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is there a "time" operator ? p & x ---> E & t ?? Rate Topic: -----

#21 7 April 2012 - 05:59 PM juanrga 


Baryon

View PostWiddekind, on 1 February 2012 - 09:18 PM, said:

By comparison to position, and momentum (proportional to the derivative of position); and the parallel to time, and energy (as the time derivative); is there a "time operator" \hat{t} \equiv t \times, having eigenstates |t_0\rangle = \delta(t-t_0) ?

If particles must be normalized w.r.t. space, s.t. \int d^3x \Psi(x) = 1; then why aren't wave-functions normalized, w.r.t. time, s.t. \int dt \Psi = 1 ? Vaguely, to tie in to relativity, would seemingly require combining time & position, into "quantum events" (t,x).


Time is an evolution parameter, not an observable. Therefore there is not time operator in QM. Yes, in RQM one would try to treat time and space in the same footing, but since time is related to causality, it cannot be given by an operator.

Precisely quantum field theory takes the contrary way, position is downgraded from observable to parameter and then both space and time are treated in the same footing.
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