# Losing a term for 3D radial schrodinger equation

I am trying to solve the Schrodinger equation For a potential $V(r)$ defined for $0<r<R$ as $$V(r)=-V_0$$ and zero everywhere else.

For wavefunction $u$ I can easily get to $$u'' =-k^2u,$$ where $$k^2 = \frac {2m}{h}(V_0 +E).$$

I understand that the general solution to a differential equation of this form is $$u=A \sin(kr) +B \cos(kr) .$$

However, in my textbook the $\cos$ term is mysteriously dropped. Why should this be?

• It's boundary condition. U(r) must be zero when r=0 and r=R. If u(0) = 0, you need to drop the cos term with equalising B=0. – aQuestion Feb 7 '15 at 12:14
• For a radial part of 3D Schrödinger's equation you are missing a first derivative term. Otherwise your equation is 1D Schrödinger's equation. And for 3D radial equation you should get spherical Bessel functions, not trigonometric ones. – Ruslan Feb 7 '15 at 14:44
• Possible duplicates: physics.stackexchange.com/q/90987/2451 , physics.stackexchange.com/q/134719/2451 and links therein. – Qmechanic Feb 7 '15 at 15:32

$u(r)=\sum_i C_i\exp\left(\lambda_ir\right)$
$\lambda^2=-k^2 \Rightarrow \lambda_{1,2}=\pm ik$
and your $u(r)$ is nothing else than a sum of $\sin(kr)$ and $\cos(kr)$.