Atomic Spectra in Astrophysics - uni-potsdam.delida/TEACH.DIR/spin_orbit.pdf · AtomicSpectrainAstrophysics LidaOskinova,HelgeTodt Astrophysik Institut für Physik und Astronomie

Atomic Spectra in Astrophysics

Lida Oskinova, Helge Todt

AstrophysikInstitut für Physik und Astronomie

Universität Potsdam

WiSe 2014/2015

L. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2014/2015 1 / 31

Application: Electron in magnetic field I

spin + charged particle →magnetic dipole:

~µ = γ~S where γ := gyromagnetic ratio (1)

Note: in classical electrodynamics: γ = q2m

in QM: γ = gJq

2m = gSq

2m ≈ 2.002 q2m

magnetic dipole in magnetic field ~B experiences torque ~M:

~M = ~µ× ~B (2)

→ torque tries to line ~µ up parallel to field (“compass needle”)Energy (and therefore Hamiltonian):

H = −~µ · ~B = −γ~B · ~S (3)


Application: Electron in magnetic field II

Larmor precessionSpin-12 particle in a homogeneous magnetic field along z-direction~B = B0~ezHamiltonian (matrix!):

H = −γB0Sz = −γB0~2

(1 00 −1

)(4)

Hamiltonian H and Spin operator Sz share same eigenstates:

χ+, energy E+ = −γB0~2 (5)

χ−, energy E− = +γB0~2 (6)

where energy is lowest for dipole moment parallel to magnetic field(otherwise: → torque tries to align)


Application: Electron in magnetic field III

→ time-dependent solution of ı~∂tχ = Hχ expressed with stationary states:

χ(t) = aχ+e−ıE+t/~ + bχ−e−ıE−t/~ =

(ae+ıγB0t/~

be−ıγB0t/~

)(7)

fix coefficients by initial conditions, e.g.

χ(t = 0) =

(ab

), where |a|2 + |b|2 = 1 (8)

Then (for real a, b) we can write a = cos(α/2) and b = sin(α/2), withconstant α, so

χ(t) =

(cos(α/2)e+ıγB0t/2

sin(α/2)e−ıγB0t/2

)(9)


Application: Electron in magnetic field IV

therefore the expectation value of the spin as a function of time:

〈Sx〉 = χ(t)†Sxχ

=(cos(α/2)e−ıγB0t/2 sin(α/2)e+ıγB0t/2

) ~2

(0 11 0

)(cos(α/2)e+ıγB0t/2

sin(α/2)e−ıγB0t/2

)= +

~2sinα cos(γB0t)

〈Sy 〉 = −~2sinα sin(γB0t)

〈Sz〉 = +~2cosα

→〈~S〉 tilted at constant angle α to z-axis, precesses about the field at

Larmor frequency of a spinning electron

ω = γB0 = 1.7606× 1011 rad s−1 T−1 · B0 → f = 28GHzT−1 · B0


Addition of angular momenta I

Now, two spin-12 particles (e.g. electron and proton in ground state ofhydrogen, so ` = 0), composite system is in a linear combination of

↑↑, ↑↓, ↓↑, ↓↓ (10)

What is the total angular momentum of the atom?We define

~S ≡ ~S (1) + ~S (2) (11)

The z-component: simply adds, each composite state is eigenstate of Sz

Szχ1χ2 = (S (1)z + S (2)

z )χ1χ2 = (S (1)z χ1)χ2 + χ1(S (2)

z χ2) (12)= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2 (13)


Addition of angular momenta II

Thus, m = m1 + m2:

↑↑ m = 1 (14)↑↓ m = 0 (15)↓↑ m = 0 (16)↓↓ m = −1 (17)

problem: two states with m = 0 → apply lowering operatorS− = S (1)

− + S (2)− to state ↑↑ to obtain the correct state:

S−(↑↑) = (S (1)− ↑) ↑ + ↑ (S (2)

− ↑) (18)= (~ ↓) ↑ + ↑ (~ ↓) = ~(↓↑ + ↑↓) (19)


Addition of angular momenta III

Therefore, the states with s = 1 in |s m〉 notation

|1 1〉 = ↑↑|1 0〉 = 1√

2(↑↓ + ↓↑)

|1 − 1〉 = ↓↓

→ triplet: s = 1 (20)

And orthogonal state |s m〉 = |0 0〉

|0 0〉 =1√2

(↑↓ − ↓↑)}→ singlet: s = 0 (21)

→ system of two spin-12 particles has total spin 1 or 0. Let us proof: tripletstates are eigenvectors of S2 with eigenvalues 2~2 and singlet state iseigenvector with eigenvalue 0:


Addition of angular momenta IV

S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2) (22)

Remember: Sxχ+ = ~2χ−, Syχ+ = − ~

2ıχ−, Szχ+ = ~2χ+, S2χ+ = 3

4~2χ+

etc.

~S (1) · ~S (2)(↑↓) = (S (1)x ↑)(S (2)

x ↓) + (S (1)y ↑)(S (2)

y ↓) + (S (1)z ↑)(S (2)

z ↓)

=

(~2↓)(

~2↑)

+

(ı~2↓)(− ı~

2↑)

+

(~2↑)(−~2↓)

=~2

4(2 ↓↑ − ↑↓) (23)

and (24)

~S (1) · ~S (2)(↓↑) =~2

4(2 ↑↓ − ↓↑) (25)


Addition of angular momenta V

So we get

~S (1) · ~S (2)|1 0〉 =~4

1√2

(2 ↓↑ − ↑↓ +2 ↑↓ − ↓↑) = −~2

4|1 0〉 (26)

~S (1) · ~S (2)|0 0〉 =~4

1√2

(2 ↓↑ − ↑↓ −2 ↑↓ + ↓↑) = −3~2

4|0 0〉 (27)

and therefore

S2|1 0〉 =

(3~2

4+

3~2

4+2

~2

4

)|1 0〉 = 2~2|1 0〉 (28)

S2|0 0〉 =

(3~2

4+

3~2

4−2~

2

4

)|0 0〉 = 0 (29)

and of course S2|1 − 1〉 = 2~2|1 − 1〉 and S2|1 1〉 = 2~2|1 1〉


Addition of angular momenta VI

Thus, answer to the general problem of combining any s1, s2 or angularmomenta:

s = (s1 + s2), (s1 + s2 − 1), . . . , |s1 − s2| (30)

and also, e.g. for hydrogen atom, net angular momentum of electron (spin+ orbital) j

j = `+12

or j = `− 12

(31)

and with proton, total angular momentum of hydrogen atom:

J = `+ 1 or J = `− 1 (32)


Addition of angular momenta VII

Therefore, state |s m〉 with total spin s and z-component m → linearcombination of composite states |s1 m1〉 |s2 m2〉:

|s m〉 =∑

m1+m2=m

C s1 s2 sm1 m2 m|s1 m1〉 |s2 m2〉 (33)

with Clebsch-Gordan coefficients C s1 s2 sm1 m2 m. E.g.

|2 1〉 =1√3|2 2〉|1−1〉+

1√6|2 1〉|1 0〉 − 1√

2|2 0〉|1 1〉 (34)

Note that the z-components have to add to m = 1→For two particles of spin 2 and spin 1 with total spin 2 and totalz-component 1: measure S (1)

z and get 2~ (probability 13), ~ (probability 1

6),or 0 (probability 1

2)


The helium atom I

Hamiltonian of helium (Z = 2), neglecting spin

H =

[− ~2

2m∇2

1 −1

4πε02e2

r1

]+

[− ~2

2m∇2

2 −1

4πε02e2

r2

]+

14πε0

e2

|~r1 − ~r2|

→without repulsion term :

ψ(~r1,~r2) = ψn`m(~r1)ψn′`′m′(~r2) (35)

with a = aH2 and E = 4EH,

E = 4(En + En′) → E0 = 8 · (−13.6 eV) = −109 eV (36)

(But measured: −79 eV = −24.6 eV− 54.4 eV, because we neglected e−e−-interaction)

with ψ0(~r1,~r2) = ψ100(~r1)ψ100(~r2) =8πa3 e

−2(r1+r2)/a (37)


The helium atom IIElectrons are identical particles, therefore

ψ(~r1,~r2) 6= ψa(~r1)ψb(~r2) but (38)ψ−(~r1,~r2) = A[ψa(~r1)ψb(~r2)− ψb(~r1)ψa(~r2)] (fermions) (39)

two electrons cannot occupy same state →Pauli exclusion principle(Proof!)ψ for fermions must be antisymmetric under exchange:

ψ(~r1,~r2) = −ψ(~r2,~r1) (40)

with ψ → ψ(~r1,~r2)χ(~r1,~r2)

as the ground state ψ0 ∼ er1+r2 of helium is symmetric → spin statemust be antisymmetric → singlet, electrons oppositely aligned

1√2

(↑↓ − ↓↑) (41)


The helium atom III

Excited helium atom:

combination ψn`mψ100(What about ψn`mψn′`′m′?)symmetric ψn`mψ100→ antisymmetric spinconfiguration (singlet) orantisymmetric ψn`mψ100→ symmetric spin configuration(triplet)thus, two different kinds ofexcited states: singlet and tripletsinglet states have sligthly higherE than corresponding triplets(because of symmetric ψ→ closer e−e− → larger Erepuls)

1

35

7 101113 1617

L

1 S 1 Po 1 D

singlet

0 1 2

2

4

6 8 912 14 15

3 S 3 Po 3 D

0 1 2

L

tripletEL. Oskinova, H. Todt (UP) Atomic Spectra in Astrophysics WiSe 2014/2015 15 / 31

The helium atom IV

IntercombinationNote that

total wavefunction must be antisymmetric by

ψ(~r1,~r2) =

{ψa(~r1,~r2)χs(~r1,~r2)ψs(~r1,~r2)χa(~r1,~r2)

(42)

the interchange operator can be written as P12 = Pψ12 Pχ12

Hamiltonian does not depend on spin → [H,Pψ12] = 0

time evolution operator U(t) = e−ı~Ht → [U,P12] = 0

if for any t0: P12|ψ(t0)〉 = −|ψ(t0), then for all times:

P12|ψ(t)〉 = P12U(t)|ψ(t0)〉 = U(t)P12|ψ(t0)〉 (43)= −U(t)|ψ(t0)〉 = −|ψ(t)〉 (44)


The helium atom V

→ symmetry of ψ(~r1,~r2) does not change with time → symmetry ofχ(~r1,~r2) does not change with time

Selection rule for multiplicityTransitions between different multiplicities are forbidden:

∆s = 0 (45)

(exact Hamiltonian depends on spin → small modification of selection rule)

Parityψ composed of ψn`mψ100 = Rn`Ym

` R10Y 00

→ parity good quantum number, odd for odd `, even for even `odd parity indicated by small letter o right from captial letter for net orbitalangular momentum → e.g. 1Po


Multi-electron atoms I

Analogously, ground-state electron configurations of other atoms describedby

H =N∑

i=1

[− ~2

2m∇2

i −1

4πε0Ze2

ri

]+∑i 6=j

14πε0

e2

|~ri − ~rj |(46)

neglect repulsion term → each electron occupies a one-particlehydrogenic state (n, `,m) → orbitaldue to Pauli exclusion principle → only two e− per orbital:n2 wavefunction for each shell n, e.g. n = 1 → two electrons, n = 2→ eight electrons, etc.Periodic system: horizontal rows ↔ filling out each shellHow to fill n = 2 (` = 0 or ` = 1) with single e−? → screening byinner electrons, favor lowest `, e.g.E `=0

n=2 < E `=1n=2, E

`=0n=4 < E `=2

n=3(larger ` → larger 〈r〉 → stronger screening of nucleus → less binding energy)


Multi-electron atoms II

Total angular momentumstate of electron represented by pair n`, where ` is a letter, e.g. s for` = 0 – m not listed, but exponent for number of electrons in `, e.g.He → 1s2 (i.e. two electrons in n = 1, ` = 0)

total orbital angular momentum indicated by capital L (letter), totalspin by captial S (multiplicity, number), and total angular momentumby capital J (number)

total configuration of atom listed as

2S+1LJ (47)

e.g. groundstate of hydrogen S = 1/2, L = S , J = 1/2 → 2S1/2

LS-coupling of multi-electrons

If spin-orbit interaction negligible: ~L =∑ ~̀i , ~S =

∑~si and ~J = ~L + ~S


Multi-electron atoms IIISpectroscopic nomenclature for atomic states

Table : Angular momenta of electrons

` 0 1 2 3 . . .letter s p d f . . .name sharp principal diffuse fundamental ?

Ground-stateelectronconfigurations offirst elements

Z El. Configuration

1 H 1s 2S1/22 He 1s2 1S03 Li (He)2s 2S1/24 Be (He)2s2 1S05 B (He)2s2 2p 2Po

1/2. . .21 Sc (Ar)4s2 3d 2D3/2


Multi-electron atoms IV

Note: Screening effect becomes larger for larger ` → overlap of shells

1s 2s 3s 4s 5s 6s 7s 8s 9s 10s2p 3p 4p 5p 6p 7p 8p 9p 10p

3d 4d 5d 6d 7d 8d 9d 10d4f 5f 6f 7f 8f 9f 10f

5g 6g 7g 8g 9g 10g6h 7h 8h 9h 10h

7i 8i 9i 10i8k 9k 10k

9l 10l10m

→ for determining the energy order of terms of a one-electron system, justgo along the diagonals, e.g. 1s 2s 2p 3s 3p 4s 3d 4p 5s.


Multi-electron atoms V

How to distribute electrons to shells?→ find configuration of minium energy

→Hund’s rules (minimum energy principle)

1 The total angular momentum J of completely filled shells (n) orsubshells (`) is 0. E.g. He: 1s2, Be: (He)2s2, Ne: (He)2s2 2p6

2 Spins of electrons are pereferably parallel, i.e. electron are distributedon subshells with m`, such that multiplicity 2S + 1 is maximal. E.g. N→ (He) 2s2 2p3: in 2p shell: ↑↑↑ → 4S .(sometimes referred to as the 1st rule or S rule)

3 For states of same S , electrons are distributed such that largestangular momentum L is achieved (larger ` → larger 〈r〉 → larger ∆xe−e−

→ less repulsion), e.g. E (3P1) < E (2S1).(sometimes referred to as the 2nd rule or L rule)


Multi-electron atoms VI

4 If outermost subshell half-filled or less: configuration with lowest totalangular momentum J = |L− S | (i.e. with smallest J) is preferred.If outermost subshell more than half-filled: configuration with highestJ = L + S is preferred.

→ reason: electron-electron interaction→ only valid for LS coupling (no spin-orbit interaction → light atoms,Z < 10)

Example: groundstate of carbon 1s2 2s2 2p2

two e− in 2p2 → 1S , 1D, or 3P? ↑↓ , ↑↓ , ↑ ↑Using S rule → largest mulitplicity S preferred, i.e. 3PJ = L + S → L = 1 and S = −1, 1,+1 → 3P0, 3P1, or 3P2Using J rule → smallest J, i.e. J = 0⇒ 3P0


Spin-orbit interaction I

When solved Schrödinger equation for hydrogen, used simple Hamiltonian

H = − ~2

2m∇2 − e2

4πε01r

(48)

But: electron orbiting around nucleus → from electrons point of view:proton orbits electron→ setting up magnetic field ~B , exerting a torque on spining electron toalign its magnetic moment ~µe along ~B

Hso = −~µe · ~B = −(− em~S)~B (49)


Spin-orbit interaction IIFrom Biot-Savart law (→ blackboard):

B =µ0I2r

(50)

where current I = e/T with L = rmv = 2πmr2/T and ~B||~L, so

~B =1

4πε0e

mc2r3~L (51)

used c = 1/√ε0µ0 to eliminate µ0. So the Spin-Orbit Interaction is

described by

Hso =12

(e2

4πε0

)1

m2c2r3~S · ~L (52)

where factor 1/2 is from Thomas precession: back transformation into restframe of proton (electron: not an inertial system, accelerated)


Spin-orbit interaction III

Now: [H,~L] 6= 0 and [H, ~S ] 6= 0 → not longer separetly conservedquantities, instead:

~J ≡ ~L + ~S (53)

commutes with H, as well as L2, S2,Jz and

J2 = (~L + ~S) · (~L + ~S) = L2 + S2 + 2~L · ~S (54)

→ ~L · ~S =12

(J2 − L2 − S2) (55)

with eigenvalues (where s = 12 in our case):

~2

2[j(j + 1)− `(`+ 1)− s(s + 1)] (56)


Spin-orbit interaction IV

For Hso we also need⟨ 1

r3

⟩:

⟨1r3

⟩=

1`(`+ 1

2)(`+ 1)n3a3(57)

Thus the energy of SO interaction is

E so = 〈Hso〉 =e2

8πε01

m2c2

(~2/2[j(j + 1)− `(`+ 1)− 34

`(`+ 12)(`+ 1)n3a3

(58)

=E 2

nmc2

n[j(j + 1)− `(`+ 1)− 34

`(`+ 12)(`+ 1)

∼ E 2n

mc2 (59)


Spin-orbit interaction V

However, only half of the truth, need correction for relativistic motion ofelectron, at least for larger Z , which is

E r = − E 2n

2mc2

[4n`+ 1

2− 3

](60)

So in total we have En + E so + E r:

Fine-structure energy levels of hydrogen

Enj = −13.6 eVn2

[1 +

α2

n2

(n

j + 12− 3

4

)](61)

→ breaks degeneracy in ` (different eigenvalues of H for same n)→m` and ms are not longer “good” quantum numbers (stationary statesnow linear combinations of states with different m`,ms)


Spin-orbit interaction VI→ n, `, s, j , mj “good” quantum numbers (write |j mj〉 as linearcombination of |`m`〉|s ms〉 with help of Clebsch-Gordan coefficents)

Finestructuresplitting ofhydrogen


Literature

D. Griffiths: Introduction to Quantum MechanicsH. Hänsel, W. Neumann: Atome · Atomkerne · ElementarteilchenH. Haken, H. C. Wolf: Atom- und QuantenphysikH. Friedrich: Theoretische Atomphysik


Documents

Atomic Spectra in Astrophysics - uni-potsdam.delida/TEACH.DIR/spin_orbit.pdf · AtomicSpectrainAstrophysics LidaOskinova,HelgeTodt Astrophysik Institut für Physik und Astronomie