Problems

Consider the center of mass system of two interacting identical fermions with spin (such as neutron-like particles without isospin.)

(a) What is the consequence of the Pauli exclusion principle on the two-particle wavefunction?

(b)  Let S1 and S2 be the spin operators of the two individual neutrons.  Show that the operators P=(1/2)(1/4) S1S2/ 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,MS>=|1,1>, |1,0>, |1,-1>, |0,0>}.

S1S2=(1/2)(S2-S12-S22).
S2|S,MS>=S(S+1)|S,MS>.
S12|S,MS>=(3/4)|S,MS>.
S22|S,MS>=(3/4)|S,MS>.
P|S,MS>=(1/21/4((1/2)
(S(S+1)-3/4-3/4))|S,MS>.

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,MS>=(1/2+1/4+1-3/4)|1,MS >=|1,MS >.
P-|1,MS>=(1/2-1/4-1+3/4)|1,MS >=0.

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=r1-r2.
Triplet state: The space wavefunction is anti-symmetic, y(r12)=-y(r21) or y(r)=-y(-r).  We therefore need l=odd. (Property of the spherical harmonics.)

(d)  For a scattering problem we have fk(r,q)=eikz+fk(q)(eikr)/r, sk(q)=|fk(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

fk(r,q)=eikz-e-ikz+[fk(q)-fk(p-q)](eikr)/r.

We therefore have sk(q)=|fk(q)-fk(p-q)|2. sk(q)=0 if q=p/2.

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

 Image778.gif (1598 bytes)

where Image779.gif (1092 bytes)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 particles in a spin-triplet state,
(e)  are identical spin particles in a spin-singlet state,
(f)  are identical unpolarized spin particles.

You may ignore any relativistic spin effects for fc(q).

Image780.gif (997 bytes)

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=r1-r2y(r) is the space part of the wavefunction.
(a)  sc(q)=|fc(q)|2 is the differential scattering cross section for detecting particle 1 in bin 1 and detecting particle 2 in bin 2.

.

(fk(r,q)=eikz+fk(q)(eikr)/r)

(b)  sc(q)=|fc(q)|2+|fc(p-q)|2 is the differential scattering cross section for detecting one particle in bin 1 and one particle in bin 2.

.

At we have q=p/2 we have sc(q)(b)= 2g2/k2. 

(c)  For the spin zero particles the space wave functions must be symmetric under the exchange of the two particles.

fk(r,q)=eikz+e-ikz+[fk(q+fk(p-q)](eikr)/r.

sc(q)=|fc(q)+fc(p-q)|2.    s c(q)=|fc(q)|2+|fc(p-q)|2+2Re[fc*(q)fc(p-q)].

Therefore sc(q)(c)=sc(q)(b)+2Re[fc(q)fc*(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.

sc(q)=|fc(q)|2-fc(p-q)|2= |fc(q)|2+|fc(p-q)|2-2Re[fc*(q)fc(p-q)].

Therefore sc(q)(d)=sc(q)(b)-2Re[fc(q)fc*(p-q)].  At q=p/2 we have destructive interference, 
s
c(p/2)(d)=0.

(e)  For the spin 1/2 particles in the singlet state the spin function is anti-symmetric under exchange, so the space wave functions must be symmetric under the exchange of the two particles.

We have sc(q)(e)=sc(q)(c).

(f)  We have a statistical distribution, 1/4 for spin singlet and 3/4 for spin triplet.

sc(q)(f)=(1/4)|fc(q)|2-fc(p-q)|2+(3/4)|fc(q)|2-fc(p-q)|2= |fc(q)|2+|fc(p-q)|2-Re[fc*(q)fc(p-q)].