(a) What is the consequence of the Pauli exclusion principle on the twoparticle wavefunction?
(b) Let S_{1} and S_{2} be the spin operators of the two individual neutrons. Show that the operators P_{±}=(1/2)±(1/4)± S_{1}×S_{2}/ are the projection operators of the triplet states and the singlet states of the spin wavefunctions.
(c) Using the Pauli exclusion principle and the symmetry properties of the spin and relative orbital angular momentum L , find the allowed values of L for any bound triplet state of the twoparticle system.
(d) Again using the Pauli exclusion principle and the symmetry properties of the space coordinate, show that the particles in a triplet state can never scatter through an angle of 90 degrees in their center of mass system.
Solution :
(a) The total state vector must be antisymmetric under the exchange of the two particles. If the two particles are in the singlet state, then the space wavefunction is symmetric, if they are in the triplet state, then the space wavefunction is antisymmetric. (b) A set of basis vectors for the spin space are {S,M_{S}>=1,1>, 1,0>, 1,1>, 0,0>}. S_{1}×S_{2}=(1/2)(S^{2}S_{1}^{2}S_{2}^{2}). For S=0: For S=1: Any state vector for the system can be written as a linear combination of the basis vectors. Therefore P_{+} is the projector onto the triplet state and P_{} is the projector onto the singlet state. (c) The relative motion of two interacting particles in their CM frame is treated by treating the motion of a fictitious particle of reduced mass m in an external potential V(r). Here r points from particle 2 to particle 1, r=r_{1}r_{2}.Triplet state: The space wavefunction is antisymmetic, y(r_{12})=y(r_{21}) or y(r)=y(r). We therefore need l=odd. (Property of the spherical harmonics.) (d) For a scattering problem we have f_{k}(r,q)=e^{ikz}+f_{k}(q)(e^{ikr})/r, s_{k}(q)=f_{k}(q)^{2}. For identical fermions the state vector must be antisymmetric. For the triplet state the spin function is symmetric, so the space function must be antisymmetric under exchange. We have f_{k}(r,q)=e^{ikz}e^{ikz}+[f_{k}(q)f_{k}(pq)](e^{ikr})/r. We therefore have s_{k}(q)=f_{k}(q)f_{k}(pq)^{2}. s_{k}(q)=0 if q=p/2. 
where and m is the reduced mass, and k is the wave number. Find the differential cross section in the center of mass if the two particles
(a) are distinguishable,
(b) are distinguishable, but the identity of the particles is not observed in either
bin,
(c) are identical spinzero particles,
(d) are identical spin ½ particles in a spintriplet state,
(e) are identical spin ½ particles in a spinsinglet state,
(f) are identical unpolarized spin ½ particles.
You may ignore any relativistic spin effects for f_{c}(q).
Solution: The relative motion of two interacting particles in their CM frame is treated by treating the motion of a fictitious particle of reduced mass m in an external potential V(r). Here r points from particle 2 to particle 1, r=r_{1}r_{2}. y(r) is the space part of the wavefunction.
