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E 1 0 0 0 0 . . . H = 0 E 2 0 0 0 . . . PowerPoint Presentation
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E 1 0 0 0 0 . . . H = 0 E 2 0 0 0 . . .

E 1 0 0 0 0 . . . H = 0 E 2 0 0 0 . . .

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E 1 0 0 0 0 . . . H = 0 E 2 0 0 0 . . .

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  1. 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

  2. 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 : :

  3. But the sub-space of angular momentum (described by just a subset of the quantum numbesrs) doesn’t suffer this complication. Angular Momentum |nlmsms…> 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

  4. 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

  5. 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, ... L2lm(,)R(r)= l(l+1)ħ2lm(,)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

  6. 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!

  7. 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

  8. 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!

  9. 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

  10. 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: Isun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

  11. 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

  12. 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, …

  13. 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, 

  14. 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

  15. 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, m1 > 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-

  16. The most common representation of angular momentum diagonalizes the Jzoperator: <jn| Jz|jm> = lmmn 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 =

  17. J±|jm> = j(j +1)-m(m±1)h |j, m1 > 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

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

  19. 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

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

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

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

  23. 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 nn Infinitesimal rotations DO commute!!

  24. 1 3 0 1 0 -2 1 0 0 R(1,2,3) = -3 1 0 0 1 1 0 1 0 20 1 0 -11 0 0 1 1 3-2 1 0 0 = -3 1 0 0 1 1 20 1 0 -11 or 1 3 0 1 0 -2 = -3 1 0 0 1 1 2-11 0 0 1 1 3-2 = -31 1 2-11

  25. 1 3-2 R(1,2,3) = -31 1 2-11 1 R(1,2,3) = 2 3 + R(1,2,3) = If we imagine building up to full rotations by applying this repeatedly N times [R(1,2,3)]N ℓim ℓim N N

  26. R(1,2,3) ℓim N Which we can re-write in the form by slightly re-writing the “vector” components: Check THIS out: We have found a “new” representation of Lx ,Ly ,Lz!!

  27. Our alternate approach to motivating where the generator was an honest-to-goodness matrix! gave representing 3-dimensional rotations with the basis: A B C which we argue A B C satisifes all the same arithmetic as Lx, Ly, Lz

  28. Notice can try diagonalizing the zth matrix: Eigenvalues of -1, 0, 1 We should be able to diagionalize s3 by aSIMILARITY TRANSFORMATION! Us3U = †

  29. Us3U =

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