- #1
ehrenfest
- 2,020
- 1
Calculate the RMSD for the odd solutions of the infinite symmetric well problem.
The odd solutions have the form:
[tex]u_n(x) = 1/\sqrt{a} \cdot sin(n \pi x/a)[/tex]
So, I should multiply u_n by x, then take the integral over the bottom of the well to get <x>. Then I should multiply u_n by x^2, take the integral over the bottom of the well to get <x^2>. From this I can get delta x.
Is all that correct?
The odd solutions have the form:
[tex]u_n(x) = 1/\sqrt{a} \cdot sin(n \pi x/a)[/tex]
So, I should multiply u_n by x, then take the integral over the bottom of the well to get <x>. Then I should multiply u_n by x^2, take the integral over the bottom of the well to get <x^2>. From this I can get delta x.
Is all that correct?