Triplet S-wave (l = 0) phase shift in n-p scattering at low energy

[wdednam]

Homework Statement

Evaluate the triplet S-wave (l = 0) phase shift in n-p scattering at low energy assuming the interaction potential is given by a square well of depth V0 = 38.5 MeV and width b = 1.73 fm. Deduce also the value of the triplet scattering length.

Homework Equations

Normalized S-wave triplet wavefunction $\varphi$ = 1/√(4∏)×u(r)/r×|S=1>|T=0>

where u(r) = A1sinKr and K² = M(E-V(r))/hbar² for r < b
and u(r) = A2sin(kr + δ) and k² = ME/hbar²

and M = mass of proton = mass of neutron and δ = phase shift

V(r) = -V0 - V0($\tau$1$\bullet$$\tau$2) for r < b
and V(r) = 0 for r > b

and $\tau$1$\bullet$$\tau$2 is the isospin operator that distinguishes between the singlet |T=0> and triplet |T=1> isospin components of the neutron-proton wavefunction and have eigenvalues:

$\tau$1$\bullet$$\tau$2|T=0> = -3|T=0>

$\tau$1$\bullet$$\tau$2|T=1> = 1|T=1>

Matching condition for radial components of wave function at r = b:

KcotKb = kcot(kb + δ)

The Attempt at a Solution

I can't use the matching condition directly to give me the triplet phase shift because that hasn't worked. So I got a hint from my lecturer to include the isopin component of the triplet wave function and solve the schrodinger equation for that. When I apply the potential operator V(r) = -V0 - V0($\tau$1$\bullet$$\tau$2) to the wave function however it gives me the dispersion relation

K² = M(E-2V0)/hbar² which when I take the limit E --> 0 has a negative number under the square root, so I can't even use it in the matching condition and solve for δ.

I am really stumped on this one and would greatly appreciate any help.

Thanks

 

[mark726]

for your question! To evaluate the triplet S-wave phase shift in n-p scattering, we can start by writing the Schrodinger equation for the system:

(-hbar^2/2m) d^2u(r)/dr^2 + [V0 + V0(\tau1\bullet\tau2)]u(r) = E u(r)

where m is the reduced mass of the system. Since we are considering the S-wave (l=0), we can use the normalized S-wave triplet wave function given in the homework equations:

\varphi = 1/√(4∏)×u(r)/r×|S=1>|T=0>

Substituting this into the Schrodinger equation and using the given potential, we get:

(-hbar^2/2m)[d^2u(r)/dr^2 + (V0 + V0(\tau1\bullet\tau2))u(r)] = E u(r)

Next, we can use the fact that the isospin operator \tau1\bullet\tau2 has eigenvalues -3 for |T=0> and 1 for |T=1>, to simplify the equation:

(-hbar^2/2m)[d^2u(r)/dr^2 + (V0 - 3V0)|T=0>u(r) + (V0 + V0)|T=1>u(r)] = E u(r)

Simplifying further, we get:

(-hbar^2/2m)[d^2u(r)/dr^2 + 2V0|T=0>u(r)] = E u(r)

Now, we can solve this equation for the S-wave triplet wave function u(r) using the boundary conditions given in the homework equations. For r < b, we have:

(-hbar^2/2m)[d^2u(r)/dr^2 + 2V0|T=0>u(r)] = E u(r)

(-hbar^2/2m)[d^2(A1sinKr)/dr^2 + 2V0|T=0|A1sinKr] = E A1sinKr

Simplifying, we get:

(-hbar^2/2m)[A1K^2sinKr - 2V0|T=0|A1sinKr] = E A