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sanitykey
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Homework Statement
Assuming that the probability density of a particle is A^2 sin^2(kx), is the particle localised in space? Using the uncertainty principle determine the degree to which the momentum of the particle is specified.
A is just a constant, k is the wavenumber, x is position.
Homework Equations
Heisenberg's Uncertainty Principle dpdx ≥ hbar i think that's the only one...
The Attempt at a Solution
I thought about what the graph would look like, like a sin(x) graph but with all the dips going below y = 0 inverted because its squared, the kx bit would shrink the graph a bit i think or something like that, but anyway the point is i thought since it looks the same, hilly, over all of x then maybe it can't be localised?
I mean the fact there are maximums means it's more likely to find the particle in certain regions looking over infinity doesn't it?
But the particle is equally likely to be at any of the maximums over infinity. So i thought dx = infinity therefore the uncertainty in momentum dp = 0 therefore the momentum can be precisely specified just like a free particle? But I'm confused because i might be wrong and this isn't a free particle is it?
I thought a free particle was one which had a constant probability density just like a straight line, equal probability of being found everywhere?
Any insight would be appreciated thanks
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