In standard quantum mechanics textbooks, the form of the momentum operator, in position space, is either given as definition, i.e. they write that the momentum operator is –id/dx (times the reduced Planck constant), or, in more advanced textbooks, like Landau & Lifshitz’s, it is derived as the generator of spatial translations.
I wonder if the form of the momentum operator, i.e. that it is a first-order differential operator, can be derived qualitatively, by means of physical arguments. In other words, does the slope of an arbitrary, i.e. non-stationary, wave function have a physical meaning?
Hardly anything in the quantum can be said to be rigorously derived. The generator of spatial translations
is found also in the quantum book of Heisenberg, is about the best you get. The wave function is in general a complex function of space and time, with meaning only in the square modulus.
Then with the observation is its noncommutivity with position, this leads to the uncertainty principle.
To relate momentum to its usual meaning, you can use say the quantum equation of motion
dx/dt = 1/i hbar [x,H], if you take say H= pp/2m, the usual kinetic energy, and calculate the commutator, with commutation rules. (Landau Lifshitz)
In this case [x, f(p)] = i hbar df/dp
Operators in QM has been derived using analogy from classical physics. See what Bohr said "For every physical quantity one can define an operator. The definition uses
formulae from classical physics replacing quantities involved by the corresponding operators"
Dear Juan Weisz and Gokaran Shukla,
thank you very much for your answers.
Perhaps it can be derived by using the least action principle, with the least quantum action equal to the reduced Planck constant, but not zero as in classical mechanics
For people who insist on some definite definition, it is defined as the dual operator to x, that is whatever operator p
whose property is [x,p]=i hbar
In the previous message I showed how to recover m dx/dt
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