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A Review of Some INVARIANCE PRINCIPLES

A Review of Some INVARIANCE PRINCIPLES. Hermitian Operators: R † = R. can be diagonalized have real eigenvalues define the physical observables. Unitary Operators: R † = R -1. represent transformations of basis (vectors/functions)

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A Review of Some INVARIANCE PRINCIPLES

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  1. A Review of Some INVARIANCE PRINCIPLES Hermitian Operators:R† = R • can be diagonalized • have real eigenvalues • define the physical • observables Unitary Operators:R† = R-1 • represent transformations of basis(vectors/functions) • can transform operators & wavefunctions by similarity • transformations that leave the Lagrangian (and therefore the • equations of motion) invariant. (include rotations/translations) AA = RAR†and   =Rso that all computed matrix elements: < |A| > = < |A | > = < |R†(RAR†)R| > = < |A| > Furthermore: If the Hamiltonian is unchanged by the similarity transformation i.e. if RHR† =Hthen RH = HR  [H,R] = 0 and the eigenvalues are constants of the equations of motion, i.e., we have conserved quantities

  2. If Both Hermitian and Unitary: R†R = R-1R = R2 = 1 Rhas eigenvalue of1 this also implies: det(R)det(R) = det(I) = 1 det(R) = 1 simple rotations SO(n) SU(n) We’ve focussed a lot on the special det(R)=+1 cases: • These are “continuous symmetries”: seamlessly connected • to the identity matrix I by infinitesimal transformations • expressible by exponential sums of infinitesimal transformations • such operators can be used to define a CURRENT • that obeys the continuity equation • which we can identify with an associated CHARGE

  3. What about the Ry = -y class of transformations? Recall then compare that to cosq sinq 0 Rz = -sinq cosq 0 0 0 1 -cosqsinq 0 +sinq cosq 0 0 0 1 or cosqsinq 0 -sinq cosq0 0 0 -1 or y y -1 0 0 0 +1 0 0 0 +1 +1 0 0 0 -1 0 0 0 +1 x x z z

  4. ALL are equivalent to a reflection (axis inversion) plus a rotation PARITY TRANSFORMATIONS y y x' x x z z y y x x z z y' -1 0 0 0 -1 0 0 0 -1 y y z' x x' x The PARITY OPERATOR on 3-dim space vectors every point is carried through the origin to the diametrically opposite location z z y'

  5. Wave functions MAY or MAY NOT have a well-defined parity (even or odd functions…or NEITHER) P P = +1 P P = -1 but the more general P

  6. However for any spherically symmetric potential, the Hamiltonian: H(-r) = H(r)  H(r) → → [ P, H ] = 0 So the bound states of such a system have DEFINITE PARITY! That means, for example, all the wave functions of the hydrogen atom!

  7. Y(r)=c(r)ml(q,f )the angular part of the solutions are theSPHERICAL HARMONICS ml(q,f ) = Pml (cosq)eimf Pml (cosq) = (-1)msinmq[()mPl (cosq)] (2l + 1)( l- m)! 4p( l + m)! d d (cosq) 1 2l l! d d (cosq) Pl(cosq) = [()l(-sin2q)l ]

  8. The Spherical Harmonics Yℓ,m(,) ℓ = 3 ℓ = 0 ℓ = 1 ℓ = 2

  9. z then note r  -r means q  p - q f  p + f q p-q y f so: eimf eimfeimp x (eip)m=(-1)m and: Pml (cosq) Pml (cos(p-q)) = Pml (-cosq) (-sin2q)l = (-1-cos2q)l butd/d(cosq)-d/d(cosq)

  10. (2l + 1)( l- m)! 4p( l + m)! ml(q,f ) = Pml (cosq)eimf Pml (cosq) = (-1)m(1-cos2q)m[()mPl (cosq)] d d (cosq) 1 2l l! d d (cosq) Pl(cosq) = [()l(-sin2q)l ] So under the parity transformation: P:ml(q,f ) =ml(p-q,p+f)=(-1)l(-1)m(-1)m ml(q,f ) = (-1)l(-1)2m ml(q,f ) )=(-1)l ml(q,f ) An atomic state’s parity is determined by its angular momentum l=0 (s-state)  constant parity = +1 l=1 (p-state)  cos parity = -1 l=2 (d-state)  (3cos2-1) parity = +1 Spherical harmonics have (-1)l parity.

  11. When acting on a vector, the parity operator gives: P P scalars have positive parity! P ??? This confusion arises from the indefinite nature of cross products… their direction is DEFINED by a convention…the right-hand rule. and reflections (parity!) change right hands  left hands! We call such derived/defined vectors PSEUDO-VECTORS (or “axial” vectors) → as opposed to POLAR VECTORS, like v.

  12. QUANTITYPARITYComments r -r position p -p momentum L +L angular momentum + intrinsic “spin” E -E electric field E = -V/r B +B B = ×A  •B + •Bmagnetic dipole moment  •E- •Eelectric dipole moment  • p - •p longitudinal polarization  •(p1×p2) + •(p1×p2) transverse polarization polar vectors axial or pseudo- vectors pseudo-vector

  13. Potential problems with this concept…or assuming its invariance… Lot’s of physics uses Right Hand Rules and cross-products… angular momentum has POSITIVE parity the Lorentz Force Law vector vectors! Just like you can’t add vectors to scalars, cannot (should not try) to add vectors and pseudovectors at least for any theory that should respect parity invariance

  14. But notice a vector crossed with a pseudo-vector: P where is a vector! → Might B really be a pseudo-vector quantity after all? the “vector” potential is a true vector! and as we have argued in quantum mechanics/high energy theory is the more fundamental field!

  15. In our Lagrangians we identify the currents as vectors. Of particular interest: J • A of electromagnetic interactions → → → → J is a vector A is a vector field i.e., the photon is a vector particle with “odd” parity! So electromagnetic interactions conserve parity. By the same argument QCD Color (the STRONG force) interactions …its Lagrangian terms all involve inner products of all 4-vectors… conserve parity!

  16. A mono-chromatic electromagnetic wave is composed of n mono-energetic photons each with In addition to energy and momentum light (the photon) also carries angular momentum (its spin!)

  17. 1909Poyntingpredicts circularly polarized electromagnetic waves carry angular momentum proposes a test: if incident upon an absorber, the absorber should rotate 1936Richard Beth detects the angular momentum of light transferred to matter. photon momentum transferred to a “half-wave plate”, a macroscopic object hung on a fiber (but with a non-vanishing torsion constant). R. A. Beth Phys.Rev.50, 115 (1936).

  18. 1964P.J.Allen updates the experiment oil drop glass bead glass fiber 1.65 cm wire P. J. Allen Am.J.Phys.34, 1185 (1964).

  19. Microwave generator beamed up cavity sets rotor in motion E LZ =  If delivered by n photons, then with means E = nħ LZ = n jZ jZ = ħ Each photon has spin 1.

  20. Energy level diagram for Hydrogen 2S 2P 2D 2F N shell M shell L shell K shell n=4 n=3 n=2 n=1 ℓ =0 ℓ =1 ℓ=2 ℓ=3 Transitions occur between adjacent angular momentum states (i.e. ℓ ±1). The parity of a state must change in such an “electric dipole transition” ℓ=±1, S=0,J=0, ±1

  21. s, d, g, … even parity p, f, h, … odd parity The atomic transitions responsible for the observed atomic spectra connect these states of different parity. When the atomic state changes, PARITY must be conserved. The ELECTRIC DIPOLE transitions (characterized by ℓ ±1) emit PHOTONS whose parity must therefore be NEGATIVE.

  22. Like s which can be singly emitted (created) or absorbed (annihilated) s are created/destroyed singly in STRONG INTERACTIONS Their “INTINSIC PARITY” is also important. To understand the role of parity in interactions, consider a system (let’s start with a single pair) of initially free (non-interacting) particles which we can describe as a product state: large P parity is a multiplicative quantum number If the pair is bound with orbital angular momentum  P

  23. If P commutes with both the free particle Hamiltonian and the full Hamiltonian with interactions the (parity) quantum numbers are conserved throughout the interactions which is obviously true for electromagnetic interactions strong interactions since the Lagrangian terms incorporating these interactions are all invariant underP

  24. With sufficient energy, collisions of hadrons can produce additional particles (frequently pions) among the final states But while p-d  nnhas been observed p-d  nnp0hasneverbeen! • Studies show the process p-d  nn • often accompanied by an X-ray spectrum • reveals the calculable excited states of a “mesonic atom” • pion orbitals around a deuteron nucleus. • This suggests • the deuteron “captures” the pion • the strong interaction proceeds only following • cascade decays to a GROUND (=0) state.

  25. p-d  nn An=0 ground state means the p-d system has J= Stotal = 1 ? So the final nn state must also have J= 1 ? =1 s=0 =0 s=1 =1 s=1 =2 s=1 Note the final state is 2 Dirac FERMIONS overall wave function must be ANTISYMMETRIC to particle exchange Space part described by the spherical harmonics (interchanging the neutrons would reverse their relative positions, i.e. r  -r P :(r,,)=(-1)(r,,) Spin part ? s = 1 ms= 1 s = 1 ms= 0 s = 1 ms= -1 s = 0 ms= 0

  26. p-d  nn Space part P :(r,,)=(-1)(r,,) Spin part ? s = 1 ms= +1 s = 1 ms= 0 s = 1 ms= -1 s = 0 ms= 0 P +1 under exchange P -1 under exchange

  27. p-d  nn Space part P :(r,,)=(-1)(r,,) Spin part ? s = 1 ms= +1 s = 1 ms= 0 s = 1 ms= -1 s = 0 ms= 0 P +1 under exchange =(-1)s+1 1/2( + ) P -1 under exchange =(-1)s+1 1/2(-) So overall must have (-1)+s+1 = -1 (-1)+s= +1 +seven =1 s=0 =0 s=1 =1 s=1 =2 s=1 Thus the parity of the final nn system Must be (-1)= -1

  28. p-d  nn Since strong interactions conserve parity we must also have P:p-d = -1 d is a bound nps-state (-1) PpPp=(+1)(+1)(+1) soP:p-d =(-1)=0 PpPd = (+1)Pp(+1) Pp = -1

  29. some Intrinsic Parity Assignments so far… p proton + n neutron + d deuteron +  charged pion -  Photon -

  30. Neutral Pions g 98.798%p 0 gg 1.198%p 0 e+e-g 3.1410-5 %p 0 e+e-e+e- g u u p 0 msTOTAL 0 2 -2 0 p 0 A B C D 0has 0 spin so can’t produce net angular momentum!

  31. p 0 A B C D Recall, under PARITY (inversion/reflections), R-handed spin→L-handed spin The sign of HELICITY reverses under PARITY. P: A→D Note: So the parity invariant states must be P: D→A A D with  parity either We can for convenience, decompose circularly-polarized photons into + right circularly polarized ms=+1 - left circularly polarized ms=+1

  32. In a direct product representation I can write: A±D=R1R2± L1L2 each component here is PLANE polarized A+D polarized planes PARALLEL polarized planesPERPENDICULAR A-D

  33. The plane defined by the trajectories of e+e- in pair production are in the plane of the E-vector of the initial gamma ray. Look at events where boths from the π0 decay both pair produce and compare the orientation of the two planes: The π0 has ODD PARITY (-1) 0 90o

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