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

[danja347]

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 $V'(x)=\epsilon$ when L/3<x<2L/3.

I get that the total ground state eigenfunction with the first order perturbation contribution is
$$u_{1}=u_{01}+{\int_{L/3}^{2L/3} {u_{02}\hat H' u_{01}dx} \over (E_{01}-E_{02})}u_{02}+{\int_{L/3}^{2L/3} {u_{03}\hat H' u_{01}dx} \over (E_{01}-E_{03})}u_{03}$$
where
$\hat H'=\epsilon}$ 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?

 

[dextercioby]

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 $V'(x)=\epsilon$ when L/3<x<2L/3.

I get that the total ground state eigenfunction with the first order perturbation contribution is
$$u_{1}=u_{01}+{\int_{L/3}^{2L/3} {u_{02}\hat H' u_{01}dx} \over (E_{01}-E_{02})}u_{02}+{\int_{L/3}^{2L/3} {u_{03}\hat H' u_{01}dx} \over (E_{01}-E_{03})}u_{03}$$
where
$\hat H'=\epsilon}$ 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.

 

[danja347]

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

 

[danja347]

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

 

[dextercioby]

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.

 

[danja347]

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