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Observables commute and time operator

  1. Nov 13, 2012 #1

    CAF123

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    I just have two questions relating to what I have been studying recently.
    1) I know that the total energy and momentum operators don't commute, while the kinetic energy and momentum operators do. Why is this the case? (explanation rather than mathematically).
    2) One form of the HUP says that we can't measure position and momentum of a particle simultaneously and when I evaluate the commuator , it gives a non zero operator. The other form of the HUP says that ## ΔEΔt ≥\frac{\hbar}{2}.##Is there a way to evaluate the commutator here - to similarly show that a non zero commutator between time and energy (if it exists) is in agreement with the HUP? (I.e do we define a time operator)?
    Many thanks.
     
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  3. Nov 13, 2012 #2

    mfb

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    Momentum determines the kinetic energy. This is not true for the total energy.
    ## ΔEΔt ≥\frac{\hbar}{2}.## is an "effective" rule - there is no time operator in quantum mechanics.
     
  4. Nov 13, 2012 #3
    1) Who says the hamiltonian doesn't commute with the momentum operator? That really depends on the hamiltonian. For example, a hamiltonian for a free particle trivially commutes with p. Most of the time, this won't happen because H will contain both position and momentum operators, which means that it won't commute with either (because p and q don't commute: HUP).


    2) As for the second question, no there's no time operator in QM and the origin of that HUP is different than the usual ones (and I think an explanation must involve QED), so you can't really evaluate a commutator for it. And keep in mind that the delta-t in that expression refers to lifetimes of certain states (and not the time you take to make a measurement). Someone might be able to elaborate further on this point.
     
    Last edited: Nov 13, 2012
  5. Nov 14, 2012 #4
    An existence of time operator with the usual commutation rule with hamiltonian implies no bound of lower energy(no ground state)
     
    Last edited: Nov 14, 2012
  6. Nov 14, 2012 #5

    dextercioby

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    This is essentially Pauli's argument in the article on wave mechanics in the <Enzyklopädie der Physik>. A careful analysis (Eric Galapon in Proc.Roy.Soc.London) shows he's quite wrong. See my blog article on this.
     
  7. Nov 15, 2012 #6
    So you mean a time operator exist.so can you tell how to get rid of the condition it implies, i mean no ground state and it's significance.
     
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