Magnetism, Angular Momentum, and Spin

Transcription

1 Magnetism, Angular Momentum, and Spin Chapter 19 P. J. Grandinetti Chem Nov 13, 2017 P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

2 In 1820 Hans Christian Ørsted discovered that electric current produces a magnetic field that deflects compass needle from magnetic north, establishing first direct connection between fields of electricity and magnetism. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

3 Biot-Savart Law Jean-Baptiste Biot and Félix Savart worked out that magnetic flux density produced at distance r away from section of wire of length dl carrying steady current I c is d B = μ 0I c 4π d l r r 3 Biot-Savart law Direction of magnetic field vector is given by right-hand rule: if you point thumb of your right hand along direction of current then your fingers will curl in direction of magnetic field. magnetic dipole µ current current radius P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

4 Example Current flowing in flat loop of wire with area A will generate magnetic field which, at distance much larger than radius, r, appears identical to field produced by point magnetic dipole with strength of μ = μ = I c A Calculate magnetic dipole moment induced by electron moving in circular orbit of radius r with linear velocity v. Solution: For e with linear velocity of v the time for one orbit is 2πr v. Taking current as charge, q e, passing point per unit time we have I c = q e v 2πr Taking area of current loop as πr 2 we obtain magnetic dipole moment of ( qe v) μ = I c A = (πr 2 ) = q e 2πr 2 r v P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

5 Microscopic Origins of Magnetism Shortly after Biot and Savart, Ampére suggested that magnetism in matter arises from a multitude of ring currents circulating at atomic and molecular scale. André-Marie Ampére P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

6 Electron Orbital Magnetic Dipole For electron with angular momentum, rearrange L = m r v, to get 1 m L = r v and then obtain μ = q e 2m e L = γ L L γ L q e (2m e ) is called orbital gyromagnetic ratio : ratio of magnetic moment to orbital angular momentum current magnetic dipole Angular momentum radius Angular momentum vector is antiparallel to magnetic dipole vector due to negative electron charge. Don t forget : current flow is defined in opposite direction of electron flow. Despite simple derivation, μ = γ L L is general and holds for non-circular orbits as long as angular momentum is conserved. It is independent of size, period, and even shape of orbit. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

7 Electron Orbit Magnetic Dipole Operator With e motion the magnetic dipole moment operator is μ l = q e L = μ B L 2m e ħ μ B is defined as Bohr magneton and given by μ B q e ħ 2m e = J/T Magnitude of dipole moment is μ l = μ B l(l + 1) Operator for z component of electron orbital magnetic dipole moment is μ z = μ B ħ L z For z (or x or y) components only observed with quantized values of μ z = μ B ħ L z = μ B ħ (m l ħ) = μ B m l P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

8 N Classical Magnetic Dipole in a Uniform Magnetic Field Energy of classical magnetic dipole in uniform magnetic field is Potential energy is V = μ B S Take kinetic energy as K = 1 2 Iω2, then total energy is E = 1 2 Iω2 μ B Magnetic dipole in magnetic field experiences torque from potential energy τ = μ B In absence of friction total energy remains constant and compass needle oscillates about direction of magnetic field. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

9 QM Dipole in Magnetic Field: Zeeman interaction For e we substitute μ l = μ B L into V = μ B and obtain ħ V = μ B L B ħ Coupling of e s magnetic dipole moment to external magnetic field is called Zeeman interaction. Taking magnetic field as along z axis, B = B z e z, expression simplifies to V = μ B ħ L z B z Zeeman interaction lifts degeneracy of m l levels of electron in H-atom. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

10 Classical Magnetic dipole with angular momentum Place classical magnetic dipole with anti-parallel angular momentum in magnetic field and magnetic moment experiences torque τ = μ B = γ L L B = γ L B L Torque causes change in angular momentum. Newton s 2nd law says Taking B = B z e z, and defining d L dt = τ = γ L B L ω 0 γ L B z we write cross product in determinant form d L dt = e x e y e z 0 0 ω 0 L x L y L z = L y ω 0 e x + L x ω 0 e y Determinant gives 3 coupled differential equations... P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

11 Classical Magnetic dipole with angular momentum dl x dt = ω 0 L y dl y dt = ω 0 L x dl z dt Coupled differential eqs. involving L x and L y can be decoupled by defining = 0 Then we find L ± = L x ± il y dl + dt = dl x dt + i dl y dt = ω 0 L y + iω 0 L x = iω 0 (L x + il y ) = iω 0 L + or dl dt = dl x dt dl ± dt i dl y dt = ω 0 L y iω 0 L x = iω 0 (L x il y ) = iω 0 L = ±iω 0 L ± giving L ± (t) = L ± (0)e ±iω 0t Converting L ± back to Cartesian and taking μ = γ L L we find... P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

12 Classical Magnetic dipole with angular momentum μ x (t) = μ x (0) cos ω 0 t μ y (0) sin ω 0 t μ y (t) = μ y (0) cos ω 0 t + μ x (0) sin ω 0 t μ z (t) = μ z (0) These equations for magnetic dipole vector components describe precession motion of vector around magnetic field direction at angular frequency of ω 0. When magnetic dipole vector has angular momentum torque does not cause dipole to align with B, but rather causes it to precess about direction of B. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

13 Classical Magnetic dipole in a Non-Uniform Magnetic Field If magnetic dipole is in non-uniform magnetic field then forces acting on 2 poles will not cancel and dipole starts to translate along magnetic field gradient direction. In this case net force is F net = F + + F = ( μ B) Dipole oriented in magnetic field gradient below is pulled to the right. N S P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

14 Classical Magnetic dipole in a Non-Uniform Magnetic Field Differently oriented dipole in same magnetic field gradient is pulled to the left. S N P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

15 Classical Magnetic dipole in a Non-Uniform Magnetic Field When dipole is perpendicular to lines of force there is no translation of dipole. N S P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

16 Classical Magnetic dipole in a Non-Uniform Magnetic Field Now with Angular Momentum Take example of magnetic field given by B( r) = gx e x + (B 0 + gz) e z g is gradient strength. Maxwell s equations require that B = 0, so we must add gradient along x to satisfy this constraint. Thus we find F net = ( μ ( gx e x + (B 0 + gz) e z ) = μ ( ( gx ex + (B 0 + gz) e z ) = μ ( g e x + g e z ) Unlike compass needle, we also want to consider magnetic dipole with angular momentum. Thus, μ precesses about magnetic field direction while translating. For this precession motion we use μ x (t) = μ x (0) cos ω 0 t μ y (0) sin ω 0 t μ y (t) = μ y (0) cos ω 0 t + μ x (0) sin ω 0 t μ z (t) = μ z (0) P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

17 Classical Magnetic dipole in a Non-Uniform Magnetic Field Now with Angular Momentum Using F net = μ(t) ( g e x + g e z ) assume B 0 gx and use μ(t) for precessing magnetic dipole moment to obtain F net (t) = g ( μ x (0) cos ω 0 t μ y (0) sin ω 0 t ) e x + gμ z (0) e z F x (t) F z Time dependent F x (t) component of force will push dipole back and forth about its original position. Time independent F z component of force will push dipole in z direction which way depends on initial orientation of dipole. Bottom line is that passing magnetic dipole through magnetic field gradient along z provides means of measuring z component of magnetic dipole moment vector. Basis for seminal experiment in 1922 by Otto Stern and Walther Gerlach. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

18 The Stern-Gerlach Experiment - First Some Background In 1896 Pieter Zeeman discovered clue about magnetic properties of yet unexplained constituents of atom when he reported broadening of yellow D-lines of sodium in flame when held inside strong magnetic field. Became known as Zeeman effect. Bohr s 1913 theory of atom predicted that e orbital angular momentum would be quantized. In 1916 Arnold Sommerfeld refined Bohr s theory to allow for elliptical orbits described by 3 quantum numbers: n, k, and m, where n = 1, 2, 3,, called the principal quantum number, k = 1,, n, called the azimuthal quantum number m = k, k + 1,, m 0,, k 1, k, called the magnetic quantum number. m described quantization of z-component of e s orbital angular momentum. Refinements allowed Sommerfeld to explain Zeeman effect through dependence of orbital magnetic dipole moment on magnetic quantum number. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

19 The Stern-Gerlach Experiment In 1922 Otto Stern and Walther Gerlach designed apparatus in attempt to measure this quantization with beam of neutral silver atoms. Since silver atom had one valence electron it was predicted in Bohr-Sommerfeld theory that ground state for this electron was n = k = 1 and m = ±1. Beam of neutral Ag atoms emitted from oven containing Ag metal vapor were collimated and sent into vacuum region of non-uniform magnetic field where atoms are pushed up or down in magnetic field gradient depending on sign of z component of atom s magnetic dipole moment. Detector was glass plate onto which atoms exiting non-uniform magnetic field were deposited. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

20 Schematic diagram of the Stern Gerlach apparatus. Measurement result N Classical prediction S Oven Atom beam Collimator Classical prediction is all values of μ z between μ and μ would be observed. Sommerfeld predicts only z components associated with m = ±1 will be observed. After 1 year of refining apparatus Stern & Gerlach observed beam splitting into 2 distinct lines. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

21 Attached is the experimental proof of directional quantization. We congratulate you on the confirmation of your theory. Postcard from Stern & Gerlach to Neils Bohr, February 8, P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

22 The Stern-Gerlach Experiment Stern-Gerlach experiment in 1922 was before De Broglie s hypothesis in 1923 and Schrödinger s wave equation in While S-G experiment successfully demonstrated quantization of electron angular momentum, Stern & Gerlach s interpretation in terms of Sommerfeld model did not allow them to appreciate full significance of their discovery. In 1927 Phipps & Taylor used S-G technique on beam of H atoms. In 1927 Schrödinger s wave equation and solutions for H atom was known. Orbital angular momentum of H atom ground state (n = 1, l = 0, and m l = 0) predicts only 1 outcome for z component of e s magnetic dipole moment, μ z = 0. N S Phipps and Taylor experiment gives 2 outcomes. At this point it became clear that Schrödinger s picture for hydrogen atom was incomplete. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

23 Electron Spin Hypothesis (now theory) that explains Stern-Gerlach (and Phipps & Taylor) result is that e has intrinsic (built-in) magnetic dipole moment, μ s, arising from intrinsic angular momentum, S, called spin. Property called spin because if e is charged ball spinning about its own axis then it would have magnetic dipole moment from this spinning motion. Physical picture of spinning charged ball doesn t hold water for various reasons. Main reason is that such a model predicts only integer values for spin angular momentum quantum number. S-G and P-T experiments give 2 outcomes from μ z, consistent with s = 1 2, with m s = ±1 2. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

24 Electron Spin and the Dirac Wave Equation In 1929 Paul Dirac ( ) proposed relativistic wave equation for e, showing it must have intrinsic angular momentum of s=1/2 and magnetic moment of μ s = g s μ B S ħ g s = is called electron g-factor. Magnitude of spin dipole moment is μ s = g s μ B s(s + 1) In same theory he also predicted existence of anti-electron (and anti-matter). P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

25 Spin Orbitals Instead of switching to Dirac equation we modify (ad hoc) Schrödinger equation solutions to be product wave function with new degree of freedom ψ( r)χ( S) χ( S) is spin part of e s wave function whose stationary states, χ ms, are given symbols α for m s = +1 2 and β for m s = 1 2. Spin part is expressed as linear combination c α and c β are complex coefficients. χ( S) = c α α + c β β Two spin states, α and β, are orthogonal, α αdσ = β βdσ = 1, α βdσ = β αdσ = 0 P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

26 Spin Orbitals H-atom stationary states modified to include e spin state ψ n,l,ml,m s (r, θ, φ, σ) = R n,l (r)y l,ml (θ, φ) χ ms and are called spin orbitals. H-atom stationary states are specified by 4 numbers n, l, m l, and m s. Write H-atom stationary states with shorthand notation. ψ n,0,0, 1 2 ψ n,1, 1, 1 2 ψ n,1,0, 1 2 ψ n,1,1, 1 2 ψ n,2, 2, 1 2 ψ n,2, 1, 1 2 ψ n,2,0, 1 2 ψ n,2,1, 1 2 ψ n,2,2, 1 2 n s α, ψ n,0,0, 1 2 n p 1 α, ψ n,1, 1, 1 2 n p 0 α, ψ n,1,0, 1 2 n p 1 α, ψ n,1,1, 1 2 n d 1 α, ψ n,2, 2, 1 2 n d 1 α, ψ n,2, 1, 1 2 n d 0 α, ψ n,2,0, 1 2 n d 1 α, ψ n,2,1, 1 2 n d 1 α, ψ n,2,2, 1 2 n s β, n p 1 β, n p 0 β, n p 1 β, n d 2 β, n d 1 β, n d 0 β, n d 1 β, n d 2 β. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

27 Spin Orbitals Occupancy of H-atom stationary states is often illustrated in form of an orbital diagram. Below is case of n = 1, l = 0, m l = 0, and m s = s 2s 2p 3s 3p 3d P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

28 Electron Spin and Magnetic Dipole Moment Operator We complete treatment of e s magnetic dipole moment in stating that total magnetic dipole moment for e includes both orbital and spin contributions, μ = μ l + μ s = g lμ B ħ L g s μ B Ŝ ħ g l = 1 and is called orbital g factor. g s = is called electron g-factor. Total magnetic dipole moment operator for e is μ = μ B ħ ( ) g l L + g sŝ and for total z component of the magnetic dipole moment operator we have μ z = μ B ħ ( gl L z + g s Ŝ z ) P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

29 Nuclear Spin Angular Momentum P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

30 Nuclear Spin In 1924 Pauli suggested existence of proton spin and magnetic dipole In 1933 Otto Stern discovered that the proton has an intrinsic spin angular momentum and associated intrinsic magnetic dipole moment. Like e, proton spin quantum number is I = 1 2 with m I = ±1 2. Magnitude of proton magnetic dipole is much smaller than e s. Like e one might expect proton magnetic dipole to be μ I = q e I 2m p Key difference is division by proton instead of e mass. Mass ratio predicts proton dipole moment is m p m e 1836 times smaller than e but it is only 656 times smaller than e s. Define nuclear magneton as μ N q eħ 2m p = J/T, and proton magnetic dipole moment is μ p = μ N P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

31 Nuclear Spin After Rutherford proposed his model of atom with e orbiting positively charged nucleus it was soon realized that there were inconsistencies between number of positive charges and mass of atom. In 1920 Rutherford suggest that nucleus also contained neutral particles with mass similar to protons, and called them neutrons. In 1932 James Chadwick produced 1st convincing evidence of neutron s existence. In 1937 Isador Rabi and his group at Columbia University developed nuclear magnetic resonance (NMR) method to measure magnetic dipole moments of atomic nuclei. Rabi determined that spin quantum number of neutron is I = 1 2 and its magnetic dipole moment of μ n = μ N Opposite in sign and little smaller than proton magnetic dipole. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

32 Nuclear Spin Total angular momentum, I, of nuclear ground state is determined by number of protons and neutrons in nucleus. Nuclear magnetic dipole moments are related to I according to μ z = γ I Î, γ I is called nuclear gyromagnetic ratio. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

33 Some NMR Active Isotopes Z N Isotope Mass/(g/mol) Abundance/% I Dipole Moment/(μ N ) H / He / Li / Be / B / C / N O / F / Ne / Na / Mg / Al / Si / P / S / Cl / K / Ca / Sc / Ti / V / Cr / Mn / Fe / Co / P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

34 Addition of Angular Momentum P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

35 Addition of Angular Momentum Discovery of electron spin angular momentum means that total electron angular momentum of H-atom becomes J = L + S In classical mechanics this would be calculated trivially, J = J x J y J z = But this requires that we know all 3 vector components of L and S simultaneously. In QM we can only know length of each angular momentum vector and one component simultaneously. We can only know sum of one component, which tells us that m j = m l + m s L x L y L z J z = L z + S z P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62 + S x S y S z

36 Addition of Angular Momentum In QM we lack information needed to calculate length of total angular momentum vector, J. But we have constraints on how big or small the total can be. Largest vector sum is when both vectors point in same direction Smallest vector sum is when both point in opposite directions In other words when we add angular momentum vectors in QM we can only constrain total angular momentum as lying between these 2 limits L + S J L S P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

37 Addition of Angular Momentum L + S J L S translates to j(j + 1) l(l + 1) s(s + 1) and constrains j to fall between 2 limits of l s j l + s with only integral or half-integral values. For each j value there will be 2j + 1 values of m j, that is, m j = j, j + 1,, j 1, j P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

38 Addition of Angular Momentum Example What are the total angular momenta resulting from the addition of l = 0 and s = 1 2? Solution: j = to gives j = 1 2 Example What are the total angular momenta resulting from the addition of l = 1 and s = 1 2? Solution: j = to gives j = 1 2 and 3 2 P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

39 Electron Magnetic Moment from Total Angular Momentum Magnetic moment associated with electron having given value of j is g j is called Landé g factor μ j = g j μ B j(j + 1) g j = j(j + 1)(g l + g s ) + [l(l + 1) s(s + 1)](g l g s ) 2j(j + 1) Total magnetic dipole moment operator for e is μ μ = g B j Ĵ ħ and for total z component of the magnetic dipole moment operator we have μ z = g j μ B ħ Ĵz P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

40 Clebsch-Gordon series expansion How do the wave functions associated with l, m l, s, and m s combine to form the wave functions associated with j and m j? They are related through Clebsch-Gordon series expansion, ψ j,mj = l s m l = l m s = s C(j, m j, l, m l, s, m s )ψ l,ml,s,m s C(j, m j, l, m l, s, m s ) are called Clebsch-Gordon coefficients. These coefficients are nonzero only when m j = m l + m s. There are recursive expressions for calculating the Clebsch-Gordon coefficients and tables and calculators are easy to find. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

41 Clebsch-Gordon coefficients d j 1j 2m1m2 j 1j 2JM = ( 1) J j 1 j 2 j 2j 1m2m1 j 2j 1JM Square-root sign is to be understood over every coefficient, e.g., for 8 15 read P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

42 Example Construct wave function with j = 1 2 and m j = 1 for p-electron in H-atom in 2 terms of wave functions associated with l, m l, s, and m s. Solution: Addition of l = 1 and s = 1 2 leads to 2 possibilities of j = 1 2 and j = 3 2. Focus on j = 1 2 ψ n,j= 1 2,m = j states and write Clebsch-Gordon series as l s m l = l m s = s ( ) C j = 1 2, m j, l = 1, m l, s = 1 2, m s ψ n,l=1,ml,s= 1 2,m s Expanding out the m j = 1 case we find six terms: 2 ( ) ( ) ψ 1 2, 1 = C 1 2 2, 1 2, 1, 1, 1 2, 1 n p 2 1 β + C 1 2, 1 2, 1, 0, 1 2, 1 n p 2 0 β ( ) ( ) + C 1 2, 1 2, 1, 1, 1 2, 1 n p 2 1 β + C 1 2, 1 2, 1, 1, 1 2, 1 n p 2 1 α ( ) ( ) + C 1 2, 1 2, 1, 0, 1 2, 1 n p 2 0 α + C 1 2, 1 2, 1, 1, 1 2, 1 n p 2 1 α Only terms where m j = m l + m s = 1 2 survive... P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

43 Example Construct wave function with j = 1 2 and m j = 1 for p-electron in H-atom in 2 terms of wave functions associated with l, m l, s, and m s. Solution: We are left with ( ) ( ) 1 ψ n, 1 2, 1 = C 2 2, 1 2, 1, 0, 1 2, 1 1 n p 2 0 β + C 2, 1 2, 1, 1, 1 2, 1 n p 2 1 α Use Clebsch-Gordon Coefficient table to find ψ n, 1 2, 1 = 1 2 n p 0 β n p 1α Using orbital diagrams we could represent j = 1 2, m j = 1 2 state as P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

44 Addition of three or more angular momenta What about addition of three or more angular momenta? For example, What about l = 1, s = 1, and i = 1? 2 2 We have 3 choices on where to start, and all are valid. First add l and s to get j = l s to l + s, then add j and i to get f = i j to i + j. First add i and s to get j = i s to i + s, then add j and l to get f = j l to j + l. First add l and i to get j = i l to i + l, then add j and s to get f = j s to j + s. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

45 Example What are total angular momenta values and number of states that result when l = 1, s = 1, and i = 1 are added? 2 2 Solution: Starting with l and s we find that j = l s to l + s gives j = 1 2, 3 2 for l = 1 and s = 1 2 then f = j i to j + i gives f = 0, 1 for j = 1 2 and i = 1 2 and f = 1, 2 for j = 3 2 and i = 1 2 Total angular momentum can be f = 2, 1, 1, 0 each with 2f + 1 values of m f Thus we have ( ) + ( ) + ( ) + ( ) = 12 states P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

46 Fine Structure of the H-Atom 1 Relativistic Correction 2 Spin Orbit Interaction P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

47 Fine Structure of the H-Atom : Relativistic Correction While classical kinetic energy expression is good starting approximation for H-atom there are slight relativistic effects that can be observed in emission spectra of H-atoms. Instead of K = p2 2m, use K = mc (v c0 ) 2 mc2 0 Derivation in notes shows relativistic correction of e kinetic energy gives (1) = p4 8m 3 c 3 0 This, in turns, gives a 1st-order correction to energy { } E (1) r = E2 n 4n m e c 2 l P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

48 Fine Structure of the H-Atom : Spin Orbit Interaction When e has orbital angular momentum the e s intrinsic spin magnetic dipole, μ s interacts with nuclear charge, Zq e, as e orbits around nucleus. nuclear rest frame electron rest frame In electron rest frame we calculate magnetic field generated at electron by orbiting nucleus and calculate spin-orbit interaction as Ṽ SO = μ s B orb Lengthy derivation in notes gives... P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

49 Fine Structure of the H-Atom : Spin Orbit Interaction Lengthy derivation in notes gives V SO = ξ(r) L Ŝ ξ(r) is spin orbit coupling constant, given by ξ(r) = 1 2m 2 e c2 0 1 dφ(r) r dr φ(r) is electrostatic potential at electron. Convention is to report spin orbit coupling constant as ζ n,l = ħ2 hc 0 ψ n,l,m l ξ(r)ψ n,l,ml dτ = ħ2 hc 0 ξ(r) ζ n,l is a wavenumber and hc 0 ζ n,l is an energy. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

50 Fine Structure of the H-Atom : Spin Orbit Interaction For H-like atom we can further calculate α is the fine structure constant α R is Rydberg constant, Note that R H = R (1 + m e m p ) α 2 R ζ n,l = Z 4 n 3 l(l + 1)(l + 1 2) q 2 e 4πε 0 ħc 0 = m e q4 e R = 8ε 2 0 h3 c 0 For 2p electron in H-atom we have ζ 2,1 = α 2 R 24 = 0.24 cm 1 or 30.2 μev, which is a small shift on E 2 = 3.4 ev, the 2p energy. Such small perturbation is observable in emission spectrum of H-atoms. V SO plays a bigger role in heavier elements since ζ n,l Z 4. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

51 Perturbation Theory : The Spin Orbit Correction Let s calculate the 1st-order energy correction from V SO using the H-atom (0) as [ ( ) (0) = ħ2 1 2m e r 2 r 2 L ] 2 r r r 2 ħ 2 Zq2 e 4πε 0 r In QM any hermitian operator that commutes with corresponds to a physical quantity that is conserved (i.e., is time independent). Hamiltonian above commutes with L, that is, [, L] = 0, so orbital angular momentum is conserved. If spin-orbit coupling didn t exist then we would find that [, Ŝ] = 0 and spin angular momentum is conserved. But once we include spin-orbit coupling to = (0) + V SO, V SO = ξ(r) L Ŝ then we find [, L] 0 and [, Ŝ] 0. In other words orbital and spin angular momentum are no longer separately conserved when spin-orbit coupling is present. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

52 Perturbation Theory : The Spin Orbit Correction [ ( ) 1 r 2 L ] 2 r 2 r r r 2 ħ Zq2 e 2 4πε 0 r (0) = ħ2 2m e + ξ(r) L Ŝ V SO Can show that (0) and V SO commute with L 2, Ŝ 2, and Ĵ = L + Ŝ Since J 2 = ( L + S) ( L + S) = L 2 + S L S it will be helpful to rewrite V SO using L Ŝ = 1 2 (Ĵ2 L 2 Ŝ 2) as V SO = 1 2 ξ(r) [ Ĵ 2 L 2 Ŝ 2] in terms of quantities Ĵ 2, L 2, and Ŝ 2, which are conserved. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

53 Fine Structure of the H-Atom : Spin Orbit Interaction Calculate 1st-order energy correction to H-atom due to V SO Starting with 1st-order energy correction is E (1) = (0) + ξ(r) L Ŝ = (0) ξ(r) [ Ĵ 2 L 2 Ŝ 2] = ψ V SO n,l,m l,m s SO ψ n,l,ml,m s dτ V = 1 [ ] [ ] 2 R (r)ξ(r)r n,l n,l(r)r 2 dr Y χ [Ĵ2 L 2 Ŝ 2] Y l,m l m s l,ml χ ms sin θdθdφ = 1 2 ξ(r) [ j(j + 1)ħ 2 l(l + 1)ħ 2 s(s + 1)ħ 2] Using ζ n,l = ħ2 hc 0 ξ(r) we obtain E (1) n,j,l,s = 1 2 hc 0ζ n,l [ j(j + 1) l(l + 1) s(s + 1) ] as 1st-order spin-orbit energy correction for H-atom. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

54 Fine Structure of the H-Atom : Spin Orbit Interaction E (1) n,j,l,s = 1 2 hc 0ζ n,l [ j(j + 1) l(l + 1) s(s + 1) ] For l = 0 (s orbitals) this correction is zero since l = 0, s = 1 j = 1 [ ] and ( 1 + 1) 0 1 ( 1 + 1) = So, s orbitals are unaffected by the spin-orbit coupling. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

55 Fine Structure of the H-Atom : Spin Orbit Interaction E (1) n,j,l,s = 1 2 hc 0 ζ [ ] n,l j(j + 1) l(l + 1) s(s + 1) For l = 1 (p orbital) there will be j = = 1 2 and j = = 3 2. So E (1) n, 3 2,1, 1 = 1 [ ] 2 hc 3 0ζ n,1 2 ( ) ( ) = 1 2 hc 0ζ n,1, for j = E (1) n, 1 2,1, 1 = 1 2 hc 0 ζ n,1 2 [ 1 2 ( ) ( ) ] = hc 0 ζ n,1, for j = 1 2 Coupling between orbital and spin angular momentum makes specification of electron state with just n, l, m l, and m s insufficient. Also need to specify j. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

56 Fine Structure of the H-Atom : Spin Orbit Interaction We can also write 1st-order energy correction from spin-orbit coupling in terms of unperturbed energy of H-atom. Combining E (1) n,j,l,s = 1 2 hc 0ζ n,l [ j(j + 1) l(l + 1) s(s + 1) ] and we obtain E (1) SO = E2 n m e c 2 0 α 2 R ζ n,l = Z 4 n 3 l(l + 1)(l + 1 2) E n = Z 2 μq 4 e 32π 2 ε 0 ħ 2 n 2 { } n[j(j + 1) l(l + 1) s(s + 1) l(l + 1 )(l + 1) 2 where E n is unperturbed energy of H-atom. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

57 Fine Structure of the H-Atom: All Together Executive Summary Adding in spin orbit and relativistic correction the total energy becomes E n,j = E (0) n + E (1) SO + E(1) r = E [ ( α2 n n 2 n 2 j )] 4 Our treatment of hydrogen atom fine structure stops here. We could go further with more energy refinements coming from couplings to nuclear degrees of freedom, and even couplings to the quantized electromagnetic field. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

58 Term Symbols P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

59 Term Symbols Atomic spectroscopists have devised approach for specifying set of states having particular set of l, s, and j with a term symbol. Given l, s, and j for electron term symbol is defined as 2s+1 {l letter} j 2s + 1 is called spin multiplicity l letter is assigned based on numerical value of l according to l = numerical value {l letter} S P D F G H I K L M O Q R T letter continuing afterwards in alphabetical order. Example What is term symbol for 2 states with j = 1 for p-electron? 2 Solution: l = 1 has symbol P, and spin multiplicity is 2s + 1 = = 2. Thus we have 2 P 1 2 (pronounced doublet P half. ) P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

60 Term Symbols and Orbital Diagrams In orbital diagram e is placed in box associated with l and m l with arrows indicating m s value. Possible orbital diagrams for 1s 1 which has l = 0, s = 1 2, and j = 1 2 are Remember that we can add only one vector component of angular momentum vector in QM conventionally it is the z component, so m j = m l + m s. Both these orbital diagrams are associated with term symbol 2 S 1 2 (pronounced doublet S half. ) P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

61 Example How are orbital diagrams for 2p 1 configuration associated with term symbols? Solution: Case of 2p 1 with l = 1 and s = 1 2 can have j = 1 2, 3 2. This leads to possible terms of 2 P 1 2 and 2 P 3 2, We can only associate 2 orbital diagrams with states j = 3 2, m j = ±3 2. Other states are part of linear combinations belonging to 2 P 1 2 and 2 P 3 2. More systematic approach uses Clebsch-Gordon series. P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62

62 Energy levels of the hydrogen atom. Taking both relativistic and spin-orbit energy corrections into account we arrive at energy level diagram of hydrogen labeled by terms P. J. Grandinetti (Chem. 4300) Magnetism, Ang. Mom., & Spin Nov 13, / 62