# Why can't we define a time related operator?

1. Mar 22, 2013

### fluidistic

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 $\hat t = \frac{d}{dt} \Psi$.

2. Mar 22, 2013

### dextercioby

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.

3. Mar 22, 2013

### fluidistic

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).

4. Mar 22, 2013

### Chopin

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 $\hat{x}$ in QFT--instead, you get observables that are functions of both time and space, such as $\phi(x)$, where $x$ 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.

5. Mar 22, 2013

### atyy

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
6. Mar 22, 2013

### Fredrik

Staff Emeritus
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.

7. Mar 23, 2013

### dextercioby

8. Mar 23, 2013

### WannabeNewton

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.

9. Mar 23, 2013

### atyy

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.

10. Mar 23, 2013

### DrDu

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.

11. Mar 23, 2013

### Fredrik

Staff Emeritus
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?

12. Mar 23, 2013

### WannabeNewton

13. Mar 23, 2013

### atyy

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.

14. Mar 23, 2013

### Staff: Mentor

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

15. Mar 23, 2013

### strangerep

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.

16. Mar 23, 2013

Staff Emeritus
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?

17. Mar 23, 2013

### strangerep

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
18. Mar 23, 2013

### strangerep

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.

19. Mar 23, 2013

### Popper

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

20. Mar 23, 2013

### strangerep

No.

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

21. Mar 23, 2013

### Staff: Mentor

Its the other way around in QFT - time and position are treated both as parameters.

From Srednicki's textbook the link to that was given previously (I have the hardcover version and have read it so know where it was located):

'Recall the axiom of quantum mechanics that says that “Observables are represented by hermitian operators.” This is not entirely true. There is one observable in quantum mechanics that is not represented by a hermitian operator: time. Time enters into quantum mechanics only when we announce that the “state of the system” depends on an extra parameter t. This parameter is not the eigenvalue of any operator. This is in sharp contrast to the particle’s position x, which is the eigenvalue of an operator. Thus, space and time are treated very diﬀerently, a fact that is obscured by writing the Schrodinger equation in terms of the position-space wave function ψ(x,t). Since space and time are treated asymmetrically, it is not surprising that we are having trouble incorporating a symmetry that mixes them up.

So, what are we to do? In principle, the problem could be an intractable one: it might be impossible to combine quantum mechanics and relativity. In this case, there would have to be some meta-theory, one that reduces in the nonrelativistic limit to quantum mechanics, and in the classical limit to relativistic particle dynamics, but is actually neither. This, however, turns out not to be the case. We can solve our problem, but we must put space and time on an equal footing at the outset. There are two ways to do this. One 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 Schrodinger 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 Xµ(τ), where X0 = T.

Relativistic quantum mechanics can indeed be developed along these lines, but it is surprisingly complicated to do so. (The many times are the problem; any monotonic function of τ is just as good a candidate as τ itself for the proper time, and this inﬁnite redundancy of descriptions must be understood and accounted for.)'

Because of that the first approach was chosen - namely to consider everything a field where position is now a parameter giving the field operator at the point and time still remains a parameter.

But of interest to this thread is that Srednicki thinks time can be made an operator. My view is without the detail of the approach he discusses its a bit hard to know the true status of his claims. From discussion I have had about it before I think he is wrong - but that's different than seeing the detail.

Thanks
Bill

Last edited: Mar 23, 2013
22. Mar 23, 2013

### atyy

I believe Srednicki has in mind the same time operator mentioned in Tong's notes http://arxiv.org/abs/0908.0333. It seems different from what dextercioby and strangerep are discussing, because it isn't conjugate to the Hamiltonian. Instead it's conjugate to some sort of "momentum".

23. Mar 24, 2013

### strangerep

Another possibility is that he was referring to other attempts (such as Stuekelberg-Horwitz) to construct many-body relativistic dynamics. Cf. Horwitz, "Time and the Evolution of States in Relativistic Classical and Quantum Mechanics", http://arxiv.org/abs/hep-ph/9606330

There's also a book: Trump & Schieve, "Classical Relativistic Many-Body Dynamics"

Arnold Neumaier gives several other references in his faq. Afaik, none of these attempts are yet satisfactory.

24. Mar 24, 2013

### fluidistic

New question

Thanks guys for the input so far. I have some related questions so I think it's appropriate to post them in this thread.
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I would like you to correct my thoughts if they are wrong:
1)I believe that there's nothing that prevent me to build any operator that depends explicitely on time such as $\hat A_{\alpha} =t \cdot \frac{d}{dt}$ (mathematically at least), however it would not correspond to any observable. The same apply for $\hat A _{\beta } =x+\frac{d}{dx} + \frac{d^2}{dx^2}$, it would not represent any observable and hence will be likely totally useless for me to do anything useful with it in QM.

2)I believe that the general consensus in QM has no operator that depends explicitely on time that correspond to an observable.
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If I'm right to both points, then the answer to my question in the title of this thread would be "nothing prevents you to define a time related operator in QM, but it would not correspond to any observable and therefore would probably be useless". Is this right guys?

25. Mar 24, 2013

### atyy

I'll ask your question in BTSM since the time operator seems standard in string theory. Maybe one of the experts can answer it.