Perturbation Theory: Calculating Ground State Eigenfunction of Particle in a Box

AI Thread Summary
The discussion revolves around calculating the ground state eigenfunction of a particle in a box with a perturbation applied in a specific region. The total ground state eigenfunction is expressed with a first-order perturbation contribution using only the first three eigenfunctions, which raises questions about the completeness of this approach. It is clarified that using only a few eigenfunctions is an approximation, as they contribute most significantly to the correction. The orthogonality of the eigenfunctions is emphasized, noting that integrals involving odd functions may yield zero results, which can affect the corrections to the eigenvector. The conversation concludes with a confirmation that the integrals should be recalculated to ensure accuracy in the perturbation analysis.
danja347
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I have a problem where I should calculate the ground state eigenfunction of a particle in the box where the potential V(x)=0 when 0<x<L and infinite everywhere else with the perturbation V&#039;(x)=\epsilon when L/3<x<2L/3.

I get that the total ground state eigenfunction with the first order perturbation contribution is
<br /> u_{1}=u_{01}+{\int_{L/3}^{2L/3} {u_{02}\hat H&#039; u_{01}dx} \over (E_{01}-E_{02})}u_{02}+{\int_{L/3}^{2L/3} {u_{03}\hat H&#039; u_{01}dx} \over (E_{01}-E_{03})}u_{03}<br />
where
<br /> \hat H&#039;=\epsilon} and u_{0n}/E_{0n}= eigenfunctions/energies of the unperturbed system.
I only need to use \{u_{01},u_{02},u_{03}\} instead of all \{u_{0n}\} when expressing the first order perturbation contribution
u_{11}=\sum_k a_{nk}u_{0k}

Is this correct?
 
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danja347 said:
I have a problem where I should calculate the ground state eigenfunction of a particle in the box where the potential V(x)=0 when 0<x<L and infinite everywhere else with the perturbation V&#039;(x)=\epsilon when L/3<x<2L/3.

I get that the total ground state eigenfunction with the first order perturbation contribution is
<br /> u_{1}=u_{01}+{\int_{L/3}^{2L/3} {u_{02}\hat H&#039; u_{01}dx} \over (E_{01}-E_{02})}u_{02}+{\int_{L/3}^{2L/3} {u_{03}\hat H&#039; u_{01}dx} \over (E_{01}-E_{03})}u_{03}<br />
where
<br /> \hat H&#039;=\epsilon} and u_{0n}/E_{0n}= eigenfunctions/energies of the unperturbed system.
I only need to use \{u_{01},u_{02},u_{03}\} instead of all \{u_{0n}\} when expressing the first order perturbation contribution
u_{11}=\sum_k a_{nk}u_{0k}

Is this correct?

Why do you use only the three vectors \{u_{01},u_{02},u_{03}\},when the theory states u should be using all the eigenvectors associated to eigenvalues different from the one (nondegenerate) u want to compute first orde corrections?

Daniel.
 
It is just an approximation and the only way I can think of an explanation to why the problem i´m supposed to solve says i only need to use these three is that they contribute most to the correction...

Daniel
 
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On the other hand... how should i do if i needed to use all \{u_{0n}\} How would i get a_{nk} in that case? Please tell me if what i have got for u_1 above is correct because then i know if its me that can't integrate because i get that the integrals are zero or if i have set it all up wrong!?

Daniel
 
Yes,the theory states that those vectors (corresponding to different energy levels) are orthogonal,and since the perturbation is a constant,then all its matrix elements between orthogonal states should annulate.
On the other hand,are u sure with the integrations??There are products of sine/cosine functions.Only in certain conditions they annulate.And besides,the eigenfunctions are orthonormalized on the domain [0,L],and yet you're integrating them on a simmetric domain wrt to the middle of the interval L/2.So if that product of eigenfunctions is an odd function,then u shouldn't be surprised the result is zero.
Check whether the products of eigenfunctions are odd.If so,the result is that the corrections to the eigenvector are identically null.
 
I calculated the integrations again and only one of them was equal to zero.
That looks a lot better since it would be strange if the eigenfunction didn´t change when adding the disturbance!

Thanks!

/Daniel
 
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