Download Chapter 7b – Electron Spin and Spin

Winter 2013 Chem 356: Introductory Quantum Mechanics Chapter 7b – Electron Spin and Spin-­‐Orbit Coupling ................................................................................. 96 H-­‐atom in a Magnetic Field: Electron Spin ............................................................................................ 96 Total Angular Momentum ................................................................................................................... 103 Chapter 7b – Electron Spin and Spin-­‐Orbit Coupling H-­‐atom in a Magnetic Field: Electron Spin !
If electron in orbital has angular momentum L , one has a magnetic moment !
m =
q !
L → 2m
−e !
L 2me
This magnetic moment can interact with a magnetic field and the interaction energy is given by ! !
V = − m ⋅ B !
B is measured in Tesla 1T = 1 Newton/(ampere meter) If we take B to be in z direction then And the total Hamiltonian would be The wavefunctions we obtained ψ n,l ,m = f n,l (r) yl (θ ,ϕ ) are also eigenfunctions of Ĥ !
V = − Bz mz =
Ĥ = Ĥ 0 +
(0)
including the magnetic field En,l ,m = En,l
+
each level n, l splits in (2l + 1) sublevels e
B L 2me z z
e
B L 2me z z
m
e
B(m!) 2me
This is what would be expected, from classical considerations. This is not what is observed! 96 Winter 2013 Chem 356: Introductory Quantum Mechanics Stern and Gerlach passed silver atoms (with electron in s-­‐orbital) through inhomogeneous magnetic field. 1922! They found this splits the beam into two. Classically one would expect (2l + 1) lines for particle with angular momentum l . Here 2l + 1 = 2 → l =
1
!! 2
This was the first evidence that electrons need another quantum number: half-­‐integer angular momentum. Jumping to what we know now, we introduce an intrinsic angular momentum operator Ŝ x , Ŝ y , Ŝ z Ŝ 2 = Ŝ x2 + Ŝ y2 + Ŝ z2 This operator is postulated to have exactly the same commutation relations as Lˆx , Lˆ y , Lˆz Hence ⎡ Ŝ x , Ŝ y ⎤ = i!Ŝ z ⎣
⎦
⎡ Ŝ z , Ŝ x ⎤ = i!Ŝ y ⎣
⎦
⎡ Ŝ y , Ŝ z ⎤ = i!Ŝ x ⎣
⎦
From out general discussion we know the possible eigenstates as Ŝ 2 s,ms = ! 2 s(s + 1) s,ms Ŝ z s,ms = ms ! s,ms Where I indicate the states as s,ms : Dirac bracket notation (page 159. MQ) The splitting of silver atom beam indicates eigenstates s=
1
1
, ms =
= α 2
2
s=
1
1
, ms = −
= β 2
2
Chapter 7b – Electron Spin and Spin-­‐Orbit Coupling 97 Winter 2013 Chem 356: Introductory Quantum Mechanics Everything we discussed on operators and angular momentum applies to spin, also the states
α
, β
are orthonormal α | α = ∫ α *(σ )α (σ ) dσ = 1 β | β = ∫ β *(σ )β (σ ) dσ = 1 α | β = ∫ α *(σ )β (σ ) dσ = 0 What are the ‘coordinates’ that we integrate over? We do not have a clue!! But, also, we do not need it: We know α , β are orthogonal because they are eigenstates of a Hermitian operator Lˆz 1
2
with different eigenvalues ± ! . Hence they must be orthogonal! (See chapter 4). We now have 2 sets of angular momentum operators Lˆx , Lˆ y , Lˆz : orbital angular momentum Sˆx , Sˆ y , Sˆz : spin angular momentum ⎡ Lˆα , Sˆβ ⎤ = 0 ⎣
⎦
The operators Lˆ2 , Sˆ 2 , Lˆz , Sˆz all commute, and they also commute with the Hamiltonian H 0 = T + V , for the H-­‐atom Then we can characterize each eigenstate through n, l , ml , S , ms 5 quantum numbers Let us facilitate the notation somewhat Note: Later on this will require modification, as we are neglecting relativistic effects, which introduce so-­‐
called spin-­‐orbit coupling. Chapter 7b – Electron Spin and Spin-­‐Orbit Coupling 98 Winter 2013 Chem 356: Introductory Quantum Mechanics We can indicate the l -­‐quantum number as s, p, d and get functions We could then write, including spin 2P1 , 2P0 , 2P−1 ml = 1,0, −1 2P1α , 2P0α , 2P−1α 2P1 β , 2P0 β , 2P−1 β In what follows we will focus on one particular n quantum number, which we can suppress. Moreover I can indicate the β -­‐spin function by an overbar. Then we get the 6 p-­‐functions Or the d-­‐functions p1 , p0 , p−1 , p1 , p0 , p−1 d 2 , d1 , d 0 , d −1 , d −2 , d 2 , d1 , d 0 , d −1 , d −2 This indicates the l , ms , S , ms quantum numbers In this way we can label the exact eigenstates of the (non-­‐relativistic) Hamiltonian. The splitting of the lines in a magnetic field would then be determined by the Hamiltonian Ĥ = Ĥ 0 +
e
ge
Bz L̂z +
B S 2me
2me z z
where g ≈ 2.0011... This factor g determining the ratio between spin and orbital interactions with the magnetic field, can be calculated using relativistic quantum field theory. (Schwinger, Tomanaga, Feynman). Far beyond our aim. It agrees to about 10 digits with the experimental value! (that is like measuring the distance from here to New York up to a millimeter!) The p1 , p1 functions are eigenfunctions of this magnetic Hamiltonian, and we can easily calculate the energy splitting. Unfortunately, this does not give correct results!! The splitting due to the magnetic field is very small. There are other corrections to the energy levels in Hydrogen atom due to relativity (think Einstein). They are of at least comparable importance, and cannot be neglected when discussing magnetic effects. The relativistic Hydrogen atom is described by the Dirac equation. This is far more complicated than we wish to discuss. Chapter 7b – Electron Spin and Spin-­‐Orbit Coupling 99 Winter 2013 Chem 356: Introductory Quantum Mechanics One can approximate effects by including so-­‐called spin-­‐orbit interaction in the Hamiltonian. Ĥ 0( R) = Tˆ + Vˆ + ξ (r) L̂ ⋅ Ŝ L̂ ⋅ Ŝ = L̂x Ŝ x + L̂y Ŝ y + L̂z S z It is called L ⋅ S coupling or spin-­‐orbit coupling Then the magnetic interaction can be included as e ˆ
ge ˆ
Hˆ = H 0( R ) +
Lz +
Sz
2me
2me Let us first examine the energy levels for the non-­‐relativistic Hamiltonian, i.e. we neglect the spin-­‐orbit or L ⋅ S coupling term. e
e
Hˆ = Hˆ 0 +
Bz Lˆz + g
Bz Sˆz 2me
2me
⎛
⎞
e
Bz ⎟ = γ and use g = 2 , then ⎝ 2me ⎠
Let us denote ⎜ !
Ĥ = H 0 +
γ
2γ
L̂z +
Ŝ !
! z
Let us consider the allowed 2 p + 1s emission lines: Normal Zeeman effect: H 0 , only include Lz include S z Chapter 7b – Electron Spin and Spin-­‐Orbit Coupling 100 Winter 2013 Chem 356: Introductory Quantum Mechanics Due to Sˆz all α -­‐levels go up by γ all β -­‐levels go down by γ . Since transitions cannot change spin, Δms = 0 , I get same transitions as without spin!! Our conclusion thus far. If one does not consider spin levels split in a magnetic field using e ˆ
Lz 2me
p → 3 equal spaced levels d → 5 equal spaced levels If we include spin, for single electron states then all α -­‐states shift up by 1 unit of γ ! , all β -­‐levels shift down by 1 unit of γ ! , and the transition energies npα → n ' sα are not affected by spin. Moreover all multiples are split by the same amount e
!B . We would not see the effects of spin. 2me z
This is what was originally observed in early experiments. It is called the ‘normal Zeeman effect’ and it was explained by Lorentz (two Dutch physicists). It appears as if only the Lz term is present. However we do observe the effects of spin in emission spectra! The story is more complicated. The complications occur already for the H-­‐atom without a magnetic field. There is a substantial correction due to what is called spin-­‐orbit interaction. A good way to think about this is as follows: We usually think of the electron as wizzing around the nucleus. From the point of view of the electron we can just as easily think that the nucleus is wizzing about the electron. This moving nucleus, with its angular momentum generates a magnetic field. This magnetic field interacts with the spin of the electron. Compare the electron with your position on the spinning earth. The sun rises and sets from our standing still point of view, and moves with incredible velocities, in this frame. For a charged particle the magnetic force would be large. It is called spin-­‐orbit coupling Chapter 7b – Electron Spin and Spin-­‐Orbit Coupling 101 Winter 2013 Chem 356: Introductory Quantum Mechanics H so =
e2 1 ˆ ˆ
L ⋅ S 2me 2 e2 r 3
Spin-­‐orbit coupling is usually said to be a relativistic effect. This is because it arises in a natural way from the fully relativistic Dirac equation. So does spin; it arises naturally. And so do particles and antiparticles, which also arise from the Dirac equation. Let me say something more about spin. The spin operators are best represented by matrices. ⎛
⎞
⎛
⎞
⎛
⎞
σ x = ! ⎜ 0 1 ⎟ , σ y = ! ⎜ 0 −i ⎟ , σ z = ! ⎜ 1 0 ⎟ ⎝ 1 0 ⎠
⎝ i 0 ⎠
⎝ 0 −1 ⎠
These matrices satisfy the commutation relations of angular momentum ⎡σ x ,σ y ⎤ = i!σ z ⎣
⎦
Moreover ⎛ 1 ⎞
⎛ 1 0 ⎞⎛ 1 ⎞
⎛ 1 ⎞
Sz ⎜
= !⎜
= !⎜
⎟
⎟
⎜
⎟
⎟ ⎝ 0 ⎠
⎝ 0 −1 ⎠ ⎝ 0 ⎠
⎝ 0 ⎠
⎛ 0 ⎞
⎛ 1 0 ⎞⎛ 0 ⎞
⎛ 0 ⎞
Sz ⎜
=
!
=
−!
⎟
⎜ 0 −1 ⎟ ⎜ 1 ⎟
⎜ 1 ⎟ ⎝ 1 ⎠
⎝
⎠⎝
⎠
⎝
⎠
Hence 2 * 2 matrices have only 2 eigenvectors. This is why we have only α , β ⎛
⎞
α = ⎜ 1 ⎟ ⎝ 0 ⎠
⎛0⎞
β = ⎜ ⎟ ⎝1⎠
The Dirac equation is a 4 * 4 matrix equation and we get (2 spin * 2 mass) solutions. The splitting due to the magnetic field is very small. There are other corrections to the energy levels in Hydrogen atom due to relativity. They are of at least comparable importance, and cannot be neglected when discussing magnetic effects. The relativistic Hydrogen atom is described by the Dirac equation. This is far more complicated than we wish to discuss. One can approximate the effects by including spin-­‐orbit interaction in the Hamiltonian. Hˆ 0( R ) = Tˆ + Vˆ + ξ (r ) Lˆ ⋅ Sˆ Lˆ ⋅ Sˆ = Lˆx Sˆx + Lˆ y Sˆ y + Lˆz Sˆz It is called L ⋅ S coupling or spin-­‐orbit coupling Then the magnetic interaction can be included as Chapter 7b – Electron Spin and Spin-­‐Orbit Coupling 102 Winter 2013 Chem 356: Introductory Quantum Mechanics e ˆ
e ˆ
Hˆ = H 0( R ) +
Lz + g
S z 2me
2me
Our original spin-­‐orbitals p1 , p0 , p−1 , p1 , p0 , p−1 are however not eigenstates of the Hˆ 0( R ) including L ⋅ S coupling We can classify the eigenstates of Hˆ ( R ) by doing a little more angular momentum theory. Let me sketch the result, as this is, finally, an accurate description. Total Angular Momentum We have Lˆx , Lˆ y , Lˆz and Sˆx , Sˆ y , Sˆz Moreover ⎡ L̂α , Ŝβ ⎤ = 0 α , β = x, y, z ⎣
⎦
This is because L̂ and Ŝ act on different coordinates Now we define total angular momentum Jˆx = Lˆx + Sˆx ; Jˆ y = Lˆ y + Sˆ y ; Jˆz = Lˆz + Sˆz Ĵ 2 = J x2 + J y2 + J z2 Jˆ x , Jˆ y and Jˆ z satisfy the usual commutation relations: ⎡ Jˆ x , Jˆ y ⎤ = ⎡ Lˆ x + Sˆx , Lˆ y + Sˆ y ⎤ = ⎡⎣ Lˆx , Lˆ y ⎤⎦ + ⎡⎣ Lˆx , Sˆ y ⎤⎦ + ⎡⎣ Sˆx , Lˆ y ⎤⎦ + ⎡⎣ Sˆx , Sˆ y ⎤⎦ = i!L̂z + 0 + 0 + i!Ŝ z = i!Ĵ z Moreover Ĵ 2 = J x2 + J y2 + J z2 = Lˆ2 + Sˆ 2 + 2L ⋅ S L ⋅ S = Lx S x + Ly S y + Lz S z And ⎡⎣ J x , L2 ⎤⎦ = ⎡⎣ Lx , L2 ⎤⎦ + ⎡⎣ Sx , L2 ⎤⎦ = 0 Likewise ⎣
⎦
⎣
⎦
=
1 2
J − L2 − S 2 ) (
2
Chapter 7b – Electron Spin and Spin-­‐Orbit Coupling 103 Winter 2013 Chem 356: Introductory Quantum Mechanics ⎡ Jˆα , Sˆ 2 ⎤ = ⎡ Lα , S 2 ⎤ + ⎡ Sα , Sˆ 2 ⎤ = 0 ⎦ ⎣
⎣
⎦ ⎣
⎦
But then also ⎡ Jα , Jˆ 2 − Sˆ 2 − Lˆ2 ⎤ = [ Jα , L ⋅ S ] = 0 ⎣
⎦
And ⎡⎣ J 2 , L2 ⎤⎦ = ⎡⎣ J 2 , S 2 ⎤⎦ = 0 Hence we can derive without too much trouble that the operators Jˆ 2 , Lˆ2 , Sˆ 2 and Jˆ z all commute. Moreover these operators commute with L ⋅ S , and also with ξ ( r ) Lˆ ⋅ Sˆ . It then follows that the angular momentum operators Jˆ 2 , Lˆ2 , Sˆ 2 , Jˆ z commute with the relativistic Hamiltonian Hˆ ( R ) . And we can classify the states with quantum numbers n, l , s, j, m j As for the non-­‐relativistic case, the angular momentum problem, defining l , s, j, m j is independent from the radial equation, and can be solved once and for all. Lˆ2 l , s, j, m j = l(l + 1)! 2 l,s, j,m j Sˆ 2 l , s, j, m j = s(s + 1)! 2 l,s, j,m j Jˆ 2 l , s, j, m j = j( j + 1)! 2 l,s, j,m j Jˆz l , s, j, m j = m j ! 2 l,s, j,m j These equations hold for the Hydrogen atom, but later on we will see that they are very similar for many-­‐electron atoms, which also have spherical symmetry. Can we deduce what the eigenstates of Ĵ 2 and Jˆ z might be? First we note that eigenfunctions of J z are easy. Ĵ z l,s,ml ,m j = ( ml + ms ) ! l,s,ml ,m j Eg. ⎛ 1⎞
Ĵ z p1 = ⎜ 1+ ⎟ !p1 ⎝ 2⎠
⎛
1⎞
Ĵ z p0 = ⎜ 0 − ⎟ !p0 2⎠
⎝
We also know that acting with Ĵ + on j , j should give 0: the highest m j value in the multiplet. It is easy to find the highest function: ml = l , ms = s Chapter 7b – Electron Spin and Spin-­‐Orbit Coupling 104 Winter 2013 Chem 356: Introductory Quantum Mechanics 3
3
, j = is the highest m j function. 2
2
1 5
In the d-­‐manifold it is d 2 = d 2α with m j = 2 + = 2 2
Hence in the p-­‐manifold p1 = p1α mj =
Acting with Ĵ − we can get the other function in the multiplet. We get (2 j = 1) in 2 ⋅
3
+ 1 = 4 functions 2
We are left with 2 functions in the p-­‐manifold (6 spin-­‐orbital in total). This creates a j =
1
multiplet. 2
1 1
1 1
, , ,− 2 functions. 2 2
2 2
How does this work, for the d-­‐ functions? l = 2 highest ml . 1 5
5
= → j = 2 2
2
5 3 1 1 3 5
m j = − , − , − , , , 2 2 2 2 2 2
mj = l j + s j = 2 +
( 2J + 1) = 6 functions In total we have 10 functions → other multiplet has 4 functions → j = l −
1
1 3
= 2 − = 2
2 2
For the Hydrogen atom we can construct always ⎛ 1⎞
j = ⎜ l + ⎟ ⎝ 2⎠
⎛ 1⎞
2 ⎜ l + ⎟ + 1 = 2l + 2 ⎝ 2⎠
Chapter 7b – Electron Spin and Spin-­‐Orbit Coupling 105 Winter 2013 Chem 356: Introductory Quantum Mechanics ⎛ 1⎞
j = ⎜ l − ⎟ ⎝ 2⎠
⎛ 1⎞
2 ⎜ l − ⎟ + 1 = 2l ⎝ 2⎠
2 ⋅ ( 2l + 1) 2 ⋅ ( 2l + 1) is the total number of spin-­‐orbitals of angular momentum l . We label the final eigenstates as 2S+1 LJ (
)
The j-­‐multiplet have slightly different energies due to the 2L ⋅ S = J 2 − L2 − S 2 coupling, which for equal L, S is seen to depend on the J quantum number. Finally, we are ready to discuss how these relativistic levels split in a magnetic field; Within a multiplet J the levels split in 2 J + 1 equally spaced levels due to a magnetic field The splitting depends on ( n, l , j ) Allowed transitions: Δl = ±1 , Δs = 0 , ΔmJ = 0, ±1 Chapter 7b – Electron Spin and Spin-­‐Orbit Coupling 106 Winter 2013 Chem 356: Introductory Quantum Mechanics All different energies. Tiny splitting, but this is how we can experimentally access degeneracies of energy levels. The splitting of energy levels in a magnetic field is a complicated subject, because relativistic effects have to be considered at the same time. Magnetic transitions using nuclei as in NMR are much simpler as we only need to consider angular momentum theory itself. Better picture of transitions including spin-­‐orbit: B = 0 10 different transitions, 10 different frequencies Selection rules Δl = ±1 , Δs = 0 , Δj = 0, ±1, Δm j = 0, ±1 2
2
2
No J = 0 → J = 0 transitions 1
2 p 3 → 2 2s1 (m j = ) 2
2
2
3 transitions Δm j = 0, ±1 1
2 p 1 → 2 2s1 (m j = ) 2
2
2
2 transitions 1
2 p 3 → 2 2s1 (m j = − ) 2
2
2
3 transitions Chapter 7b – Electron Spin and Spin-­‐Orbit Coupling 107 Winter 2013 Chem 356: Introductory Quantum Mechanics 2
1
2 p 1 → 2 2s1 (m j = − ) 2
2
2
2 transitions 10 transitions in total To good approximation one can evaluate the shifts in magnetic field from Neglecting spin-­‐orbit coupling. e ˆ
2e ˆ
Lz +
S z 2me
2me
Chapter 7b – Electron Spin and Spin-­‐Orbit Coupling 108