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Understanding observables in QM

  1. Jan 9, 2008 #1
    Hi community,

    In the past weeks I'm busy ordering my thought on quantum mechanics. Somehow I really like to understand this part of physics, I don't want to just make a good exam. That's why I started writing down everything I know on quantum mechanics (from a mathematical point of view) and try to fill in the gaps. Most questions I had could be answered by studying my books or reading up on the internet. But I have never really understood how operators are formed.

    All books and articles I found simply give the operators for momentum and location in momentum space or position space. Or they postulate that observables are represented by self-adjoint operators, from which I can't see how the actual operators are formed.

    So could someone of more knowledge then me shed some light on this subject? Thanks in advance.

    Hendrik
     
    Last edited: Jan 9, 2008
  2. jcsd
  3. Jan 9, 2008 #2

    olgranpappy

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    can you tell us which books and then we can recommend some more?

    In terms of vectors (representing the state of the system), an operator acts on a vector and returns another (probably different) vector. I.e., it's a function of a vector whose value is a vector.

    An observable is *defined* as a self-adjoint operator whose eigenvectors form a basis for the space.
     
  4. Jan 9, 2008 #3

    f95toli

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    Maybe I am wrong, but I think the OP is asking how we can know that the observables in QM are related to quantities we actually measure.
    While it is true that observables are easy to define mathematically there is -as far as I know- no "algorithm" that can be used to derive an operator for a given measurable quantity (all observables can of course be measured in principle, but that is not the same thing as knowing that apparatus A will measure the expectation value of a certain operator).
    I have seen a few "derivations" but they mainly focus on using physical "intuition" (e.g. drawing analogies etc) and then showing that the operator has a set of useful properties. However, sometimes this is far from obvious and there are quite a few papers where the authors argue for or agains a certain operator being the "correct" one for a certain measurement.
     
  5. Jan 9, 2008 #4
    First of all thanks for the reply, but it isn't exactly what I was looking for.

    Lets start of with the books I've been using:
    *Introductory quantum mechanics (fourth edition) by R.L. Liboff (the book used in the course I'm following).
    *Quantum physics by S. Gasiorowicz
    *Quantum mechanics by C. Cohen-Tannoudji, B Diu and F. Laloe (not in my possession at the moment).

    The answer you give is what I've been hearing over and over, sorry if I was unclear in my initial post. What I was really interested in is why the momentum operator in position space looks the way it does:
    [tex] -i \hbar \frac {\partial} {\partial x} [/tex]


    Ah, I just received an e-mail with the new reply from f95toli. Also thanks for the reply.
    That is indeed what I was asking about. Could you perhaps point me to such a derivation? I understand that it won't be of much help, but I would like to have some pointer at least.
     
  6. Jan 9, 2008 #5

    olgranpappy

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    in fact, there is an algorithm:

    take the classical quantity which you are interested in measuring (which is a function of 'x' and 'p') and symmetrize it wrt to 'x' and 'p' and then promote 'x' and 'p' to operators.

     
  7. Jan 9, 2008 #6

    f95toli

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    That method usually works quite well, but there are cases when it is not as straigtforward. One example would be the fact that it us not always possible to replace a classical current with the Heisenberg current operator, the reason is that the latter does not commute with itself at different times. This is a problem when e.g. studying noise since we then need to calculate the autocorrelation function*.

    *one way to avoid the problem is to instead use a symmetrized product, this is often used in various books but might in fact be wrong; it can be argued that measurements of e.g. excess noise is in fact be measuring the non-symmetrized product.
    See. e.g. Gasih et al
    http://arxiv.org/abs/cond-mat?papernum=0211646
     
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