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A Operators in Quantum Mechanics

  1. May 12, 2017 #1
    Hey guys,
    Am facing an issue, we know that x and y operators take the same form due to isotropy of space, but sir if we destroy the isotropy, then what form will it take?
    Can u pleases throw some light on this!
    Thanks in advance
     
  2. jcsd
  3. May 12, 2017 #2

    vanhees71

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    Where does this come from? For me the question is so dark that I've no clue how to shed light on it :-(.
     
  4. May 12, 2017 #3
    Actually I was thinking like we say that the x-axis y-axis and the z-axis are identical due to the isotropy of space, we know that in momentum space the position operator takes some functional form so when we represent the x operator or the y operator, their functional natures are the same. Now if we break the isotropy of space then it is expected that the functional dependencies change, so how do we account for the changes in the operator algebra!
     
  5. May 12, 2017 #4
    can you give an example of a non isotropic space?
     
  6. May 12, 2017 #5

    vanhees71

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    Space-time around a Kerr black hole? SCNR.
     
  7. May 12, 2017 #6

    atyy

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    There are two sorts of x and y operators in elementary quantum mechanics: spin operators and position operators.

    Spin operators: In curved spacetime treated as a fixed, classical background, space is locally Lorentzian. The spin operators are local operators, so they remain the same. However, they must be formulated using the tetrad formalism for general relativity. See eg. https://arxiv.org/abs/1108.3896, Localized qubits in curved spacetimes, by Matthew C. Palmer, Maki Takahashi, Hans F. Westman.

    Position: In curved spacetime treated as a fixed, classical background, there are no particles except in a very approximate sense, and one must use fields and field observables, eg. quantum probabilities for particle detection along the worldline of a detector, eg. p17 of https://arxiv.org/abs/gr-qc/0308048, Introduction to Quantum Fields in Curved Spacetime and the Hawking Effect, by Ted Jacobson. Matthias Blau has links to many more good references in http://www.blau.itp.unibe.ch/QFTCST/.

    However, if spacetime is treated as curved, but not fixed, then the problem is about observables in quantum gravity. As far as I understand, there are good informal arguments that there are no local observables in quantum gravity: https://diracseashore.wordpress.com/2008/10/09/observables-in-quantum-gravity/, Observables in quantum gravity, by Moshe Rozali.
     
    Last edited: May 12, 2017
  8. May 13, 2017 #7
    Like say, at the edge of a crystal
     
  9. May 17, 2017 #8
    Operators don't take the form they take because of the "isotropy of space". Rather the isotropy of space implies that matrix elements of those operators must transform in well defined rules when you rotate your coordinates.

    I think this is covered pretty well in most graduate level books of quantum mechanics, so I'll refer to Sakurai, chapter on symmetries etc... Basically, the isotropy of space implies that Rotation operators must satisfy composition rules of the rotation group, etc.. In quantum mechanics there's an additional subtlety of the phase ambiguity, you can read about "projective representation" which is the technical term for that.
     
    Last edited: May 17, 2017
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