- #1

- 343

- 33

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

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary, there is no operator of time in quantum mechanics because time is not an observable. While all normal observables have properties that can be measured through interactions with the physical system, time can be measured independently through a clock. This means that time can be treated as a classical numerical parameter in quantum mechanics. Additionally, there is only one time parameter in quantum mechanics, measured by a clock, rather than two separate time variables.

- #1

- 343

- 33

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

Physics news on Phys.org

- #2

- 1,766

- 63

referframe said:central reason why there is no "Time" operator in QM?one

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

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

- 343

- 33

meopemuk said: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.

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

(1) The

(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

- 1,766

- 63

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

(1) Theof time (thelengthbetween 2 system times) for which a quantum state is stationary determines the sharpness of that quantum state's energy. (Energy-Time Uncertainty).difference

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

referframe said:(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.

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.

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

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

Eugene.

- #5

- 6,724

- 429

meopemuk said: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.

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

- 1,766

- 63

bcrowell said: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.

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

bcrowell said:(2) The whole argument would seem to apply to position just as well as it applies to time.

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

- 14,108

- 6,610

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.referframe said:Is therecentral reason why there is no "Time" operator in QM?one

For more details see

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

- #8

- 32

- 0

meopemuk said: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.

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.

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.

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

- 32

- 0

- #10

Homework Helper

- 2,176

- 8

- #11

- 1,766

- 63

dpackard said: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.

Suppose that you have defined an Hermitian operator of time with eigenvalues t and eigenvectors [tex]| t \rangle[/tex]. 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.

dpackard said: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).

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

- 14,108

- 6,610

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.dpackard said:

- #13

- 12

- 0

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

Staff Emeritus

Science Advisor

Gold Member

- 10,877

- 422

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.referframe said:Is therecentral reason why there is no "Time" operator in QM?one

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:

- #15

- 14,108

- 6,610

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?Fredrik said:As we all know, time is measured by a clock, and a clock doesn't interact with the system in the way described above.

- #16

- 1,766

- 63

Demystifier said: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.

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.

Demystifier said:In addition, if you think that time is fundamentally different from the position in space, then does it mean that Lorentz invariance is violated?

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

Eugene.

- #17

- 14,108

- 6,610

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?meopemuk said: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.

- #18

- 1,766

- 63

Demystifier said: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?

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

Staff Emeritus

Science Advisor

Gold Member

- 10,877

- 422

An ideal measurement is supposed to start with a process of the typeDemystifier said: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?

[tex]\bigg(\sum_i c_i|s_i\rangle\bigg)|a\rangle\rightarrow\sum_i c_i |s_i\rangle|a_i\rangle[/tex]

where the [itex]|s_i\rangle[/itex] are eigenstates of the measured observable and [itex]|a_i\rangle[/itex] is the state of the measuring device that indicates that the result was [itex]s_i[/itex]. 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

- 14,108

- 6,610

See e.g. the classic paperFredrik said:Nothing like this happens when you measure time.

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.

See also

Sec. 12.7. of A. Peres, Quantum Theory: Concepts and Methods

- #21

- 14,108

- 6,610

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.meopemuk said: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.

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

- 684

- 5

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

[tex]\int |\psi(x,y,z,t)|^2 d^3xdt = 1[/tex]

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 [tex]\int |\phi(x,y,z,t)|^2 d^3x \neq constant[/tex] (or, another way to put is there is no conserved current for [tex]|\phi(x,y,z,t)|^2[/tex] ) possibly this is because under relativistic conditions we can no longer ignore the dispersion of the time-eigenstates. We need to treat [tex]|\phi(x,y,z,t)|^2[/tex] (or whatever the correct quantity is) as a pdf on 4-space.

- #23

- 1,766

- 63

Demystifier said: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.

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

- 14,108

- 6,610

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 paperpellman said: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 [tex]\int |\phi(x,y,z,t)|^2 d^3x \neq constant[/tex]

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

- 684

- 5

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

- #26

- 684

- 5

meopemuk said: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

We don't have to necessarily treat the clock as a quantum object. We make position measurements without treating rulers as quantum objects.

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.

This is true. In regular QM there

- #27

- 1,766

- 63

pellman said:This is true. In regular QM thereisno time operator. But why is the theory formulated that way? Making a time measurement really is a measurement like any other. In fact, it is very often (always?) inferred from a position measurement. So how can we get away with ignoring that?

My point was that formalism of quantum mechanics should be discussed only in the context of a specific experiment. Tell me which experiment you have in mind (what is the observed system and what is the measuring apparatus) and I will be able to answer how you should describe the clock.

If, for example, the observed system is an unstable particle, and you are studying its decay probability as a function of time, then the clock is not a part of the quantum-mechanical system. It is a part of the measurement setup. Clock's readings are classical parameters.

If your observed system is the arm of the clock, and you are measuring the arm's position. Then the Hilbert space you are dealing with is the space of states of the arm. The clock is not used to measure time, there should be some other device to do this job in your laboratory.

I understand that this sounds as a contradiction (different theoretical descriptions of the same object depending on the question asked). But this is the fundamental unavoidable contradiction of the QM formalism. This contradiction results from the fact that QM is not a complete theory of everything. QM can be applied to specific well-defined individual experiments. It cannot be applied to the world as a whole.

Eugene.

- #28

- 14,108

- 6,610

That's where my point of view significantly differs from yours.meopemuk said:This contradiction results from the fact that QM is not a complete theory of everything. QM can be applied to specific well-defined individual experiments. It cannot be applied to the world as a whole.

- #29

- 1,766

- 63

Demystifier said:That's where my point of view significantly differs from yours.

I hope you agree that in the QM formalism the observed system is represented by a state vector, while the measuring apparatus (or corresponding observable) is represented by a Hermitian operator. This fact alone should give a clue that QM implies a clear separation boundary between the system and the measuring device. Dynamical observables of the measuring apparatus are not considered in the QM formalism. So, this formalism is fundamentally incomplete.

However, this incompleteness is totally understandable and natural. In science we are not allowed to speak about things that cannot be experimentally verified. Since, be definition, dynamical variables of the measuring apparatus cannot be measured, then it is pointless to talk about them. Of course, we can decide to choose our measuring device to be our observed system, but this means a complete change of the whole experimental setup, choice of a new measuring apparatus, and repeat of the same logic in new circumstances.

Eugene.

- #30

- 4,078

- 593

meopemuk said:Time "measured" by clocks is not a quantum-mechanical observable. It is just a numerical parameter.

While I think your description is a quite fair description of normal QM, I think it does not come with a very insightful view of time, and particularly the arrow of time, in a way that I think will be necessary to also merge this with GR.

Your reasoning relies heavily of "laboratory contexts" where we have clocks without uncertainty. I think from an information theoretic perspective, this is a clear idealisation and I am not satisfied with it.

In a conditional information perspective, the IMO more natural decomposition is

1) what the observer thinks he knows for sure

2) what he can distinguishable that he doesn't know, given "posable questions"

I think the clock must then be encoded in (1), and if you think that the observer can only encode finite amounts of information, as you envision larger and larger systems like you mention in (system+measuringdevice) etc, then at some point the encoding capacity is full, and there is no "room" for "certain" clocks.

So either we must constrain ourselves to study only relative to our own complexity, SMALL subsystems, or we must find a generalisation of QM formalism.

/Fredrik

- #31

- 684

- 5

meopemuk said:My point was that formalism of quantum mechanics should be discussed only in the context of a specific experiment. Tell me which experiment you have in mind (what is the observed system and what is the measuring apparatus) and I will be able to answer how you should describe the clock.

Take any experiment (such as the 2-slit) in which particles are detected by hitting a "screen" where they make a "mark". I put these terms in quotes because I don't really know how these things are done in practice.

Now the screen is principally a 3-d object. It extends in 2 spatial dimensions and in time. And just as we treat the screen as theoretically spatially infinite, we can treat it as being fixed in place through all time.

A measurement is made and the result is a mark on the screen. The mark is approximately a 1D object. No spatial extension but it begins at some time and endures indefinitely. That point where it begins is an event (x,y,t) on the 2+1 dimensional screen. From a quantum theory perspective the "measurement" occurs at the appearance of the mark. Or rather, the appearance of the mark at such and such position and such and such time on the 2+1 D screen

If, for example, the observed system is an unstable particle, and you are studying its decay probability as a function of time, then the clock is not a part of the quantum-mechanical system. It is a part of the measurement setup. Clock's readings are classical parameters.

In my experiment the clock is part of the measurement setup, true. And equally so is the ruler.

So what about this situation allows us to single out time as a parameter while the position is an observable?

I suspect is that it is non-relativistic approximation, similar to being able to define simultaneity for all observers in classical mechanics (and hence treat time as a parameter).

- #32

- 1,225

- 75

As a classic treatise of this problem W. Pauli, in Handbuch der Physik, edited by S. Flugge, Springer, Berlin (1958) Vol.5/1,p.60, showed that a "time operator" T conjugate to a Hamiltonian H could not exist if the spectrum of H is bounded.

Regards.

- #33

- 1,766

- 63

pellman said:So what about this situation allows us to single out time as a parameter while the position is an observable?

In order to measure position (to see marks on the screen) you need to have a physical system. You must release an electron in the double-slit setup and find its mark on the screen. Then you can say: electron's coordinates were x and y. So, position is definitely an attribute or property that depends on the state of the electron - the observed physical system. In other words, position is an observable.

You do not need to have a physical system in order to "measure time". You can turn off the electron gun in the double-slit experiment and you can still tell what the time is by looking at the clock. The clock readings do not depend on the state of the observed system. They remain the same even if there is no system to observe (the electron gun is turned off). So, it is not appropriate to call time an observable.

Eugene.

- #34

- 1,766

- 63

sweet springs said:

As a classic treatise of this problem W. Pauli, in Handbuch der Physik, edited by S. Flugge, Springer, Berlin (1958) Vol.5/1,p.60, showed that a "time operator" T conjugate to a Hamiltonian H could not exist if the spectrum of H is bounded.

Regards.

You are absolutely right. Introduction of the "time operator" leads to all kinds of paradoxes and inconsistencies.

Eugene.

- #35

- 684

- 5

meopemuk said:You do not need to have a physical system in order to "measure time". You can turn off the electron gun in the double-slit experiment and you can still tell what the time is by looking at the clock. The clock readings do not depend on the state of the observed system. They remain the same even if there is no system to observe (the electron gun is turned off). So, it is not appropriate to call time an observable.

Eugene, how is this different than saying we can choose an origin, lay out a ruler and read off the positions x and y even if there is no system to observe?

Share:

- Replies
- 11

- Views
- 804

- Replies
- 3

- Views
- 709

- Replies
- 12

- Views
- 822

- Replies
- 2

- Views
- 473

- Replies
- 2

- Views
- 143

- Replies
- 8

- Views
- 696

- Replies
- 3

- Views
- 883

- Replies
- 3

- Views
- 964

- Replies
- 1

- Views
- 203

- Replies
- 27

- Views
- 2K