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Time operator?

  1. Mar 6, 2006 #1
    Just wondering if there exist a "time operator" in quantum mechanics why or why not?
  2. jcsd
  3. Mar 6, 2006 #2
    And what would it do exactly?

    Translate along the time axis?- Hamiltonian/time evolution
    Reverse direction of time?- Time reversal operator

    You mean an operator that creates time? That would be time evolution, it gives states a time dependance.
  4. Mar 6, 2006 #3


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    It's not an observable if that's what you mean. You don't measure the 'time of a particle'.
  5. Mar 6, 2006 #4
    I know there are two formulations of QM where one of them the operators are themselves derived as functions of time in the Hamiltonian, which I believe makes the mathematics much more difficult, but the conceptual understanding easier; where as, we usually learn it as the operators are independent of time ... well something along these lines. Does anyone know what I'm referring to, and maybe shed a little more light on it?
  6. Mar 6, 2006 #5


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    You're talking about the Heisenberg and Shrödinger pictures. In the Shrödinger picture the operators don't depend on time, but the state evolves.
    The Heisenberg picture is more like classical mechanics. The observables change in time for a given system. It's not really more difficult per se. The relation between the two is such so any measurable prediction is the same in both pictures.
  7. Mar 6, 2006 #6
    Yes thats what I meant, an observable with eigenvalues and functions.

    This is where I'm confused, the momentum operator is basically a deriviative with respect to position, but classically, momentum equals mass times velocity, or m * dx/dt, then where does a component of a particle's velocity come from in quantum mechanics since time is not incorporated in the momentum operator?

    Also, since the position operator is defined, why not time? It would be like a measurement of a particles time would place it in a state at that particluar time?
  8. Mar 6, 2006 #7
    thanks galileo, that's exactly what i'm thinking of, it's been a while since i've played in the qm realm...should go back and play, i remember how much fun it was. in fact, i just did the ladder operators for the sho on a random problem the other day. all in all, good times with the qm.
  9. Mar 6, 2006 #8
    have you considered the commutation relationships between the corresponding operators? also, remember, the relationship between observables, such as position, momentum must observe the uncertainty principle.
  10. Mar 7, 2006 #9


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    Here's the trick:

    Time needn't be the evolution parameter[/B] in ordinary (nonrelativistic, purely classical) mechanics. Think of that [itex] \tau ^[/itex] variable which appears when one tries to prove the Noether theorem for systems with a finite no. of degrees of freedom...

    Why do we choose "time" for evolution in nonrelativistic QM ...? Simply because of habit. We could choose [itex] tau [/itex], but it still wouldn't be quantized...

    Here's the case for special relativity ,QM and QFT:

    In ordinary QM displacement along "x" is a self-adjoint operator [itex] \hat{X} [/itex], but time is just a scalar parameter which labels Schrödinger state vectors or Heisenberg operators such as [itex] \hat{X}\left(t\right) [/itex]. This doesn't sit well with special relativity, which places displacement & time on equal footing.

    If time were an operator [itex] \hat{T} [/itex], then it would be canonically conjugated to the Hamiltonian operator, [itex] \hat{H} [/itex].

    The commutation relation

    [tex] \left[\hat{T},\hat{H}\right]_{-}=i\hat{1} [/itex]

    implies that

    [tex] \exp\left(-i\epsilon \hat{T}\right) \hat{H} \exp\left(+i\epsilon \hat{T}\right) =\hat{H}-\epsilon \hat{1} [/tex]

    for any real constant \epsilon. The thing above means that the operator [itex] \exp\left(i\epsilon\hat{T}\right) [/itex] when applied to an eigenstate [itex] |E\rangle [/itex] of the Hamiltonian with eigenvalue E produces a new eigenstate with energy E-\epsilon. This means that a continuous energy spectrum [itex] -\infty <E<+\infty [/itex] occurs contrary to the requirement that the energy of a physical system be bounded from below.

  11. Mar 7, 2006 #10


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    There is an operator called the evolution operator, which 'translates the state over time'. Leaving the math details behind, you can integrate the S.E (assuming H doesn't depend on time):
    [tex]i\hbar \frac{\partial}{\partial t}\Psi=\hat H \Psi[/tex]
    to give:
    [tex]\Psi(t)=\exp(-i\hat H(t-t_0)/\hbar)\Psi(t_0)[/tex].

    [itex]U(t,t_0)=\exp(-i\hat H(t-t_0)/\hbar)[/itex] is called the evolution operator. But this is not an observable (For one thing, it isn't Hermitian). What sense does it make to measure the time of a particle? Time is the parameter on which the dynamical depend.
    Last edited: Mar 7, 2006
  12. Nov 28, 2011 #11
    I'm not sure why there has to be a focus on translation... the operator [itex]\hat{x}[/itex] isn't responsible for translation (except in p-space), so why would we want/expect [itex]\hat{t}[/itex] to translate through time...?

    I seems fine to me to have an operator [itex]\hat{t}[/itex] that draws out the coordinate [itex]t[/itex]. We constantly have an arbitrary origin for [itex]\hat{x}[/itex] so do that same thing for [itex]\hat{t}[/itex]. Then the eigenvalue would simply be the time coordinate, the time on someones watch.

    Why would such an operator be forbidden?
  13. Nov 28, 2011 #12


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    There are several proposed and acceptable models of a time operator embedded in the standard Hilbert space formulation of QM.

    The so-called <time of occurance> operator has been proposed. A review is made by Srinivas <The time of occurance in Quantum Mechanics>.
  14. Nov 28, 2011 #13
    Thanks dexter :smile:
  15. Nov 29, 2011 #14


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    If we define a classical state as a point in phase space and classical observables as functions on this space, then <time> is also not an observable in classical mechanics. Given a point in phase space, I can't calculate the corresponding <time> from that. Or am I overlooking something?
  16. Nov 29, 2011 #15

    A. Neumaier

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    No. Your observation is right to the point. Classically and quantum mechanically, time is not (and should not be) an observable on the basic Hamiltonian or Lagrangian level.

    When we measure time, we measure - seen from a fundamental point of view - a pointer position or a counter state.
  17. Nov 29, 2011 #16

    A. Neumaier

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    Whether measured or not, a state is _always_ a state at a particular time. Thats why we write psi(t), and have a dynamical equation to tell how the state changes with time.

    A position operator is defined only for a system consisting of a single particle alone in the universe, in an observer-dependent coordinate system. (The observer must be outside this mini universe.)

    For an N-particle system, one has N position operators. If time were like position, each particle would have its own time, which would make the concept of time meaningless.

    Note that in a relativistic theory, position becomes like time rather than time like position: Instead of trajectories depending on time we have fields depending on space and time. And particle positions for multiparticle systems don't make relativistic sense - position becomes an intrinsically smeared concept, even classically!

    What one can have consistently in relativity is only relative positions of one particle with respect to
    a frame attached to a particular particle and its world line - in the case of the GPS this ''particle'' is the earth. These positions are again time-dependent.

    Thus there is an intrinsic asymmetry between observed position and time - even in the classical relativistic case!
  18. Nov 29, 2011 #17
    Isn’t time the fourth part of position? Don’t we already assign particles with their own time, just like the emission of a photon their time began upon their own emission? Relative to my present all atoms were formed at the same “time” and still exist as my present today. How does this make the concept of time meaningless?
  19. Nov 30, 2011 #18

    A. Neumaier

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    I didn't say that time is meaningless but ''... would make time meaningless''.

    In relativity, time is the 0-component of 4-position. But this gives 4-position the same status as time has in the nonrelativistic case - not time the same status as position has in the nonrelativistic case. The latter is the nongeneric case, meaningful only when considering a single particle alone in the world.
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