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As far as I understand, to normalize the eigenfunctions, corresponding to the continuous spectrum, we use Dirac delta function:

$\langle \psi_\lambda \mid \psi_{\lambda'} \rangle = \delta(\lambda - \lambda')$

However, I don't get how does that work if the eigenvalue $\lambda$ is degenerate. This is the case for the free particle Hamiltonian $-\frac{d^2}{dx^2}$ (I am omitting the constants like $\hbar$ and $m$): for every energy value $E > 0$ there are two linearly independent wavefunctions $\psi_+(x) = e^{i \sqrt{E} x}$ and $\psi_-(x) = e^{-i \sqrt{E} x}$. Intuitively, they should be orthogonal in some sense, however, there is no way of getting zero in the right hand side of the Dirac orthonormality condition (since $\delta(\lambda - \lambda) = \delta(0)$).

What am I missing?

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As you found, the two wavefunctions aren't orthogonal; what makes you think they should be? In general, degenerate eigenvectors of a symmetric matrix are not orthogonal.

Note that since the two wavefunctions are linearly independent, the Gram–Schmidt process can be used to form an orthogonal basis for the eigenspace corresponding to the energy value $E$.

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Okay, but how do I understand they are not orthogonal then? I mean, since the integral $\int\limits_{i = -\infty}^\infty \psi_+(x) \psi_-^*(x) dx = \int\limits_{i = -\infty}^\infty e^{2 i \sqrt{E} x} dx$ does not exist, what kind of intuition or formal calculations should I use to prove non-orthogonality? Also, the Gram-Schmidt process relies on calculating the inner product $\langle \psi_+ \mid \psi_- \rangle$, which involves this integral again, how should I interpret it for non-normalizable states? – karlicoss Apr 10 '14 at 22:14
Note that $\langle \psi_+ | \psi_+ \rangle$ does not exist either, indicating that the wave functions are not even normalizable for your chosen inner product. The simplest way to resolve this issue is to artificially trap the wave functions into a box by changing the limits of integration to finite values. – Shivam Sarodia Apr 10 '14 at 23:30
I was able to reason that $\psi_+$ and $\psi_-$ are actually orthogonal. Not sure if my reasoning is correct, but: We can formally express $\int\limits_{-\infty}^\infty e^{-i k x} = 2 \pi \delta(k)$ (using the inverse Fourier transform for the constant function). Now, $\langle \psi_+ \mid \psi_- \rangle = \int\limits_{-\infty}^\infty e^{ikx} (e^{-ikx})^* dx = \int\limits_{-\infty}^\infty e^{2 i k x} dx = \int\limits_{-\infty}^\infty \frac{1}{2} e^{i k x} dx = \frac{1}{2} 2 \pi \delta(k)$, which is zero for any positive energy (since $k$ is proportional to $\sqrt{E}$). – karlicoss Apr 11 '14 at 21:53
I meant Fourier transform, not its inverse. – karlicoss Apr 11 '14 at 22:12

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