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Let's consider the following Schrödinger equation: $$\psi''(x)+k^2\psi(x)=0$$ with the following boundary condition $$\psi(0)+a\psi'(0)=0$$ $k$ is supposed to be larger that $0$.

This equation is considered in the region $x<0$. For the sake of simplicity I will suppose that $a=1$.

On the first glance, this problem may seem a bit "artifitial", but I faced it dealing with a special semiconducting structure.

This equation can be easily solved: $$\psi_k (x)=c(k)\left(\frac{ik-1}{\sqrt{k^2+1}}e^{ikx}+\frac{ik+1}{\sqrt{k^2+1}}e^{-ikx} \right)$$ The problems appears when one wants to normalize this solution in a common way for the states of continous spectra.

It is known that for continous spectra it is always possible to normalize the states in the following way: $$\int \psi_{k'}^*(x)\psi_k (x)dx=\delta (k-k')$$ However, the explicit evaluation of this integral for the example above seems a bit inconsistent with this formula in a sense that the states with different $k$ are not even orthogonal.

Let's define $$e^{-i\phi_k}=\frac{ik+1}{\sqrt{k^2+1}}$$ Then the noramlization integral looks as follows $$c(k)c(k')^*\int_{-\infty}^{0}dx\left( e^{i\phi_{k'} +ik'x}-e^{-i\phi_{k'} -ik'x}\right)\left(- e^{i\phi_{k} +ikx}+e^{-i\phi_{k} -ikx}\right) $$ The interals can be easily calculated. For this purpose it is necessary to add a small $\epsilon$ times $x$ in every exponent. $$c(k)c(k')^* \mathrm{lim}_{\epsilon \rightarrow 0_+} \left(\frac{e^{i(\phi_k-\phi_k')}}{\epsilon+i(k-k')}+\frac{e^{-i(\phi_k-\phi_k')}}{\epsilon-i(k-k')}-\frac{e^{i(\phi_k+\phi_k')}}{\epsilon+i(k+k')}-\frac{e^{-i(\phi_k+\phi_k')}}{\epsilon-i(k+k')}\right)=$$ $$=c(k)c(k')^* \mathrm{lim}_{\epsilon \rightarrow 0_+} \left(\frac{(\epsilon-i(k-k'))e^{i(\phi_k-\phi_k')}}{\epsilon^2+(k-k')^2}+\frac{(\epsilon+i(k-k'))e^{-i(\phi_k-\phi_k')}}{\epsilon^2+(k-k')^2}-\frac{(\epsilon-i(k+k'))e^{i(\phi_k+\phi_k')}}{\epsilon^2+(k+k')^2}-\frac{(\epsilon+i(k+k'))e^{-i(\phi_k+\phi_k')}}{\epsilon^2+(k+k')^2}\right)=$$ $$=c(k)c(k')^*(2\pi\delta(k-k')\cos(\phi_k-\phi_k')-2\pi\delta(k+k')\cos(\phi_k+\phi_k')+2\frac{\sin(\phi_k-\phi_k')}{k-k'}+2\frac{\sin(\phi_k+\phi_k')}{k+k'})$$ $\delta(k+k')$ can be thrown away. What's clear is that there are terms that are large as $k\rightarrow k'$. However, it is also clear that this expression doesn't equal 0 for $k\neq k'$.

Are eigenstates that correspond to the continuos spectra always orthogonal? Apperently, I don't understand this, or there is a mistake in the evaluation above somewhere. Could you please take a look and say what's wrong? What is the proper way to noramlize eigenstates of the continuous spectra?

I have found that the problem like this one rises in the other problems, too. For example, are scattering states in 1D scattering problem on the delta-function potential orthogonal?

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There is something wrong with your evaluation. Let us simplify the wave functions to $$ \Psi_k(x) = {\bar c}_k\left[ (ik-1)e^{ikx} + (ik+1)e^{-ikx}\right] $$ Then what follows for $k'\neq k$ after introducing the small $\epsilon>0$, doing the integrals, and taking the limit $\epsilon \rightarrow 0$ is $$ \int_{-\infty}^0{dx \;\Psi^*_{k'}(x)\Psi_k(x)} \sim $$ $$- \int_{-\infty}^0{dx \left[ (ik'+1)e^{-ik'x} + (ik'-1)e^{ik'x}\right] \left[ (ik-1)e^{ikx} + (ik+1)e^{-ikx}\right] } \rightarrow $$ $$ \rightarrow i\frac{(ik'+1)(ik-1)}{k-k'} - i\frac{(ik'+1)(ik+1)}{k+k'} + i\frac{(ik'-1)(ik-1)}{k+k'} - i\frac{(ik'-1)(ik+1)}{k-k'} = $$ $$ i\frac{(ik'+1)(ik-1) - (ik'-1)(ik+1)}{k-k'} + i\frac{(ik'-1)(ik-1) - (ik'+1)(ik+1)}{k+k'} = $$ $$ i\frac{-k'k -ik'+ik -1 + k'k-ik' +ik +1}{k-k'} + i\frac{-k'k-ik'-ik+1 +k'k-ik'-ik-1}{k+k'} = $$ $$ -2\frac{k-k'}{k-k'} + 2\frac{k+k'}{k+k'} = 0 $$ Reevaluating for $k=k'$ produces the $\left(k^2+1\right)$ factor, so eventually $$ \Psi_k(x) = \frac{(ik-1)e^{ikx} + (ik+1)e^{-ikx}}{\sqrt{k^2 +1}} $$ and $$ \int_{-\infty}^0{dx \;\Psi^*_{k'}(x)\Psi_k(x)} = \delta(k'-k) $$

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