Saturday, July 08, 2006

Is there an observable of time?

We already discussed here, here, and here that the principle of relativity does not require the symmetry between space and time coordinates alleged in Einstein's special relativity. Here I would like to suggest another argument for the deep division between time and position. The point is that position is an observable, while time is not.

When we measure position of a particle or any other physical system, we know that the result of measurement will depend on the state of the observed system.
In quantum mechanics there is an Hermitian operator corresponding to position, and observations of the position are normally associated with uncertainties that are characteristic to measurements of quantum observables.

Time is different. Time is "measured" by simply looking at the wall clock in the laboratory. The value of time does not depend on what physical system is actually observed and what is the state of the system. The value of time would be just the same if we haven't observed any physical system at all. So, time is simply a classical parameter associated with the act of measurement. We perform a measurement of some observable and we attach to this measurement a numerical label called "time". This label belongs to the same class as nine other labels that should be attached to each measurement. These are labels that uniquely specify the reference frame in which the measurement was made: its position, orientation, and velocity. So, time is just one of the 10 parameters of the reference frame or laboratory, and time has nothing to do with the properties of the physical system that may be (or may be not) present in the laboratory.

So, it is not surprising that there is no operator of time in quantum mechanics. (Some published attempts to introduce such an operator do not look convincing.) This observation is just another reason to stop using the Minkowski space-time formalism in relativistic theories.

14 Comments:

Anonymous Chris Oakley said...

Eugene,

I agree that the first problem one has to overcome in trying to make quantum mechanics relativistic is the nature of the time operator. in NRQM one can successfully use Hermitian 3-position operators for each of the particles being considered, but time runs into the immediate problem that <t'|t> must be zero if T is a Hermitian operator and t'≠t, which means that every amplitude for every real process is zero.

As you will probably have gathered, I do not think that the answer is to abandon explicit covariance.

My reasoning is as follows: the equations of classical mechanics and classical electrodynamics may or may not be usable at the microscopic level. The assumption is that they are, subject to the vagaries of quantization and the introduction of concepts, such as spin, which have no classical analogue, but the fact is that whereas it is essential that microscopic physics gives macroscopic physics as the classical limit, the converse is not true - there may be all sorts of weird and wonderful things going on at the quantum level that never show up in classical mechanics or electrodynamics.

Not so Special Relativity. SR is so bold that you cannot apply it piecemeal, any more than you can be a little bit pregnant. If it is indeed the case that the laws of physics are invariant under Lorentz transformations, then this must apply as much to microscopic physics as it does to macroscopic physics.

There is also the issue of aesthetics. The axioms of QM plus SR lead to Wigner's irreducible representations of the Poincare group. The helicity one solutions are just the source-free Maxwell equations: a lot of output for little input. There is also, of course, the vast amount of bubble chamber data that could not be explained without SR.

Although I don't accept that the usual decomposition H = H_0 + V makes any sense in a relativistic theory - mainly because one cannot construct H_0 once the interactions are turned on, I do accept the fact that we currently have no clue about how to calculate anything unless we are allowed to make this decomposition, so the path my own work has followed is to calculate invariant amplitudes and then look for the correspondence with NR time-dependent perturbation theory amplitudes through the singular contributions to the former. However, I would not rule out the possibility that it might be possible to formulate a completely covariant method that bypasses this final step. But I do not view the idea of doing everything non-relativistically with the hope of a relativistic theory emerging, as an attractive prospect.

Saturday, July 8, 2006 at 2:13:00 PM PDT  
Blogger Eugene Stefanovich said...

Chris:

I am glad that we are in agreement about the uselessness of the time operator in quantum mechanics. Now, if we also agree that the observable of position should be represented by a Hermitian operator even in the relativistic theory, then we have a sweet paradox between QM and the 4D space-time unification.


But I do not view the idea of doing everything non-relativistically with the hope of a relativistic theory emerging, as an attractive prospect.

One message that I probably failed to convey (though I tried ) is that what I am doing is a perfectly relativistic theory. The question is how to distinguish a relativistic theory from a non-relativistic one.

I don't think that manifestly covariant tensor transformation laws of all ingredients is a sign of a good relativistic theory. The manifest covariance does not follow logically from the principle of relativity and the invariance of the speed of light. The manifest covariance is, in fact, an additional assumption.

My definition of a relativistic quantum theory is the same as yours and Wigner's: there should be an unitary representation of the Poincare group in the Hilbert space of the system. This definition works fine for elementary systems as shown by Wigner [1], and for compound ones as shown by Dirac [2]. The same defintion was used by Weinberg to prove the relativistic invariance of QED [3]. The point I am trying to make is that this definition neither implies nor requires the "manifest covariance". Some observables (e.g., the total momentum-energy) may transform as tensors. Transformations of other observables may be not that trivial. For example, the transformations of momentum-energy of individual particles are interaction-dependent.

[1] E. P. Wigner, "On unitary representations of the inhomogeneous Lorentz group", Ann. Math., 40, (1939)149.

[2] P. A. M. Dirac, "Forms of relativistic dynamics", Rev. Mod. Phys., 21, (1949)392.

[3] S. Weinberg, "Photons and gravitons in perturbation theory: Derivation of Maxwell's and Einstein's equations", Phys. Rev., 138 , (1965) B988.

Saturday, July 8, 2006 at 4:07:00 PM PDT  
Blogger Eugene Stefanovich said...

Chris,

I am puzzled by your statement:

Although I don't accept that the usual decomposition H = H_0 + V makes any sense in a relativistic theory - mainly because one cannot construct H_0 once the interactions are turned on...

What do you mean by that? We can easily write the non-interacting Hamiltonian H_0 in the Fock space. We can also write an infinite number of interaction Hamiltonians V. There is a minor problem that this interaction should obey the Poincare commutators with the interacting boost operator K = K_0 + W and other generators. But this is a technical issue, and there are well-established ways how to do that. Why do you say that the sum H = H_0 + V does not make sense?

Sunday, July 9, 2006 at 1:59:00 AM PDT  
Anonymous Chris Oakley said...

Well, for example, the matrix element

<p';k|p>

where p and p' are electron momenta and k is a photon momentum, at equal times, should be zero if H_0 exists, since the existence of H_0 enables one to construct an orthonormal basis of particle states. In the presence of interactions, I find this not to be the case.

Sunday, July 9, 2006 at 3:07:00 AM PDT  
Blogger Eugene Stefanovich said...

nigel,

I had to delete your comments, because they were completely off-topic.

Sunday, July 9, 2006 at 3:22:00 PM PDT  
Blogger Eugene Stefanovich said...

Chris,

why do you say that the matrix element between states "one electron + one photon" and "one electron" can be non-zero? These states are eigenstates of the particle number operator with different eigenvalues, so they are orthogonal. They are orthogonal at time t=0 and at any other time, because the time evolution operator is unitary. Are you saying that the orthogonality of these states contradict observations or any physical principle? What is this principle?

Sunday, July 9, 2006 at 3:33:00 PM PDT  
Anonymous Chris Oakley said...

I don't believe that there is a particle number operator in an interacting theory (except for overall charge conservation). What I get for this matrix element is a transient, i.e. it goes to zero as t→∞ which shows that one would not expect a single electron to spontaneously emit a photon, but it contains order e contributions which are non-vanishing even when t=0. For "real" processes, e.g. e+e- annihilation, one gets a pole in (time) Fourier space, plus transients. A Laurent expansion yields the usual contribution to the Feynman amplitude via time-dependent P.T. plus non-singular parts that give no contribution as t→∞.

I should qualify all of this by saying that I have yet to persuade anyone that anything I have done in this regard is reasonable. But on reflection I think that it is an inevitable consequence of relativity: anything one sets up using H_0, one should also be able to set up with γ(H_0+βp_x0). In a free theory this is not a problem as time or space translation causes only phase rotation. But an interacting theory is a different matter altogether.

Monday, July 10, 2006 at 1:38:00 AM PDT  
Blogger Eugene Stefanovich said...

Chris,

I don't believe that there is a particle number operator in an interacting theory (except for overall charge conservation).

Here we disagree. I do believe that there is a particle number operator, and its form is exactly the same for interacting and non-interacting theories.

What I get for this matrix element is a transient, i.e. it goes to zero as t→∞ which shows that one would not expect a single electron to spontaneously emit a photon, but it contains order e contributions which are non-vanishing even when t=0.

Here we disagree again. I think that a single electron cannot emit a photon neither at infinite time nor at finite time. Such an spontaneous emission would completely disagree with experiment. So, my understanding is that the matrix element you wrote is identically zero.

Monday, July 10, 2006 at 12:32:00 PM PDT  
Blogger Eugene Stefanovich said...

Chris,

this is true that in traditional QED, a single (bare) electron can emit/absorb photons. This is because bare particles of QED have very little in common with the real physical particles. Physical electron is an infinite linear combination of bare particle states.

The idea of the "dressed particle" approach is to abandon the misleading bare particle picture, and to have everything expressed in terms of physical particles. In this physical particle representation, a free electron cannot emit/absorb photons at any time, which is physically correct.

Monday, July 10, 2006 at 1:25:00 PM PDT  
Anonymous Chris Oakley said...

Eugene,

A non-zero but transient <eγ|e> is not necessarily a "disagreement with experiment". Think of it like the LSZ formalism, where one looks to the asymptotic behaviour to derive the physical interpretation. One difference between LSZ and my formalism, though, is that I do not require asymptotic states to be free.

Tuesday, July 11, 2006 at 2:39:00 AM PDT  
Blogger Eugene Stefanovich said...

Chris,

I believe that the absence of photon emission by free electrons is a well-established experimental fact, and I base my entire approach on this fact. If future experiments will demonstrate that this assumption is wrong, the I am ready to admit that my approach is complete nonsense.

Tuesday, July 11, 2006 at 10:43:00 AM PDT  
Anonymous Chris Oakley said...

the absence of photon emission by free electrons is a well-established experimental fact

You seem to be suggesting that I don't agree on this.

Tuesday, July 11, 2006 at 1:00:00 PM PDT  
Anonymous IdoCocaine said...

Hi! I think you are totally wrong (special relativity aside).
The true evolution (assumed in Schroedinger's equation and its elaborations) of quantum states might be better modeled by discontinuous processes (and it is only incidental that in certain limits it behaves as Schroedinger's eqn predicts).
Why shouldn't alternative geometries for the parameter/s ("time"/s) of the dynamics' equations find success? The only argument for a unidirectional time is the 2nd law of thermodynamics (which doesn't even prove homogeneity since the rate at which net entropy increases can probably vary) which gets violated at time scales smaller than Planck time anyway. e.g. time/change inside the Planck-length might be modeled by equations which are parameterized by multidimensional time, whatever that may entail for potential "observables". I know this for a fact because I do a lot of cocaine.
-Joey

Friday, August 1, 2008 at 2:17:00 AM PDT  
Blogger Eugene Stefanovich said...

Hi Joey,

unfortunately I (and nobody) know what happens at the Planck scale. My best guess is that nothing special happens there. Time remains a continuous parameter. However, your guess is as good as mine. If you can build a predictive theory based on your guess, then more power to you! I don't do cocaine.

Eugene.

Friday, August 1, 2008 at 11:22:00 AM PDT  

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