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Rigged Hilbert Space Φ ⊂ H ⊂ Φ'

  1. Jan 21, 2015 #1
    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.
  2. jcsd
  3. Jan 21, 2015 #2


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    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'##.
  4. Jan 21, 2015 #3
    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?
  5. Jan 21, 2015 #4


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    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'##.
  6. Jan 21, 2015 #5


    Staff: Mentor

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

    But a background in analysis is required:

    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.

    Last edited by a moderator: May 7, 2017
  7. Jan 21, 2015 #6


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    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.
  8. Jan 21, 2015 #7
    Thanks, Dr.Du, bhoba and devtercioby. With your advise I am going to learn RHS.
  9. Jan 30, 2015 #8
    I am still reading the paper. The space [tex]\Phi[/tex] 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 [tex]A\Phi \subset \Phi[/tex]. The paper says when eigenvalue is discrete we do not need [tex]\Phi^*[/tex] which is the linear functional space of [tex]\Phi[/tex].

    I think of energy eigenstate of bound state of square well potential. It should belong to [tex]\Phi[/tex] but expectation value of [tex]P^6[/tex] or higher order diverges. What's wrong?
  10. Jan 30, 2015 #9


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    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(?).
  11. Jan 31, 2015 #10
    Thanks for commentn strangerep.

    When we replace V(x) be similar, mathematically smooth so more "physical" one, e.g.
    [tex]V(x)=V_0 (tanh(\frac{a-x}{d})+tanh(\frac{x-b}{d}))[/tex]
    ,where [tex]0<V_0\ \ \ a<b, \ \ \ 0<d << b-a[/tex], 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?
    Last edited: Jan 31, 2015
  12. Jan 31, 2015 #11


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    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'##.
  13. Jan 31, 2015 #12


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    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...
    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.)
    Last edited: Jan 31, 2015
  14. Jan 31, 2015 #13
    Thanks for your interest, strangerep.

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


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

    Wave function of the ground state in coordinate representation is:
    [tex]\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}).[/tex]

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


    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.

    Last edited: Jan 31, 2015
  15. Jan 31, 2015 #14


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    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.)

    [...] 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.
    Last edited: Feb 1, 2015
  16. Feb 1, 2015 #15


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    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.
  17. Feb 1, 2015 #16
    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.

    Last edited: Feb 1, 2015
  18. Feb 1, 2015 #17


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    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.
    Last edited: Feb 1, 2015
  19. Feb 1, 2015 #18
    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).
  20. Feb 1, 2015 #19


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    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.
    Last edited: Feb 1, 2015
  21. Feb 1, 2015 #20
    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.

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