EE201/MSE207 Lecture 10 Angular momentum

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1 EE20/MSE207 Lecture 0 Angular momentum (A lot of mathematics; I will try to explain as simple as possile) Classical mechanics r L = r p (angular momentum) Quantum mechanics The same, ut p x = iħ x p = m v L x = y p z z p y L y = z p x x p z L z = x p y y p x Therefore L x = iħ y z z y z x y p y = iħ y L y = iħ z x x z p z = iħ z L z = iħ x y y x To find quantized values of angular momentum, we can solve 3D differential equations to find eigenvalues and eigenstates. However, there is a simpler way, similar to the trick for oscillator.

2 Commutation relations Simple to show: (check yourself) L x, L y = iħ L z L y, L z = iħ L x L z, L x = iħ L y Most of the theory of angular momentum (and spin) can e derived from these commutation relations. Will not write hats elow for revity First oservation: L x, L y, L z are all incompatile (do not commute), so if L x has a definite value, then L y and L z do not have definite values (except when all are zero, σ Lx σ Ly ħ L 2 z ). However, L 2 = L 2 x + L 2 y + L 2 z commutes with each component. Simple to show: L 2, L x = 0 L 2, L y = 0 L 2, L = 0 L 2, L z = 0 Therefore, L 2 (square of total angular momentum) and one component can have definite values at the same time (usually z-component is chosen)

3 Eigenvalues of L 2 and L z We want to find eigenvalues of operators L 2 and L z (they are the quantized values for the angular momentum). Need to solve: L 2 f = λ f L z f = μ f (f is not very important) Trick of ladder operators Let us introduce operators Will show (next goal): L ± = L x ± il y λ = ħ 2 l l + μ = ħ m l = 0, 2,, 3 2, m = l, l +, l These operators are not Hermitian (unphysical), while L x, L y, L z are of course Hermitian [L z, L ± ] = L z, L x ± i L x, L y = iħl y ± i iħ L x = ±ħl ± Thus L z, L ± = ±ħ L ± [L 2, L ± ] = 0

4 Lemma If f is an eigenvector of oth L 2 (with eigenvalue λ) and L z (with eigenvalue μ), then L ± f are also eigenvectors, with the same λ, ut μ μ ± ħ. Proof L 2 L ± f = L ± L 2 f = λl ± f L z, L ± = ±ħ L ± [L 2, L ± ] = 0 L z L ± f = L z, L ± f + L ± L z f = ±ħl ± f + μl ± f = μ ± ħ L ± f Now the same logic as for oscillator: apply L + many times, the process should stop at some point (since L z 2 cannot exceed L 2 ), therefore L + f top = 0. L z f top = ħlf top (so far l is aritrary) f top (L + ) 2 f L + f f L 2 f top = λf top Let us show λ = ħ 2 l(l + ) (eigenvalues are related!) Use formula L 2 = L L + + L z 2 + ħl z (simple to derive explicitly), then λ f top = ħl 2 f top + ħ 2 lf top = ħ 2 l l + f top

5 Similarly, applying L many times, we should eventually get L f ottom = 0. Then similarly L z f ottom = ħl f ottom (with aritrary l ), and can show the relation etween eigenvalues: λ = ħ 2 l (l ) Comparing these two equations, we otain l = l. But they are related y integer numer (integer numer of applying L ± ). Therefore, l is integer or half-integer. Thus, result: Eigenvalues of L 2 (i.e., possile measured values) are λ = ħ 2 l(l + ), where l = 0,,, 3, 2,, 2 2 eigenvalues of L z (i.e., possile measured values) are μ = ħ m, where m = l, l +, l, l If we denote the corresponding eigenvectors y f l m, then L 2 f l m = ħ 2 l l + f l m L z f l m = ħm f l m Interesting that L z 2 ħ 2 m 2 ħ 2 l 2 is always smaller than L 2 ħ 2 l l +. So, z-component cannot e parallel to L. (If it would, then L x = L y = 0, impossile.)

6 Eigenvectors of L 2 and L z We found eigenvalues of L 2 and L z : What are the eigenvectors f l m? L 2 f l m = ħ 2 l l + f l m L z f l m = ħm f l m This is not simple. Discuss only the result (not the derivation). If the eigenvector f l m is represented y ψ-function in real space, ψ(r, θ, φ), then l is integer (not half-integer) and is the eigenfunction, L 2 ψ = ħ 2 l l + ψ L z ψ = ħm ψ ψ r, θ, φ = R r Y l m (θ, φ) Thus, azimuthal quantum numer l means that square of the total angular momentum is L 2 = ħ 2 l(l + ) (s, p, d, f oritals). Magnetic quantum numer m means that L z = ħm (degeneracy 2l + ). any However, half-integer l values really exist: spin (consider next). spherical harmonics

7 Spin Existence of half-integer values of l is experimental fact (Stern-Gerlach, 922: neutral silver atoms travel though inhomogeneous magnets field, split into two eams) Cannot explain as ψ(r, θ, φ) something else. It is an angular momentum, which does not correspond to motion in (x, y, z) intrinsic angular momentum (spin). The theory of spin is essentially the same as theory of angular momentum, just no ψ(r, θ, φ), only operator formalism. S x, S y S y, S z S z, S x Total spin: = iħ S z = iħ S x = iħ S y Useful relation: s = 0, 2,, 3 2, Denote eigenstates of S 2 and S z as s, m This is jargon, actually More correctly, z-component: m = s, s +, s total spin z-component S 2 = ħ s(s + ), S z = ħm S 2 s, m = ħ 2 s s + s, m S z s, m = ħm s, m S ± s, m = ħ s s + m(m ± ) s, m ± s = 0: pi-mesons, Higgs oson, s = 2: electron, proton, neutron, quarks, s = : photon, W, Z, s = 3 2: deltas, omega, s = 2: graviton, etc.

8 S 2 = ħ 2 2 The case s = /2 (spin one-half) (simplest and most important) 3 2 = 3 4 ħ2 S z : mħ = ± ħ 2 What is the wavefunction? It cannot depend on x, y, z. Let us introduce something that satisfies formalism (commutation relations). Spinor χ = a = a Spin operators S z = ħ 2 S x = ħ 2 S y = ħ 2 spin-up χ 0 0 = ħ 2 σ z 0 0 = ħ 2 σ x 0 i i 0 = ħ 2 σ y σ x, σ y, σ z are called Pauli matrices spin-down χ (χ is chosen to resemle ψ) Normalization: a = These operators satisfy desired commutation relations: S x, S y = iħ S z, S y, S z = iħ S x, S z, S x = iħ S y Also, as desired S 2 = S 2 x + S 2 y + S 2 z = ħ2 4 (σ x 2 + σ 2 y + σ 2 z ) = ħ2 0 = 3 4 ħ2

9 Eigenvectors and eigenvalues We know that S z has eigenvalues ±ħ/2; what are the eigenvectors? S z a ± ħ 2 = λ a S z = ħ 2 χ = 0 Check: S z 0 = ħ (spin-up) 0 = ħ 2 0 Similarly S z = ħ 2 χ = 0 (spin-down) However, z-direction is not different from other directions, therefore the same eigenvalues for S x and S y. S x a ± ħ 2 = λ a Check: S x / 2 / 2 = ħ 2 S x = ħ 2 χ = / 2 / 2 = 2 S x = ħ 2 χ = / 2 / 2 = / 2 / 2 = ħ 2 / 2 / 2 (similar for ħ/2)

10 not included into this course Similarly find eigenvectors for S y S y a = λ a S y = ħ 2 χ = / 2 i/ 2 = 2 i ± ħ 2 S y = ħ 2 χ = / 2 i/ 2 = 2 i Actually, S x and S y correspond to new ases of eigenstates, and everything can also e considered in these new ases a (z) a i 2 a + i 2 (y) a + 2 a 2 (x) (we will consider only the standard z-asis)

11 χ = a Measurement of spin (in standard z-representation) Measure S z : get S z = get S z = ħ 2 with proaility P z+ = a 2, after that χ = 0 ħ 2 with proaility P z = 2, after that χ = 0 Measure S x : get S x = S x = P x+ = 2 ħ 2 with P x = 2 ħ 2 with proaility a 2 2 = a + 2 after that χ = a 2 = 2 a 2 then χ = 2 Measure S y : get S y = ħ 2 with proaility P y+ = i a = a i 2 then χ = 2 2 i S y = ħ 2 with P y = +i a = a + i 2 then χ = 2 2 i

12 Experimental measurement of spin (Stern-Gerlach experiment, 922) (not included into this course) N inhomogeneous magnetic field produces force onto a magnetic moment neutral atoms S H = γb S angular momentum (or spin) magnetic field gyromagnetic ratio If B is inhomogeneous (not constant), then F = H = γ (B S) So, force depends on S. If B x, y, z = B 0 + αz k αx i then F = γα S z k S x i F z = γαs z spin-up is deflected down (γ < 0), spin-down is deflected up (for magnetic field B x not important + B y + B z x y not important (oscillates ecause of Larmor precession aout z-axis, so zero on average) z = 0), (particle should e neutral ecause otherwise charge will circle in magnetic field)