- #1
angass
- 1
- 0
Hi,
I´m not sure if my way of tackling a question, probably it's a trivial problem, but it's important for me to get it right so any help will be greeted.
The question is as follows:
Problem: consider a particle in a one-dimensional system. The wave function ψ(x) is as follows:
ψ(x)= 0 for (-∞<x<0),
ψ(x)= 1/√a for (0<x<a)
ψ(x)= 0 for (a<x<∞)
i) if the kinetic energy is measured what is the most probable value?
ii) which values of the momentum can never be found when measuring it?
My reasoning:
i) I use the operator for kinetic energy: K= -[itex]\frac{h^2}{2m}[/itex][itex]\frac{∂^2}{∂x^2}[/itex]
which when applied: ∫ψ(x)*Kψ(x)dx gives me zero.
If this was right I assume the most probable value of the kinetic energy is zero.
ii) I have no clue whatsoever what the question means.
Thanks and forgive my bad english, regards from Spain!
I´m not sure if my way of tackling a question, probably it's a trivial problem, but it's important for me to get it right so any help will be greeted.
The question is as follows:
Problem: consider a particle in a one-dimensional system. The wave function ψ(x) is as follows:
ψ(x)= 0 for (-∞<x<0),
ψ(x)= 1/√a for (0<x<a)
ψ(x)= 0 for (a<x<∞)
i) if the kinetic energy is measured what is the most probable value?
ii) which values of the momentum can never be found when measuring it?
My reasoning:
i) I use the operator for kinetic energy: K= -[itex]\frac{h^2}{2m}[/itex][itex]\frac{∂^2}{∂x^2}[/itex]
which when applied: ∫ψ(x)*Kψ(x)dx gives me zero.
If this was right I assume the most probable value of the kinetic energy is zero.
ii) I have no clue whatsoever what the question means.
Thanks and forgive my bad english, regards from Spain!