# Why Not a Time Operator?

1. Dec 24, 2009

### referframe

Is there one central reason why there is no "Time" operator in QM?

I know this question has been asked before, but I thought I would try to stimulate some fresh thinking. Thanks in advance.

2. Dec 24, 2009

### meopemuk

There is no operator of time, because time is not an observable.

Generally, observable is a property or attribute of the physical system. You measure the observable by bringing your measuring device in contact with the physical system and you can obtain different values of the observable depending on the state of the system. All normal observables (position, momentum, mass, spin, etc) are covered by this definition.

On the other hand, you cannot measure "time of the physical system". In fact, you can say what time is it without even having a physical system. You can simply look at the clock hanging on your laboratory's wall. The reading of the clock does not depend on what kind of physical system you are studying in the laboratory. This reading is the same even you don't have any system to study. When you do a measurement of any true observable in your physical system you simply attach the time label (obtained by looking at the wall clock) to this measurement. This means that it is OK to treat time as a classical numerical parameter in quantum mechanics.

Eugene.

3. Dec 24, 2009

### referframe

Ok. Let's call your above time parameter the "system time".

(1) The length of time (the difference between 2 system times) for which a quantum state is stationary determines the sharpness of that quantum state's energy. (Energy-Time Uncertainty).
(2) The system time, t, and the energy, E, are related via the Fourier Transform in the same way that x and p are related in the wave function.

It's almost as if there are two time variables: The system time and the time-interval related to Energy sharpness.

Make sense?

4. Dec 24, 2009

### meopemuk

If quantum state is stationary then it remains stationary for infinite time.

Yes there is analogy between E/t and p/x. But this analogy is not perfect, and you shouldn't push it too far. For example, the lifetime of an unstable state is inversely proportional to its energy spread. However, this doesn't mean that there exists a "wave function in the time representation" which is a Fourier transform of the energy-representation wave function.

There is only one time parameter - the one measured by the laboratory clock.

Eugene.

5. Dec 24, 2009

### bcrowell

Staff Emeritus
This seems to me to be unlikely to be a correct explanation, for a couple of reasons. (1) I don't see where, in this whole argument, there is any reference to any specific properties of quantum mechanics, as opposed to classical physics. (2) The whole argument would seem to apply to position just as well as it applies to time.

Here http://math.ucr.edu/home/baez/uncertainty.html is an argument that at least passes these two tests: it specifically invokes properties of QM, and it doesn't apply to position.

6. Dec 24, 2009

### meopemuk

This is true. Time plays similar roles in both quantum and classical theories. Why do you see that as a disadvantage?

No, this argument does not apply to particle position. Position is a particle's property, whose measurement depends on particle's state. In different states the measurement of position would yield different results. So, position is a true observable.

On the other hand, "measurements" of time do not depend on the state of the observed physical system. These "measurements" can be performed even if there is no physical system to observe. So, time is a specific quantity, which is an attribute of the reference frame or laboratory rather than attribute of the observed physical system.

Eugene.

7. Dec 25, 2009

### Demystifier

Yes there is. Because, by DEFINITION, a wave function is a vector in the Hilbert space of functions of points in SPACE. If you extend that definition, i.e., if you introduce an enlarged Hilbert space in which a wave function is a vector in the Hilbert space of functions of points in SPACETIME, then time operator can be naturally defined.

For more details see
http://xxx.lanl.gov/abs/0811.1905 [ Int. J. Quantum Inf. 7 (2009) 595]

8. Dec 25, 2009

### dpackard

This is unclear. There is no a priori reason we cannot speak of time as a property that belongs to a particle in the same way you speak of space. In principle, we could define a set a time states that would correspond to the particle existing at each instant of time. You're simply restating the fact that it ISN'T done, but not explaining why.

I think I understand what you're trying to say, but you muddy it with the false distinction you made above. It is NOT true that we can measure time without observing some physical system; be it the tiny gear mechanisms turning in a pocket watch or the nuclear decay of an atom, we cannot tell time by ourselves (unless we're counting our heartbeats, but that still counts as observing a system).

The difference is an empirical one. We can measure the particle in different positions corresponding to different states because we have the ability to access the same position states multiple times, whereas time is unidirectional to us. As far as I can tell this is why it is not done - we cannot do very much with the set of time basis states, since they're constantly changing.

9. Dec 25, 2009

### dpackard

Hmm, the bit in that paper about the time operator producing unphysical eigenstates in standard QM is strange. Wish I understood that. How could the time operator have unphysical eigenstates? Superpositions of time states?

10. Dec 25, 2009

### diazona

For what it's worth, I've heard that certain people have tried to develop a version of quantum field theory in which time is an operator, along with position. It turned out to be exceedingly difficult. But that's just hearsay; I've never actually looked at any of that work.

11. Dec 25, 2009

### meopemuk

Suppose that you have defined an Hermitian operator of time with eigenvalues t and eigenvectors $$| t \rangle$$. Then there should exist a particle (e.g., electron) state
coinciding with this eigenvector. In such a state the probability of finding the electron at time t is non-zero, while the probability of finding the electron at any other time is zero. This state violates all known conservation laws, in particular the laws of conservation of energy and charge. This is just another reason why the "operator of time" is not a good idea.

Of course clocks (wall clocks, pocket watches, human heart, etc.) are examples of physical systems. They are made of atoms, molecules, etc. However, when they are used as devices for measuring time in physics, they should not be treated as physical systems. For example, their quantum properties (e.g., position-momentum uncertainty) are not interesting for physicists. If the laboratory clock exhibits some quantum uncertainties, then it is simply a bad clock and should be replaced by another one, which ticks at regular predictable intervals. This property of ticking at regular intervals is the only clock property that is important for physics.

So, clocks should not be regarded as physical systems. They should be regarded as parts of the experimental setup. The same is true for other laboratory equipment. For example, each laboratory should have 3 mutually orthogonal measuring rods, which allow us to measure x-y-z positions of particles. These rods should not be regarded as physical systems. They are measuring devices, and their quantum-mechanical theoretical counterpart is the operator of position, rather than any state vector.

Any physical experiment requires the presence of three components:

1. The physical system - the object whose properties are measured. In QM, states of the physical system are described by unit vectors (or densitry operators) in the Hilbert space.

2. Measuring devices (such as x-y-z measuring rods). In QM, these objects are described by Hermitian operators of observables.

3. A clock, whose readings are described by a c-number parameter in QM.

Of course, you may decide to treat your laboratory clock as a physical system and to perform some measurements (of position, velocity, mass, energy, etc.) on this clock. But then you are considering a completely different experimental setup in which the roles of different components have changed. Quantum-mechanically, this setup should be described in a different Hilbert space, and the role of clock should be played by a different device.

Eugene.

12. Dec 26, 2009

### Demystifier

There is a classical analogy of that. A classical particle is local in space but nonlocal in time. It lives for a long time and you cannot localize it in time. In fact, this is so because particles are slower than light. If they were faster than light (tachyons), the roles of space and time would be exchanged. Photons moving with the velocity of light are in the middle of these two types of behavior.

13. Dec 30, 2009

### yuanyuan5220

two papers
arXiv: 0710.1128 Supplementary material to Heavy electrons and the symplectic symmetry of spin
arXiv: 0810.5144 Symplectic N and time reversal in frustrated magnetism

14. Jan 1, 2010

### Fredrik

Staff Emeritus
Demystifier's answer in #7 is a complete answer for wave mechanics (the non-relativistic quantum theory of a single particle), but I think Meopemuk's answer in #2 is better, because it applies to all quantum theories. An observable is an equivalence class of measurement devices. When we allow the system that we want to perform a measurement on to interact with the measurement device, the first thing that happens is that states of the measured system become correlated with the states of the measurement device. The composite system (measured system + measurement device) is still in a pure state, but interactions with the environment quickly turn it into a very good approximation of a mixed state. This can be taken as a definition of a measurement device.

As we all know, time is measured by a clock, and a clock doesn't interact with the system in the way described above. Also note that the above definition of an "observable" already relies on the concept of time. Observables are measured, and measurements take time.

Last edited: Jan 1, 2010
15. Jan 3, 2010

### Demystifier

Why do you think so? I think that a clock is a measuring apparatus like any other. It involves entanglement between many degrees of freedom, decoherence at the macroscopic level which provides the emergence of clasicallity, etc. In addition, if you think that time is fundamentally different from the position in space, then does it mean that Lorentz invariance is violated?

16. Jan 3, 2010

### meopemuk

Yes, but in order to do its job a clock does not need to interact with the studied physical system. In this sense clocks are different from all other measuring devices. A clock does not provide any information about the system. It simply provides a numerical label that can be attached to any true measurement performed on the system.

You may find interesting the ongoing discussion of the Lorentz invariance in the thread https://www.physicsforums.com/showthread.php?t=365817

Eugene.

17. Jan 4, 2010

### Demystifier

I disagree. Suppose that you want to measure time elapsed between two quantum events. Obviously, the clock that measures this time must interact with the quantum systems describing these events. Otherwise, how can clock "know" that the elapsed time has anything to do with these events?

18. Jan 4, 2010

### meopemuk

Let me specify in more detail how do we "measure time elapsed between two quantum events". In order to do this measurement we need to have

1) a physical system in which the two events are happening
2) a set of inertial ("instantaneous") observers O(t) parameterized by the quantity t (which is the reading of the clock associated with this sequence of observers). Observers O(t) are related to each other by time translation transformations. The set O(t) can be as trivial as a person sitting in a chair and holding a watch in his hand.

Each observer from the set O(t) measures all relevant observables in the system (positions of particles x, their momenta p, spins s, etc), and combining all these data one can obtain the time dependencies of these observables x(t), p(t), s(t),... Then, according to our definition of the event (e.g., a coincidence of coordinates of two particles in the system) we can find two values of t (e.g. t_a and t_b) at which the two events have happened. Then the time elapsed between the two events is simply t_b-t_a. In this situation there is absolutely no interaction between the clock (the watch in observer's hand) and the observed physical system.

Eugene.

19. Jan 4, 2010

### Fredrik

Staff Emeritus
An ideal measurement is supposed to start with a process of the type

$$\bigg(\sum_i c_i|s_i\rangle\bigg)|a\rangle\rightarrow\sum_i c_i |s_i\rangle|a_i\rangle$$

where the $|s_i\rangle$ are eigenstates of the measured observable and $|a_i\rangle$ is the state of the measuring device that indicates that the result was $s_i$. Nothing like this happens when you measure time.

This view of measurements is described in section 2.5.1 of Schlosshauer (here) and also in chapter 9 of Ballentine, if I remember correctly.

20. Jan 4, 2010

### Demystifier

See e.g. the classic paper
H. Salecker and E. P. Wigner, Phys. Rev. 109, 571 (1958).
It demonstrates how the measurement of time reduces to a measurement of an ordinary quantum observable.
Sec. 12.7. of A. Peres, Quantum Theory: Concepts and Methods

21. Jan 4, 2010

### Demystifier

I think that the parameter t above has, a priori, nothing to do with a reading of a clock. Take for example a mechanical clock with a needle. What one calls "time" measured by this clock is actually the POSITION of the needle. And the position is, of course, a quantum observable like any other.
My point is that even though the external parameter t is not an operator in standard QM, the time measured by an actual CLOCK is described by an operator.

22. Jan 4, 2010

### pellman

I'm not sure where this fits into the discussion, but I think it is something to keep in mind. If we introduced time as an observable, then it would pertain to a specific particle just as position does. So where for N particles we have N observables $$x_i$$ that describe their positions on the x-axis, we would have N time observables $$t_i$$ for each particle's "position" along the time-axis.

Also, as Demystifier points out in his paper, with time on an equal footing, the full wave function would be normalized in some fashion such as

$$\int |\psi(x,y,z,t)|^2 d^3xdt = 1$$

Schrodinger theory may be a "time-eigenstate" approximation to such a theory. That may have something to do with explaining the objection made by Fredrik in post 19. I suspect that since a system (described in the full relativistic time-as-variable way) starts out in a nearly time-eigenstate, since the dispersion under short time periods is very small, it is still in a near time-eigenstate when a measurement is made, and we don't notice any reduction due to measurement. Which would be the very reason that the "time eigenstate" approximation (time as parameter) works.

I'm probably not doing a very good choosing terms here. I hope you see what I mean.

Also, neglecting this aspect may be the very reason why we can't identify a Klein-Gordon pdf in the position representation. That reason being that $$\int |\phi(x,y,z,t)|^2 d^3x \neq constant$$ (or, another way to put is there is no conserved current for $$|\phi(x,y,z,t)|^2$$ ) possibly this is because under relativistic conditions we can no longer ignore the dispersion of the time-eigenstates. We need to treat $$|\phi(x,y,z,t)|^2$$ (or whatever the correct quantity is) as a pdf on 4-space.

23. Jan 4, 2010

### meopemuk

When speaking about measurements in QM we should be extremely careful with terminology. What is physical system? what is measuring apparatus? what is measured? how is it measured? Some think that this "measurement problem" in QM is dreadful. I think it is quite simple. Let me suggest you my vision of these things.

In experimental physics there is always a clear separation between observed physical system and measuring apparatus. Every experimentalist can tell you exactly where the boundary is. As quantum mechanics is designed to describe actual experiments, the same separation is present in the QM formalism. QM uses very different descriptions for the physical system and for measuring apparatuses. The physical system is described by its Hilbert space, where different possible (pure) states of the system are represented by unit vectors. On the other hand (different) measuring apparatuses are described by (different) Hermitian operators. The apparatus measuring position (the ruler) is represented by the operator of position. The Stern-Gerlach apparatus is represented by the operator of spin, etc. The act of measurement is *not* described as a dynamical interaction between two systems (the physical system and the measuring apparatus). In QM, the act of measurement of observable F is described as a series of rather formal steps:

1. find eigensubspaces (with eigenvalues f) of the operator F in the Hilbert space of the observed system
2. expand the state vector of the system in the above basis
3. Squares of the expansion coefficients are probabilities for measuring values f in this state.

One may say than unequal treatment of the physical system and measuring devices makes quantum theory incomplete. This is true. QM is not designed to provide a complete description of the entire world. It is focused on specific experimental situations only (where the separation "physical system/measuring device" always exists).

Of course, one may say: "I now consider my chosen physical system as comprising both the observed object and the measuring apparatus together". Fine, no problem. Then one needs to define what is the super-measuring-apparatus, which makes observations on this super-physical-system, build a new (bigger) Hilbert space, and do all QM calculations in this new Hilbert space. This new theoretical setup reflects the new experimental setup.

In addition to various measuring devices, each laboratory has a specific device, which is not used to do measurements on the physical system. This device is the clock. It's role is to provide all measurements with the classical number-label called "time". In QM clock is not a quantum object. The same is true in experiments. If the clock used in experiments shows some quantum behavior (uncertainties), then it is simply a bad clock and it should be replaced with a more stable one.

As before, you can always decide to treat the clock quantum-mechanically and include it in the observed physical system, and add its degrees of freedom to the QM Hilbert space. Then, as I explained earlier, you've moved the boundary separating the physical system from the laboratory. So, in the new laboratory you will need to choose a new device that will play the role of the clock.

So, the conclusion is that clocks are *not* treated as quantum systems in QM. Clocks are *not* treated as measuring devices either (they do not make any contact with the observed physical system). Clocks are just separate beasts. Time "measured" by clocks is not a quantum-mechanical observable. It is just a numerical parameter.

Eugene.

24. Jan 5, 2010

### Demystifier

It is interesting to note that this quantity IS a constant when phi is a superposition of positive frequencies only (or negative frequencies only). The proof is presented in the Appendix of my paper
http://lanl.arxiv.org/abs/0804.4564 [Found.Phys.38:869-881,2008]
which was discovered by another member (not me) of this forum.

25. Jan 5, 2010

### pellman

That little fact is particularly interesting to me, D. Thank you.