- #26

- 1,789

- 4

Thanks for your reply DrDu. That does clear things up a bit. But how does one get a transcendental equation for [itex]\phi[/itex] which 'cannot be solved exactly'? I seem to have found one which can be solved exactly for [itex]\phi[/itex]. They seem to want something like [itex]\phi = f(\phi)[/itex]. Do you see something wrong in my reasoning?How shall I say? For a potential of infinite range, the statement only holds asymptotically. But for a potential of finite range d, if A exp (ikx)+B exp(ikx) =C sin (kx+phi) holds asymptotically (i.e. abs(A)=abs(B)) then it holds down to x=d.