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Why can't we define a time related operator?

  1. Mar 22, 2013 #1

    fluidistic

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    I asked my professor if we could define time as an operator and he said no.
    I've read on the web that time isn't an observable because "you don't measure the time of a particle". Yet, to me at least, the sentence "you don't measure the time of a particle" is similar to "you don't measure the position of a particle".
    When I give the position of a particle it's always according to a reference frame ##\vec x _0##. Why can't I give the "time of a particle" with respect to a reference time ##t_0##?
    So if I have an external clock to the system I'm considering and that I had set that clock to 0 at ##t_0## then I could give the time and position of the particle at any moment.
    Position like time is not an intrinsec property of particles as far as I know. So why a position operator but no time operator?

    My professor wrote on the blackboard that for stationary states, ##\Psi _t =e^{-iEt/\hbar }## and that ## \langle \Psi _t , \hat A \Psi _t \rangle = \langle e^{-iEt/\hbar }\Psi, \hat A e^{-iEt/\hbar } \Psi = e^{iEt/\hbar } \cdot e^{-iEt/\hbar } \langle \Psi, \hat A \Psi \rangle = \langle \Psi , \hat A \Psi \rangle ## and then he concluded that for any self-adjoint linear operator ##\hat A##, the expected values of all the observables are time independent. I follow the math and his conclusion, but it's true if ##\hat A## does not depend on time. Why can't we define an observable ##\hat A## that depends on time? Such as ##\hat t \Psi =t\cdot \Psi## or ##\hat t = \frac{d}{dt} \Psi##.
     
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  3. Mar 22, 2013 #2

    dextercioby

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    There's a long history behind a time operator. You can google this easily. Well, since it's not universally accepted, it's not textbook material after all.
     
  4. Mar 22, 2013 #3

    fluidistic

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    I see, thank you.
    It's over my head (I just started QM). I'll assume for now and on that I can't build any operator that depends on time until I get more knowledge to understand what's strange about time.
    Meanwhile, what do you think about the document http://organizations.utep.edu/portals/1475/hilgevoord-time in quantum mechanics.pdf ? He more or less says that the argument given by Pauli which forbids the time operator is wrong (3rd page, or page 303).
     
  5. Mar 22, 2013 #4
    This problem becomes a big deal when you try to unify QM with relativity, where you have to consider time and space to be one and the same. It may interest you to know that in Quantum Field Theory, which is the commonly-accepted answer to the problem, you handle it by going the opposite direction from your suggestion--instead of trying to create a time operator, you get rid of the position operator.

    So there is no such thing as [itex]\hat{x}[/itex] in QFT--instead, you get observables that are functions of both time and space, such as [itex]\phi(x)[/itex], where [itex]x[/itex] is a relativistic four-vector. By doing that, and putting time and space back on the same footing, you can construct a Lorentz-invariant quantum theory that obeys all of the normal laws of relativity.
     
  6. Mar 22, 2013 #5

    atyy

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    Srednicki's draft version of his QFT text discusses the time operator briefly. He says a time operator is possible but complicated.

    http://web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf (bottom of p25)
     
    Last edited: Mar 22, 2013
  7. Mar 22, 2013 #6

    Fredrik

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    You measure position with a particle detector. You measure time with a clock. A clock measures time just fine without interacting the particle, so you can hardly think of a time measurement as a determination of a property of the particle.
     
  8. Mar 23, 2013 #7

    dextercioby

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  9. Mar 23, 2013 #8

    WannabeNewton

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    Really? Ballentine summarizes Pauli's argument in his text too, but doesn't really say anything about its possible inaccuracies (page 344). After that paragraph he talks about what Fredrik mentioned before regarding how time is actually measured via say a clock.
     
  10. Mar 23, 2013 #9

    atyy

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    I'm not sure this is rigourous, but in addition to the comment in Srednicki's text, Tong's notes explicitly mention a time operator. http://arxiv.org/abs/0908.0333

    If this is correct, then Pauli's argument can't be right without additional assumptions.
     
  11. Mar 23, 2013 #10

    DrDu

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    I also heard the following argument:
    Time and energy are canonically conjugate variables. But the energy spectrum is bounded from below. This makes it impossible to construct a conjugate variable. Similar problems occur when one tries to construct a phase operator conjugate to particle number or a radial momentum operator conjugate to the radial coordinate.
     
  12. Mar 23, 2013 #11

    Fredrik

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    A question for those of you who have studied one or more definitions of something that can be considered a "time operator": Is there a measuring device associated with the operator?
     
  13. Mar 23, 2013 #12

    WannabeNewton

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  14. Mar 23, 2013 #13

    atyy

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    It looks like the time operator that dexter defines there is different from Tong's. dexter defines it by commutation with the Hamiltonian, but Tong (p28) defines it with commutation with position and conjugate momenta.
     
  15. Mar 23, 2013 #14

    bhobba

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    I dont think there is anything that actually forbids a time operator. Indeed in Quantum Field Theory the difficulty it addresses is that relativity treats time and space on the same footing yet bog standard QM has time a parameter and position an operator. To get around this QFT has both time and position as parameters. Evidently, the other way to overcome it was also tried - time and position as operators. But evidently the mathematics was horrid and it couldn't be made to work. I had read in Srednicki's textbook that while very difficult it could be made to work and led to the same math as QFT, but people more knowledgeable about such things posted that wasn't correct - doesn't matter how hard it was tried it couldn't be made to work.

    Thanks
    Bill
     
  16. Mar 23, 2013 #15

    strangerep

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    Careful. That's a misleading summary of what Hilgevoord is trying to communicate.

    To others responding in this thread: I get the feeling not everyone read that Hilgevoord paper before replying. (Am I right?) The important point in that paper (afaict) is that the notions of time-as-spacetime-coordinate and time-as-dynamical-variable are not interchangeable. In QM, the latter tends to be more important since we often quantize a classical Hamiltonian dynamics. This distinction is also part of the reason why other puzzles about time in QM can sometimes be resolved by appealing to some kind of clock apparatus/process which can participate in quantum-level interactions and correlations with other subsystems.
     
  17. Mar 23, 2013 #16

    Vanadium 50

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    Before going into the details of how one constructs a "time operator", one should think about what exactly one is: I have a hydrogen atom in its ground state (just to pick a random, specific example): what exactly should <psi | T | psi> return?
     
  18. Mar 23, 2013 #17

    strangerep

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    Well, time as an operator canonically-conjugate to a Hamiltonian operator with spectrum bounded-below is untenable. (This was also part of Pauli's argument, iirc.) The commutation relation ##[T,H] = i## leads to
    $$\def\eps{\epsilon} H' ~:=~ e^{-i\eps T} \, H \, e^{+i\eps T} ~=~ H - \eps$$ where ##\eps## can take any real value. So if the time-translated operator ##H'## is applied to an energy eigenstate ##|E\rangle## it implies the existence of another eigenstate ##|E-\eps\rangle##, and so on.

    Actually, the recurring "time operator" questions are yet another reason why it's better to approach QFT from the perspective of Poincare unireps: one introduces the field operators in 4-momentum space, but restricted to the +ve energy mass hyperboloid (i.e., physical field operators are 0 off that hyperboloid). Since the latter has only 3 degrees of freedom, Fourier transforming back to 4-space parameter manifold then yields a more reliable picture. There can only be 3 independent degrees of freedom in the Fourier-transformed field, regardless of first impressions that may suggest the contrary.
     
    Last edited: Mar 23, 2013
  19. Mar 23, 2013 #18

    strangerep

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    I get your point.
    But, (extending your point), what should the ##\langle 0| X |0\rangle## return? The vacuum state here corresponds to a system which is not spatially bounded. :biggrin:
     
  20. Mar 23, 2013 #19
    Your professor is quite right. Time is a parameter in quantum mechanics whereas position and momentum of a particle is not. We don't really measure time but note it. That is to say when a particle hits one of many in an array of detectors and we here a click regestering its detection then we look at a clock and note the time. The clock isn't doing anything but ticking. It will not change. It's a macroscopic system, not a quantum one, designed to be independant of how our instruments work and when they detect a particle. The same thing is true in classical mechanics.

    I don't know field theory yet but I understand that there is time operator there since you have to treat space and time on the same footing in relativistic field theory.

    Given what I just said, however, there is such a thing called a "Time of Evolution Operator." This is quite different than a time operator.

    You may find this interesting Time as an Observable, by J. Oppenheim, B. Reznik, W.G. Unruh. To be published in Proceedings of the 10th Max Born Symposium, eds. Ph. Blanchard, A. Jadczyk, Wroclaw - Sept., 1997, Springer-Verlag, Lecture Notes in Physics.

    It's online at - http://arxiv.org/abs/quant-ph/9807058
     
  21. Mar 23, 2013 #20

    strangerep

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    No.

    BTW, it's usually wise to read through the entire thread before responding to the 1st post. :wink:
     
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