Physics 195bProblem set number 13 – Solution to Problem 64Due 2 PM, Thursday, January 30, 2003READING: Read sections 1-5 of the “Approximate Methods” course note.PROBLEMS:61. Exercise 15 of the Angular Momentum course note.62. Exercise 17 of the Angular Momentum course note.63. More on decay angular distributions: Do exercise 19 of the AngularMomentum course note.64. Let us discuss some issues relevant to the proof you gave of Yang’stheorem (by the way, the original reference is: C. N. Yang, PhysicalReview 77 (1950) 242) in exercise 10 of the Angular Momentum coursenote.(a) Let us consider the parity again, and try to establish the connec-tion between how I expected you to do the homework, and anequivalent argument based on the identical boson symmetry.In general, the parity of a system of two particles, when their stateis an eigenstate of the parity operator, may be expressed in theform:P = ηintrinsicηspatial,where P2=1,ηintrinsicrefers to any intrinsic parity due to theinterchange of the positions of the particles, and ηspatialrefers tothe effect of parity on the spatial part of the wave function. Givena system of two identical particles, and considering the action ofparity on YL0(θ, φ), determine the parity of the system for givenorbital angular momentum L.For a two-photon system, such as we considered in our discussionof Yang’s theorem, explicitly consider boson symmetry (i.e.,thatthe wave function is invariant under interchange of all quantumnumbers for two identical bosons) to once again determine theeffect of parity of the states:|↑↑ , |↓↓ , |↑↓ + |↓↑ , |↑↓ − |↓↑ 38Also, give the allowed possibilities for the orbital angular momen-tum for these states, at least at the level of our current discussion.You may now wish to go back to your original derivation of Yang’stheorem, and determine where you implicitly made use of theboson symmetry.Solution: The action of parity on a system of given orbital an-gular momentum L is:PYL0(θ, φ)=YL0(π − θ, π + φ) (133)=(−)LYL0(θ, φ), (134)since YL0does not depend on φ, and the effect of the tranformationon θ is determined by the even/odd properties of the Legendrepolynomials. Note that it is sufficient to consider YL0, since ifM = 0, we can always rotate to a basis in which M =0,and[J,P] = 0. Thus, the parity of a system of two identical particlesof orbital angular momentum L is (−)L.The | ↑↑ is symmetric under interchange of the spins. Therefore,it must be symmetric under interchange of spatial coordinatesin order for the total wave function to be symmetric under in-terchange of the two photons. Thus, the parity of this state iseven. The same argument applies for | ↓↓ and |↑↓+ ↓↑ .The| ↑↓ − ↓↑ state is odd under spin interchange, hence must be oddunder space interchange; the parity of this state is odd.The even parity states can have even orbital angular momenta,and the odd parity state can have odd orbital angular momenta.(b) Continuing our discussion of Yang’s theorem, there may be someconcern about the total spin angular momentum of the two photonstates, and whether the appropriate values are possible to give theright overall angular momentum when combined with even or oddorbital angular momenta. Using a table of Clebsch-Gordan coef-ficients or otherwise, let us try to alleviate this concern. Thus,decompose our four 2-photon helicity states (with Jzvalues in-dicated by |↑↑ , |↓↓ ,1√2[|↑↓ + |↓↑ ],1√2[|↑↓ − |↓↑ ], where thephotons are travelling along the + and −z axis) into states oftotal spin angular momenta and spin projection along the z-axis:| S, Sz . Hence, show that a JP=0+particle may decay into39two photons with relative orbital angular momentum L = 2 or 0,and a JP=0−particle may decay into two photons with relativeangular momentum L =1.Solution: We are asked to combine the spins of the two photonsto determine the given states in terms of the total spin and itsprojection along the z axis. Using a table of Clebsch-Gordancoefficients, we find:| ↑↑ = |22 (135)| ↑↓ =1√6|20 +1√2|10 +1√3|00 (136)| ↓↑ =1√6|20 −1√2|10 +1√3|00 (137)| ↓↓ = |2 − 2 . (138)Hence, we also have the desired linear combinations:1√2(| ↑↓ + | ↓↑ )=1√3|20 +23|00 (139)1√2(| ↑↓ −| ↓↑ )=|10 . (140)The JP=0+state, with even parity, can only decay to the evenparity state | ↑↓ + | ↓↑ . Note that we must have L = S,inorderto couple to angular momentum 0. We see that this is possiblewith L = 0, coupling to the |00 spin state, or L = 2, coupling tothe |20 spin state. The JP=0−state, with odd parity, can onlydecay to the odd parity state | ↑↓ − | ↓↑ . This is possible onlywith L =1.65. Do exercise 1 of the Approximate Methods course