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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?

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    $\begingroup$ 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. $\endgroup$ – aQuestion Feb 7 '15 at 12:14
  • $\begingroup$ 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. $\endgroup$ – Ruslan Feb 7 '15 at 14:44
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/90987/2451 , physics.stackexchange.com/q/134719/2451 and links therein. $\endgroup$ – Qmechanic Feb 7 '15 at 15:32
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It's just an Ansatz and as long as it fulfills the given homogeneous ordinary differantial equation it's OK to use it. In the end you'll get the same results with the more general formula you gave because at one point (e.g. when inserting initial or boundary conditions) your B turns out to be 0.

The most general (correct me if I'm wrong) Ansatz would be something like:

$u(r)=\sum_i C_i\exp\left(\lambda_ir\right)$

Then your problem transforms to:

$\lambda^2=-k^2 \Rightarrow \lambda_{1,2}=\pm ik$

and your $u(r)$ is nothing else than a sum of $\sin(kr)$ and $\cos(kr)$.

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