Sommerfeld (1920) noted energy levels of Li deduced from spectroscopy looked like H, with slight adjustment of principal quantum number:

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1 Spin. Historical Spectroscopy of Alkali atoms First expt. to suggest need for electron spin: observation of splitting of expected spectral lines for alkali atoms: i.e. expect one line based on analogy of alkali atom with Hydrogen, see two closely spaced ones instead. Naive argument alkali atoms: Argue valence electron in alkali atom feels mostly Coulomb force due to extra proton in nucleus analogous to hydrogen: Z=3: Lithium=Helium + proton + electron Z=: Sodium= Neon + proton + electron Sommerfeld (90) noted energy levels of Li deduced from spectroscopy looked like H, with slight adjustment of principal quantum number:

2 E n Rydberg (n + p l ) () due to the fact that once the valence electron gets inside the He-like core electrons, it will feel a higher nuclear charge, potential falls faster than /r see figure. Correction term p l < 0 gets small in magnitude for large l, however, since these orbits are outside the core with high probability. Expt l. observations:. Some of the spectral lines for Li appear in close doublets (pairs of closely spaced energies). Sommerfeld said must imply influence of two possible nuclear or core states. Pauli argued two-valuedness came from two possible quantum states of valence electron.. Doublets observed to move farther apart in applied magnetic field, other lines where splitting too small to see in B = 0 turn into doublets when B 0. Zeeman effect. To explain splitting in applied mag. field, Uhlenbeck and Goudschmidt (95) suggested electron has spin angular momentum s = /, therefore two states m s = ±/. Predicted observed splitting, with E = µ B µ = e m S energy of electron magnetic moment () in applied magnetic field B relation of magnetic moment to electron spin operator (3)

3 S = (S x, S y, S z ) = electron spin operator (4) S has e value h s(s + ) = 3 h /4 (5) S z has e value hm s = (±/) h (6) Q: Should we have expected spin on classical grounds? Electron might be spinning in analogy to planet revolving around sun total a.m. J is sum of orbital a.m. (L r p) around sun plus its spin angular momentum S Iωˆn. (Bogus argument: quantum mechanical spin is intrinsic and permanent. Electron is as far as we know point particle.) It is interesting to note that if we had a classical charge e orbiting in a circle of radius a with angular velocity ω, the ratio of the magnetic moment to the angular momentum (gyromagnetic ratio) would be µ l = Ja Iω = (e ω/(π))(πa ) (ma )ω = e m, where J = eω/(π) is the current carried by the orbiting particle. Note that the radius cancels, so we can in fact discuss the classical gyromagnetic ratio of any spinning object using this formula since it s just made up of many such charge el- 3

4 ements orbiting around the same axis. Note also that the gyromagnetic ratio for the quantum mechanical spin determined by fits to the field dependence of the doublet splitting is a factor of two different from the classical expectation.. Free spin: Pauli Spin Matrices Recall we showed that operators which obeyed the algebra [ ˆL i, ˆL j ] = i hɛ ijk ˆLk (7) had eigenvalues of total ˆL i ˆL i restricted to h l(l + ), with l =,, 3,, 5,.... (Orbital ang. mom. ˆL only took on values with l =integer since we showed additionally that ˆL z = m h, l m integer l.) So let s demand that the components of spin angular momentum obey same comm. rels.: because [Ŝi, Ŝj] = i hɛ ijk Ŝ k (8). by analogy with orbital angular momentum, looking for a quantum operator which has exactly two discrete e values to explain spectral doublets, recall a.m. algebra allows this.. will use algebra (Eq. (8)) later to show S generates rotations of spins. Spin algebra (8) implies [Ŝ, Ŝz] = 0 as before, so we can find simultaneous eigenstates of Ŝ and Ŝz, with eigenvalues h s(s + ) and m s h, with s = / and m s = ±s as before. 4

5 Write two allowed spin states of a particle as ( spin up ) and, so we now need to specify the spin index σ of a wave function, by ψ σ, i.e. ψ (r) = ψ (9) ψ (r) = ψ (0) Of course the operator S is completely specified by the matrix elements m s Ŝi m s. Calculate: Ŝz: or, in this basis, m s Ŝz m s = m s hδ ms m s () S z = h 0 0 () (Now for S x and S y we ll need ladder operators: Ŝ + = Ŝx + iŝy (3) Ŝ = Ŝx iŝy (4) with the usual conditions Ŝ+ = 0 and Ŝ = 0 to terminate the ladder. The proportionality constant given as usual (i.e., as for the orbital a.m. ladder ops): Ŝ ± m s = h s(s + ) m s (m s ± ) m s ± (5) i.e. for s = and m s = ± 5

6 Ŝ + = h and Ŝ = h (6) So, using Ŝx = (Ŝ+ +Ŝ )/ and Ŝy = (Ŝ+ Ŝ )/(i) can now calculate (check it!) Ŝx: m s Ŝx m s = h m s = h Ŝ + + Ŝ m s (7) 3 4 m s(m s + )δ ms,m + s+ h 3 4 m s(m s )δ ms,m (8) s So and similarly Ŝy: S x = S x = h S y = S y = h i i 0 (9) (0) Notation: 3 dimensionless matrices 0 0 i σ x ; σ 0 y i 0 ; σ z 0 0 () are called Pauli spin matrices, sometimes written in vector form, σ = (σ x, σ y, σ z ), so that S = ( h/) σ. 6

7 Mathematical aside: Rank unitary matrices of determinant form a group in the mathematical sense, called SU(). The Pauli matrices are a representation of the generators of SU(), meaning any matrix of determinant can be written as i exp σ. () ˆn.3 Two spin-/ particles Consider two electrons, and suppose that only dynamical variables of interest for moment are spins, Ŝ and Ŝ. and refer to two different particles, so Ŝ and Ŝ commute. So we have Total spin operator [Ŝi, Ŝj] = i hɛ ijk Ŝ k (3) [Ŝi, Ŝj] = i hɛ ijk Ŝ k (4) Ŝ i, Ŝj] = 0. (5) S = S + S (6) Ŝ = Ŝ + Ŝ + S S (7) As in Prob. we did in homework, can specify two distinct sets of commuting observables: ) Ŝ, Ŝz, Ŝ, Ŝz (8) ) Ŝ, Ŝz, Ŝ, Ŝ (9) 7

8 Note, e.g. Ŝ can t be added to set ) because of Eq. (7): S S won t commute with Ŝz, e.g. Assume both particles spin /, then all state vectors are eigenstates of Ŝ and Ŝ with eigenvalues, e.g. h S (S + ) = 3 h /4, same for spin. Drop S labels, complete set of eigenstates for this problem is then m, m, m = ±, m = ± (30) or, could choose as our complete set Completeness = S, m, m S with (3) Ŝ S, m = h S(S + ) S, m and (3) Ŝ z S, m = hm S, m (33) S, m = m, m m, m S, m (34) m,m Find expansion coefficients m, m m, m S, m : Use spin version of favorite identities: Ŝ = Ŝ Ŝ+ + S z + hŝz (35) Ŝ = Ŝ+Ŝ + S z hŝz (36) As usual start with state with max. ang. momentum, Check: m = /, m = / (37) 8

9 where good use was made of So clear we can write Ŝ = h, (38) Ŝ z = h (39) Ŝ z = Ŝz + Ŝz (40) S =, m = = m = /, m = / (4) Now play usual game and apply lowering op. Ŝ = Ŝ +Ŝ to find (check!):, 0 = + (4) and, = (43) (Note I dropped S =, m = and m =, m labels since the distinction is clear: S, m are always integers for this problem.) Finally we need remaining state for combination of two spins with total S = 0. Must be orthogonal to all 3 states. Could construct formally, but easy to guess, and verify orthonormality: 0, 0 = (44) 9

10 singlets and triplets. The three S = states (4-43) are referred to together as triplet states, whereas the S = 0 combination Eq. (44) is called a singlet..4 Pauli principle Pauli: two identical particles in quantum mechanics have symmetric (antisymmetric) wavefunctions under interchange of all particle labels if they have integer (half-integer) spin. Particles with integer (half-integer) spin are called bosons (fermions). So bosons ψ(, ) = ±ψ(, ) (45) fermions Remarks on Pauli principle:. Immediate consequence for atomic physics: no two electrons can be in same single-particle quantum state. (Then wave function would be φ()φ(), manifestly symmetric no good.). Two electrons are in -particle wavefctn. which consists of product of spatial part (dependence on e coordinates r and r ) and spin part. Could write ψ(, ) = R(r, r )χ(, ) (46) Note if spatial part has definite sign under interchange, spin part must have opposite sign to make total wavefctn antisymmetric! 0

11 Example. Suppose R = R( r r ) symmetric under, so must have χ(, ) = χ(, ). System is in spin singlet state! ( ( ) Example. Suppose R (r r ) ẑ (relative p-wave state). Then must have χ(, ) = +χ(, ) lin. comb. of triplet states. ( ( +, ), or ).5 Spin / in external magnetic field Spin was assumed by Goudschmidt and Uhlenbeck to give rise to magnetic moment just as orbiting charged particle does. Small difference: orbiting charge e has moment (check!) µ = e m L (47) whereas Goudschmidt and Uhlenbeck found they needed to assume for the electron µ = e m S (48) in order to explain experiment. They assumed then the coupling to the external field was just as in classical electromagnetism: H Zeeman = µ B = e m S B (49) µ 0 σ B (50) where µ 0 e h/(m) is called the Bohr magneton. Time evolution of spin states. Ignore all parts of Hamiltonian except H Zeeman for now, and take B = B 0 ẑ. Time evolution

12 operator is exp iht/ h, so amplitude for spin state m at time t is χ m (t) = m ψ(t) = m e iht/ h ψ = e iµ 0mB 0 t/ h χ m (5) Note if, at t = 0, ψ is in an eigenstate of Ŝz, it stays that way, as must be since [Ŝz, H] = 0! (Anything which commutes with the Hamiltonian is conserved...) On other hand, if it is in another state, spin precesses in general. Example: start spin off in eigenstate of Ŝx: or, h 0 0 χ χ }{{}}{{} = h = h χ χ (5) S x ψ (53) χ (54) χ In other words, χ = χ for this state at t = 0. Using normalization and (49) we have ψ = ; ψ(t) = e iµ 0B 0 t/ h e iµ 0B 0 t/ h (55) Note ψ(t) generally not eigenstate of Ŝx anymore. However at t = T = π h/(µ 0 B), ψ(t ) = (56) again an eigenstate of Ŝx with eigenvalue h/! ( spin points along ˆx ).

13 Quick calculation shows (check) that at T/4, ψ(t/4) = i + i (57) which is an eigenstate of Ŝy now with eigenvalue h/ ( spin points along ŷ axis). At T/, ψ(t/) = i (58) which is an eigenstate of Ŝx again but with eigenvalue h/ ( spin points along ˆx ). And so on... see that spin is rotating in xy plane with period T! Alternative viewpoint on same problem: Q: System is in state ψ(t). What is probability you will obtain a result h/ when you measure Ŝx? A: E state of Ŝx with e value h/ is φ = as we saw. is (59) Therefore probability of obtaining h/ when measuring Ŝx P = φ ψ(t) (60) = [ ] e 0B 0 t/ h e iµ 0B 0 t/ h = cos πt/t (6) consistent with previous picture of a rotating spin! 3

14 Or, calculate expectation value of Ŝx in ψ(t) : ψ(t) Ŝx ψ(t) = [e iµ 0B 0 t/ h e iµ 0B 0 t/ h ] e iµ 0B 0 t/ h e iµ 0B 0 t/ h = cos πt/t (6).6 Magnetic resonance Put electron in large static applied field B 0 ẑ, assume spins initially in eigenstate of Ŝ z corresponding to h/ ( = spin down ). (Note this is equilibrium config. since µ likes to align with field, but µ S for an electron!) Apply oscillating field B cos ωt ˆx to it as shown. Hamiltonian is then H = µ 0 σ B (63) 0 0 = µ 0 B 0 + µ 0 0 B cos ωt (64) 0 4

15 If we once again put ψ = χ χ Then Schrödinger eqn. has two components:, (65) i h χ = µ 0 B 0 χ + µ 0 B cos ωtχ (66) t i h χ = µ 0 B 0 χ + µ 0 B cos ωtχ (67) t Exact soln. possible but not enlightening. Suppose B B 0, assume initial conditions χ = 0, χ = at t = 0. So for small t, can assume χ is small, and S.-eqn is approx. i h χ µ 0 Bχ, so t χ e iµ 0B 0 t/ h and substituting into (64) we get i h χ t µ 0B 0 χ + µ 0 B cos ωt e iµ 0B 0 t/ h Rewrite as i h ( χ e iµ 0B 0 t/ h ) = µ 0 B cos ωte iµ 0B 0 t/ h t (68) (69) (70) 5

16 Integrate, use cos x = (e ix +e ix )/, and define ω 0 = µ 0 B 0 / h: i hχ e iµ 0B 0 t/ h = µ 0 B t 0 dt cos ωt e iµ 0B 0 t / h = µ 0B e i(ω+ω 0)t i(ω + ω 0 ) + ei(ω 0 ω)t i(ω 0 ω) (7) From (69) we find by ) noting that for ω ω 0 nd term is much bigger, then ) taking complex magnitude χ µ 0 B sin (ω ω 0 )t/ (7) h (ω ω 0 ) Recap: At t = 0, spin is down. At time t, χ is probability that spin is up. Clearly prob. is max. when ω = ω 0, resonant frequency. Note resonance occurs when hω = µ 0 B 0, i.e. difference between two magnetic field split levels. Consistent with picture that photon of energy just equal to splitting is causing spin flip. (Not quite kosher since we are treating B- field classically!) 6