a) The uncertainity principle goes like that:
So I think the right answer is: "The observables are simultaneously measurable if mean is zero."
I have the following problems:
a) Under which conditions we can simultaneously measure (with arbitrary precision) two observables which operators does not commute?
b) Does energy and coordinates of particle's momentum are in general simultaneously measurable. Give the conditions under which this physical quantities are simultaneously measurable. Give an example of quantum system having such property.
c) Are there any conditions under which position and square of momentum are simultaneously measurable. Find an example of such conditions.
Let denote differentiation over i-th coordinate. We have one particle with mass m. The energy operator should be like that:
Momentum operator (coordinate k):
I will focus on the commutator:
I shall assume :
Let's assume the wavefunction belongs to class (mathematiclly very restrictive assumption but I consider it as "physical")
As I remember there was some theorem which states that we can change sequence of differentiation (I guess the assumption was that the second derivative is continuous). It was important theorem but I don't remember it's name .
So under this assumptions for wavefunction we should get:
So the commutator is zero
I guess it is obvious that:
Which means that operators , commute.
Thus we can simulataneously measure momentum and energy with infinite precision.
It seems to me a little strange because the question "Give the conditions under which this physical quantities are simultaneously measurable. Give an example of quantum system having such property." suggests that they should not commute (otherwise giving the example is trivial - any system has such proprety). Could you check if my reasoning is correct?