The behavior of the cross section near a resonance

Let us assume low energy scattering off a square well potential such that ka<<1 and only s-wave scattering is important.  At the resonance energy E_r, the phase shift increases through p/2.  The scattering amplitude is

The phase shift d₀ is a function of E. Near E=E_r let us expand sind₀ and cosd₀ .

since

2/G=dd₀/dE.

Therefore

The total cross section is now given by

This is the Breit-Wigner resonance formula.  G is the width of the resonance curve.

Let us assume low energy scattering off a square well potential such that ka<<1 and only s-wave scattering is important.  Assume that the potential has enough depth so that a bound state with a small binding energy exists.

For a bound state we have:

We need to be continuous at r=a. This yields

and

, or

Since E is small we replace

We then have

If k₀a is close to p/2, i.e. a bound state barely exists, we have -r/k₀<<1, tan(p/2-k₀a)<<1, or, approximately, -r/k₀=(p/2-k₀a), k₀a=(p/2+r/k₀).

Now let us look at a continuum state in the same potential.  For a low energy continuum state we have:

Since k²<<U₀ we approximately have , and we write

or

Now

We also have tanka»ka. Therefore

Since ra<<1, we have kcotd₀=-r+ak²=-(1/a)+(1/2)r₀k².  The quantity a is defined as the scattering length, and the quantity r₀ is defined as the effective range.  As k®0 we have kcotd₀®-(1/a), 1/d₀®-(1/ak), d₀®-ak, f₀®-a, s_T®4pa². Since a=1/r we we have established a connection between the energy of a bound state and the low-energy scattering cross sections for potentials that barely support 1 bound state.

At very low energy the Schroedinger equation for r>a becomes and the asymptotic solution is a linear function of r. 

As k®0 we have rf_k(r,q,f)=0 when r=a.  The scattering length can be interpreted as the r intercept of u_k0(r) since f_k(r,q,f)=0 where u_k0=0.

Since we can infer the energy of a loosely bound state by performing a near zero energy scattering experiment.  Wigner first attempted to apply this to neutron-proton scattering.

Let us use what we know about the deuterium and try to predict the near zero energy scattering cross section.  With a binding energy of -2.23 MeV and a reduced mass of 940 MeV we find We have a=(ra+1)/r.  Taking a=1.2´10^-13cm yields a=5.5´10^-13cm and s_T=4´10^-24cm².  The measured cross section however is s_T=20´10^-24cm².  This discrepancy is mainly due to the spin dependence of the scattering cross section.  In the triplet state the neutrons scatter different than in the singlet state.  The bound state of the deuterium is the triplet state, the singlet state is not bound.  We should expect s_T=(3/4)s_t+(1/4)s_s (statistical population).  Our prediction is for the triplet state.  For the singlet state we then expect s_s=4s_T-3s_t= 68´10^-24cm².  This is a huge cross section and it suggests a resonance.  The singlet state potential just misses to produce a bound state near zero energy.  The singlet state wavefunction u_k0(r) just misses to turn over and have a node and therefore the scattering length is large and negative.

Laboratory to center of mass frame transformations

We study the scattering of particles m₁ and m₂ with interaction energy V(r₁-r₂) by studying the scattering of a fictitious particle of reduced mass  by a potential (energy) V(r).  The vector r(r,q,f) describes the relative position of the two particles.  It points from particle 2 to particle 1.  The vector dr(r,q,f)/dt describes the relative velocity of the two particles, and its direction is the direction of the velocity of particle 1 in the CM frame.   We therefore calculate the cross section s(q,f) in the CM frame of the two particles.  In most experimental situations, however, particle 1, with velocity v, approaches particle 2, at rest in the laboratory.  We measure the laboratory scattering angles q₀ and f₀ of particle 1.  We need the to relate the angles and the cross sections in the two frames.

Let v’’ be the velocity of particle 1 in the center of mass frame, v₁ its velocity in the laboratory frame, and v’ the velocity of the center of mass frame with respect to the laboratory frame.  Then, as can be seen from the figure,

Here g=v'/v''=m₁/m₂.  The ratio v'/v'' is equal to the ratio m₁/m₂.

[In the laboratory particle 1 approaches particle 2 with speed v. In the CM frame particle 1 approaches the CM with speed  from the left and particle 2 approaches the CM with speed from the right.  Their relative speed is v and their total momentum is zero. The speed of the CM in the laboratory is v’.]

The total number of particles scattered in to a detector is independent of the reference frame of the observer.  Therefore We can therefore write

[

Using  we can evaluate

Using  we can evaluate We therefore can find  ]

We can now compare the cross sections measured in the laboratory frame with the cross sections calculated in the center of mass frame.

Dependence on g

g=m₁/m₂.  For g<1, m₁<m₂, v’<v’’, q₀ increases monotonically from 0 to p as q increases from 0 to p.  For g=1, v’=v’’, q₀=½q and varies from 0 to ½p as q varies from 0 to p.  No particles appear in the backward direction. For g>1, m₁>m₂, v’>v’’, q₀ increases from zero to a maximum value sin^-1(1/g) as q increases from 0 to cos^-1(-1/g);  q₀ then decreases to 0 as q increases to p. s₀(q,f) is usually infinite at the maximum value of q₀.  However this singularity gives a finite contribution to the total cross section.  No particles appear beyond the maximum value of q₀ in the laboratory.  [In the figure the maximum value of q₀ occurs when v₁ is tangent to the circle.]