write the radial equation for a particle with mass m and angular momentum l=0 which is under the influence of the following potential:
write all the conditions for the solution of the problem.
Hamiltonian: H=p/2m +V = pr/2m+L^2/2mr^2+V(r)
The Attempt at a Solution
since the angular momentum is zero, the radial equation appears as:
the conditions I can think of are:
1) continuity of u
2) u(infinity)= 0 (for u to be square integrable)
3) from integration of Schroedinger's equation on the interval [R-epsilon, R+epsilon] the jump in the first derivative of u at r=R should be -2mau(R)/hbar^2
but there is another condition according to the answers, that is, u(0)=0.
where does this condition come from?