Rigged Hilbert Space Φ ⊂ H ⊂ Φ'

[sweet springs]

Hi,
I am reading the paper http://arxiv.org/abs/quant-ph/0502053 listed in the reference of Wikipedia Rigged Hilbert Space. I have a question about the relation, Φ ⊂ H ⊂ Φ', where H is Hilbert space, Φ is its subspace and Φ' is dual space of Φ.
Φ⊂H and Φ⊂Φ' are obvious. How can we say H ⊂ Φ' ? Φ' has nothing to do with the elements of H　that is not in Φ, I assume. Your advice is appreciable.

 

[DrDu]

As H is it's own dual, it must also be included in the dual of $\Phi$, which is a subset of H. Hence H is also a subset of $\Phi'$.

 

[sweet springs]

Thanks. From your post I assume there is a theorem Φ⊂H ⇒ H' ⊂ Φ' where A' means dual space of A. Am I on the right track?

 

[DrDu]

You can make a theorem out of everything. However, it should be obvious. The definition of H includes that you have a scalar product defined between all elements. Hence each element of H is also an element of H', or H=H'. As $\Phi \subset H$ $\langle h| \phi \rangle$ is well defined for all $h\in H$ and all $\phi \in \Phi$. Hence H must be a subset of $\Phi'$.

 

[bhobba]

If you want to really delve into Rigged Hilbert Spaces this doctoral dissertation is a good reference:
http://physics.lamar.edu/rafa/webdis.pdf

But a background in analysis is required:
http://fa.its.tudelft.nl/~haase/Dokus/afa-lectures1-14.pdf

In my early days of QM I got caught up in this stuff and it was a long detour. I was fortunate in having done a math degree that included analysis. I came out the other end with a good understanding of things like nuclear spaces etc but in hindsight it would have been better to simply accept some things and leave it for later. Ballentine's textbook on QM provides a good enough overview to get started.

Thanks
Bill

 

[dextercioby]

Hi,
I am reading the paper http://arxiv.org/abs/quant-ph/0502053 listed in the reference of Wikipedia Rigged Hilbert Space. I have a question about the relation, Φ ⊂ H ⊂ Φ', where H is Hilbert space, Φ is its subspace and Φ' is dual space of Φ.
Φ⊂H and Φ⊂Φ' are obvious. How can we say H ⊂ Φ' ? Φ' has nothing to do with the elements of H　that is not in Φ, I assume. Your advice is appreciable.

The Φ ⊂ H ⊂ Φ' is a shorthand for the full version: Φ ⊂ H ≅H'⊂ Φ', where that middle isomorphism is a restatement of the Riesz-Fiescher lemma.

 

[sweet springs]

Thanks, Dr.Du, bhoba and devtercioby. With your advise I am going to learn RHS.

 

[sweet springs]

I am still reading the paper. The space $$\Phi$$ is the maximal invariant subspace of the algebra generated by Q, P
and H, which is the largest subdomain of the Hilbert space that remains invariant under the action of any power of Q, P or H, say A $$A\Phi \subset \Phi$$. The paper says when eigenvalue is discrete we do not need $$\Phi^*$$ which is the linear functional space of $$\Phi$$.

I think of energy eigenstate of bound state of square well potential. It should belong to $$\Phi$$ but expectation value of $$P^6$$ or higher order diverges. What's wrong?

 

[strangerep]

I suspect the short answer is that there are difficulties if the usual momentum operator is naively applied in the square well potential example (i.e., there are issues to do with the boundary conditions). There was a thread about this a while back, iirc(?).

 

[sweet springs]

Thanks for commentn strangerep.

When we replace V(x) be similar, mathematically smooth so more "physical" one, e.g.
$$V(x)=V_0 (tanh(\frac{a-x}{d})+tanh(\frac{x-b}{d}))$$
,where $$0<V_0\ \ \ a<b, \ \ \ 0<d << b-a$$, the bound energy eigenstate belongs to the invariant subsapce in Hilbert space thus any expectation value <p^n> does not diverge, as suggested by this paper?

 

[Fredrik]

Thanks. From your post I assume there is a theorem Φ⊂H ⇒ H' ⊂ Φ' where A' means dual space of A.

If $\Phi$ is a proper subset of $H$, then $H'$ and $\Phi'$ are actually disjoint, because elements of $H'$ are functions with domain $H$ and elements of $\Phi'$ are functions with domain $\Phi$. But there's an obvious way to map $H'$ into $\Phi'$. For each $f\in H'$, the restriction of $f$ to $\Phi$ is an element of $\Phi'$. So the set of these restricted functions is a subset of $\Phi'$.

 

[strangerep]

Thanks for commentn strangerep.

When we replace V(x) be similar, mathematically smooth so more "physical" one, e.g.
$$V(x)=V_0 (tanh(\frac{a-x}{d})+tanh(\frac{x-b}{d}))$$
,where $$0<V_0\ \ \ a<b, \ \ \ 0<d << b-a$$, the bound energy eigenstate belongs to the invariant subsapce in Hilbert space thus any expectation value <p^n> does not diverge, as suggested by this paper?

I haven't studied that tanh well potential in detail, so I'd better not comment. Did you get that example from another source?

BTW, I just realized that you were most likely talking about a finite square well, whereas I was talking about an infinite square well. These cases are deceptively different -- iirc, the latter cannot be reached (rigorously) by taking a limit of the former, although some insight can be gained by proceeding in that direction. [Cf. Ballentine section 4.5.]

In fact,... it's worthwhile to explore all this more carefully, so let's go back to your earlier post...

I think of energy eigenstate of bound state of square well potential. It should belong to
$$\Phi$$but expectation value of$$P^6$$or higher order diverges. What's wrong?

This is better addressed by working through the detail. Which textbook are you working from? (I was vaguely thinking of working through the detail of Ballentine section 4.5, but he doesn't delve into the $P^6$ case.)

 

[sweet springs]

Thanks for your interest, strangerep.

I cannot find detailed description in textbooks or web contents, so I will show you the result of my calculation that was not difficult at all.

****************

Square well potential:
$$V(x)= -D\ for\ -w/2 < x < w/2,\ 0\ otherwise.$$

Wave function of the ground state in coordinate representation is:
$$\psi(x)= B e^{\kappa (x + \frac{w}{2})} \theta (-x-\frac{w}{2})+ C cos (kx)\theta (-x+\frac{w}{2})\theta (x+\frac{w}{2}) + B e^{-\kappa (x - \frac{w}{2})}\theta (x-\frac{w}{2}).$$

By Fourier transfromation of this equation we can get the wave function in momentum representation
$$\psi(p) \equiv \frac{1}{\sqrt{h}}\int_{-\infty}^{+\infty} \psi(x) e^{\frac{-ipx}{\hbar}} dx= 2cosec\ \alpha\sqrt{\frac{\frac{w}{2}}{h\alpha (\alpha + cot\ \alpha)}} \frac { cos(\alpha \rho)-\rho sin(\alpha \rho)}{(1-\rho)(1+\rho)(1+\rho^2 cot^2\ \alpha)} ,$$
where
$$\theta (x) \equiv \frac {sgn(x) + 1}{2}$$
$$sgn(x)=-1\ for\ x<0, \ 0 for\ x=0, 1\ for\ x>0.$$
Other coefficents are given by α[0,π/2],
$$D=\frac{\hbar^2}{2m} (\frac{2}{w})^2 \alpha^2 sec^2 \ \alpha,$$
$$C= (\frac{2}{w})^{1/2} (1+\frac{cot\ \alpha }{\alpha})^{-1/2},$$
$$B= (\frac{2}{w})^{1/2} (1+\frac{cot\ \alpha }{\alpha})^{-1/2} cos\ \alpha,$$
$$k=\frac{2}{w} \alpha,$$
$$\kappa =\frac{2}{w} \alpha \ tan \ \alpha,$$
Scaled non dimensional momentum
$$\rho \equiv \frac{\frac{w}{2}}{\hbar\alpha}p,$$　　　
Energy measured from the bottom of the well
$$E=\frac{\hbar^2}{2m} (\frac{2}{w})^2 \alpha^2.$$

****************

You can find the momentum wave function decreases as p^-3. |ψ(p)|^2 decreases as p^-6. Thus expectation value of p^6 or higher even power diverges.

For α→π/2-0, the infinite depth well, it decreases as p^-2. |ψ(p)|^2 decreases as p^-4. Thus expectation value of p^4 or higher even power diverges.

 

[strangerep]

I cannot find detailed description in textbooks or web contents,

Do you have a copy of Ballentine? If so, we could adapt his section 4.5, starting on p107. (His potential differs from yours simply by having the energy zero at the bottom of the well.)

so I will show you the result of my calculation that was not difficult at all.

[...] I feel a need to refresh my memory on the gory details, so I'm going to reproduce Ballentine's calculation from scratch for myself.

Back later, or maybe tomorrow.

 

[strangerep]

Back now,...

OK... I'm surprised how much of the detail had been archived to offline storage by my brain. I'm going to solve the entire problem for myself, not just a piece of it like Ballentine does. Hopefully, that will bring it all back into RAM.

Afaict, your solution is not general, and assumes a particular range of E. The general method should deduce what range(s) of E are physically allowed, and the solutions in each case.

Might take me a while to find enough spare time to solve the general case carefully.

 

[sweet springs]

Afaict, your solution is not general, and assumes a particular range of E.

Ya, this was about the ground state as I wrote. D and w determines α and thus all the coefficients and E.

I will be glad if you would verify or correct my result.

Best.

 

[strangerep]

Square well potential:
$$V(x)= -D\ for\ -w/2 < x < w/2,\ 0\ otherwise.$$
Wave function of the ground state in coordinate representation is:
$$\psi(x)= B e^{\kappa (x + \frac{w}{2})} \theta (-x-\frac{w}{2})+ C cos (kx)\theta (-x+\frac{w}{2})\theta (x+\frac{w}{2}) + B e^{-\kappa (x - \frac{w}{2})}\theta (x-\frac{w}{2}).$$

How did you get the $e^{\kappa (x + \frac{w}{2})}$ and $e^{-\kappa (x - \frac{w}{2})}$ ?
I would have thought they should be of the form $e^{-\kappa |x|}$ .

I also don't really follow how you conclude that $\langle P^6 \rangle$ diverges. Maybe try doing it in position representation? $P^2 \psi \propto \psi$ for an energy eigenstate, so something is wrong. (Did you enforce $d\psi/dx$ to be continuous at the well boundaries?)

Also, you didn't say whether you can access a copy of Ballentine. His treatment is far less messy than yours, hence easier to follow.

 

[sweet springs]

How did you get the e κ(x+w2 ) e^{\kappa (x + \frac{w}{2})} and e −κ(x−w2 ) e^{-\kappa (x - \frac{w}{2})} ?
I would have thought they should be of the form e −κ|x| e^{-\kappa |x|} .

The exponentilal damping of wave function outside the well starts from the edges of the well(x=-w/2,w/2), not from the center of the well(x=0).

 

[strangerep]

The exponentilal damping of wave function outside the well starts from the edges of the well(x=-w/2,w/2), not from the center of the well(x=0).

That's not the right way to think about it. Just solve the TISE as if the potential was constant and then restrict the solution accordingly. Then match up the various pieces of the solution at the well boundaries.

Edit: I suppose one could simply absorb $e^{\kappa\omega/2}$ into the factor, so maybe it doesn't matter.

 

[sweet springs]

I also don't really follow how you conclude that ⟨P 6 ⟩ \langle P^6 \rangle diverges. Maybe try doing it in position representation? P 2 ψ∝ψ P^2 \psi \propto \psi for an energy eigenstate, so something is wrong. (Did you enforce dψ/dx d\psi/dx to be continuous at the well boundaries?)

Momentum wave function is convenient and easy way when we calculate expectation values of momentum function, e.g. p^n.
The result does not rely on which representations or wave functions we use. We can do it in coordinate wave function. As you have assumed before higher differentiation of delta functions at the edges cause the divergence.

Now I can read web text of Ballentine, you can refer equations or explanations there in your comments.

Best

 

[strangerep]

As you have assumed before higher differentiation of delta functions at the edges cause the divergence.

I realized later that if both $\psi$ and $d\psi/dx$ are continuous at the well boundaries, then any theta/delta function contributions should cancel out.

Now I can read web text of Ballentine, you can refer equations or explanations there in your comments

OK -- read section 4.5. The square well stuff starts on p107.

 

[sweet springs]

That's not the right way to think about it. Just solve the TISE as if the potential was constant and then restrict the solution accordingly. Then match up the various pieces of the solution at the well boundaries.

Please go your way. With the connection at the edges you will get the same result as mine. Please check it and let me know.

Edit: After up this message I noticed your edit.

 

[sweet springs]

any theta/delta function contributions should cancel out.

should? How do you deal with d^2ψ/dx^2, d^4ψ/dx^4, ...　that appear in calculation of <p^n>?

 

[strangerep]

should? How do you deal with d^2ψ/dx^2, d^4ψ/dx^4, ...　that appear in calculation of <p^n>?

If $\psi$ is an energy eigenstate, then [EDIT:] $d^2 \psi/dx^2$ can be expressed in terms of $\psi$ multiplied by something involving just theta functions.

 

[sweet springs]

If ψ \psi is an energy eigenstate, then [EDIT:] d 2 ψ/dx 2 d^2 \psi/dx^2 can be expressed in terms of ψ \psi multiplied by something involving just theta functions.

Along the line,
d 3 ψ/dx 3 can be expressed in terms of ψ multiplied by something involving delta functions,
d 4 ψ/dx 4 can be expressed in terms of ψ multiplied by something involving delta^(1) functions,
d 5 ψ/dx 5 can be expressed in terms of ψ multiplied by something involving delta^(2) functions, ... right?.

 

[strangerep]

Along the line,
d 3 ψ/dx 3 can be expressed in terms of ψ multiplied by something involving delta functions,
d 4 ψ/dx 4 can be expressed in terms of ψ multiplied by something involving delta^(1) functions,
d 5 ψ/dx 5 can be expressed in terms of ψ multiplied by something involving delta^(2) functions, ... right?.

Let's write the full solution like this:
$$\psi(x) ~=~ \Big(1 - \theta(x+\omega/2) \Big)\psi_L(x)
~+~ \Big(\theta(x+\omega/2) - \theta(x-\omega/2)\Big)\psi_M(x)
~+~ \theta(x-\omega/2) \, \psi_R(x) ~,$$where the ``L,M,R'' subscripts denote "Left", "Middle", and "Right".

Now compute $\psi'(x)$, i.e., $d\psi(x)/dx$. There will be a term like $-\delta(x+\omega/2)\psi_L(x)$ and also a term like $\delta(x+\omega/2)\psi_M(x)$. But at $x= -\omega/2$, we have $\psi_L(-\omega/2) = \psi_M(-\omega/2)$ so these $\delta(x+\omega/2)$ terms cancel.

A similar cancellation happens at $x = +\omega/2$.

So, unless I've made a mistake, we get:
$$\psi'(x) ~=~ \Big(1 - \theta(x+\omega/2) \Big)\psi'_L(x)
~+~ \Big(\theta(x+\omega/2) - \theta(x-\omega/2)\Big)\psi'_M(x)
~+~ \theta(x-\omega/2) \, \psi'_R(x) ~.$$
Similar cancellations happen for $\psi''(x)$, provided we have ensured that the $\psi'(x)$ pieces match at the boundaries.

I'll leave it to you to work out $\psi'''(x)$ explicitly...

 

[sweet springs]

ψ′(x) = (1−θ(x+ω/2))ψ′L(x) + (θ(x+ω/2)−θ(x−ω/2))ψ′M(x) + θ(x−ω/2)ψ′R(x) .​

Similar cancellations happen for ψ′′, provided we have ensured that the ψ′(x)\psi'(x) pieces match at the boundaries.

No similar cancellations. Provided ψ′(x) pieces match at the boundaries, in the formula of ψ’’(x) , delta functions cancel, so

ψ'''(x) = (1−θ(x+ω/2))ψ''L(x) + (θ(x+ω/2)−θ(x−ω/2))ψ''M(x) + θ(x−ω/2)ψ''R(x)

Since ψ''L(-w/2) ≠ ψ''M(-w/2) or so, ψ'''(x) has gaps at the potential boundaries. Differentiation of gap makes delta function appear in ψ^(3) and so on.

 

[strangerep]

We can use the Schrodinger equation to help with $\psi'''(x)$. We have:
$$-\frac{\hbar^2}{2m}\, \psi''_L(x) ~=~ E \psi_L(x) ~,~~~~~~~ \mbox{and similarly for}~ \psi_R(x) ~.$$Also,
$$-\frac{\hbar^2}{2m}\, \psi''_M(x) ~=~ (E-D) \psi_M(x) ~.$$
So, in $\psi'''(x)$, replace each $\psi''_?$ by the corresponding expression involving $\psi_?$ . E.g., replace $\psi''_L$ by $-\frac{2mE}{\hbar^2}\psi_L$, and similarly for the middle and right parts.

Then, since (e.g.,) $\psi_L(-\omega/2)=\psi_M(-\omega/2)$, there is some cancellation at $x=\pm\omega/2$, but we're left with a term like $$-\frac{2mD}{\hbar^2} \, \delta(x+\omega/2)\, \psi_M(x) ~,$$and a similar one at $x=+\omega/2$. These remaining deltas do not make $\langle \psi|\psi'''\rangle$ diverge, as far as I can tell.

I need to think a bit more about the higher derivatives. Back later.

 

[sweet springs]

I'll leave it to you to work out ψ′′′(x) explicitly...

$$\psi^{(2)}(x) ~=~ \Big(1 - \theta(x+w/2)\Big)\psi^{(2)}_L(x) ~+~ \Big(\theta(x+w/2)-\theta(x-w/2)\Big)\psi^{(2)}_M(x) ~+~ \theta(x-w/2) \psi^{(2)}_R(x) ~$$

$$\psi^{(3)}(x) ~= \Big(\psi^{(2)}_M(-w/2) -\psi^{(2)}_L(-w/2)\Big) \delta(x+w/2)+\Big(\psi^{(2)}_R(w/2) - \psi^{(2)}_M(w/2) \Big)\delta(x-w/2) \\ +~ \Big(1 - \theta(x+w/2)\Big)\psi^{(3)}_L(x) ~+~ \Big(\theta(x+w/2)-\theta(x-w/2)\Big)\psi^{(3)}_M(x) ~+~ \theta(x-w/2) \psi^{(3)}_R(x) ~$$

Expectation value of p^6
$$(ψ,P^6 ψ) = -\hbar^6 (ψ,ψ^{(6)})=\hbar^6 (ψ^{(3)}, ψ^{(3)})=\hbar^6 ||ψ^{(3)}||^2=\infty$$
because ψ^(3), which contains delta functions, is not normalizable. Thus we showed the expectation value of p^6 diverges in coordinatie representation as well as in momentum representation.

PS Sorry for inconvenience caused by cancelling my previous #28 and uploading this new #29 for my Latex problem. Best.

 

[strangerep]

Expectation value of p^6
$$(ψ,P^6 ψ) = -\hbar^6 (ψ,ψ^{(6)})=\hbar^6 (ψ^{(3)}, ψ^{(3)})=\hbar^6 ||ψ^{(3)}||^2=\infty$$

In cases like this, your manipulation: $$-(ψ,ψ^{(6)}) =(ψ^{(3)}, ψ^{(3)})$$ might be unreliable, since $P$ is only self-adjoint on a suitably well-behaved space of functions. I'm not sure that $ψ^{(3)}$ is one of those functions.

We need to evaluate $P^6 \psi$ directly to be sure, but I'm still thinking about $P^3\psi$ and $P^4\psi$.

 

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

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

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

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

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

 

[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