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So as an exercise in using this notation let’s look at ( x ) g ( y ) ( x) ( y ) x g g x y y The indices indicate very specific matrix or vector components/elements. These are not matrices themselves, but just numbers, which we can reorder as we wish. We still have to respect the summations over repeated indices! g g ?? x y i.e. ( x) ( y ) x y x y And remember we just showed (g) = g x y y x All dot products are INVARIANT under Lorentz transformations. even for ROTATIONS 0 0 as an example, consider rotations about the z-axis 1 2 1 2 ( x) ( y) x y ( x cos x sin )( y cos y sin ) ( x cos x sin )( y cos y sin ) x y 2 1 2 1 3 3 x1 y1 cos2 x1 y 2 sin cos x 2 y1 sin cos x 2 y 2 sin 2 3 3 x y 2 2 2 x y 1 2 2 1 1 1 2 x y cos x y sin cos x y sin cos x y sin 0 0 0 0 1 1 2 2 3 3 x y x x x x x y x y The relativistic transformations: E px ( p x ) c E E E ( E cp x ) ( p x ) c c suggest a 4-vector E that also p ( ; p) ( p) p transforms by c so p p should be an invariant! 2 E 2 2 p c In the particle’s rest frame: E = ?mc2 pp = ?m2c2 px = ?0 In the “lab” frame: E px ( p x ) (0 mc) = mv c E E ( p x ) = E = mc c c c so 2 2 2 2 2 2 2 ( p) ( p) m c m c (1 )m c 2 2 2 m c 2 2 2 Limitations of Schrödinger’s Equation 1-particle equation 2 2 i ( x , t ) ( x, t ) V ( x ) ( x, t ) 2 t 2m x1 2-particle i ( x1 , x2 , t ) ( x , x , t ) ( x1 , x2 , t ) 1 2 2 2 equation: t 2m2 x1 2m2 x2 2 2 2 2 V ( x1 ) V ( x2 ) V ( x1 , x2 ) ( x1 , x2 , t ) mutual interaction np+e++e But in many high energy reactions the number of particles is not conserved! n+p n+p+3 e+ p e+ p + 6 + 3 Sturm-Liouville Equations a class of differential equations that include: Legendre's equation i.e. a class of differential eq's to which Schrodinger's equations all belong! the associated Legendre equation Bessel's equation the quantum mechanical harmonic oscillator whose solutions satisfy: (*n m ) n* mdr3 0 for different eigenfunctions, If we adopt the following as a definition of the "inner product" * dr3 * v compare this directly to the vector "dot product" n un n then notice we have automatically * n Recall: any linear combination of simple solutions to a differential equation is also a solution, and, from previous slide: m n m n mn eigenvalues are REAL and different eigenfunctions are "orthogonal" Thus the set of all possible eigenfunctions (basic solutions) provide an "orthonormal" basis set and any general solution to the differential equation becomes expressible as an n where an n n * dr3 n any general solution will be a function in the "space" of all possible solutions (the solution set) sometimes called a Hilbert Space (as opposed to the 3-dimensional space of geometric points. What does it mean to have a matrix representation of an operator? of Schrödinger’s equation? H n En n H n En n where n represents all distinguishing quantum numbers (e.g. n, m, ℓ, s, …) m H n En m n mn En Hmn m H Em m H n En n since † H H H= E1 0 0 E2 0 0 0 0 :. 0 0 0 0 E3 0 0 E4 0 0 0 0 ... ... ... ... with the “basis set”: 1 0 0 0 1 0 0 0 1 ... : : : · , · , · , This is not general at all (different electrons, different atoms require different matrices) Awkward because it provides no finite-dimensional representation That’s why its desirable to abstract the formalism 32 Hydrogen Wave Functions 1 z zr / a 100 e a 32 1 Z Zr Zr / 2 a 200 1 e 2a 2a 52 1 Z Zr / 2 a i 211 re sin e 8 a 52 1 Z Zr / 2 a 210 cos re 2a 0 1 0 0 0 :: 0 0 1 0 0 :: 52 1 Z Zr / 2 a 211 sin ei re 8 a 1 Z Zr Z 2 r 2 Zr / 3a 300 27 18 2 2 e a a 81 3 a 32 1 0 0 0 0 :: 0 0 0 1 0 :: 0 0 0 0 1 :: 0 0 0 0 But the sub-space of angular momentum (described by just a subset of the quantum numbers) doesn’t suffer this complication. Angular Momentum |lmsms…> l = 0, 1, 2, 3, ... Lz|lm> = mh|lm> for m = l, l+1, … l1, l L2|lm> = l(l+1)h2|lm> Sz|lm> = msh|sms> for ms = s, s+1, … s1, s S2|lm> = s(s+1)h2|sms> Of course |nℓm> is dimensional again can measure all the spatial(x,y,z) components (and thus L itself) of L r mv not even possible in principal ! L r i , , ir x y x So, for example Lz i x y i x y azimuthal angle in polar coordinates nlm … Angular Momentum l Measuring Lx alters Ly (the operators change the quantum states). The best you can hope to do is measure: l = 0, 1, 2, 3, ... L2lm(,)R(r)= l(l+1)ħ2lm(,)R(r) Lz lm(,)R(r) = mħ lm(,)R(r) for m = l, l+1, … l1, l States ARE simultaneously eigenfunctions of BOTH of THESE operators! We can UNAMBIGUOULSY label states with BOTH quantum numbers ℓ=2 mℓ = 2, 1, 0, 1, 2 2 ℓ=1 mℓ = 1, 0, 1 1 1 0 0 L2 = 1(2) = 2 |L| = 2 = 1.4142 L2 = 2(3) = 6 |L| = 6 = 2.4495 Note the always odd number of possible orientations: A “degeneracy” in otherwise identical states! Spectra of the alkali metals (here Sodium) all show lots of doublets 1924: Pauli suggested electrons posses some new, previously un-recognized & non-classical 2-valued property Perhaps our working definition of angular momentum was too literal …too classical perhaps the operator relations Lx Ly Ly Lx iLz Ly Lz Lz Ly iLx Lz Lx Lx Lz iLy may be the more fundamental definition Such “Commutation Rules” are recognized by mathematicians as the “defining algebra” of a non-abelian (non-commuting) group [ Group Theory; Matrix Theory ] Reserving L to represent orbital angular momentum, introducing the more generic operator J to represent any or all angular momentum J x J y J y J x iJ z J y J z J z J y iJ x study this as an algebraic group J z J x J x J z iJ y Uhlenbeck & Goudsmit find actually J=0, ½, 1, 3/2, 2, … are all allowed! quarks leptons spin 12 p, : n, e, , , e , , , u, d, c, s, t, b the fundamental constituents of all matter! ms = ± 1 2 spin “up” spin “down” s = 32 ħ = 0.866 ħ sz = |nlm>| 1 2 1 2 >= 1 2 ħ ( )nlm 1 0 “spinor” the most general state is a linear expansion in this 2-dimensional basis set () () () = 1 + 0 0 1 with 2 + 2 = 1 SPIN ORBITAL ANGULAR MOMENTUM fundamental property of an individual component relative motion between objects Earth: orbital angular momentum: rmv plus “spin” angular momentum: I in fact ALSO “spin” angular momentum: Isunsun but particle spin especially that of truly fundamental particles of no determinable size (electrons, quarks) or even mass (neutrinos, photons) must be an “intrinsic” property of the particle itself Total Angular Momentum nlm sm j… l = 0, 1, 2, 3, ... l s Lz|lm> = mħ|lm> for m = l, l+1, … l1, l L2|lm> = l(l+1)ħ2|lm> Sz|lm> = msħ|sms> for ms = s, s+1, … s1, s S2|lm> = s(s+1)ħ2|sms> In any coupling between L and S it is the TOTAL J = L + s that is conserved. Example J/ particle: 2 (spin-1/2) quarks bound in a ground (orbital angular momentum=0) state Example Either spin-1/2 electron in an l=2 orbital. Total J ? 3/2 or 5/2 possible BOSONS spin 1 FERMIONS spin ½ p, n, e, Nuclei (combinations of p,n) can have J = 1/2, 1, 3/2, 2, 5/2, … BOSONS FERMIONS mesons spin 0 “psuedo-scalar” spin ½ 0 0 quarks and leptons e, , , u, d, c, s, t, b, Baryon “octet” p, n, mediators spin 1 Force spin 3/2 “vector”bosons: ,W,Z Baryon “decupltet” D, S, X, W , , , K ,K ,K “vector” mesons r, , , J/, spin 2 : spin 5/2 : Combining any pair of individual states |j1m1> and |j2m2> forms the final “product state” |j1m1>|j2m2> What final state angular momenta are possible? What is the probability of any single one of them? Involves “measuring” or calculating OVERLAPS (ADMIXTURE contributions) or forming the DECOMPOSITION into a new basis set of eigenvectors. S j1j2 |j1m1>|j2m2> = z j j1 j2; m m1 m2 | jm> j=| j1j2 | Clebsch-Gordon coefficients Matrix Representation for a selected j J2|jm> = j(j+1)h2| j m > Jz|jm> = m h| j m > for m = j, j+1, … j1, j J±|jm> = j(j +1)m(m±1) h | j, m1 > The raising/lowering operators through which we identify the 2j+1 degenerate energy states sharing the same j. adding J+ = Jx + iJy J = Jx iJy subtracting 2Jx = J+ + J Jx = (J+ + J )/2 2iJy = J+ J Jy = i(J J+)/2 The most common representation of angular momentum diagonalizes the Jz operator: <jn| Jz |jm> = mmn 1 0 0 = 0 0 0 0 0 -1 (j=1) Jz (j=2) Jz = 2 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 -2 J±|jm> = j(j +1)m(m±1) h | j, m1 > < 1 0 | J | 1 1 > = 2|10> < 1 -1 | J | 1 0 > = 2 | 1 -1 > 0 J = 0 0 2 0 0 0 2 0 J = 0 0 0 J | 1 -1 > = 0 < 1 0 | J | 1 -1 > = 2|10> < 1 1 | J | 1 0 > = 2|11> J | 1 1 > = 0 2 0 0 2 0 0 For J=1 states Jx a matrix representation of the angular momentum operators 0 1 2 2 0 0 1 J y i 2 2 0 Jz 2 0 2 i 2 0 i 2 1 0 0 0 1 0 0 0 1 0 2 0 0 i 0 0 1 2 0 1 2 0 1 2 1 2 0 0 0 2 i 2 0 i 2 0 i 2 i 2 0 0 Which you can show conform to the COMMUTATOR relationship you demonstrated in quantum mechanics for the differential operators of angular momentum [Jx, Jy] = iJz i 0 2 Jx J y Jy Jx = 0 0 i 0 2 i 2 i 2 0 0 i i 2 2 0 0 0 i 2 1 0 0 0 i 0 0 0 i 0 0 1 2 = iJz J J 1 / 2 z 0 1/ 2 1/ 2 0 J J 1z 1 0 0 0 1 0 0 0 1 J J 3 / 2 z 0 0 0 3/ 2 1/ 2 0 0 0 0 0 1/ 2 0 0 0 3/ 2 0 J J 2 z 2 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 2 0 R(1,2,3) = 1 0 0 0 cos1 sin 1 0 sin cos 1 1 z z′ 1 y′ 1 y x =x′ R(1,2,3) = z′ z′′ 2 1 z cos 2 0 sin 2 0 0 0 sin 2 1 1 0 0 cos1 sin 1 0 cos 2 0 sin 1 cos1 2 y′ =y′′ 1 y x =x′ 2 x′′ R(1,2,3) = z z′ z′′′ = z′′ 2 1 cos 3 sin 3 0 cos 2 sin cos 0 0 3 3 0 0 1 sin 2 0 0 0 sin 2 1 1 0 0 cos1 sin 1 0 cos 2 0 sin 1 cos1 3 y′′′ 3 2 y′ =y′′ 1 y x =x′ 2 x′′ 3 x′′′ R(1,2,3) = about about x-axis about y′-axis z′′-axis 1st cos 3 sin 3 0 cos 2 sin cos 0 0 3 3 0 0 1 sin 2 0 0 0 sin 2 1 1 0 0 cos1 sin 1 0 cos 2 0 sin 1 cos1 2nd 3rd These operators DO NOT COMMUTE! but as nn Recall: the “generators” of rotations are angular momentum operators and they don’t commute! Infinitesimal rotations DO commute!! 1 3 3 1 0 2 2 0 1 0 2 1 3 0 1 3 1 0 3 1 0 3 1 0 1 0 0 1 0 2 2 0 1 3 0 1 1 1 0 0 3 0 0 1 2 1 0 2 0 2 1 0 0 1