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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!. And remember
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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! And remember we just showed (g) = g i.e. All dot products are INVARIANT under Lorentz transformations.
as an example, consider rotations about the z-axis even for ROTATIONS
The relativistic transformations: suggest a 4-vector that also transforms by so should be an invariant!
In the particle’s rest frame: px = ? 0 mc2 pp = ? m2c2 E = ? In the “lab” frame: = -mv E c = = mc so ?
Limitations of Schrödinger’s Equation 1-particle equation 2-particle equation: mutual interaction np+e++e n+p n+p+3 e-+ p e-+ p + 6 +3g But in many high energy reactions the number of particles is not conserved!
Sturm-Liouville Equations • a class of differential equations that include: • Legendre's equation • the associated Legendre equation • Bessel's equation • the quantum mechanical harmonic oscillator i.e. a class of differential eq's to which Schrodinger's equations all belong! 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" then notice we have automatically
Recall: any linear combination of simple solutions to a differential equation is also a solution, and, from previous slide: eigenvalues are REAL and different eigenfunctions are "orthogonal" 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 solutionwill be a function in the "space" of all possible solutions (the solution set)sometimes called a Hilbert Space (as opposed to the 3-dimensional spaceof 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, …) Hmn † since
E1 0 0 0 0 . . . H = 0 E2 0 0 0 . . . 0 0 E3 0 0 . . . 0 0 0 E4 0 . . . : . 1 0 0 : · 0 0 1 : · 0 1 0 : · 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 1 0 0 0 0 0 1 0 0 0 : : 0 0 1 0 0 : : 0 0 0 1 0 : : 0 0 0 0 1 : : : : 0 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, … l-1, l L2|lm> = l(l+1)h2|lm> Sz|lm> = msh|sms> for ms = -s, -s+1, … s-1, s S2|lm> = s(s+1)h2|sms> Of course |nℓm> is dimensional again
Classically can measure all the spatial (x,y,z) components (and thus L itself) of Quantum Mechanically not even possible in principal ! azimuthal angle in polar coordinates So, for example
Angular Momentum nlml… 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, … l-1, 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 ℓ = 1 mℓ= -1, 0, 1 2 1 0 1 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:Paulisuggested 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 Such “Commutation Rules” are recognized by mathematicians as the “defining algebra” of a non-abelian (non-commuting) group may be the more fundamental definition [ 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 & Goudsmitfind actually J=0, ½, 1, 3/2, 2, … are all allowed!
quarks 3 2 1 2 1 2 1 2 1 2 1 2 leptons spin : p, n,e, , , e , , ,u, d, c, s, t, b the fundamental constituents of all matter! spin “up” spin “down” s = ħ = 0.866 ħ ms = ± sz = ħ ( ) 1 0 | n l m > | > = nlm “spinor” ( ) ( ) ( ) the most general state is a linear expansion in this 2-dimensional basis set 1 0 0 1 = + with a 2 + b 2 = 1
ORBITAL ANGULARMOMENTUM SPIN 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 l = 0, 1, 2, 3, ... Lz|lm> =mħ|lm> for m = -l, -l+1, … l-1, l L2|lm> = l(l+1)ħ2|lm> Sz|lm> = msħ|sms> for ms = -s, -s+1, … s-1, s S2|lm> = s(s+1)ħ2|sms> nlmlsmsj… 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, m 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 : : quarks and leptons e, m, t, u, d, c, s, t, b, n “psuedo-scalar” mesons p+, p-, p0, K+,K-,K0 Baryon “octet” p, n, L Force mediators “vector”bosons: g,W,Z Baryon “decupltet” D, S, X, W “vector” mesons r, w, f, J/y,
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. j1+j2 S |j1m1>|j2m2>= zj j1 j2;mm1m2| jm> j=| j1-j2 | Clebsch-Gordon coefficients
Matrix Representation for a selected j J2|jm> =j(j+1)h2|j m> Jz|jm> = mh|j m> for m = -j, -j+1, … j-1, j J±|jm> = j(j +1)-m(m±1)h |j, m1 > The raising/lowering operators through which we identify the 2j+1degenerate energy states sharing the same j. J+ = Jx + iJy J- = Jx- iJy subtracting adding 2Jx = J+ + J- Jx = (J+ + J-)/2 Jy = i(J-- J+)/2 2iJy = J+- J-
The most common representation of angular momentum diagonalizes the Jzoperator: <jn| Jz|jm> = lmmn 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 1 0 0 0 0 0 0 0 -1 (j=1) (j=2) Jz = Jz =
J±|jm> = j(j +1)-m(m±1)h |j, m1 > J-|1 1> = <1 0| |1 0> 0 0 0 0 0 0 0 J-|1 0> = J- = <1 -1| |1 -1> J-|1 -1> = 0 J+|1 -1> = |1 0> <1 0| 0 0 0 0 0 0 0 J+|1 0> = J+ = |1 1> <1 1| J+|1 1> = 0
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 [Jx, Jy] = iJz JxJy-JyJx = = iJz
R(1,2,3) = z z′ 1 y′ 1 y x =x′
R(1,2,3) = z z′ z′′ 1 2 2 y′ =y′′ 1 y x =x′ 2 x′′
R(1,2,3) = z z′ z′′ 1 z′′′ = 2 3 y′′′ 3 2 y′ =y′′ 1 y 3 x =x′ 2 x′′ x′′′
R(1,2,3) = about x-axis about y′-axis about z′′-axis 1st 2nd 3rd These operators DO NOT COMMUTE! Recall: the “generators” of rotations are angular momentum operators and they don’t commute! but as nn Infinitesimal rotations DO commute!!