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Squared operators and sums of operators in practice

  1. Sep 6, 2014 #1
    Consider a one dimensional harmonic oscillator.

    We have:
    $$\hat{n} = \hat{a}^{\dagger} \hat{a} = \frac{m \omega}{2 \hbar} \hat{x}^2 + \frac{1}{2 \hbar m \omega} \hat{p}^2 - \frac{1}{2}$$
    And:
    $$\hat{H} = \hbar \omega (\hat{n} + \frac{1}{2})$$

    Let's say we want to measure the total energy. We can, using the number operator, but, apparently, this requires (by the definition above) that we measure location and momentum twice consecutively in two separate identical systems.

    This raises questions:

    1. After measuring one observable, immediately measuring it again should not change the value measured nor the function. So, what is the meaning of this in practice?

    2. The apparent requirement of interacting with two independent systems in order to calculate/measure the number of quanta in a system has to be wrong. ?

    So finally: How does one measure total energy or number of quanta in reality?
     
  2. jcsd
  3. Sep 6, 2014 #2

    atyy

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    The assumption is wrong. In "pure theory without conservation laws" we can just assume that all observables can be measured by a single apparatus.

    In "pure theory with conservation laws" not all observables can be measured.
    http://arxiv.org/abs/1012.4362

    But for the actual process in real life of measuring, we don't always know how to measure every observable, and even if we did, it might be too expensive. However, for some systems, we do know how to measure all observables.
    http://arxiv.org/abs/quant-ph/0408011
    http://arxiv.org/abs/quant-ph/0512227v1
     
  4. Sep 8, 2014 #3
    atyy, Thanks a lot for these enlightening references.
     
  5. Sep 9, 2014 #4

    tom.stoer

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    A self-adjoint operator, which defines a "measurable quantity", i.e. an "observable" tells you what you can measure in principle, but not how to measure it in practice; the obervable is silent about the construction of the apparatus and about the measurement process.
     
  6. Sep 9, 2014 #5

    DrDu

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    An operator like H which is the sum of x^2 and p^2 is a new operator. It does not mean that you have to measure both x and p at the same time. There will be other operators containing both x and p whose variance will become very large on the eigenstates of H.
     
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