"Continuous eigenstates" vs "discrete eigenstates"


by giova7_89
Tags: discrete eigenstates
giova7_89
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#19
May9-11, 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!
dextercioby
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#20
May9-11, 11:02 AM
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Quote Quote by A. Neumaier View Post
But the inner product can be partially extended to this extended space, too, in our case for all psi such that the limit
<E|psi>:= lim_{j\to inf} <\phi_j|psi>
exists and is independent on the approximating sequence phi_j --> |E>. This can be made precise more generally with the notion of a partial inner product space (mainly promoted by by Antoine).

In this extension (and such an extension must be used to make sense at all of the question of the OP), the scattering states themselves are orthogonal to the bound states.
Sorry, but I'm not familiar with PIPs. I only presented my version based on what I read and understand on the RHS approach, as I continously try to make sense of Rafael de la Madrid's PHD thesis of 2001.
dextercioby
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May9-11, 11:23 AM
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Quote Quote by giova7_89 View Post
I take advantage of the presence of people who seem to know a lot about maths to ask this: if I have an Hamiltonian with both discrete and continuous spectrum does the resolution of the identity with the eigenstates of such an Hamiltonian have to take in account both continuous and discrete eigenstates?
Yes, actually the so-called <resolution of identity> is a consequence of the general spectral theorem for a self-adjoint operator.

Quote Quote by giova7_89
And if I have a physical state vector (which is normalizable) which I want to expand with the eigenstates of this Hamiltonian, do I have to use both kind of states (discrete and continuous)?
Yes, the interesting situation arises if a physical state vector (as you say, normalizable) needs to be expanded with respect to the continuum of states of an operator which doesn't have discrete spectrum (let's choose the coordinate operator for the free particle). The decomposition is possible, it's expressible through an integral which contains as 'nucleus' the so-called <Schroedinger wave function>.

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

Quote Quote by giova7_89
For example, the Hamiltonian for the Hydrogen atom has both kind of spectrums. What should I do to expand a general state vector?
See Arnold's comment above.
A. Neumaier
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May9-11, 11:26 AM
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Quote Quote by dextercioby View Post
Sorry, but I'm not familiar with PIPs. I only presented my version based on what I read on the RHS approach, as I continously try to make sense of Rafael de la Madrid's PHD thesis of 2001.
PIP is a refinement of the RHS, keeping an infinite scale of spaces between the extremes.
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 bra-ket notation work correctly in the case of a mixed spectrum.
A. Neumaier
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#23
May9-11, 11:29 AM
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Quote Quote by A. Neumaier View Post
Yes, you need both. If the dissociation threshold has zero energy then in Dirac's notation, with appropriate scaling of the continuum states,
[tex] |\psi\rangle=
\sum_n |n\rangle\langle n|\psi\rangle + \int_0^\infty dE |E\rangle\langle E|\psi\rangle.
[/tex]
Actually, this is the formula for a single oscillator only. For hydrogen, the continuous spectrum is degenerate and one needs to integrate also over two additional labels complementing the energy.
dextercioby
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May9-11, 11:57 AM
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Quote Quote by A. Neumaier View Post
PIP is a refinement of the RHS, keeping an infinite scale of spaces between the extremes.
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 bra-ket notation work correctly in the case of a mixed spectrum.
Unfortunately, I don't have access to non-free articles.

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 so-called <tight rigging> of the H-space, he could miss the existence of Gamov vectors which are used in scattering theory.
giova7_89
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#25
May9-11, 12:50 PM
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I know that this can be regarded as off-topic, 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 time-dependent 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 <E|n_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...
dextercioby
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May10-11, 02:26 AM
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Quote Quote by giova7_89 View Post
I know that this can be regarded as off-topic, 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 time-dependent 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 <E|n_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...
Regarding your dilemma, I'm pretty sure that there's no published mathematically rigorous version of what you're trying to do (but I may be wrong). I have said it in another thread, the non-stationary perturbation theory in most textbooks is developed from an unperturbed Hamiltonian which has only bound (elements of a Hilbert space) states. The discussion one finds, e.g. in Cohen & Tannoudji (2.volume), on the <Fermi golden rule>, even though uses eigenstates for continous spectrum, is not put in a mathematical rigorous fashion, namely using rigged Hilbert spaces. Apparently, no textbook does. I would be interested in placing that discussion in terms of RHS, then tackling the Dyson series.

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.
strangerep
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#27
May10-11, 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 time-dependent 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 (time-dependent) interaction Hamiltonian, and the psi's can
be any of the eigenstates of the free Hamiltonian.
A. Neumaier
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#28
May10-11, 06:23 AM
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Quote Quote by dextercioby View Post
Unfortunately, I don't have access to non-free articles.
You can find some articles online at http://www.fyma.ucl.ac.be/FymaPubli

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.

Quote Quote by dextercioby View Post
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 so-called <tight rigging> of the H-space, he could miss the existence of Gamov vectors which are used in scattering theory.
I don't understand. Gamov vectors do not correspond to the real portion of the spectrum, so why should they _not_ be missing?
giova7_89
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#29
May10-11, 06:26 AM
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Quote Quote by strangerep View Post
giova7_89,

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 time-dependent perturbation theory".
I read the article in the link you posted and it considers an Hamiltonian with only discrete eigenstates, but so close to each other that their discrete index can be approximated to a continuous index. There's nothing wrong with me, I just wanted to know if that was the way it's done. This evening (I live in Italy so it'll be afternoon in America!) when I'll be able to use my computer with LaTex I'll write down some formulas in a .pdf and I'll post it. Thank you very much for the help, I'll try to make it as soon as possible!
giova7_89
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#30
May10-11, 04:21 PM
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I wrote a short .pdf with my calculations and I'll post it here. Let me know what you think. Thank you!
Attached Files
File Type: pdf Dyson series.pdf (106.9 KB, 9 views)
strangerep
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#31
May10-11, 10:06 PM
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Quote Quote by giova7_89 View Post
I wrote a short .pdf with my calculations [...]
In your eq(4), you haven't allowed for the possibility that perhaps

[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.
jfy4
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#32
May10-11, 11:30 PM
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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.
giova7_89
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#33
May11-11, 04:11 AM
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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
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#34
May12-11, 12:18 PM
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Quote Quote by A. Neumaier View Post
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
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#35
May13-11, 08:38 AM
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Quote Quote by dextercioby View Post
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....


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