Rigged Hilbert Space Φ ⊂ H ⊂ Φ'

[sweet springs]

−(ψ,ψ(6))=(ψ(3),ψ(3))​

-(ψ,ψ^{(6)}) =(ψ^{(3)}, ψ^{(3)}) might be unreliable, since P is only self-adjoint on a suitably well-behaved space of functions.

P does not appear in this equation but differentiation of wave function do. Partial integration works here because any ψ^(n) is well damped far away.
Anyway, I will be interested in your evaluation.

 

[sweet springs]

Almost mechanically,
<br /> \psi^{(3)}(x) ~=<br /> \Big(\psi^{(2)}_M(-w/2) -\psi^{(2)}_L(-w/2)\Big) \delta(x+w/2)+\Big(\psi^{(2)}_R(w/2) - \psi^{(2)}_M(w/2) \Big)\delta(x-w/2) \\<br /> +~ \Big(1 - \theta(x+w/2)\Big)\psi^{(3)}_L(x)<br /> ~+~ \Big(\theta(x+w/2)-\theta(x-w/2)\Big)\psi^{(3)}_M(x)<br /> ~+~ \theta(x-w/2) \psi^{(3)}_R(x),

\psi^{(4)}(x) ~=<br /> \Big(\psi^{(2)}_M(-w/2) -\psi^{(2)}_L(-w/2)\Big) \delta^{(1)}(x+w/2)+\Big(\psi^{(2)}_R(w/2) - \psi^{(2)}_M(w/2) \Big)\delta^{(1)}(x-w/2) \\+\Big(\psi^{(3)}_M(-w/2) -\psi^{(3)}_L(-w/2)\Big) \delta(x+w/2)+\Big(\psi^{(3)}_R(w/2) - \psi^{(3)}_M(w/2) \Big)\delta(x-w/2)<br /> \\+~ \Big(1 - \theta(x+w/2)\Big)\psi^{(4)}_L(x)<br /> ~+~ \Big(\theta(x+w/2)-\theta(x-w/2)\Big)\psi^{(4)}_M(x)<br /> ~+~ \theta(x-w/2) \psi^{(4)}_R(x),

\psi^{(5)}(x) ~=<br /> \Big(\psi^{(2)}_M(-w/2) -\psi^{(2)}_L(-w/2)\Big) \delta^{(2)}(x+w/2)+\Big(\psi^{(2)}_R(w/2) - \psi^{(2)}_M(w/2) \Big)\delta^{(2)}(x-w/2)<br /> \\+\Big(\psi^{(3)}_M(-w/2) -\psi^{(3)}_L(-w/2)\Big) \delta^{(1)}(x+w/2)+\Big(\psi^{(3)}_R(w/2) - \psi^{(3)}_M(w/2) \Big)\delta^{(1)}(x-w/2)<br /> \\+\Big(\psi^{(4)}_M(-w/2) -\psi^{(4)}_L(-w/2)\Big) \delta(x+w/2)+\Big(\psi^{(4)}_R(w/2) - \psi^{(4)}_M(w/2) \Big)\delta(x-w/2)<br /> \\+~ \Big(1 - \theta(x+w/2)\Big)\psi^{(5)}_L(x)<br /> ~+~ \Big(\theta(x+w/2)-\theta(x-w/2)\Big)\psi^{(5)}_M(x)<br /> ~+~ \theta(x-w/2) \psi^{(5)}_R(x),

\psi^{(6)}(x) ~=<br /> \\ \Big(\psi^{(2)}_M(-w/2) -\psi^{(2)}_L(-w/2)\Big) \delta^{(3)}(x+w/2)+\Big(\psi^{(2)}_R(w/2) - \psi^{(2)}_M(w/2) \Big)\delta^{(3)}(x-w/2)<br /> \\+\Big(\psi^{(3)}_M(-w/2) -\psi^{(3)}_L(-w/2)\Big) \delta^{(2)}(x+w/2)+\Big(\psi^{(2)}_R(w/2) - \psi^{(3)}_M(w/2) \Big)\delta^{(3)}(x-w/2)<br /> \\+\Big(\psi^{(4)}_M(-w/2) -\psi^{(4)}_L(-w/2)\Big) \delta^{(1)}(x+w/2)+\Big(\psi^{(4)}_R(w/2) - \psi^{(4)}_M(w/2) \Big)\delta^{(1)}(x-w/2)<br /> \\+\Big(\psi^{(5)}_M(-w/2) -\psi^{(5)}_L(-w/2)\Big) \delta(x+w/2)+\Big(\psi^{(5)}_R(w/2) - \psi^{(5)}_M(w/2) \Big)\delta(x-w/2)<br /> \\+~ \Big(1 - \theta(x+w/2)\Big)\psi^{(6)}_L(x)<br /> ~+~ \Big(\theta(x+w/2)-\theta(x-w/2)\Big)\psi^{(6)}_M(x)<br /> ~+~ \theta(x-w/2) \psi^{(6)}_R(x) ,

and so on. In general,
\psi^{(q)}(x) ~\\<br /> = \sum^{\{m+n+1=q\}}_{m,n=0\ or\ positive\ integer}\Big(\Big(\psi^{(m)}_M(-w/2) -\psi^{(m)}_L(-w/2)\Big) \delta^{(n)}(x+w/2)+\Big(\psi^{(m)}_R(w/2) - \psi^{(m)}_M(w/2) \Big)\delta^{(n)}(x-w/2) \Big)\\+~ \Big(1 - \theta(x+w/2)\Big)\psi^{(q)}_L(x) ~+~ \Big(\theta(x+w/2)-\theta(x-w/2)\Big)\psi^{(q)}_M(x) ~+~ \theta(x-w/2) \psi^{(q)}_R(x)

 

[strangerep]

I'll introduce the following definitions to simplify things.
$$a ~:=~ \frac{2mE}{\hbar^2} ~,~~~~ b ~:=~ \frac{2mD}{\hbar^2} ~.$$
Then, from the Schrodinger eqn, $$\psi''_L = - a \psi_L(x) ~,~~~~~
\psi''_M(x) = - (a+b) \psi_M(x) ~,~~~~~
\psi''_R(x) = - a \psi_R(x) ~.$$So, $\psi^{(3)}$ simplifies to
$$\psi^{(3)}(x) ~=~ a \Big(\theta(x+1) - 1 \Big) \psi'_L(x)
~-~ (a+b) \Big(\theta(x+1) - \theta(x-1) \Big) \psi'_M(x)
~-~ a \theta(x-1) \psi'_R(x) \\~~~~~~~~~~~~~~~~~~
~-~ b \Big(\delta(x+1) - \delta(x-1) \Big) \psi_M(x) ~.$$
Therefore, inner products like $\left\langle \phi|\psi^{(3)}(x)\right\rangle$ remain finite, but the norm $\left\|\psi^{(3)}(x)\right\|$ is ill-defined since the integral involves products of delta functions.

Similar behaviour occurs for higher derivatives. E.g, inner products like $\left\langle \phi|\psi^{(6)}(x)\right\rangle$ remain finite, even though norm $\left\|\psi^{(6)}(x)\right\|$ is undefined.

Important: This is what I meant before (in post #30) about $P$ not being self-adjoint when applied to $\psi^{(3)}(x)$ and higher derivatives.

So, FINALLY, we can get back to rigged Hilbert spaces: the above shows that the $\psi^{(3)}(x)$ and higher derivatives can be regarded as elements of a larger space of distributions (since things like $\left\langle \phi|\psi^{(3)}(x)\right\rangle$ are finite for $\Phi\in H$).

 

[sweet springs]

Important: This is what I meant before (in post #30) about PP not being self-adjoint when applied to ψ(3)(x)\psi^{(3)}(x) and higher derivatives.

Thanks for confirming difficulties of square well bound state eigenfunction. Many textbook describe the solution but now we know that it leads to silly results like <p^6> diverges.

As I wrote in #10, I assume this difficulty results from not realizable square shape.

Best.

 

[sweet springs]

So, FINALLY, we can get back to rigged Hilbert spaces: the above shows that the ψ(3)(x) and higher derivatives can be regarded as elements of a larger space of distributions (since things like ⟨ϕ|ψ(3)(x)⟩ are finite for Φ∈H).

The author of the paper introduces Rigged Hilbert Space for continuous specrum of eigenvalue.
He says that the states of discrete eigenvalue remains in Hilbert space.(ref my #8)
We have been looking for the gournd state of square well potential which has discrete energy eigenvalue. You state that the ground state does not belong to Φ but in the space larger than Hilbert space.

How do you think this contradiction takes place?

I think that Hamiltonian should be carefully conditioned for the theories to be applied properly. When we choose wrong Hamiltonians, the discrete eigenstates do not belong to Φ yet. From the case of square well potential we shoule learn that potential V(x) should be C^∞ function of x, at least. Failed model V(x) results a failed answer. (ref my#10, #20)

Best

 

[strangerep]

Thanks for confirming difficulties of square well bound state eigenfunction. Many textbook describe the solution but now we know that it leads to silly results like <p^6> diverges.

Whether or not such a result is "silly" or "expected", depends on one's point of view.

As I wrote in #10, I assume this difficulty results from not realizable square shape.

I'm not sure that is the fundamental reason. What if $V(x)=0$ everywhere (i.e., a free particle)? The plane wave solutions are not normalizable.

When we construct a Hilbert space from solutions of a particular Schrodinger equation, we construct a representation of the dynamical symmetry group (i.e., the group which maps solutions into solutions). There is no guarantee that other operators are necessarily well-represented on that Hilbert space. We're seeing an example of this in the square well problem, but even with smooth potentials, there's no guarantee that arbitrary operators are well-represented on the Hilbert space of solutions.

 

[strangerep]

The author of the paper introduces Rigged Hilbert Space for continuous specrum of eigenvalue. He says that the states of discrete eigenvalue remains in Hilbert space.(ref my #8)
We have been looking for the ground state of square well potential which has discrete energy eigenvalue.

But what about all the other bound states?? What values of $E$ are allowed by the Schrodinger equation? You must find the entire spectrum before we can say the problem is "completely solved".

You state that the ground state does not belong to Φ but in the space larger than Hilbert space.

That's not what I said. I said $\psi^{(3)}$ is not in the Hilbert space (since its norm is undefined).

I think that Hamiltonian should be carefully conditioned for the theories to be applied properly. When we choose wrong Hamiltonians, the discrete eigenstates do not belong to Φ yet. From the case of square well potential we should learn that potential V(x) should be C^∞ function of x, at least. Failed model V(x) results a failed answer. (ref my#10, #20)

I think you have missed the point. To understand why, you must first answer the question: "What values of $E$ are allowed for the bound states in the current square well problem?"

 

[sweet springs]

After the post I found the relating discription in the paper as follows.
"However, it is not clear that one can always
find a nuclear space that remains invariant under the action of the observables.
Nevertheless, Roberts has shown that such exists when the potential is infinitely
often differentiable except for a closed set of zero Lebesgue measure [19]."
I have not understood it fully but infinitely differentiability of potential seems important.

Best

 

[strangerep]

After the post I found the relating discription in the paper as follows.
"However, it is not clear that one can always find a nuclear space that remains invariant under the action of the observables.
Nevertheless, Roberts has shown that such exists when the potential is infinitely often differentiable except for a closed set of zero Lebesgue measure [19]."
I have not understood it fully but infinitely differentiability of potential seems important.

The square well potental in $C^\infty$ everywhere, except at the isolated points $x = \pm\omega/2$. An isolated point has zero Lebesgue measure, iiuc.

Anyway, we already know that a nuclear space exists for the current problem -- it gives the usual Schwarz theory of distributions.

 

[sweet springs]

Anyway, we already know that a nuclear space exists for the current problem -- it gives the usual Schwarz theory of distributions.

I wonder how we can mediate your above line and the irregularity of ||ψ＾(3)|| for square well potential.
Best

 

[strangerep]

I wonder how we can mediate your above line and the irregularity of ||ψ＾(3)|| for square well potential.

I already told you: $\psi^{(3)}$ belongs to the larger space $\Phi'$ of distributions.

 

[sweet springs]

So back to my #8, you say that energy eigenstate of bound state of square well potential ψwhich has discrete enregy eigenvalue does not belong to Φ, the maximal invariant subspace of the algebra generated by Q, P
and H, because P^3ψ does not come back to Φ? Some eigenstates of discrete eigenvalues also need linear functional space Φ' as well as eigenstates of continuous eigenvalues do, right?
Best

 

[strangerep]

[...] you say that energy eigenstate of bound state of square well potential ψ which has discrete energy eigenvalue does not belong to Φ,

I did not say "discrete energy eigenvalue".

I think we cannot make further progress until you at least try to answer my earlier question about the spectrum of the energies of bound states. So... one (and only 1) more time: what values of E does the Schrodinger equation allow for bound states in the square well problem?

 

[sweet springs]

I have already showed my solution of the Shroedinger equation in #13. What can I do more?

 

[strangerep]

I have already showed my solution of the Shroedinger equation in #13.

In your post #13, I see only a solution for the ground state.

EDIT: Looking a bit closer, I see that although you have called it "ground state", it actually covers all the bound states. So... since you have an expression for E, what range of numerical values can E take?

 

[sweet springs]

You will find in #13 that first the depth D and the width w defines α=α(D,w) , then all the other coefficients and parameters including energy are expressed by α and w, thus D and w.
Best

 

[sweet springs]

Surely we can get solutions for the 1st, 2nd, .. nth excited bound states with discrete energy eigenvalues in similar way as in #13. And we can get solutions of non-binding states of continuous energy eigenvalues. How could these full set solutions matter with the difficulty of high order derivative of the ground state Ψ? May continuous spectrum part do harm on discrete spectrum part? Best

 

[strangerep]

I have become confused about what your question is. The bound states (E<0) have a discrete spectrum (though it is hard to find analytically, as it involves a transcendental equation). The scattering (E>0) states have continuous spectrum.

Please restate your question.

 

[sweet springs]

I think we cannot make further progress until you at least try to answer my earlier question about the spectrum of the energies of bound states. So... one (and only 1) more time: what values of E does the Schrodinger equation allow for bound states in the square well problem?

I just tried to answer your above question in #46, #47 and #48. Now is it OK?

 

[strangerep]

OK, so,... is this thread finished now, or is there more to discuss?

 

[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.