So as an exercise in using this notation let’s look at 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 And remember we just showed  i.e. All dot products are INVARIANT under Lorentz transformations.

even for ROTATIONS as an example, consider rotations about the z-axis

The relativistic transformations: suggest a 4-vector that also transforms by so should be an invariant!

EcEc In the particle’s rest frame: p x = ? E = ? p  p  = ? 0 mc 2 m2c2m2c2 In the “lab” frame: =   mv =  =  mc so

Limitations of Schrödinger’s Equation 1-particle equation 2-particle equation: mutual interaction But in many high energy reactions the number of particles is not conserved! n  p+e + + e n+p  n+p+3  e  + p  e  + p + 6  + 3 

i.e. a class of differential eq's to which Schrodinger's equations all belong! Sturm-Liouville Equations then notice we have automatically a class of differential equations that include:  Legendre's equation  the associated Legendre equation  Bessel's equation  the quantum mechanical harmonic oscillator whose solutions satisfy: for different eigenfunctions,  n 0 If we adopt the following as a definition of the "inner product" compare this directly to the vector "dot product"

eigenvalues are REAL and different eigenfunctions are "orthogonal" Recall: any linear combination of simple solutions to a differential equation is also a solution, and, from previous slide:  mn 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 where 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? where n represents all distinguishing quantum numbers (e.g. n, m, ℓ, s, …) H mn since †

E H =0 E E E :. 100 : ·100 : · 010 : · 010 : · 001 : ·001 : ·,,,... with the “basis set”: 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

Hydrogen Wave Functions : : : : : : : : : : : :

Angular Momentum |lmsm s …> l = 0, 1, 2, 3,... L z |lm> = mh|lm> for m =  l,  l+1, … l  1, l L 2 |lm> = l(l+1)h 2 |lm> S z |lm> = m s h|sm s > for m s =  s,  s+1, … s  1, s S 2 |lm> = s(s+1)h 2 |sm s > Of course |n ℓ m> is  dimensional again But the sub-space of angular momentum (described by just a subset of the quantum numbers) doesn’t suffer this complication.

can measure all the spatial ( x,y,z ) components (and thus L itself) of not even possible in principal ! So, for example azimuthal angle in polar coordinates

Angular Momentum nlm l … L z  lm ( ,  )R(r) = mħ  lm ( ,  )R(r) for m =  l,  l+1, … l  1, l L 2  lm ( ,  )R(r)= l(l+1)ħ 2  lm ( ,  )R(r) l = 0, 1, 2, 3,... Measuring L x alters L y (the operators change the quantum states). The best you can hope to do is measure: States ARE simultaneously eigenfunctions of BOTH of THESE operators! We can UNAMBIGUOULSY label states with BOTH quantum numbers

ℓ = 2 m ℓ =  2,  1, 0, 1, 2 L 2 = 2(3) = 6 |L| =  6 = ℓ = 1 m ℓ =  1, 0, 1 L 2 = 1(2) = 2 |L| =  2 = 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 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 study this as an algebraic group Uhlenbeck & Goudsmit find actually J=0, ½, 1, 3 / 2, 2, … are all allowed!

m s = ± 1212 spin “up” spin “down” s = ħ = ħ 3232 s z = ħ 1212 | n l m > | > =  nlm ( ) “spinor” the most general state is a linear expansion in this 2-dimensional basis set  1 0  0 1 ( ) =  +  ( ) with  2 +  2 = 1 spin : 1212 p, n, e, , , e, , , u, d, c, s, t, b leptons quarks the fundamental constituents of all matter!

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: I sun  sun 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 l sm s j… l = 0, 1, 2, 3,... L z |lm> = mħ|lm> for m =  l,  l+1, … l  1, l L 2 |lm> = l(l+1)ħ 2 |lm> S z |lm> = m s ħ|sm s > for m s =  s,  s+1, … s  1, s S 2 |lm> = s(s+1)ħ 2 |sm s > 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 spin-1/2 electron in an l =2 orbital. Total J ? Either 3/2 or 5/2 possible

BOSONSFERMIONS spin 1 spin ½  e,  p, n, Nuclei (combinations of p, n ) can have J = 1/2, 1, 3/2, 2, 5/2, …

BOSONSFERMIONS spin 0 spin ½ spin 1 spin 3 / 2 spin 2 spin 5 / 2 : : “psuedo-scalar” mesons             quarks and leptons e, ,  u, d, c, s, t, b, Force mediators “vector”bosons: ,W,Z “vector” mesons  J  Baryon “octet” p, n,  Baryon “decupltet” 

Combining any pair of individual states |j 1 m 1 > and |j 2 m 2 > forms the final “product state” |j 1 m 1 >|j 2 m 2 > What final state angular momenta are possible? What is the probability of any single one of them? Involves “measuring” or calculating OVERLAPS (ADMIXTURE contributions) |j 1 m 1 >|j 2 m 2 > =  j j 1 j 2 ; m m 1 m 2 | j m >  j=| j 1  j 2 | j1j2 j1j2 Clebsch-Gordon coefficients or forming the DECOMPOSITION into a new basis set of eigenvectors.

Matrix Representation for a selected j J 2 |jm> = j(j+1)h 2 | j m > J z |jm> = m h| j m > for m =  j,  j+1, … j  1, j J ± |jm> =  j(j +1)  m(m±1) h | j, m  1 > The raising/lowering operators through which we identify the 2j+1 degenerate energy states sharing the same j. J + = J x + iJ y J  = J x  iJ y 2J x = J + + J  J x = (J + + J  )/2 J y = i(J   J + )/2 adding 2iJ y = J +  J  subtracting

The most common representation of angular momentum diagonalizes the J z operator: = m  mn J z = (j=1) J z = (j=2)

J  | 1 1 > = J ± |jm> =  j(j +1)  m(m±1) h | j, m  1 > J  | 1 0 > = J  | 1 -1 > = J  | 1 0 > = J  | 1 -1 > = J  | 1 1 > = | 1 0 > | 1 -1 > 0 | 1 0 > | 1 1 > 0 J  = J  = < 1 0 | < 1 -1 | < 1 0 | < 1 1 |

For J=1 states a matrix representation of the angular momentum operators

Which you can show conform to the COMMUTATOR relationship you demonstrated in quantum mechanics for the differential operators of angular momentum [J x, J y ] = iJ z J x J y  J y J x = = iJ z

x y z z′ R (  1,  2,  3 ) = 11 y′ 11 =x′

x y z z′ R (  1,  2,  3 ) = 11 y′ 11 =x′ 22 22 22 x′′ z′′ =y′′

x y z z′ R (  1,  2,  3 ) = 11 y′ 11 =x′ 22 22 22 x′′ z′′ =y′′ 33 y′′′ z′′′ = x′′′ 33 33

R (  1,  2,  3 ) = These operators DO NOT COMMUTE! about x-axis about y ′ -axis about z ′′ -axis 1 st 2 nd 3 rd Recall: the “generators” of rotations are angular momentum operators and they don’t commute! but as  n  n Infinitesimal rotations DO commute!!