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Eigenvalues of the position operator

  1. Mar 21, 2013 #1

    fluidistic

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    I'm new to QM, but I've had a linear algebra course before. However I've never dealt with vector spaces having infinite dimension (as far as I remember).
    My QM professor said "the eigenvalues of the position operator don't exist". I've googled "eigenvalues of position operator", checked into wikipedia's and wolfram's pages but they seem to only talk briefly on eigenfunctions of that operator.
    I understand that the basis of the vector space spanned by all possible wave functions ##\Psi##'s has infinite dimension so I expect that if I want to write the position operator ##\hat x## under matrix form, it would be an infinite matrix. But I don't think this implies that the eigenvalues don't exist (I guess they are infinite?).
    Why are the eigenvalues non existing and what does that mean exactly?
    Thanks...
     
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  3. Mar 21, 2013 #2

    fzero

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    I would guess he means that the eigenfunction of position (in the position basis) is a distribution rather than a function (as in http://en.wikipedia.org/wiki/Position_operator#Eigenstates). This is a statement about the wavefunction ##\langle x' | x \rangle ## and not the eigenvalue, as you note. There's no problem defining the operator, or its eigenvalue/eigenstate. The eigenfunction can even be well-defined in the position basis, where we just get a plane wave solution.
     
  4. Mar 21, 2013 #3

    fluidistic

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    I see, thank you.
    1)Let's take a particular example. I choose the strange example of a particle in 1 dimension whose position is "perfectly" known, i.e. just like in classical mechanics. Mathematically, ##\Psi (x ,t )=\delta ( x - a )##. I'm not really sure that in this case ##|\Psi (x,t)|^2## means something though. But I'll assume for now that the particle is at x=a with probability 1.
    Now I wonder, in such case what would be the eigenfunction(s) and eigenvalue(s).
    So I take the position operator, ##\hat x \Psi = x \Psi = \lambda \Psi##. But eigenvectors/functions are different from 0 by definition so I'm left with ##a\Psi=\lambda \delta (0)##. The delta does not make any sense outside an integral... I guess my example is a hard one. I don't know how to get the answer to my question.

    2)Also I wonder when looking at http://scienceworld.wolfram.com/physics/PositionOperator.html, I see that the position operator can be written down under matrix form as if the elements of the basis of the vector space spanned by all possible ##\Psi##'s were countable.

    It's like, if I take the general wave that is a superposition of plane waves: ##\Psi (\vec k ,t )=\int _{\text{range of k}}A(\vec k ) e^{i\vec k \cdot \vec x - \omega (\vec k ) t} dk##, the index k is continuous and goes from - to + infinity. Every time I fix "k", I get an element of the basis of the vector space spanned by all possible psi's. The possible k's are, I believe, not countable (since it's a continuous index). Thus, how is it possible to write down the position operator under an explicit matrix form, as if the elements of its basis were countable?
     
  5. Mar 21, 2013 #4

    fzero

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    The matrix in eq (3) is for the case of a SHO. I believe that if you were to attempt to diagonalize this matrix, you'd find that the eigenvectors involve an infinite sum over the SHO states. They are some sort of coherent states that are peaked at a particular value of position.

    It's probably hard unless we have a nice enough basis in the momentum representation. The harmonic oscillator is an example where we can, because the Hamiltonian is the square of the ladder operators, which are linear in the momentum and position operators. For another, rather unphysical, example, we could consider a free particle in the momentum basis. If we restrict the particle to a hard momentum band (analogous to the infinite square well), then we would find that the position was quantized. The position eigenstates are given by trig functions subject to the specific boundary conditions.
     
  6. Mar 21, 2013 #5

    strangerep

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    Dirac once said that "the mathematicians got it wrong" when they tried to make his QM rigorous by formulating it in ordinary Hilbert space. Cases involving continuous spectrum and unbounded operators, are better dealt with by the "rigged Hilbert space" formulation of QM.
    See, e.g., http://arxiv.org/abs/quant-ph/0502053

    Ballentine also gives a (shorter) introduction to RHS in ch1 of this textbook "QM -- A Modern Development."
     
  7. Mar 22, 2013 #6

    DrDu

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    I don't know whether this is true. The treatment of unbounded operators and of their spectrum in ordinary Hilbert space is very nicely laid out in the classic book by J. von Neumann, Mathematical foundations of quantum mechanics.
    One may argue that it is more physical to discuss only wavepackets of finite extensions instead of going forcedly to delta function type of distributions.
    However, this is largely a matter of taste.
     
  8. Mar 22, 2013 #7

    fluidistic

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    Ok thanks guys.
    If I understand well the matrix representation of the position operator depends on the system of the particle(s) I'm considering. If I consider a plane wave (thus not well localized), is there any matrix representation of the position operator?
     
  9. Mar 23, 2013 #8

    DrDu

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    Only a subclass of all bounded operators allows for a matrix representation, the "trace class" operators.
     
  10. Mar 23, 2013 #9

    dextercioby

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    Not true, think about the harmonic oscillator and the basis formed by eigenstates of the Hamiltonian. X and P have infinite matrix representations, while being obviously unbounded in l2(R).
     
  11. Mar 23, 2013 #10

    DrDu

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    Of course you can formally do so, but the trace of these infinite matrices depends on the choice of the basis vectors. So you can't just use the formulas valid for finite matrices.
     
  12. Mar 24, 2013 #11

    Jano L.

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    Can you be more specific? I have some doubts about using matrices for X,P too, but it seems that they have been used with some success. I am thinking of Heisenberg's matrix mechanics, or the more recent approaches to model dissipative evolution via quantum Langevin equation. Some scientists just replace position x by operator X and use the same differential equation, Do you think there might be some problem with that?
     
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