε0

m2c2r3~S · ~L

H ′so ≈

8πε0

m2c2r3~S · ~L

where ge ≈ 2 + α/π = 2.002

H ′ = −~µ · ~B =gee

4πε0

mc2r3~S · ~L

8πε0

m2c2r3~S · ~L

H ′so ≈

8πε0

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 angular

momentum ~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 ≈

8πε0

m2c2r3~S · ~L

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