- #1
fluidistic
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I asked my professor if we could define time as an operator and he said no.
I've read on the web that time isn't an observable because "you don't measure the time of a particle". Yet, to me at least, the sentence "you don't measure the time of a particle" is similar to "you don't measure the position of a particle".
When I give the position of a particle it's always according to a reference frame ##\vec x _0##. Why can't I give the "time of a particle" with respect to a reference time ##t_0##?
So if I have an external clock to the system I'm considering and that I had set that clock to 0 at ##t_0## then I could give the time and position of the particle at any moment.
Position like time is not an intrinsec property of particles as far as I know. So why a position operator but no time operator?
My professor wrote on the blackboard that for stationary states, ##\Psi _t =e^{-iEt/\hbar }## and that ## \langle \Psi _t , \hat A \Psi _t \rangle = \langle e^{-iEt/\hbar }\Psi, \hat A e^{-iEt/\hbar } \Psi = e^{iEt/\hbar } \cdot e^{-iEt/\hbar } \langle \Psi, \hat A \Psi \rangle = \langle \Psi , \hat A \Psi \rangle ## and then he concluded that for any self-adjoint linear operator ##\hat A##, the expected values of all the observables are time independent. I follow the math and his conclusion, but it's true if ##\hat A## does not depend on time. Why can't we define an observable ##\hat A## that depends on time? Such as ##\hat t \Psi =t\cdot \Psi## or ##\hat t = \frac{d}{dt} \Psi##.
I've read on the web that time isn't an observable because "you don't measure the time of a particle". Yet, to me at least, the sentence "you don't measure the time of a particle" is similar to "you don't measure the position of a particle".
When I give the position of a particle it's always according to a reference frame ##\vec x _0##. Why can't I give the "time of a particle" with respect to a reference time ##t_0##?
So if I have an external clock to the system I'm considering and that I had set that clock to 0 at ##t_0## then I could give the time and position of the particle at any moment.
Position like time is not an intrinsec property of particles as far as I know. So why a position operator but no time operator?
My professor wrote on the blackboard that for stationary states, ##\Psi _t =e^{-iEt/\hbar }## and that ## \langle \Psi _t , \hat A \Psi _t \rangle = \langle e^{-iEt/\hbar }\Psi, \hat A e^{-iEt/\hbar } \Psi = e^{iEt/\hbar } \cdot e^{-iEt/\hbar } \langle \Psi, \hat A \Psi \rangle = \langle \Psi , \hat A \Psi \rangle ## and then he concluded that for any self-adjoint linear operator ##\hat A##, the expected values of all the observables are time independent. I follow the math and his conclusion, but it's true if ##\hat A## does not depend on time. Why can't we define an observable ##\hat A## that depends on time? Such as ##\hat t \Psi =t\cdot \Psi## or ##\hat t = \frac{d}{dt} \Psi##.