Problem:
Consider the center of mass system of two interacting identical fermions with
spin ½.
(a) What is the consequence of the Pauli exclusion principle on the
two-particle wave function?
(b) Let S1 and S2 be the spin operators of
the two individual neutrons.
Show that the operators P± = ½ ± ¼ ± S1∙S2/ħ2
are the projection operators of the triplet states and the singlet states of the
spin wave functions.
(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:
- Concepts:
Indistinguishable particles, identical fermions
- Reasoning:
A state of many identical particles is totally antisymmetric if they
are fermions.
- Details of the calculation:
(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 wave function is symmetric, if they are in the triplet state, then the
space wave function is anti-symmetric.
(b) A set of basis vectors for the spin space are {|S,MS> =
|1,1>, |1,0>, |1,-1>, |0,0>}.
S1∙S2 = ½(S2 - S12
- S22).
S2|S,MS> = S(S + 1)ħ2|S,MS>.
S12|S,MS> = ¾ħ2|S,MS>.
S22|S,MS> = ¾ħ2|S,MS>.
P±|S,MS> = (½ ± ¼ ± (½(S(S + 1)/ħ2 - ¾ -
¾))|S,MS>.
For S = 0, the singlet state:
P+|0,0> = (½ + ¼ - ¾)|0,0> = 0.
P-|0,0> = (½ - ¼ + ¾)|0,0> = |0,0>.
For S = 1, the triplet state:
P+|1,MS> = (½ + ¼ + 1 - ¾)|1,MS > = |1,MS
>.
P-|1,MS> = (½ - ¼ - 1 + ¾)|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 U(r). Here r points from particle 2 to
particle 1, r = r1-r2.
Triplet state: The space wave function is anti-symmetic,
ψ(r12) = -ψ(r21) or ψ(r) = -ψ(-r).
We therefore need l = odd. (Property of the spherical harmonics.)
(d) For a scattering problem we have Φk(r,θ) = eikz +
fk(θ)(eikr)/r, σk(θ) = |fk(θ)|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
Φk(r,θ) = eikz - e-ikz + [fk(θ)
- fk(π-θ)](eikr)/r.
We therefore have σk(θ) = |fk(θ) - fk(π-θ)|2.
σk(θ) = 0 if θ = π/2.
Problem:The Coulomb scattering amplitude for two non-relativistic
particles in the center of mass system is
fc(θ) = -exp(iδ0)
γ exp[-iγ ln(sin2(θ/2))]/(2ksin2(θ/2)),
where
γ
= -μe2Z1Z2/(ħ2k), μ 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.
Evaluate each at θ = π/2.
You may ignore any relativistic spin effects for fc(θ).
Solution:
- Concepts:
Scattering, indistinguishable particles
- Reasoning:
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 U(r).
Let r points from particle 2 to
particle 1, r = r1 - r2 and
let ψ(r) be the space part of the wave function.
In a central potential, stationary state solutions of the eigenvalue
equation HΦk(r) = EkΦk(r)
with the asymptotic form
Φk(r) = eikz + fk(θ) eikr/r
exist.
The differential scattering cross section is σk(θ) = |fk(θ)|2
- Details of the calculation:
(a) Let Φk(r,θ) = eikz + fk(θ)(eikr)/r
as r --> ∞. Let fk(θ) = fc(θ).
σc(θ)a = |fc(θ)|2 is the differential scattering cross section for
detecting particle 1 in bin 1 and detecting particle 2 in bin 2.
σc(θ)a = γ2/(4k2sin4(θ/2)).
At θ = π/2 we have σc(θ)b = γ2/k2.
(b) σc(θ)b = |fc(θ)|2 + |fc(π -
θ)|2 is the differential scattering cross section for
detecting one particle in bin 1 and one particle in bin 2.
σc(θ)b = γ2/(4k2)[1/sin4(θ/2)
+ 1/sin4((π - θ)/2)].
At θ = π/2 we have σc(θ)b = 2γ2/k2.
(c) For the spin zero particles the space wave functions must be symmetric under
the exchange of the two particles.
Φk(r) = eikz + e-ikz + [fk(θ) +
fk(π - θ)](eikr)/r as r --> ∞.
σc(θ) = |fc(θ) + fc(π - θ)|2.
σc(θ) = |fc(θ)|2 + |fc(π - θ)|2
+ 2Re[fc(θ) fc*(π - θ)].
Therefore σc(θ)c = σc(θ)b +
2Re[fc(θ) fc*(π - θ)].
fc(θ) fc*(π - θ) = [γ2/(4k2sin2(θ/2) sin2((π -
θ)/2)] exp[-iγ (ln(sin2(θ/2)) - ln(sin2((π - θ)/2)].
Re[fc(θ) fc*(π - θ)] = [γ2/(4k2sin2(θ/2) sin2((π -
θ)/2)] cos[γ ln{sin2(θ/2)/sin2((π - θ)/2)})].
At θ = π/2 we have Re[fc(θ) fc*(π - θ)] = γ2/k2
and σc(θ)c = 4γ2/k2.
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.
Φk(r) = eikz - e-ikz + [fk(θ)
- fk(π - θ)](eikr)/r as r --> ∞.
σc(θ)d = |fc(θ)|2 + |fc(π -
θ)|2 - 2Re[fc(θ) fc*(π - θ)] = σc(θ)b
- 2Re[fc(θ) fc*(π - θ)].
At θ = π/2 we have
destructive interference, σd(π/2) = 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 σc(θ)e = σc(θ)c.
(f) We have a statistical distribution, 1/4 for spin singlet
and 3/4 for spin triplet.
σc(θ)f = (1/4)σc(θ)e + (3/4)σc(θ)f
= |fc(θ)|2 + |fc(π - θ)|2 - Re[fc(θ) fc*(π -
θ)].
At θ = π/2 we have Re[fc(θ) fc*(π - θ)] = γ2/k2
and σc(θ)c = γ2/k2