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To find the allowed energies for a system, I can find the spectrum of the Hamiltonian $\hat{H}_{\psi}$ given a wavefunction $\psi$ representing the state of the system. 3 cases might happen: either there are boundary conditions that will lead to full quantization of the spectrum, or the spectrum can be partially quantized, partially continuous, or last case, it can be fully continuous, if there are no boundary conditions (the system is free). However, the eigenvalues I find are not necessarily "physical": in the case of the free system in particular, the eigenstates $\psi_k$ might not be normalizable, therefore you can't compute $E_k=\langle\psi_k|\hat{H}|\psi_k\rangle$ (the denominator blows up). So, do I have to insist on the corresponding eigenstate being normalizable to declare that I have found a possible energy for the system? (i.e. it's not enough to find the spectrum of $\hat{H}$)

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If $\lambda$ is a value in the spectrum of $H$ there always exists a state $\omega$ such that $\lambda =\omega(H)$ and $\Delta_\omega(H)=0$, so every point in the spectrum of $H$ is a possible outcome of an experiment. What can fail to happen is that, in a particular representation, like the Schrödinger representation, the state $\omega$ is a vector state. So either you allow distribution to come into play through the rigged Hilbert space, where Dirac's bras and kets actually live, or you content yourself with approximate eigenvectors.

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You say

"To find the allowed energies for a system, I can find the spectrum of the Hamiltonian $\hat H_{\psi}$ given a wavefunction $\psi$ representing the state of the system."

No, the spectrum of the Hamiltonian you find not for a certain state, you have to find the eigenvalues of the Hamiltonian. For each such eigenvalue there exists an eigenstate.

Also, a system may have a fully continuous spectrum without being completely free, i.e. a potential well may be present, limited by a potential barrier. But the bottom energy of the well should be non-negative (or only slightly negative of a small absolute value that doesn't admit bound states), and with a potential barrier finite in height and length. Such situations are found in nuclei.

Such functions are called scattering states or generalized Fourier functions because they come from $-\infty$ and go to $\infty$, as the Fourier functions. Such waves cover all the space, from $-\infty$ to $\infty$ and normalize to $\delta$ Dirac,

$$\langle \psi_{\ \vec k}|\psi_{\ \vec k'}\rangle = \delta(\vec k - \vec k'). \tag{i}$$

In other words, the integral of their absolute intensity over all the space is infinite. This is why they are considered non-physical, because they cannot be normalized to 1.

What we do with such functions is, first of all, find their form and eigenvalues (energies). This is what the book did. Next, we construct normalizable wave-packets

$$\Psi(\vec r, t) = \int_{-\infty}^{\infty} A(\vec k) \psi (\vec r, t; \vec k) \ \text d \vec k. \tag{ii}$$

As each value of $|\vec k|$ implies a corresponding energy, $\hbar^2 k^2/(2m)$, the wave-packet comprises many energies. An example of such a wave-packet, constructed from Fourier functions, is the Gaussian wave-packet.

"So, do I have to insist on the corresponding eigenstate being normalizable to declare that I have found a possible energy for the system?"

No, you shouldn't falsify the eigenfunctions of the continuousm spectrum.

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  • $\begingroup$ I am confused - take Griffiths, 2nd ed, page 60. At the bottom of the page, he says that the free particle system under study doesn't admit definite energies, although he just found some $\psi_k$ corresponding to some $E$ - I think he means that these eigenstates are not physically meaningful, and therefore the corresponding $E$ are not physically meaningful either. Hence the idea that once you hold an $E_k$ and $\psi_k$ eigen pair from working on the Hamiltonian, you still need to validate that $\psi_k$ makes physical sense. $\endgroup$ – Frank Mar 4 '15 at 23:45

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