Why time is not an operator in quantum mechanics?
We've had similar discussions on this very topic here lately. You can search for those posts/threads.
In short, there's no universally accepted reason for the rejection of a time observable in <ordinary> quantum mechanics. There are very many articles in peer-reviewed literature arguing in favor of putting a time operator in the formalism.
Here is one (recent) such paper:
http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595]
I think a more interesting question is "why is there a position operator?". Energy, momentum and spin can all be defined as generators of Galilean transformations, but position requires a more complicated definition. (I don't know that definition myself, but I think Ballentine covers it).
Are you saying that an operator should be a generator of a transformation? Well, if that's what you are saying, then position operator can be viewed as a generator of the translation in the momentum space.
Anyway, I don't think that it helps much to answer the surena's question. Or maybe it does?
I would imagine that its because you don't go out and measure the time...observable quantities correspond to operators. In ordinary run of the mill non relativistic quantum mechanics, time is a parameter that describes the evolution of states. In a relativistic quantum theory I would expect time to be an observable/operator and proper time or the invariant interval to be the parameter describing state vector evolution. I also think its funny when we run across apparent conflicts between relativity and quantum mechanics...its like people have forgotten that things like Bell's Inequality are derived using a non relativistic quantum theory.
I'm not sure what I'm saying actually, or if it helps answer the question. I like to think of QM as being defined by the usual axioms about Hilbert space and the probability rule (but not the Schrödinger equation). An important class of theories in that framework can be defined simply by specifying a symmetry group that includes translations in time, and by specifying which irreducible repesentations we're interested in. (We can do what Weinberg did in chapter 2 of his QFT book, with another symmetry group. The most important group are of course the connected parts of the Poincaré and Galilei groups). Translations in time need to be included because we get the Schrödinger equation from the properties of the time translation subgroup.
The reason I'm a bit confused here is that I don't know how position and mass enters this picture, but as I said, I think Ballentine has some good stuff about that. I just haven't read it yet. I have too many things I'd like to read and understand.
...and observables correspond to measuring devices, like...clocks. So the question is certainly valid.
It isn't. And there seems to be no way to define a position operator for arbitrary quantum systems, in particular massless particles. (An operator called the Newton-Wigner position operator does the job in a lot of cases, but not all).
The standard is to use the coordinate time of some inertial frame.
The differences between relativistic and non-relativistic QM are much smaller than you think. The derivations of Bell inequalities that I've seen are all valid in relativistic QM too.
I guess the point I was trying to make is that suppose you have some region of space and then measure the position of some particle in some state...the particle may or may not be located within the region you've chosen...but the particle always exists at any particular instant. Its not like you'll make a "time measurement" and find that the particle isn't anywhere in the universe at that time. Its not like the particle can be found only at 2pm plus or minus 6 minutes. That means time isn't an observable.
As for relativistic quantum theory...unfortunately my education ends at the Klein-Gordon equation. I'm actually quite suprised that time isn't an operator in modern RQM! I always figured that the EPR paradox wouldn't be an issue in RQM too so I'm suprised to hear that you can still derive Bell's Inequality using it.
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