Continuous eigenstates vs discrete eigenstates

In summary: A scattering eigenstate is (if considered a state at all) a singular state in some dual space, which is representable as a limit state of regular states in the space spanned by all scattering states. All these are orthogonal to the bound state, so the same holds for the limiting state.
  • #1
giova7_89
31
0
"Continuous eigenstates" vs "discrete eigenstates"

There's this thing that's bothering me: if I have an Hamiltonian with a discrete and continuous spectrum, every book I read on quantum mechanics says that eigenvectors of discrete eigenvalues are orthogonal in the "Kronecker sense" (their scalar product is a Kronecker delta) and that eigenvectors of continuous eigenvalues are orthogonal in the "Dirac sense" (their scalar product is a Dirac delta). But what about the scalar product between a continuous eigenstate and a discrete eigenstate? Is it 0 or not?
 
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  • #2


There's no real scalar product in the space in which the eigenstates <live>. I'll get back on this.
 
  • #3


You mean in the space where the continuous eigenstates live?
 
  • #4


The continuous eigenstates (linear functionals) live in one space, while the discrete ones (ordinary vectors) in another. Surely, one can move the vectors to the functionals' space, but that doesn't make the functionals' space a genuine scalar product space. A scalar product defines a norm, computing the norm of a continuous eigenstate yields infinity, because they are not properly orthogonal.
 
  • #5


The scalar product between a continuous eigenstate and a discrete eigenstate is simply zero, no delta function involved, since they cannot be the same state.

Consider one-dimensional scattering from a potential well deep enough to have bound states. You can write ∫ψ1*(x)ψ2(x) dx for any two of the states. For two scattering states it will be a Dirac delta function, for two bound states it will be a Kronecker delta, and for one scattering state and one bound state it will be zero.
 
  • #6


Ok I too thought about the example of one dimensional scattering. I'm sorry I can't appreciate the explanation of dextercioby because in my course of QM there wasn't any deeper hint to mathematics than what I wrote in the original post. Thank you both!
 
  • #7


The bottom line is that such a mathematical expression

[tex] \langle \psi_{\mbox{scattering}} |\psi_{\mbox{bound state}}\rangle [/tex]

can be made rigorous, but does not define a scalar product between the 2 psi's.
 
  • #8


dextercioby said:
The bottom line is that such a mathematical expression

[tex] \langle \psi_{\mbox{scattering}} |\psi_{\mbox{bound state}}\rangle [/tex]

can be made rigorous, but does not define a scalar product between the 2 psi's.

Not quite. The physical Hilbert space is the direct sum of the part spanned by the bound states and the part spanned by the scattering states. in this Hilbert space, superpositions of bounds states and superpositions of scattering states are always orthogonal.
 
  • #9


I don't agree, Arnold. What you call <physical Hilbert space> is not a scalar product space at all, not to mention a complete scalar product space. So, even if one can define the bra(c)ket <scattering|bound>, it doesn't define a scalar product in the mathematical way, by reasons explained in post #4.
 
  • #10


dextercioby said:
I don't agree, Arnold. What you call <physical Hilbert space> is not a scalar product space at all, not to mention a complete scalar product space. So, even if one can define the bra(c)ket <scattering|bound>, it doesn't define a scalar product in the mathematical way, by reasons explained in post #4.

You reasons are not cogent.

Consider the Morse oscillator http://en.wikipedia.org/wiki/Morse_potential
It has finitely many bound states n> with energies E_n<0 and in-scattering states |E> with arbitrary positive energy E. As one can easily show even without explicit formulas for the spectrum, the dissocation states psi=integral dE psi_E |E> (with |psi_E| decaying fast enough) are orthogonal to the bound states, and the physical Hilbert space L^2(R), on which the Morse Hamiltonian is defined, is a direct sum of the invariant subspaces C^N and L^2(R_+). Volume 3 of Thirring's book extends this to much more general multiparticle situations.
 
  • #11


Consider the Morse oscillator http://en.wikipedia.org/wiki/Morse_potential
It has finitely many bound states n> with energies E_n<0 and in-scattering states |E> with arbitrary positive energy E. As one can easily show even without explicit formulas for the spectrum, the dissocation states psi=integral dE psi_E |E> (with |psi_E| decaying fast enough) are orthogonal to the bound states, and the physical Hilbert space L^2(R), on which the Morse Hamiltonian is defined, is a direct sum of the invariant subspaces C^N and L^2(R_+). Volume 3 of Thirring's book extends this to much more general multiparticle situations.

I understand your point but you're talking about superpositions of eigenstates: my original question was about the "scalar product" between a single scattering state (I used the word "continuous state") and a single bound state.
 
  • #12


giova7_89 said:
I understand your point but you're talking about superpositions of eigenstates: my original question was about the "scalar product" between a single scattering state (I used the word "continuous state") and a single bound state.

A scattering eigenstate is (if considered a state at all) a singular state in some dual space, which is representable as a limit state of regular states in the space spanned by all scattering states. All these are orthogonal to the bound state, so the same holds for the limiting state.

There is nothing ill-defined, although the scattering state is not in the original Hilbert space (but in a Sobolev space with a smoothing inner product).
 
  • #13


A. Neumaier said:
You reasons are not cogent.

Consider the Morse oscillator http://en.wikipedia.org/wiki/Morse_potential
It has finitely many bound states n> with energies E_n<0 and in-scattering states |E> with arbitrary positive energy E. As one can easily show even without explicit formulas for the spectrum, the dissocation states psi=integral dE psi_E |E> (with |psi_E| decaying fast enough) are orthogonal to the bound states, and the physical Hilbert space L^2(R), on which the Morse Hamiltonian is defined, is a direct sum of the invariant subspaces C^N and L^2(R_+). Volume 3 of Thirring's book extends this to much more general multiparticle situations.

Yes, the dissociation states are indeed orthogonal to the bound states, simply because they live in the Hilbert space L^2 (R) where a scalar product can be defined. But I was talking about the pure scattering states, which don't live in L^2 (R), but into some extension of its topological dual.

You said and I quote: <The physical Hilbert space is the direct sum of the part spanned by the bound states and the part spanned by the scattering states. in this Hilbert space, superpositions of bounds states and superpositions of scattering states are always orthogonal.>

I just said that wasn't true and your example confirms me.
 
  • #14


Hm, I always thought that the idea of eigenvectors to continuous "eigenvalues" made no sense anyway (there are no continuous eigenvalues. the position operator for example has no eigenvalues at all. its spectrum is the real numbers, but it has no eigenvalues).
in a mathematical sense, one has to use the spectral theorem for noncompact operators and that only says that you can find appropriate projections to decompose the 1-operator, not that those projections correspond to eigenvectors. so i agree with dextercioby. the idea of a scalar product between the two is wrong...
 
  • #15


QuantumCosmo said:
Hm, I always thought that the idea of eigenvectors to continuous "eigenvalues" made no sense anyway (there are no continuous eigenvalues. the position operator for example has no eigenvalues at all. its spectrum is the real numbers, but it has no eigenvalues).
in a mathematical sense, one has to use the spectral theorem for noncompact operators and that only says that you can find appropriate projections to decompose the 1-operator, not that those projections correspond to eigenvectors. so i agree with dextercioby. the idea of a scalar product between the two is wrong...

Eigenstates for the continuum are well-defined in the so-called rigged Hilbert space setting. http://en.wikipedia.org/wiki/Rigged_Hilbert_space

This setting must be used to interpret Dirac's formalism mathematically. In this context the question about the inner product makes sense and one finds that the inner product is zero.
 
  • #16


dextercioby said:
Yes, the dissociation states are indeed orthogonal to the bound states, simply because they live in the Hilbert space L^2 (R) where a scalar product can be defined. But I was talking about the pure scattering states, which don't live in L^2 (R), but into some extension of its topological dual.

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.
 
  • #17


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? 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)? For example, the Hamiltonian for the Hydrogen atom has both kind of spectrums. What should I do to expand a general state vector?
 
  • #18


giova7_89 said:
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? 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)? For example, the Hamiltonian for the Hydrogen atom has both kind of spectrums. What should I do to expand a general state vector?

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]
This decomposition needs the fact (discussed before, though disputed by dextercioby) that bound states and scattering states are orthogonal in any mathematical framework where Dirac's notation makes sense.

In case of a pure point spectrum (no dissociation), the integral is absent, while in case of a pure continuous spectrum (no bound states), the sum is absent.
 
  • #19


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


A. Neumaier said:
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.
 
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  • #21


giova7_89 said:
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.

giova7_89 said:
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]

giova7_89 said:
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.
 
  • #22


dextercioby said:
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.
 
  • #23


A. Neumaier said:
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.
 
  • #24


A. Neumaier said:
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.
 
  • #25


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...
 
  • #26


giova7_89 said:
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 continuous 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.
 
  • #27


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.
 
  • #28


dextercioby said:
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.

dextercioby said:
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?
 
Last edited by a moderator:
  • #29


strangerep said:
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!
 
  • #30


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

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  • #31


giova7_89 said:
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.
 
  • #32


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. :smile:
 
  • #33


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


A. Neumaier said:
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.
 
  • #35


dextercioby said:
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...
 

1. What is the difference between continuous eigenstates and discrete eigenstates?

Continuous eigenstates refer to a set of eigenstates that form a continuous spectrum, while discrete eigenstates refer to a set of eigenstates that form a discrete or countable spectrum.

2. How are continuous eigenstates and discrete eigenstates used in quantum mechanics?

Continuous eigenstates are used to describe systems with a continuous spectrum, such as the position of a particle, while discrete eigenstates are used to describe systems with a discrete spectrum, such as the energy levels of an atom.

3. Can a system have both continuous and discrete eigenstates?

Yes, a system can have both continuous and discrete eigenstates. For example, the energy levels of a particle in a box are discrete, while the position of the particle can take on any value within the box, making it continuous.

4. How are continuous eigenstates and discrete eigenstates related to the uncertainty principle?

The uncertainty principle states that the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa. In the case of continuous eigenstates, the position is known precisely, but the momentum is uncertain, while in the case of discrete eigenstates, the energy is known precisely, but the position is uncertain.

5. Can continuous eigenstates and discrete eigenstates be used interchangeably?

No, continuous eigenstates and discrete eigenstates cannot be used interchangeably as they represent different types of spectra and have different mathematical properties. It is important to use the appropriate type of eigenstates depending on the system being studied.

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