Dispersion in for a free particle

[marcelnv]

Hey,
I am trying to calculate the dispersions [itex]\Delta[/itex]P and [itex]\Delta[/itex]X for a free particle with definite momentum [itex]\Phi[/itex](x) = Aexp(ikx) where A is a normalization constant. This i get
[itex]\Delta[/itex]P = 0
[itex]\Delta[/itex]X = [itex]\infty[/itex]
What physical interpretation can I give to these results? Is is that the momentum is well specified and can always be obtained with certainty and that the position can be anywhere?
Also,
I am trying to solve the same problem for a free particle in the state
[itex]\Phi[/itex](x) = A*Cos(kx).
I get integrals involving sin and cos to be evaluated at [itex]\infty[/itex]. Using taylor expansion, i argue that
[itex]\Delta[/itex]P = [itex]\infty[/itex]
[itex]\Delta[/itex]X = [itex]\infty[/itex]

Am I correct?
The function [itex]\Phi[/itex](x) = A*Cos(kx) seems not to be normalizable to me, so should not be used to represent the state of a quantum mechanical system. What do you think?

Thanks

Marcel

 

[LightHero]

Yes, your calculations are correct. The physical interpretation of the delta P and delta X values is that the momentum is well specified and can always be obtained with certainty, but the position can be anywhere. This is true for any free particle with a definite momentum. For the \Phi(x) = A*Cos(kx) state, it is not normalizable and hence cannot be used to represent the state of a quantum mechanical system. Normalizability is a requirement for any valid wavefunction.