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In Quantum Mechanics, position is an observable, but time may be not. I think that time is simply a classical parameter associated with the act of measurement, but is there an observable of time? And if the observable will exist, what is an operator of time?

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marked as duplicate by Qmechanic Nov 7 '13 at 6:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
possible dup physics.stackexchange.com/questions/12287/… –  Yrogirg Aug 15 '12 at 18:24
    
Possible duplicate: physics.stackexchange.com/q/6584/2451 –  Qmechanic Aug 15 '12 at 18:38
    
In the first chapter of Srednicki's book on QFT he states that one route to QFT is to promote time to an operator on an equal footing with position. He says this is viable but complicated so in general we do QFT by demoting position to a label on an equal footing with time. I don't know more about this but hope it may be of interest. –  Mistake Ink Aug 15 '12 at 18:53
    
The first link above is a related but different question. the second link is more or less the same question, but the answers there are quite different from the answers below. –  Arnold Neumaier Aug 15 '12 at 19:02
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The problem of extending Hamiltonian mechanics to include a time operator, and to interpret a time-energy uncertainty relation, first posited (without clear formal discussion) in the early days of quantum mechanics, has a large associated literature; the survey article

P. Busch. The time-energy uncertainty relation, in Time in quantum mechanics (J. Muga et al., eds.), Lecture Notes in Physics vol. 734. Springer, Berlin, 2007. pp 73-105. doi:10.1007/978-3-540-73473-4_3, arXiv:quant-ph/0105049.

carefully reviews the literature up to the year 2000. (The book in which Busch's survey appears discusses related topics.) There is no natural operator solution in a Hilbert space setting, as Pauli showed in 1958,

W. Pauli. Die allgemeinen Prinzipien der Wellenmechanik, in Handbuch der Physik, Vol V/1, p. 60. Springer, Berlin, 1958. Engl. translation: The general principles of quantum mechanics, p. 63. Springer, Berlin 1980.

by a simple argument that a self-adjoint time operator densely defined in a Hilbert space cannot satisfy a CCR with the Hamiltonian, as the CCR would imply that $H$ has as spectrum the whole real line, which is unphysical.

Time measurements do not need a time operator, but are captured well by a positive operator-valued measure (POVM) for the time observable modeling properties of the measuring clock.

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In QM, the temporal variable $t$ is not an observable in the technical sense (i.e., in the same sense that position and momentum are). In order to be an observable, it should have to exist a linear self-adjoint operator $\hat T$ whose eigenvalues $t$ were the outcomes of measurements. But then (at least in the most naive way and according with the Schr. equation) the Hamiltonian and the temporal operator should be non-compatible observables with canonical commutation relations like position and momentum. And this is not possible because in a quantum theory the Hamiltonian must be bounded from bellow and this would imply that its conjugate (time operator) were no self-adjoint.

However, there exist mean lifetimes which are quantum-mechanically computable (they are inverses of probabilities per unit of time) and have units of time. In some sense (arguably vague sense), this is a quantum notion of time.

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What you give is just Pauli's argument alluded to in my answer. - Note that the mean lifetimes are not the eigenvalues of natural operators, hence don't have the same status as observables represented by Hermitian operators. - This also shows that there is something seriously amiss with the traditional notion of a quantum observable as ''defined'' by the Born interpretation. –  Arnold Neumaier Aug 15 '12 at 18:59
    
Yes, it is exactly the same argument. I started to write my answer before yours shows up. I did not read yours before answering, I'm not the fastest answering questions. I think it is clear in my answer that there is not time operator and mean lifetimes are related to probabilities and are not eigenvalues. I just wanted to point out that mean lifetimes give us some notion of time. –  drake Aug 15 '12 at 19:07
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