(a) What is the consequence of the Pauli exclusion principle on the two-particle 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 two-particle 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 anti-symmetric 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 anti-symmetric.

(b) A set of basis vectors for the spin space are {|

*S,M*>=|1,1>, |1,0>, |1,-1>, |0,0>}._{S}**S**_{1}×**S**_{2}=(1/2)(*S*^{2}-*S*_{1}^{2}-*S*_{2}^{2}).

*S*^{2}|*S,M*>=_{S}*S(S+1*)|*S,M*>._{S}

*S*_{1}^{2}|*S,M*>=_{S}*(*3/4*)*|*S,M*>._{S}

*S*_{2}^{2}|*S,M*>=_{S}*(*3/4*)*|*S,M*>._{S}

*P*_{±}*|S,M*>=_{S}*(*1/2±1/4±((1/2)(*S(S+1*)-3/4-3/4))|*S,M*>._{S}For S=0:

*P*>=_{+}|0,0*(*1/2+1/4-3/4)|*0,0*>=0.

*P*>=_{-}|0,0*(*1/2-1/4+3/4)|*0,0*>=|*0,0*>.For S=1:

*P*1_{+}|*,M*>=_{S}*(*1/2+1/4+1-3/4)|1*,M*>=|1_{S}*,M*>._{S}

*P*1_{-}|*,M*>=_{S}*(*1/2-1/4-1+3/4)|1*,M*>=_{S}*0.*Any state vector for the system can be written as a linear combination of the basis vectors. Therefore

(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*P*is the projector onto the triplet state and_{+}*P*is the projector onto the singlet state._{-}*V(*. Here**r**)points from particle 2 to particle 1,**r**.**r**=**r**_{1}-**r**_{2}

Triplet state: The space wavefunction is anti-symmetic,*y*(**r**_{12})=-*y*(**r**_{21}) or*y*()=-**r***y*(-). We therefore need*r**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 anti-symmetric. For the triplet state the spin function is symmetric, so the space function must be anti-symmetric under exchange. We have

f

_{k}(r,q)=e^{ikz}-e^{-ikz}+[f_{k}(q)-f_{k}(p-q)](e^{ikr})/r.We therefore have

*s*0 if_{k}(q)=|f_{k}(q)-f_{k}(p-q)|^{2}. s_{k}(q)=*q=p/2*.

The Coulomb scattering amplitude for two non-relativistic particles in the center of mass system is

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 spin-zero particles,

(d) are identical spin ½ particles in a spin-triplet state,

(e) are identical spin ½ particles in a spin-singlet 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(*. Here**r**)points from particle 2 to particle 1,**r**.**r**=**r**_{1}-**r**_{2}*y*() is the space part of the wavefunction.**r**- (a)
*s*is the differential scattering cross section for detecting particle 1 in bin 1 and detecting particle 2 in bin 2._{c}(q)=|f_{c}(q)|^{2}.

(f

_{k}(r,q)=e^{ikz}+f_{k}(q)(e^{ikr})/r) - (b)
*s*is the differential scattering cross section for detecting one particle in bin 1 and one particle in bin 2._{c}(q)=|f_{c}(q)|^{2}+|f_{c}(p-q)|^{2}.

At we have

*q=p*/2 we have*s*_{c}(q)_{(b)}= 2g^{2}/k^{2}. - (c) For the spin zero particles the space wave functions must be symmetric under
the exchange of the two particles.
f

_{k}(r,q)=e^{ikz}+e^{-ikz}+[f_{k}(q+f_{k}(p-q)](e^{ikr})/r.s

_{c}(q)=|f_{c}(q)+f_{c}(p-q)|^{2}. s_{ c}(q)=|f_{c}(q)|^{2}+|f_{c}(p-q)|^{2}+2Re[f_{c}^{*}(q)f_{c}(p-q)].Therefore

*s*2Re_{c}(q)_{(c)}=s_{c}(q)_{(b)}+*[f*_{c}(q)f_{c}^{*}(p-q)]..

.

.

At

*q=p*/2 we have . The cross section is enhanced through constructive interference. - (d) For the spin 1/2 particles in the triplet state the spin function is symmetric
under exchange, so the space wave functions must be anti-symmetric under the exchange of
the two particles.
*s*_{c}(q)=|f_{c}(q)|^{2}-f_{c}(p-q)|^{2}= |f_{c}(q)|^{2}+|f_{c}(p-q)|^{2}-2Re[f_{c}^{*}(q)f_{c}(p-q)].Therefore

*s*_{c}(q)_{(d)}=s_{c}(q)_{(b)}-*2*Re*[f*At_{c}(q)f_{c}^{*}(p-q)].*q=p/2*we have destructive interferenc*e,*

s_{ c}(p/2)_{(d)}=0. - (e)
We have

*s*._{c}(q)_{(e)}=s_{c}(q)_{(c)} (f) We have a statistical distribution, 1/4 for spin singlet and 3/4 for spin triplet.

*s*1/4)_{c}(q)_{(f)}=(*|f*(3/4)_{c}(q)|^{2}-f_{c}(p-q)|^{2}+*|f*_{c}(q)|^{2}-f_{c}(p-q)|^{2}= |f_{c}(q)|^{2}+|f_{c}(p-q)|^{2}-Re[f_{c}^{*}(q)f_{c}(p-q)].

- (a)