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Continuous eigenstates vs discrete eigenstatesby giova7_89
Tags: discrete eigenstates 
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#19
May911, 09:33 AM

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Ok that's exactly what I thought one should do. I hope to be able to learn more about the mathematical framework as I proceed in my studies!



#20
May911, 11:02 AM

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#21
May911, 11:23 AM

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[tex] \varphi = \int dx \, x\rangle \, \varphi(x) [/tex] 


#22
May911, 11:26 AM

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Like a scale of Sobolev spaces. See, e.g., @article{antoine1980partial, title={{Partial inner product spaces. IV. Topological considerations}}, author={Antoine, J.P.}, journal={Journal of Mathematical Physics}, volume={21}, pages={2067}, year={1980} } and other work by Antoine that you can easily find with Google. Independent of that, your remarks on the resolution of unity (and my explicit form of it in the mixed case) show that orthogonality between bound states and scattering states must be assumed to make the Dirac braket notation work correctly in the case of a mixed spectrum. 


#23
May911, 11:29 AM

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#24
May911, 11:57 AM

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I managed though to find a review (<Quantum Mechanics beyond Hilbert space>) by Antoine in 1998 in a book edited among others by Arno Bohm which, however, prompted me to the concept of <tight riggings> (not fully emphasized by de la Madrid's analysis mentioned above) which is really interesting from my perspective. It turns out that, if one is interested in finding only the real portion of the observables' spectra by making a socalled <tight rigging> of the Hspace, he could miss the existence of Gamov vectors which are used in scattering theory. 


#25
May911, 12:50 PM

P: 31

I know that this can be regarded as offtopic, but I'll ask anyway: the reason I asked this question was this. When one has a Dyson series (which is an operator series) and an unperturbed Hamiltonian with both continuous and discrete spectrum (assuming for simplicity that they're not degenerate), the thing one does is insert "a lot" of resolutions of the identity with the eigenstates of the unperturbed Hamiltonian and then obtains a formula that is the generalization of the formula of timedependent perturbation theory that is found on many books. This way one can make any initial state evolve frome time 0 to time t. I'll take the initial state to be a bound eigenstate and I'll look for the probability density to find at time t the initial state in an interval of energy in the continuous spectrum of the unperturbed Hamiltonian. That is, I evaluate the scalar product <En_t> (where E> is a "continuous eigenstate" and n_t> is the initial bound state evolved at time t) and take the modulus square of this complex number. From what I know about standard quantum mechanics, this is the approach used to derive Fermi's golden rule. The thing is that if I use the resolution of the identity with both bound and scattering states this probability density comes off as 0 (due to the fact that, as you told me, scattering states are orthogonal to bound states), while in all of the books I read the authors use the resolution of the identity ONLY with bound states even if they say that the unperturbed Hamiltonian possesses scattering staets as well. I don't understand where my reasoning is wrong...



#26
May1011, 02:26 AM

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Bottom line, I don't know how to put your calculations in a rigorous way. Try to attach a scanned version of your work, to see if, at least at a formal level, you're doing something wrong. 


#27
May1011, 03:40 AM

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giova7_89,
Like Dex said, post at least some of your work so people can see what you're doing. Problems like you describe should be straightforward, (at least at low orders of perturbation). E.g., a basic derivation of the Fermi golden rule is given in one of the external links at the bottom of this Wiki page: http://en.wikipedia.org/wiki/Fermi_golden_rule I.e., follow the like titled "Derivation using timedependent perturbation theory". The cohabitation of discrete and continuous eigenstates is reasonably straightforward at the practical level. You just need to calculate something like [tex] \langle \psi_a  H'(t)  \psi_b \rangle\> [/tex] where H' is the (timedependent) interaction Hamiltonian, and the psi's can be any of the eigenstates of the free Hamiltonian. 


#28
May1011, 06:23 AM

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http://streaming.ictp.trieste.it/preprints/P/85/232.pdf is free. http://www.hindawi.com:80/journals/amp/2010/457635/ also seems to be free. 


#29
May1011, 06:26 AM

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#30
May1011, 04:21 PM

P: 31

I wrote a short .pdf with my calculations and I'll post it here. Let me know what you think. Thank you!



#31
May1011, 10:06 PM

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[tex] \langle n  V  E \rangle ~\ne~ 0 ~. [/tex] 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. 


#32
May1011, 11:30 PM

P: 647

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 [itex]\xi^r[/itex] and [itex]\xi^s[/itex] to denote discrete eigenvalues and [itex]\xi'[/itex] and [itex]\xi''[/itex] to denotes continuous eigenvalues, we have the set of equations [tex]\langle\xi^r\xi^s\rangle=\delta_{\xi^r\xi^s}[/tex] [tex]\langle\xi^r\xi'\rangle=0[/tex] [tex]\langle\xi'\xi''\rangle=\delta(\xi'\xi'')[/tex] as the generalizations of [tex]\langle\xi'\xi''\rangle=\delta_{\xi'\xi''}[/tex] and [tex]\langle\xi'\xi''\rangle=\delta(\xi'\xi'').[/tex] 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. 


#33
May1111, 04:11 AM

P: 31

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!



#34
May1211, 12:18 PM

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#35
May1311, 08:38 AM

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Unless you also relax your interest.... 


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