Continuous eigenstates vs discrete eigenstates

[strangerep]

I wrote a short .pdf with my calculations [...]

In your eq(4), you haven't allowed for the possibility that perhaps

<br /> \langle n | V | E \rangle ~\ne~ 0 ~.<br />

I.e., you've assumed that the interaction V does not mix the discrete
and continuous subspaces of the total (rigged) Hilbert space. Hence
you get a null result.

Try a simple specific potential (e.g., Morse, as Arnold Neumaier suggested),
which has an exactly solvable discrete and continuous spectrum.

 

[jfy4]

This has already been resolved for the OP, I think, but I thought I would give another reference since it always helps to see multiple angles.

Here from Dirac's Principles of Quantum Mechanics page 65, he lays out three specific rules for descrete and continuous eigenvalues.

"Using \xi^r and \xi^s to denote discrete eigenvalues and \xi&#039; and \xi&#039;&#039; to denotes continuous eigenvalues, we have the set of equations

\langle\xi^r|\xi^s\rangle=\delta_{\xi^r\xi^s}

\langle\xi^r|\xi&#039;\rangle=0

\langle\xi&#039;|\xi&#039;&#039;\rangle=\delta(\xi&#039;-\xi&#039;&#039;)

as the generalizations of

\langle\xi&#039;|\xi&#039;&#039;\rangle=\delta_{\xi&#039;\xi&#039;&#039;}

and

\langle\xi&#039;|\xi&#039;&#039;\rangle=\delta(\xi&#039;-\xi&#039;&#039;).

These equations express that the basic vectors are all orthogonal, that those belonging to discrete eigenvalues are normalized and those belonging to continuous eigenvalues have their lengths fixed..."

Hope this helps in some way.

 

[giova7_89]

Thank you strangerep I realized my mistake! When I used the resolution of the identity on the right and on the left I forgot, as you say, to take into account the "mixing" between discrete and continuous eigenstates. Now my problem is solved!

 

[dextercioby]

I don't understand. Gamov vectors do not correspond to the real portion of the spectrum, so why should they _not_ be missing?

Depends on your goal. Normally, they don't occur, except for when you search for them specifically. For that, you need to <relax> a little the topology, the Hamiltonian becomes symmetric and that way its spectrum can be complex.

 

[A. Neumaier]

Depends on your goal. Normally, they don't occur, except for when you search for them specifically. For that, you need to <relax> a little the topology, the Hamiltonian becomes symmetric and that way its spectrum can be complex.

But you assumed in

if one is interested in finding only the real portion of the observables' spectra by making a so-called <tight rigging> of the H-space, he could miss the existence of Gamov vectors

that one is interested only in the real portion. Because of this assumption, one _should_ miss the complex spectrum in the deformation!

Unless you also relax your interest...