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Chapter 7B – Electron Spin and Spin-‐Orbit Coupling

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Chapter 7B – Electron Spin and Spin-‐Orbit Coupling 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 ! q ! −e ! m = L → L 2m 2m e 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 e V = −B m = B L z z 2m z z e And the total Hamiltonian would be e Hˆ = Hˆ + B L 0 2m z z e The wavefunctions we obtained ψ = f (r)ym (θ,ϕ) are also eigenfunctions of Hˆ n,l,m n,l l e including the magnetic field E = E(0) + B(m!) n,l,m n,l 2m e each level nl, splits in (2l + 1)sublevels 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 1 angular momentum l . Here 212l += → l = !! 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 Sˆ ,Sˆ ,Sˆ Sˆ 2 = Sˆ 2 + Sˆ 2 + Sˆ 2 x y z x y z ˆˆˆ This operator is postulated to have exactly the same commutation relations as LLLxyz,, Hence ⎡Sˆ ,Sˆ ⎤ = i!Sˆ ⎣ x y ⎦ z ⎡Sˆ ,Sˆ ⎤ = i!Sˆ ⎣ z x ⎦ y ˆ ˆ ˆ ⎡S y ,Sz ⎤ = i!Sx ⎣ ⎦ From out general discussion we know the possible eigenstates as Sˆ 2 s,m = !2s(s +1) s,m s s Sˆ s,m = m ! s,m z s s s Where I indicate the states as s,m : Dirac bracket notation (page 159. MQ) s The splitting of silver atom beam indicates eigenstates 11 sm===, α 22s 11 sm==, −=β 22s 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: ˆ We know α , β are orthogonal because they are eigenstates of a Hermitian operator Lz 1 with different eigenvalues ± ! . Hence they must be orthogonal! (See chapter 4). 2 We now have 2 sets of angular momentum operators ˆˆˆ LLLxyz,, : orbital angular momentum ˆˆˆ SSSxyz,, : spin angular momentum ⎡⎤LSˆ ,0ˆ = ⎣⎦αβ ˆˆ22ˆˆ The operators LS,,, LSzz all commute, and they also commute with the Hamiltonian HTV0 =+, for the H-atom Then we can characterize each eigenstate through nlm,,ls , Sm , 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 2P , 2P , 2P m 1, 0, 1 1 0 −1 l =− We could then write, including spin 2P1α , 2P0α , 2P−1α 2P1β , 2P0 β , 2P−1β 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 p , p , p , p , p , p 1 0 −1 1 0 −1 Or the d-functions d2 , d1 , d0 , d−1, d−2 , d2 , d1 , d0 , d −1, d−2 This indicates the lm,,,ss Sm 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 e ge Hˆ = Hˆ + B Lˆ + B S where g ≈ 2.0011... 0 2m z z 2m z z e e 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 p , p functions are eigenfunctions of this magnetic Hamiltonian, and we can easily calculate the 1 1 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. Hˆ ( R) Tˆ Vˆ (r)Lˆ Sˆ 0 = + + ξ ⋅ Lˆ ⋅ Sˆ = Lˆ Sˆ + Lˆ Sˆ + Lˆ S x x y y z z It is called LS⋅ coupling or spin-orbit coupling Then the magnetic interaction can be included as ege HHˆˆ=+()R L + Sˆ 0 22mmzz ee Let us first examine the energy levels for the non-relativistic Hamiltonian, i.e. we neglect the spin-orbit or LS⋅ coupling term. ˆˆee ˆ ˆ HH=+0 BLgzz + BS zz 22mmee ⎛ e ⎞ Let us denote ! B = γ and use g = 2, then ⎜ 2m z ⎟ ⎝ e ⎠ ˆ γ ˆ 2γ ˆ H = H0 + Lz + Sz ! ! Let us consider the allowed 21ps+ emission lines: Normal Zeeman effect: H 0 , only include Lz include Sz Chapter 7b – Electron Spin and Spin-Orbit Coupling 100 Winter 2013 Chem 356: Introductory Quantum Mechanics ˆ Due to Sz 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. e ˆ If one does not consider spin levels split in a magnetic field using 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 np n' s are not affected by spin. Moreover all ! αα→ e multiples are split by the same amount !B . We would not see the effects of spin. 2m z e 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 2 e 1 ˆ ˆ HLSso =22 3 ⋅ 2mere 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. ⎛ 0 1 ⎞ ⎛ 0 −i ⎞ ⎛ 1 0 ⎞ , , σ x = !⎜ ⎟ σ y = !⎜ ⎟ σ z = !⎜ ⎟ ⎝ 1 0 ⎠ ⎝ i 0 ⎠ ⎝ 0 −1 ⎠ These matrices satisfy the commutation relations of angular momentum ⎡σ ,σ ⎤ = i!σ ⎣ x y ⎦ z Moreover ⎛ 1 ⎞ ⎛ 1 0 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ S z ⎜ ⎟ = !⎜ ⎟ ⎜ ⎟ = !⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 −1 ⎠ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎛ 0 ⎞ ⎛ 1 0 ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ S z ⎜ ⎟ = !⎜ ⎟ ⎜ ⎟ = −!⎜ ⎟ ⎝ 1 ⎠ ⎝ 0 −1 ⎠ ⎝ 1 ⎠ ⎝ 1 ⎠ ⎛ 1 ⎞ ⎛⎞0 Hence α = β = ⎜ 0 ⎟ ⎜⎟1 ⎝ ⎠ ⎝⎠ 2*2 matrices have only 2 eigenvectors. This is why we have only α , β 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.
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