The question in the title has been asked many times on this site before, of course. Here's what I found:

  1. Time as a Hermitian operator in QM? in 2011. Answer states time is a parameter.
  2. Is there an observable of time? in 2012. Answer by Arnold Neumaier cites surveys in the literature, and very briefly summarizes Pauli's 1958 proof: if a time operator satisfied a CCR with the Hamiltonian, it would imply spectrum of Hamiltonian is unbounded below.
  3. Why there is no operator for time in QM? in March 2013. Closed as duplicate without answer.
  4. What are the Time Operators in Quantum Mechanics? in Nov 2013. Answer confusingly states there is no time operator, then describes a time operator. Then closed as duplicate.
  5. What is the correct way to treat operators that has "time" in QM? in Feb 2016. Answer describes an explicit time operator, but points out that multiplication by $t$ does not yield an $L^2(\mathbb{R})$ function since norm is preserved in time.
  6. Let me also link Why isn't the time-derivative considered an operator in quantum mechanics?, Why $\displaystyle i\hbar\frac{\partial}{\partial t}$ can not be considered as the Hamiltonian operator?, Time in special relativity and quantum mechanics, and Why isn't the Heisenberg uncertainty principle stated in terms of spacetime? which address slightly different questions, but contain related discussions.
  7. We also have Position operator in QFT, where Valter Moretti points out that a sensible construction of coordinate operators in relativistic QFT yields the Newton-Wigner operators, but only for the spatial coordinates, since none exists for the timelike coordinate by Pauli's theorem.
  8. And off of physics.se, a great resource is John Baez's website, where he has a note The Time-Energy Uncertainty Relation which mentions that by the Stone-von Neumann a time operator satisfying a CCR with the Hamiltonian would necessitate the Hamiltonian have unbounded spectrum.

Should Pauli's theorem (which is probably a corollary of Stone-von Neumann theorem), as it's referred to in some of the answers, be taken as the definitive answer? Most of the sources seem to agree.

But some of these answers seem to be ignoring an elephant: in a relativistic theory, time and space are treated on an equal footing. If there are spatial operators, surely there must be a time operator. Indeed, some of the comments in the above questions refer to chapter 1 of Srednicki's book. Let me quote the relevant passage:

One [option to combine quantum mechanics and relativity] is to demote position from its status as an operator, and render it as an extra label, like time. The other is to promote time to an operator.

Let us discuss the second option first. If time becomes an operator, what do we use as the time parameter in the Schrödinger equation? Happily, in relativistic theories, there is more than one notion of time. We can use the proper time τ of the particle (the time measured by a clock that moves with it) as the time parameter. The coordinate time $T$ (the time measured by a stationary clock in an inertial frame) is then promoted to an operator. In the Heisenberg picture (where the state of the system is fixed, but the operators are functions of time that obey the classical equations of motion), we would have operators $X^\mu(τ)$, where $X^0 = T$. Relativistic quantum mechanics can indeed be developed along these lines, but it is surprisingly complicated to do so.

In a subsequent paragraph, he points out that promoting all the coordinates to operators, including time, is exactly what string theory is, though with the additional complication that there is an additional parameter.

So how do we reconcile these two camps? Is Pauli's theorem incompatible with relativistic considerations? Is it really a problem if your Hamiltonian's spectrum is not bounded below? Why is there no timelike Newton-Wigner operator? Does there in fact exist a relativistic quantum theory which includes an operator whose eigenvalues are time coordinates?

I feel like the existing answers on the site did not address this issue, so I'm hoping for some new answers which look at it from that perspective.

  • $\begingroup$ If they're all saying the same thing (that time isn't an operator), what's the question? $\endgroup$
    – Kyle Kanos
    Nov 26, 2015 at 16:35
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    $\begingroup$ @KyleKanos: The physics.se answers and Pauli's theorem are contradicted by Srednicki's answer, as well as the explicit example of string theory. $\endgroup$
    – ziggurism
    Nov 26, 2015 at 16:37
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    $\begingroup$ String theory is a red herring - yes, the target space "coordinates" are operators, but the QFT lives on the worldsheet, not on the target space. There isn't any "worldsheet time operator", the worldsheet coordinates are just labels. There is no contradiction. $\endgroup$
    – ACuriousMind
    Nov 26, 2015 at 16:53
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    $\begingroup$ String theory has a 1+1 CFT with a "time operator" for a completely unrelated 10D target space. That you can have in any QFT, but that's not what people mean when they say a time operator is forbidden - they mean an operator for the time coordinate of the d+1 spacetime the QFT lives on is forbidden $\endgroup$
    – ACuriousMind
    Nov 26, 2015 at 17:09
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    $\begingroup$ Just how about: $\hat{t}=i \hbar \frac{\partial}{\partial E}$ and $\left[\hat{t}^2, E\right]=0$ (doi.org/10.1016/j.aop.2006.10.007)? $\endgroup$ Feb 22 at 15:13

2 Answers 2


Just open any string text which has a discussion of the relativistic point particle.

http://arxiv.org/abs/0908.0333 - Section 1 for example or Green, Schwartz, Witten Volume 1


1) Time can be introduced as an operator but you need to introduce a 'proper time' parameter for which the system evolves with. In doing this you introduce a gauge redundancy which needs to be accounted for when doing the path integral

2) The proper-time Hamiltonian vanishes. So all the "Schroedinger equation" says is that the wavefunctions don't depend on propertime.

3) The canonical momentum you quantize $p_\mu = \frac{\partial L}{\partial \dot{X}^\mu} \rightarrow -i \partial_\mu$ aren't independent: $p^2 + m^2 = 0$

4) Imposing that constraint on the wavefunction gives you the Klein-Gordan equation. If you try to solve it you get 'negative energy' solutions. This is the unbounded spectrum issue. The correct interpretation is that these modes represent 'anti-particles': positive energy solutions travelling backwards in time.

Is that last part surprising? I don't think so. You demanded a formalism which treats time and space equally and got solutions which can travel backwards and forwards in space. Seems reasonable to expect you get things that travel backwards and forwards in time. This doesn't let you break causality. In fact this is a feature of relativity and to remove it breaks causality.

From here you could do the same kind of thing for spinful particles or start including interactions but in all honesty the single particle formalism isn't as useful to describe interactions. Physics demands a many-particle/field description to describe interactions. Of course you'll be able to recover this, and non-relativistic quantum mechanics, in the appropriate limits.

  • $\begingroup$ Those notes by Tong didn't say what the CCR were, and didn't discuss whether a time operator was compatible with the result of Pauli or Stone-von Neumann, which is what I'm looking for. My guess is that it's ok for the operator $i\partial_t$ to have unbounded spectrum in a relativistic theory (negative energy states or something?), therefore we can also have a time operator conjugate to $\partial_t$. $\endgroup$
    – ziggurism
    Nov 26, 2015 at 21:52
  • $\begingroup$ Quantizing $p_\mu \rightarrow -i \partial_\mu$ is so that $[x_\mu , p_\nu] = i \delta_{\mu \nu}$ and yeah $p_0$ will have unbounded spectrum: anti-particles. $\endgroup$
    – SM Kravec
    Nov 26, 2015 at 22:36
  • $\begingroup$ Replace $\delta$ with $\eta$ my brain is fuzzy from thanksgiving $\endgroup$
    – SM Kravec
    Nov 26, 2015 at 22:47
  • $\begingroup$ And so all those people running around saying there cannot be a time operator in quantum mechanics, are really only talking about a particular simpler kind of quantum mechanics, without the reparametrization invariance, where we need $i\partial_t=H$ to be semibounded. $\endgroup$
    – ziggurism
    Nov 27, 2015 at 15:06
  • $\begingroup$ Yeah they're talking about non-relativistic quantum where there's one universally agreed upon notion of time and its a parameter as opposed to an operator. That's usually what people ask about with these SE questions. Notice there that $i \partial_t \neq H$ as time isn't an operator on operator on the Hilbert space $\endgroup$
    – SM Kravec
    Nov 27, 2015 at 18:50

This is one of the open questions in Physics.

J.S. Bell felt there was a fundamental clash in orientation between ordinary QM and relativity. I will try to explain his feeling.

The whole fundamental orientation of Quantum Mechanics is non-relativistic. Even though, obviously, QM can be made relativistic, it goes against the grain to do so, because the whole concept of measurement, as developed in normal QM, falls to pieces in relativistic QM. And one of the reasons it does so is that there is no time operator in ordinary QM, time is not an observable that gets measured in the same sense as position can. Yet, as you and others have pointed out, in a truly relativistic theory, time should not be treated differently than position.

I presume Srednicki is has simply noticed this problem and has asked for an answer. This problem is still unsolved. There is a general dissatisfaction with the Newton-Wigner operators for various reasons, and the relativistic theory of quantum measurement is not nearly as well developed as the non-relativistic theory.

It is notable that QFT, the most accepted and useful way to unify relativity and quantum theories, really does not possess the same logical structure as ordinary QM: the quantum fields are described by functions on space-time rather than as wave-functions of a many-particle system on a polarisation in a phase space. This is due to the whole philosophical difference between relativity and ordinary QM: relativity gives a privileged position to the space-time coordinates. But QM uses phase space, a very high-dimensional one, and then picks half the variables (e.g., only the positions of all the particles) for its variables. (This is called, by e.g. Souriau and Vergne, a polarisation.) QFT abandons this QM technique by using occupation numbers or operators instead of wave functions, but thus gets away from the whole structure of ordinary QM's measurement theory involving observables etc.

  • $\begingroup$ I really like your response, can you clarify where QM only "picks half the variables". Is this due to the representation as a complex analytic signal or that we project onto a <ket| that measures position and not position+momentum when getting a "measurable" quantity? $\endgroup$
    – Mikhail
    Nov 26, 2015 at 20:03
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    $\begingroup$ Yes. Example: The classical phase space for a two particle system has twelve dimensions: three position variables for the first particle, three more for the second. Three momentum variables for the first particle, three more for the second. A polarisation picks one half of these: for example x position and y position and z-momentum for the first particle, and x position and y momentum and z-momentum for the second particle. A stupid choice, but a good example. A more usual example: pick the momentum variables for both particles. $\endgroup$ Nov 26, 2015 at 20:13
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    $\begingroup$ Can you give some references for your last paragraph? Is Souriau and Vergne a book or paper or just names of two people? $\endgroup$
    – Curiosa
    Oct 26, 2016 at 23:36

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