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How can I solve the following ecercise: Use perturbation theory to calculate the energy levels of a particle in a box with a co...Asked by Juan Manuek
How can I solve the following ecercise: Use perturbation theory to calculate the energy levels of a particle in a box with a cosine function botto. Let the box extend from x=a to x=a and let the perturbing potential be H' = V0[1+cos(2*Pi*m*x/a)]. This potential oscilates between 2V0 y . The number of oscilations is determined by m. Discuss restrictions on the possible values of k+1 and k-1 in the integrals H'kl
Answers
Answered by
drwls
What is a botto?
Your sentence:
<< This potential oscilates between 2V0 y .>>
appears to be incomplete.
I don't think any of us are going to be able to answer this question.
Your sentence:
<< This potential oscilates between 2V0 y .>>
appears to be incomplete.
I don't think any of us are going to be able to answer this question.
Answered by
Juan Manuek
bottom
2V0 or 2V is the potential in x=0
2V0 or 2V is the potential in x=0
Answered by
Juan Manuek
How can I solve the following ecercise: Use perturbation theory to calculate the energy levels of a particle in a box with a cosine function bottom. Let the box extend from x=a to x=a and let the perturbing potential be H' = V0[1+cos(2*Pi*m*x/a)]. This potential oscilates between 2V0 y 0. The number of oscilations is determined by m. Discuss restrictions on the possible values of k+1 and k-1 in the integrals H'kl
Answered by
Count Iblis
I would shift x so that the box is from x = 0 to x = 2 a. The unperturbed states are then:
psi_n(x) = a^(-1/2) sin[n pi x/)(2 a)]
The perturbation is:
H' = V0[1+cos(2*Pi*m*(x-a)/a)] =
V0[1+cos(2*Pi*m*x/a)]
The first order correction is:
E1 = <psi_n|H'|psi_m> =
Integral from x = 0 to 2 a of
|psi_n|^2 H'(x) dx =
1/a Integral from x = 0 to 2 a of sin^2[n pi x/)(2 a)]*
V0[1+cos(2*Pi*m*x/a)] dx
which is straightforward to evaluate.
psi_n(x) = a^(-1/2) sin[n pi x/)(2 a)]
The perturbation is:
H' = V0[1+cos(2*Pi*m*(x-a)/a)] =
V0[1+cos(2*Pi*m*x/a)]
The first order correction is:
E1 = <psi_n|H'|psi_m> =
Integral from x = 0 to 2 a of
|psi_n|^2 H'(x) dx =
1/a Integral from x = 0 to 2 a of sin^2[n pi x/)(2 a)]*
V0[1+cos(2*Pi*m*x/a)] dx
which is straightforward to evaluate.
Answered by
drwls
Thanks Count Iblis, for understanding and answering this question. I should not have assumed none of us could answer it.
I still don't understand the << This potential oscilates between 2V0 y 0. >>
statement. Does the "y" represent Spanish "and" ?
I still don't understand the << This potential oscilates between 2V0 y 0. >>
statement. Does the "y" represent Spanish "and" ?
Answered by
Count Iblis
Yes, it looks like "y" = "and". I think Juan also needs to consider the second order correction, because the question mentions the matrix elements H'kl, which occur in the second order corrections.
Answered by
Juan Manuek
H'(x=0) = 2V0
H'(X = a/2) = 0
H'(x= a) = 2 V
(y is and) in spanish
and I don't know wath does mean k+1 and k-1. It refers to l=1?
H'(X = a/2) = 0
H'(x= a) = 2 V
(y is and) in spanish
and I don't know wath does mean k+1 and k-1. It refers to l=1?
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