Coulomb Klein Gordon: Where does e^(-iEt) come from?

  • #1

Main Question or Discussion Point

Hi everyone,

I've been reading about the Klein Gordon equation with the Coulomb Potential. The full solution can be found here:

http://wiki.physics.fsu.edu/wiki/index.php/Klein-Gordon_equation#Klein-Gordon_equation_with_Coulomb_potential

I'm confused near the beginning of this. I understand that the solution is going to look like

$$\Phi(r,t) = R(r)Y(\theta,\phi)T(t)$$

since this is radially symmetric and a second order linear homogeneous partial differential equation, so it is separable. What I don't understand is why all the solutions I look at assume

$$T(t)=e^{-iEt/\hbar}.$$

I've looked up a number of solutions and they all make this assumption without explanation, so I'm assuming I'm missing something obvious. Can anyone explain it to me?

Thanks!
 

Answers and Replies

  • #2
hilbert2
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It's an eigenfunction of ##\left(i\hbar\frac{\partial}{\partial t}\right)^2##, isn't it? In this kind of solutions, the stationary states are looked for, meaning that the time dependence has the form of a trivial phase factor.
 

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