in relativistic quantum mechanics(quantum field theory) we take co-ordinate time as a time observable to bring space and time on equal footing ...why can't we take proper time as a time observable ? by this approach we may overcome the problem, called "renormalization."
Preprint PLAYING DICE WITH PROPER TIME
Proper time assumes that we measure the time for a particle at rest, and that is impossible from the quantum mechanical point of view.
@ Daniel Baldomir
my dear Daniel Baldomir do you mean moving particle don't have proper time?
Dear Hussainsha,
What I wanted to say is that Quantum mechanics tell us that as much as you know the energy of a particle you ignore its associated time. A particle at rest has very well known its energy and therefore you cannot have a kind of proper time because this variable is going to fluctuate without allowing to have a fixed value.
@Daniel Baldomir
my dear Daniel Baldomir does it mean you should not take it (proper time) as a time observable?
ya you are right according to uncertainty principle if you know energy, time fluctuate and if you know position, momentum fluctuate......here it does not mean that if the momentum of a particle is known, position is not a observable.....
therefore here my intuition is; why can't we take proper time as a time observable by ignoring energy associated with it rather than taking co-ordinate time?
@ Jan-Martin Wagner
my dear Jan-Martin Wagner where would you end up if you ignored momentum?
Dear Hussainsha,
Obviously you cannot ignore the energy of the system. Notice that at this quantum scale you have not its conservation because the system doesn't need to have time traslations as a symmetry. But what is clear is that one rest observer associated to a quantum particle is without meaning.
@ Daniel Baldomir
my dear Daniel Baldomir i replied you with a word 'neglecting' under the context of uncertainty principle ....
please i am appealing you to understand my question with utmost concentration...once again repeating my question as follows......
i want to measure time therefore according to uncertainty principle it is impossible to calculate energy ,,,,,,,, here i want to take proper time as a observable,,,,,,,,,
finally you are not answered my question...... moving particle don't have proper time ? this answer tells, one rest observer associated to a quantum particle is without meaning or not.
One can consider time as coordinate of clock pointer. See
Article Formulation of quantum mechanics with dynamical time
Dear Hussainsha,
I don't know how can tell you clearer:
1. Time cannot be an operator.
2.Proper time only can be defined for a particle at rest. This is not possible to do it in quantum mechanics.
Dear Hussainsha,
Let me try to be more explicit and try to explain a little bit more why time is not an operator in quantum mechanics.
1. Time in quantum mechanics transforms as in classical mechanics, thus it is an absolute quantity, a label more than a variable.
2. Several states of particles have the same time and therefore it is impossible to assing an only eigenvalue to one of them associated to the imagined time operator.
3. In quantum field theory in principle time has one operator well defined, although it is not quite useful and never is used.
@ Daniel Baldomir
1) my dear Daniel Baldomir is is not possible to change our thinking?
2) is it not illogical........how can be Several states of particles have the same time at the surface of the earth and at the top of mount Everest?
3) in quantum field theory co-ordinate time is operator , although it is not quite useful and never is used ....why can't we take proper time as a operator so it is quite useful?
s h s> why can't we take proper time as a time observable?
The proper time of which objects, distributed throughout the universe? Given a space-time geometry defined by the line element
c2 dτ2 = gμν(x) dxμdxν
it may be possible to choose a time coordinate such that g00 = 1, so that the proper time for objects at rest in these coordinates is equal to coordinate time. However, this would in most cases be a very bad idea, because then the other components of gμν become horribly ugly.
s h s> ... in quantum field theory co-ordinate time is operator
No, in quantum field theory ("second quantisation") all space-time coordinates play the role of parameters. In non-relativistic quantum mechanics time acts as a parameter, while the position coordinates are operators. It is possible to construct a kind of "relativistic quantum mechanics" where all space-time coordinates behave like operators.
In special relativity, if one chooses Minkowski coordinates (t,x,y,z) in a frame of reference S, then the time coordinate t is the physical time that elapses in S and hence, in the equations of motion, it is this t that appears in the derivatives. Proper time depends on the state of the particle. For example, if a particle is moving with speed v, then the proper time interval in its rest frame is dt.sqrt (1-v^2/c^2). Clearly, the differential eqs. that describe the laws of motion cannot be based on proper time, otherwise it will become state specific.
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