Rigged Hilbert Space Φ ⊂ H ⊂ Φ'

[sweet springs]

Answering your question, I was expecting you would show me something wonderful as outcome of it but, OK, Now I come back to my question #8 and start again.

In the paper the author says

"[37] The reason why the derivatives of ϕ(x) must vanish at x = a, b is that we want to be able to
apply the Hamiltonian H as many times as we wish. Since repeated applications of H to ϕ(x)
involve the derivatives of V (x)ϕ(x), and since V (x) is discontinuous at x = a, b, the function
V (x)ϕ(x) is infinitely differentiable at x = a, b only when the derivatives of ϕ(x) vanish at
x = a, b. For more details, see Ref. [19]. The vanishing of the derivatives of ϕ(x) at x = a, b must
be viewed as a mathematical consequence of the unphysical sharpness of the discontinuities of
the potential, rather than as a physical consequence of Quantum Mechanics. Note also that in
standard numerical simulations, for example, Gaussian wave packets impinging on a rectangular
barrier, one never sees that the wave packet vanishes at x = a, b. This is due to the fact that on
a Gaussian wave packet, the Hamiltonian (3.1) can only be applied once."

Obviously the ground state of square well potential with discrete energy eigenvalue ψ in #13 does not belong to Φ or S(R-{a,b}) in spite of #38 and #39 ( I am not yet able to understand it rigorously though).

We may have two different interpretations:

It's OK. Some eigenstates with discrete eigenvalues are not in Φ but in Φ', linear functional space of Φ.(ref #41) We should follow mathematics. Physical or unphysical does not make sense here.

or

It's not OK. If some eigenstates with discrete eigenvalues are not in Φ but in Φ', it means that their Hamiltonian are unphysical. We know the discrete energy eigenstate lies here at laboratory. It really is in Φ. No extraoridinary divergence is happening here. We should replace Hamiltonian of our system with more real one.

I prefer the latter (See #10) I wish C^∞ function potential saves us.

Eigenstates of 1D rectangular barrier discussed in this paper has the same discontinuity at x=a,b but Φ' is required by spectral continuity anyway.

Best

 

[strangerep]

Answering your question, I was expecting you would show me something wonderful as outcome of it

Sorry if I disappointed you. But,... I see several (very) important features and lessons in the square-well example. If you list here the features/lessons you think are/were important, I'll try to add some more.

 

[strangerep]

[...]
Obviously the ground state of square well potential with discrete energy eigenvalue ψ in #13 does not belong to Φ or S(R-{a,b}) in spite of #38 and #39

http://arxiv.org/abs/quant-ph/050205 discusses the rectangular barrier potential, not the square well potential. You keep trying to apply results/statements from the rectangular barrier case to the square well case. But that's invalid. Here's why...

For the square well, we found a set of bound (normalizable) states (E<0) with discrete energy spectrum, and a set of scattering (non-normalizable) states (E>0) with continuous energy spectrum. We have only been talking about the bound states, and for these an ordinary Hilbert space formulation is adequate for doing physics -- because the E<0 eigenstates are normalizable. Hence we don't need the more complicated machinery of rigged Hilbert space for the bound states (though we do need it for the scattering states).

In contrast, for the rectangular barrier there are no bound states. The energy spectrum is $[0,\infty)$, i.e., continuous. In this case, rigged Hilbert space is useful, since the energy eigenstates are non-normalizable.

So,... if we're going to talk about that paper, we should talk in terms of the rectangular barrier example, not the square well example.

 

[sweet springs]

Thanks for your suggestion.

because the E<0 eigenstates are normalizable. Hence we don't need the more complicated machinery of rigged Hilbert space for the bound states (though we do need it for the scattering states).

I would like to support you, but as for the ground state of square well Ψ, as you said

I already told you: ψ＾(3) belongs to the larger space Φ′of distributions.

P^3 Ψ leaves Hilbert space for the space of distribution Φ'. How can we mediate these two ?

 

[strangerep]

P^3 Ψ leaves Hilbert space for the space of distribution Φ'. How can we mediate these two ?

To do quantum physics, we need essentially 2 things:

1) A dynamical symmetry Lie algebra applicable to the system (i.e., the generators of the Lie group which maps solutions of the equations of motion into other solutions).

2) A representation of this dynamical group in which probabilities (and a concept of measurement) can be sensibly defined. This means we need a representation in which the important generators of the dynamical algebra are self-adjoint (so that they have real eigenvalues), and we need normalizable states.

The difficulty for (e.g.,) systems like the rectangular barrier potential is that the energy eigenstates are not normalizable. Nevertheless, we can construct linear combinations of those eigenstates which are normalizable, and therefore live in a Hilbert space (even though the energy eigenstates live in the larger space $\Phi'$). Therein lies the central point of the rigged Hilbert space construction: the nuclear spectral theorem shows that the physical states can be decomposed in terms of the unphysical (non-normalizable) states in $\Phi'$.

Then,... regarding $P^3 \psi$ in the square well case: a crucial point here is that $P^3$ by itself is not part of dynamical symmetry algebra of the square well. One can see this from the fact that the system is not translation-invariant: if we apply a finite translation to $\psi(x)$ like this:$$\psi(x) ~\to~ \psi(x+q) = e^{-iqP}\psi(x) ~,$$we get a bad answer sometimes, e.g., if we translate a point outside the well to the inside where the solution is very different.

In contrast, the Hamiltonian (i.e., the operator on the left hand side of the TISE) is a good dynamical symmetry generator for this system. (I think there aren't any more, but I'm not 100% sure -- that needs some calculation.) In the bound state (E<0) case, the energy eigenstates are normalizable, hence we don't need a larger space $\Phi'$. The ordinary Hilbert space gives us what we need. But for the scattering states (E>0) this is no longer true, and we do need the more sophisticated RHS construction.

 

[sweet springs]

Then,... regarding P3ψP^3 \psi in the square well case: a crucial point here is that P3 by itself is not part of dynamical symmetry algebra of the square well. One can see this from the fact that the system is not translation-invariant: if we apply a finite translation to ψ(x)\psi(x) like this:

ψ(x) → ψ(x+q)=e−iqPψ(x) ,​

we get a bad answer sometimes, e.g., if we translate a point outside the well to the inside where the solution is very different.

Lie albebra teaches us the physical quantity conserves or not. Whether Symmetric or not the system is, we can apply momentum operator to the states.
I do not think the symmetry or Lie algebra is a crucial part of our problem. For example, harmonic oscillator V(x)=ax^2 has only discrete energy eigenstates and its derivatives show no divergence. What's the difference?

 

[strangerep]

Lie albebra teaches us the physical quantity conserves or not.

No. Only those Lie symmetries which commute with the Hamiltonian correspond to conserved quantities. A Lie symmetry which does not commute with the Hamiltonian does not correspond to a conserved quantity.

Whether Symmetric or not the system is, we can apply momentum operator to the states.

Perhaps you should study more about quantum dynamical symmetries before making such assertions.

I do not think the symmetry or Lie algebra is a crucial part of our problem. For example, harmonic oscillator V(x)=ax^2 has only discrete energy eigenstates and its derivatives show no divergence. What's the difference?

The harmonic oscillator's dynamical symmetry group is the (so-called) Schrodinger--Niederer group, which contains the ordinary Galilei group as a subgroup, and one of its generators is indeed the usual spatial translation operator, which corresponds to the familiar momentum operator $-i\hbar d/dx$.

In general, it is more likely that a well-behaved potential will yield more Lie symmetries (since Lie symmetries are continuous symmetries). But precisely which Lie symmetries apply to a particular case depends on the details of that case.

 

[sweet springs]

Well,

Then,... regarding P3ψin the square well case: a crucial point here is that P3P^3 by itself is not part of dynamical symmetry algebra of the square well.

So you say that P or P^2 is part of dynamical symmetry algebra but P^3 is not, right?

 

[sweet springs]

Another idea to remove for the difficulty is modification of energy eigenfunction;

Square well potential:

V(x)=−D for −w/2<x<w/2, 0 otherwise.​

Wave function of the ground state in coordinate representation is:

ψ(x)=Beκ(x+w2)θ(−x−w2)+Ccos(kx)θ(−x+w2)θ(x+w2)+Be−κ(x−w2)θ(x−w2).​

.

for x: (-∞, -w/2-ε), (-w/2+ε, w/2-ε),(w/2+ε, ∞) and

ψ(x)=0 for [ -w/2-ε, -w/2+ε], [ w/2-ε, w/2+ε] thus ψ^(n)(x)=0, though there is concern below shown.

0<ε→0

I try to follow the prescription of S(R-|a,b}) in the paper (4.1). Thus δand δ^(n) so divergence disappear.

I find two concerns in my prescription;
-Differentiation at x= -w/2-ε, -w/2+ε, w/2-ε, w/2+ε where only right or left side differentiation is allowed. If distributions appear here also there is no merit of this change.
-Momentum wave function by Fourier transformation of the coordination wave function now clearly shows divergence of <p^6>. Does momentum wave function also change with the chnage of the coordinate wave function above? I am not sure Fourier transformation care aout such a subtle modification of the coordinate wave function.Best

 

[strangerep]

First, I must correct something I said earlier:

The harmonic oscillator's dynamical symmetry group is the (so-called) Schrodinger--Niederer group, [...]

My statement was misleading, so I must correct it...

The harmonic oscillator algebra $ho(1)$ is merely isomorphic to the Schrodinger--Niederer algebra. The relationship involves a fractional-linear coordinate transformation. The usual spatial translation operator is not a generator of $ho(1)$. The original reference is:

U. Niederer,
"The Maximal Kinematical Invariance Group of the Harmonic Oscillator",
Helv. Phys. Acta, vol 46, 1973, p191.
Accessible here.

[...] So you say that P or P^2 is part of dynamical symmetry algebra but P^3 is not, right?

I think not. I.e., I think that $P =i\hbar d/dx$ is not one the dynamical symmetries. You can check by the following exercise:

Suppose $\psi(x)$ is a solution of the (square-well) TISE for energy $E_\psi$. I.e., $\psi(x)$ satisfies:
$$H(d/dx,\,x) \psi(x) ~=~ E_\psi \psi(x) ~.$$Then let
$$\phi(x) ~:=~ e^{q\frac{d}{dx}}\psi(x) ~\equiv~ \psi(x+q) ~.$$
If you can prove that $\phi(x)$ satisfies
$$H\left(\frac{d}{dx},\,x+q\right) \phi(x) ~=~ E_\phi \phi(x) ~,$$ where $E_\phi$ is one of the allowed energy eigenvalues in this problem, then $d/dx$ generates a symmetry: it maps solutions to solutions. If this is not so, then $d/dx$ is not a symmetry of that case.

 

[strangerep]

Another idea to remove for the difficulty is modification of energy eigenfunction; [...]

It is usually more profitable to study thoroughly what other people have already achieved before constructing one's own theory.

In this vein, it may be helpful to study distribution theory more deeply. I found Appendix A of Nussenzveig's book on "Causality and Dispersion Relations" quite helpful.

 

[sweet springs]

Thanks for your comment.
I am excited to know that the square well potential that appears n many QM textbooks seems to have something more to be explored.
I agree with you that I should learn more.