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How to understand operators representing observables are Hermitian?

  1. Dec 13, 2013 #1
    As we know, all operators representing observables are Hermitian. In my undersatanding, this statement means that all operators representing observables are Hermitian if the system can be described by a wavefunction or a vector in L2. For example, the momentum operator p is Herminitian, because for any states described by Ψ and Φ, we have

    <Ψ|p|Φ> = ∫Ψ*(-ihdΦ/dx) dx = Ψ*(-ih)Φ + ∫(-ihdΨ/dx)*Φdx = <pΨ|Φ>

    where Ψ*(-ih)Φ = 0 is employed at -∞ and +∞.

    But the eigen function of p-operator (2πh)-1/2exp(ipx/h) does not satisfy = 0 at -∞ and +∞, and the eigen function of the momentum operator, (2πh)-1/2exp(ipx/h), is not a wave function describing quantum state. Am I right?
     
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  3. Dec 14, 2013 #2

    WannabeNewton

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    We had the exact same question being asked earlier today except it was with regards to the eigenfunctions of the position operator instead of those of the momentum operator. Regardless, see here: http://arxiv.org/pdf/quant-ph/0502053v1.pdf
     
  4. Dec 14, 2013 #3
    Thanks, so the Hilbert space is replaced by rigged Hilbert space. It is a good paper and I will read it. One more question about the momentum operator.

    In the 1D infinite well, 0<x<a, the momentum eigen equation is given by

    -ih*dψ/dx=p*ψ, with ψ(0)=ψ(a)=0

    and its solution is ψ=A*exp(ipx/h) which only has one constant "A" for the two boundary conditions ψ(0)=ψ(a)=0, and thus ψ=0 must hold. Does that mean the momentum operator p has no eigen wavefunctions in such a case? But according to the QE hypothesis, all operators representing observables are Hermitian, and their eigenfunctions constitute complete systems. How to understand this?
     
  5. Dec 14, 2013 #4

    bhobba

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    It's the Rigged Hilbert Space formalism. The physically realizable states are taken as test functions (and they do the nice stuff like vanish at infinity, continuously differentiable etc etc) but for mathematical convenience it is extended by means of the Gelfland Triple to functions like the above and even weirder stuff like the Dirac Delta function:
    http://homepage.univie.ac.at/roza.aceska/testfunctions.pdf [Broken]
    http://en.wikipedia.org/wiki/Rigged_Hilbert_space

    Test functions are all that are considered physically realizable, but can approximate the functionals that are the superset of the Hilbert space in the Gelfland triple to any desired degree of accuracy. For example a wave that continues to infinity is not physically realizable, but is a good approximation to a wave that vanishes a long distance away. A position that is a Dirac Delta function doesn't exist but if it's known to a high degree of accuracy is a good approximation.

    Thanks
    Bill
     
    Last edited by a moderator: May 6, 2017
  6. Dec 14, 2013 #5

    WannabeNewton

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    For a more detailed analysis of this system using the Rigged Hilbert space formalism see section 3 (p.10) of the paper.

    EDIT: And see here for even more details: http://arxiv.org/pdf/quant-ph/0407195v1.pdf :wink:

    Yes. More precisely we want them to be self-adjoint so that the spectral theorem applies. To ensure this we must restrict the domains of operators (that correspond to dynamical variables) to appropriate dense subspaces of ##L^2## so that said domains agree with those of the adjoints.

    For bounded operators there is really no issue with regards to any of this but we must be extremely careful with domains when it comes to unbounded operators; the property of being Hermitian is in a sense attached solely to the unbounded operator but the property of being self-adjoint is very sensitive to the domain because the domain of the adjoint has to match and so we must be cautious when dealing with unbounded operators (and by unbounded here I mean those operators which do not have a continuous extension from their dense domain of definition to all of ##L^2##).

    No. This is not part of the postulates of QM. Self-adjoint operators over infinite dimensional spaces can easily fail to have a complete set of eigenfunctions; the existence of a complete set of eigenfunctions for self-adjoint operators is not as fundamental as the spectral theorem. The spectral theorem is the main resort.
     
    Last edited: Dec 14, 2013
  7. Dec 14, 2013 #6
    Thanks a lot. You corrected me a wrong idea. So in the 1D inifite well (discrete eigenvalue problem), the momentum operator p does not have eigenfunction, because its solution is ψ=A*exp(ipx/h) which only has one constant "A" for the two boundary conditions ψ(0)=ψ(a)=0, and thus ψ=0 must hold. Right?
     
  8. Dec 14, 2013 #7

    WannabeNewton

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  9. Dec 14, 2013 #8

    dextercioby

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    Due to the very strong boundary conditions, yes, the 0 vector is the only eigenfunction, thing which is trivial, as any operator when applied on the 0 vector turns the 0 vector times an arbitrary constant. So that for any operator, the 0 vector is a trivial eigenvector.

    No, all observables are represented by self-adjoint operators. Nothing is being said on their spectral properties. For example, for the free galilean particle in 1D, the position operator is an observable, because it's represented on L^2(R) as a self-adjoint operator, however it has no eigenfunctions in L^2(R).
     
  10. Dec 14, 2013 #9
    But how to understand the following statement in the book by Griffiths, [D. Griffiths, "Introduction to quantum mechanics", (Prentice Hall, NJ, 1995), p. 106, Chapter 3]:

    "As we have seen, in the finite-dimensional case the eigenvectors of a Hermitian operator always span the space."

    In my understanding, "always span the space" means the eigenvectors are complete. The 1D infinite-well case is of a "finite-dimensional case", and the momentum operator is Hermitian, but it has no eigenfunction. Seems not consistent with the above statement. Did I miss something? Thanks a lot.
     
  11. Dec 14, 2013 #10

    WannabeNewton

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    The dimensionality there refers to the dimensionality of the state space, not that of the physical system. For example, the energy eigenstates in the coordinate representation of a particle confined to a 1D infinite square well with the usual boundary conditions are sinusoidal functions restricted to the width of the square well. These are clearly square integrable functions residing in ##L^2## which is infinite dimensional.

    All operators over finite dimensional spaces are bounded hence being Hermitian is equivalent to being self-adjoint; furthermore over finite dimensional spaces all self-adjoint operators have a complete set of eigenvectors due to the finite dimensional spectral theorem as you probably know from your elementary linear algebra course(s). Over infinite dimensional spaces such as ##L^2## the situation is much more complicated as already noted above.
     
  12. Dec 14, 2013 #11

    dextercioby

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    Well, you chose your username a misspelling of Schrödinger, but the connection you look for resides however in the matrix mechanics invented by Born, Heisenberg and Jordan in 1925. There are no finite matrices in a formulation of the 1D infinite well case for any of the operators at stake: position, momentum, energy, so that your bolded comment has nothing to do with the quote from Griffiths (side note: there are probably at least 5 better textbooks to learn QM than Griffiths). So the red part above is true.

    Fact: any finite dimensional complex Hilbert space is isomorphic to C^n, for some n a natural number. C^n is the setting for the theory of square matrices with n2 entries.
     
  13. Dec 14, 2013 #12
    WannabeNewton and dextercioby, I got it, thanks a lot.
     
  14. Dec 15, 2013 #13
    I still have a question. The momentum operator p does have any eigenfunctions in the 1D inifite well, since zero vector cannot be counted as a eigenvector. No eigenvectors, no eigenvalues. Then how to understand the postulate: A measurement of observable p is certain to return one of the eigenvalues of the operator p?
     
  15. Dec 16, 2013 #14

    dextercioby

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    Extremely simple: if the wavefunction is required to be nil at the 2 points forming the boundary of the box, then momentum is not an observable, hence no contradiction with the measurement postulate.
     
  16. Dec 16, 2013 #15
    Sorry, I got confused a little bit.

    (1) In principle, the momentum operator is Hermitian, observable operator. Why not observable?

    (2) In the book by Griffiths [D. Griffiths, Introduction to quantum mechanics, (Prentice Hall, NJ, 2005), 2ed, p. 38, Chapter 2, Problem 2.5 (d) ], there is an exercise for calculating the expectation value of p, <p>, in the 1D infinite well; the answer is <p>=(8h/3a)sin(3ωt). This also indicates that p should be observable. Am I right?
     
  17. Dec 16, 2013 #16

    dextercioby

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    For the point 1), well said 'in principle'. That leaves the door open for exceptions such as the momentum operator in a box = 1D infinite square well.
    2) I don't see how that follows. You're merely computing <psi, p psi> for a given psi. How does that tell you it's an observable ?
     
  18. Dec 16, 2013 #17

    WannabeNewton

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    This is a very nice question my friend :smile:!

    Take a look at example 4 in page 6 of this paper, see if you can answer the question yourself, and then go to page 39 of said paper for the resolution: http://arxiv.org/pdf/quant-ph/9907069.pdf

    The moral of the story is that Dirac's braket notation, if taken more seriously than being a mere calculational facility, will lead to a lot of mathematical contradictions in functional analysis.
     
  19. Dec 16, 2013 #18
    Physically, in my undersatanding, <p> means that many measurements of p for an ensemble of the wells, we get an average value, <p>. Is that right?
     
  20. Dec 16, 2013 #19
    Interesting! I am asking the same question as the paper does. But I don't understanding some words:

    (1) "P does not admit any eigenvalues. Nevertheless, the spectrum of P is the entire complex plane [8] and P does not represent an observable."
    The first statement seems not consistent with the second.

    (2) "Thus, we established that P is Hermitian, but not self-adjoint."
    In my understanding, "Hermitian" means "self-ajoint", <Ψ|p|Φ> = <pΨ|Φ>.
     
  21. Dec 16, 2013 #20

    WannabeNewton

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    It's very hard to explain these things if you don't know basic functional analysis. When we are talking about operators over infinite dimensional spaces, the spectrum of the operator contains more than just the eigenvalues of the operator. See here: http://en.wikipedia.org/wiki/Decomposition_of_spectrum_(functional_analysis)

    Hermitian does not mean self-adjoint and I've explained why previously in the thread. An operator can be Hermitian but if it's domain does not agree with the domain of it's adjoint then clearly it cannot be self-adjoint. See post #5.

    If you want, I can recommend some functional analysis texts to get you started.
     
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