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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?

Masahiko Yamaguchi
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  • possible dup http://physics.stackexchange.com/q/12287/ – Yrogirg Aug 15 '12 at 18:24
  • Possible duplicates: http://physics.stackexchange.com/q/6584/2451 and links therein. – Qmechanic Aug 15 '12 at 18:38
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    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.

Emilio Pisanty
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Arnold Neumaier
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  • However, wouldn't it make more sense not to consider the commutation w.r.t the Hamiltonian operator, but rather the (now a bona fide operator since we are effectively talking about a wave function in space-time, not just space, if we're going to promote time to equal status as position) energy operator $\hat{E} := i\hbar \frac {\partial}{\partial t}$ instead? The Hamiltonian and energy are conceptually not the same thing, not even in classical Hamiltonian mechanics, which is exemplified by the fact there exist situations in CM where $H \ne K + U = E_\mathrm{mech}$! – The_Sympathizer Mar 13 '19 at 20:44
  • If we do that, and take $\hat{t} := [(x, t) \mapsto t]$, analogous to the position operator, then $[\hat{E}, \hat{t}] = i\hbar$. Where does this scheme run afoul? – The_Sympathizer Mar 13 '19 at 20:51
  • @The_Sympathizer: It changes the space of wave functions to a much bigger space - even a qubit' state space becomes infinite-dimensional. Moreover, this bigger space has no natural inner product. – Arnold Neumaier Mar 15 '19 at 06:23
  • Sure. However, not every possible wave function in that space would be physically valid, any more than that an arbitrary path of a classical particle drawn all over classical space-time represents a valid evolution for some given set of forces (or even at all, when we go to relativity and consider the limitations of Minkowski spacetime, or even causality alone!). If we're going to unite both space and time, then we necessarily have to accept some restrictions on the form things can take within both together, since that's what having dynamics/physics is all about. – The_Sympathizer Mar 15 '19 at 06:35
  • That said, the loss of inner product may be a bigger problem - although, how does that work, exactly? – The_Sympathizer Mar 15 '19 at 06:36
  • Also, regarding the qbit - that situation is no different from the same thing that happens if we bring time into the description of an ordinary, classical bit. You get now an infinite number of possible spatio-temporal histories of that classical bit, where it may or may not change abruptly at various, arbitrary, times. Again, that doesn't invalidate doing so, or describing classical bits in a unified space-time picture like that of special relativity. – The_Sympathizer Mar 15 '19 at 07:42
  • @The_Sympathizer: But a history is something very different from a state. Operators in quantum mechanics operate on state space, not on history space. Unlike states, histories are heavily constrained. – Arnold Neumaier Mar 15 '19 at 11:03
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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.

Diego Mazón
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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. – Diego Mazón Aug 15 '12 at 19:07