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I am stuck with a problem which effectively boils down to this: Given the eigenstates of a Hamiltonian with a step potential in the x direction

[itex]H=-\hbar^2/2m \nabla^2 + V_0 \Theta(x)[/itex]

[itex]\psi(q)_{in}=cos(qx)-\frac{\sqrt{K_{V_0}^2-q^2}}{q}sin(qz) \qquad x<0[/itex]

[itex]\psi(q)_{out}=e^{-\sqrt{K_{V_0}^2-q^2}x} \qquad x>0[/itex]

where [itex]K_\xi[/itex] refers to [itex]\sqrt{2m \xi /\hbar^2}[/itex] and q is the momentum quantum number labelling the eigenstate with eigenenergy [itex]\frac{\hbar^2q^2}{2m}[/itex].

How do I normalize this? It seems to me that I should be able to get normalized states of the form

[itex]\int_{-\infty}^{0}\psi(q)_{in}\psi(q')_{in}^* dx+\int_{0}^{∞}\psi(q)_{out}\psi(q')_{out}^* dx=\delta(q-q')[/itex]

but I cannot get anything resembling a delta function out of the left hand side when I plug the eigenfunctions into this equation.

Where am I going wrong? Shouldn't it be possible to get orthonormal states for this scenario?

Thanks...

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# Normalization of step potential eigenfunction

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