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I Generating a Hilbert space representation of a wavefunction

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  1. Nov 1, 2017 #1

    SeM

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    Hello, I Have a particle with wavefunction Psi(x) = e^ix

    and would like to find its Hilbert space representation for a period of 0-2pi. Which steps should I follow?

    Thanks!
     
  2. jcsd
  3. Nov 1, 2017 #2

    strangerep

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    Hmm. Is this homework? (If so, you should be posting this over in one of the HW forums.)

    Anyway,... the answer should follow immediately if you list the defining properties of "Hilbert space" (which are easily obtainable from Wikipedia, or by googling).
     
  4. Nov 1, 2017 #3

    bhobba

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    Do you know what a Hilbert space is? If so does what you wrote belong to it?

    And if you don't know what it is then its a rather hard question for us to answer - the answer is it isn't even a member of a Hilbert Space - but any detail beyond that requires you to know what one is.

    Thanks
    Bill
     
  5. Nov 2, 2017 #4

    strangerep

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    Heh, that depends on whether the OP meant ##0## to ##2\pi## as a domain... :oldwink:
     
  6. Nov 2, 2017 #5

    bhobba

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    True - we really need more info.

    Thanks
    Bill
     
  7. Nov 2, 2017 #6

    SeM

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    I think this explains what I am wondering about:
    https://physics.stackexchange.com/questions/246647/why-do-we-need-both-hamiltonian-and-hilbert-space-to-specify-a-quantum-system/246650
    However, I am not sure which steps one should follow to prove the solutions to a Hamiltonian in a Hilbert space. So far, I have only used a Cartesian system with 2 dimensions, and got a solution to the Hamiltonian, the eigenstate f(x) = e^ikx. However, this is for a cartesian system. How can I analyse this solution and the Hamiltonian in a N-finite Hilbert space for the interval of 0 to 2pi? Are there 3-4 steps one can stick to in order to interpret the existing Hamiltonian and its Cartesian solution (for one particle) in a Hilbert space for N particles?
     
  8. Nov 2, 2017 #7

    bhobba

    Staff: Mentor

    Take the Hamiltonian of a free particle, Solve it. Are they in a Hilbert space? What you wrote is one such solution (p=1) in certain units. Is it in a Hilbert space?

    Again I mention if you do not know what a Hilbert space is (all functions in a Hilbert Space are Lebesque square integrable as you should either know or look up in your investigations). Is the function you wrote square integrable - multiply it by its conjugate - that's the squaring bit, then integrate from -∞ to ∞. What do you get? Does that imply its in a Hilbert space? Indeed how do you apply the Born Rule to such a function?

    Please post at least an attempt to answer these questions - its very very important and led no less a person than the great Von-Neumann to exasperation and to also no less a person than the equally great mathematician, Grothendieck, along with others such as Schwartz and Gelfland to sort out - and it took time as well - it didn't happen overnight.

    Now I could tell you the answer but will not for three reasons. One - what you figure out for yourself you understand better. And two it will ensure you fully understand concepts like Hilbert space etc. Many beginning students see it mentioned in texts but don't grasp what it is. Its wise to rectify that. And thirdly you develop self reliance - really your question should have been whats going on with the solution you wrote and states form a Hilbert space, detailing the exact issue..

    Thanks
    Bill
     
  9. Nov 3, 2017 #8

    SeM

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    Dear Bill, thanks for this extensive explanation. Answer to your question, the Hamiltonian its solved for a Cartesian space, not Hilbert space.

    I will first try your advice, and then get back to you here or via PM, is that OK? It will be next week.

    First I check if it is Hermitian, then I do a square integration over - to + infinity with its complex conjugate and normalize it to 1. At that stage I apply the Born rule to it, then I get back to you!

    Thanks!

    Have a good weekend
     
  10. Nov 3, 2017 #9

    bhobba

    Staff: Mentor

    Nope - the solution is an element of a Hilbert space, or to be absolutely precise it is the representation of a member of a Hilbert space, but we will not be that carefull to start with.

    I would rather you post in this thread.

    First though can you tell us what textbook you are using?

    Try that and see what happens - but I will this time outline it first - multiply e^ix by its conjugate and what you get is 1. Take the integral from -∞ to ∞ and what do you get? That depends on your definition of such improper integrals - in some texts you take the limit as l → ∞ of the integral from -l to l in which case you get zero. In others you break it into two bits and you get -∞ + ∞. What that means again depends on how you interpret such things. Some may say its undefined while others again say it's zero - we will take zero again. To normalize your wave-function divide it by zero - well again depending on your view of such things that is either undefined or infinite. Either way you are up the creek without a paddle.

    Whats going on? That solution to the Schrodinger equation is NOT a member of a Hilbert space, its not normalizeable - as you would expect physically anyway since its cos x + i sine x for x -∞ to ∞. It's not zero at infinity which is physical nonsense - in fact the limit at infinity of cos x or sine x, since its an oscillating function, is undefined - but it does not go to zero. You have obtained an nonphysical solution.

    Once you understand that I can explain the resolution which depends on your mathematical sophistication.

    Thanks
    Bill
     
  11. Nov 4, 2017 #10

    SeM

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    Dear Bill, the wavefunction is actually quite more complex than this, I used this simple version to see what to do in the first place. Now I see that:

    1. Check hermiticiy
    2. Integrate the complex conjugate with the wavefunction over infinite space, normalize to 1, then check for it towards the Born Rule.
    3. Get back to you.

    If there is something else I coud do, was thinking of generating the Product of the WFN with N, so

    Psi1* Psi2*...Psi

    I will do this on Monday, because its too important to be done on a half way on a Sunday.

    Sergio
     
  12. Nov 4, 2017 #11

    bhobba

    Staff: Mentor

    Yes the actual solution for Schrodinger's equation is more complex and like you said you should really do separation of variables to solve it.

    Do all those things - but the same issue will arise - for the Schrodinger equation with no potential term you get terms of the form you wrote and exactly the same issues arrise when you try to find its norm then calculate the normalized solution - you will not be able to do it.

    Get back here when you have done it and I will explain the solution.

    It will be greatly facilitated if you do a bit of study before into whats called distribution theory:
    http://orion.lnu.se/pub/education/course/electrical/4ED044/Distributions.pdf

    Thanks
    Bill
     
  13. Nov 5, 2017 #12

    SeM

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    Dear Bill, that is all correct which you wrote. The potential term is not included because there is no proper potential form, as these are quite free particles without the conventional potential form. I have therefore omitted that term, and look only at the kinetic term. I will look at all you have wrote, and get back in some days. This may take some more time.

    Thanks for your help!
     
  14. Nov 5, 2017 #13

    bhobba

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    The wave-functions of free particles do not belong to a Hilbert space - they are non-physical as I explained.

    However one can, in a sense I will not go into (called weak convergence) find a sequence of functions, fn, that are zero outside a certain interval (called functions of compact support) such that fn → f where f is a function like you got. Since the functions are of compact support they are of course square integergrable. So all you do is consider the solution you got simply as an approximation to fn when n is so large for all practical purposes its the same as f - the function you got. The interval where its not zero could be the length of the universe - just as long as its way outside what can be measured. That's all that required to resolve it - physically that is. Mathematically - the solution is much harder and requires Rigged Hilbert Spaces:
    https://arxiv.org/pdf/quant-ph/0502053.pdf

    Thanks
    Bill
     
  15. Nov 6, 2017 #14

    vanhees71

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    There are plenty of wave functions for free particles that are square integrable. You can start, e.g., with a Gaussian wave packet, and it's easy to calculate that with time it stays a Gaussian wave packet with its "center" moving with constant velocity and its "width" getting broader and broader with time. It's a good exercise to calculate the corresponding solution of the Schrödinger equation from scratch!
     
  16. Nov 6, 2017 #15

    SeM

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    Dear Bill, the function does however satisfy the given condition:

    \begin{equation}
    \int_{0}^{2\pi}|\psi|^2 dx < \infty,
    \end{equation}




    which is satisfied, one can define the inner product of two such functions by the formula

    \begin{equation}
    \langle f,g \rangle = \int_{0}^{2\pi} f g dx ,
    \end{equation}

    How does this NOT coincide with a Hilbert "Geometry" ?
     
  17. Nov 6, 2017 #16

    bhobba

    Staff: Mentor

    Because the axioms of QM say inner products etc are taken over infinity.

    Maddening isn't it.

    You can try what Vanhees said as well - but that is not an eigenfunction of momentum - still that maybe is all that you require - a solution that is a member of the Hilbert space - not necessarily an eigenfunction of momentum - I cant comment because I do not know your intent.

    Added later
    In light of Stragerep's correct comment the position observable X has eigenvalues from -∞ to +∞ which implies the expansion in terms of position eigenvectors |x> is a function f(x) such that |u> = ∫f(x)|x> where since the eigenvalues can go from -∞ to +∞ so does the integral and the inner product.

    But in QM all that is required is a compete set of commuting observables, and one does not have to be the position observable.

    Thanks
    Bill
     
    Last edited: Nov 6, 2017
  18. Nov 6, 2017 #17

    SeM

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    Yes, a solution member of the Hilbert space is indeed sufficient. I will generate the wavepackets, as this has also been advised from another person.

    Thanks so far.
     
  19. Nov 6, 2017 #18

    strangerep

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    Huh? Which axioms are you using?

    (One can derive useful quantum features from a purely abstract Hilbert space if an algebra of observables is given. Cf. the derivation of the quantum angular momentum spectrum in Ballentine sect 7.1.)
     
  20. Nov 6, 2017 #19

    bhobba

    Staff: Mentor

    The ones in Von-Neumann based on the Hilbert Space of square integrateable functions.

    But I take your point. This was in relation to normal wave-functions of position. All, of course, one needs is a complete set of commuting observables which is one of Dirac's axioms.

    Thanks
    Bill
     
  21. Nov 7, 2017 #20

    SeM

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    Hi Bill, given that the wavefunction at hand here in this thread is non-orthogonal, are there are classifications of non-orthogonal wavefunctions that can be made? For instance, if it obeys some inequality or some integrals which separate it from other non-orthogonal functions? In other words, except for the following rigged Hilbert space definition, defined by

    \begin{equation}
    \int_{0}^{2\pi}|ψ|^2dx<\infty
    \end{equation}

    can I add some more classifications? I have generated a Hilbert Transform of it for instance, which is quite interesting. However, I am currently not aware of other "classifications" of non-orthogonal wavefunctions, before I go to wavepackets.
     
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