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Spin the Evidence of Intrinsic Angular Momentum Or Spin and Its Spin The evidence of intrinsic angular momentum or spin and its associated magnetic moment came through experiments by Stern and Gerlach and works of Goudsmit and Uhlenbeck. The spin is called intrinsic since, unlike orbital angular momentum which is extrinsic, it is carried by point particle in addition to its orbital angular momentum and has nothing to do with motion in space. The magnetic moment ~µ of silver atom was measured in 1922 experiment by Stern and Gerlach and its projection µz in the direction of magnetic fieldz ^ B (through which the silver beam was passed) was found to be just −|~µj and +j~µj instead of continuously varying between these two as limits. Classically, magnetic moment is proportional to the angular momentum (12) and, assuming this proportionality to survive in quantum mechanics, quan- tization of magnetic moment leads to quantization of corresponding angular momentum S, which we called spin, q q µ = g S ) µ = g ~ m (57) z 2m z z 2m s as is the case with orbital angular momentum, where the ratio q=2m is called Bohr mag- neton, µb and g is known as Lande-g factor or gyromagnetic factor. The g-factor is 1 for orbital angular momentum and hence corresponding magnetic moment is l µb (where l is integer orbital angular momentum quantum number). A surprising feature here is that spin magnetic moment is also µb instead of µb=2 as one would naively expect. So it turns out g = 2 for spin { an effect that can only be understood if linearization of Schr¨odinger equation is attempted i.e. make it consistent with special theory of relativity. The algebra for spin angular momentum is considered very similar to the orbital angular momentum algebra, 2 2 S js; msi = s(s + 1)~ js; msi Sz js; msi = ms~ js; msi; −s ≤ ms ≤ +s p S± js; msi = (s ∓ ms)(s ± ms + 1) ~ js; ms ± 1i; (58) where S± = Sx ± iSy. Any component of spin angular momentum S has 2s + 1 eigenvalues and therefore we expect ms to have 2s + 1 disctinct values between −s ≤ ms ≤ s. However, unlike orbital angular momentum quantum number l, the spin angular momentum quantum number can take both integer and half-integer values, s = 0; 1=2; 1; 3=2; 2;:::. In fact, the experimental result of Stern and Gerlach showing only 2 distinct µz can explained if s = 1=2 is considered bacause ms = ±1=2. Every elementary particle has specific value of spin s: π has spin 0, electron spin is 1/2, photons have spin 1, deltas' spins are 3/2 and so on. Matrix representation for spin S can be derived using the equations in (58), the same way as L, and they are, 3 1 0 1 1 0 S2 = 2 ;S = ; 4 ~ 0 1 z 2 ~ 0 −1 1 0 1 1 0 −i S = ;S = ; x 2 ~ 1 0 y 2 ~ i 0 0 1 0 0 S = ;S = : (59) + ~ 0 0 − ~ 0 1 We may write this representation as 1 S = σ (60) 2 ~ 1 where, 0 1 0 −i 1 0 σ = ; σ = ; σ = (61) x 1 0 y i 0 z 0 −1 are the Pauli spin matrices. The commutation and anti-commutation relations obeyed by Pauli matrices are, [σi; σj] = 2i ijk σk fσi; σjg = 2 δij: (62) A few other properties of Pauli spin matrices are, 2 2 2 y σx = σy = σz = 1; σx σy = i σz etc:; σj = σj Tr (σj) = 0; det (σj) = −1: (63) These relations are peculiar to spin 1/2 representations and do not hold for l = 1 matrices. The eigenstates of S for spin 1/2 particles will be represented by a two-component column matrix, called spinor: 1 9 spin − up : χ = > + 0 => a χ = a χ + b χ = ; and jaj2 + jbj2 = 1: (64) 0 + − b spin − dn : χ− = > 1 ; As we have discussed before, there is nothing special about the z-component of angular momentum operator and just because the σz is diagonal the spin-states ±~=2 are easy to see, otherwise any other directions are just fine. To illustrate this, letus try to figure out the eigenvalues and eigenspinors of Sx. To find the eigenvalues, diagonalize the Sx, (x) (x) (x) Sx jχ i = λ jχ i ! (Sx − λ) jχ i = 0 −λ ~=2 2 2 ) = 0 ) λ = (~=2) ! λ = ±~=2 (65) ~=2 −λ which says the spin eigenvalues are the same as that for Sz as it should be. However, the space in which Sx is diagonal will obviously be different and they are obtained as, α 0 1 α α β α S = ~ = ±~ ) = ± ; (66) x β 2 1 0 β 2 β α β i.e. β = ±α. Hence the eigen spinors of Sx are, α 1 1 χ(x) = = p e − val : ~ + α 2 1 2 α 1 1 χ(x) = = p e − val : − ~: (67) − −α 2 −1 2 2.
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