Conceptual problem with perturbation theory

[eljose]

-Ok..Let,s be the Hamiltonian H=H_0 +W in one dimension where W is a "weak" term so we can apply perturbation theory.

-The "problem" comes when we need to calculate the eigenvalues and eigenfunction of H0 of course we set the system in an "imaginary potential well of width L" so we have the set of eigen-values-functions:

E_n =P^{2}/2m p=(n\pi \hbar)/L \Phi(x) =e^{in\pi x/L}

the problem is...what is the value of L?...so when doing calculations..what,s the value of the width of our "imaginary" well..if we set L--->oo then the Energies and Wave functions tend all to 0.

 

[Perturbation]

-Ok..Let,s be the Hamiltonian H=H_0 +W in one dimension where W is a "weak" term so we can apply perturbation theory.

-The "problem" comes when we need to calculate the eigenvalues and eigenfunction of H0 of course we set the system in an "imaginary potential well of width L" so we have the set of eigen-values-functions:

E_n =P^{2}/2m p=(n\pi \hbar)/L \Phi(x) =e^{in\pi x/L}

the problem is...what is the value of L?...so when doing calculations..what,s the value of the width of our "imaginary" well..if we set L--->oo then the Energies and Wave functions tend all to 0.

For one, when you set L\rightarrow\infty, \Phi_n does not tend to zero, it tends to 1, because that's a complex exponential,

e^{in\pi\tfrac{x}{L}}=\cos\left(n\pi\tfrac{x}{L}\right)+i\sin\left(n\pi\tfrac{x}{L}\right).

Since each \Phi_n(x) =e^{in\pi x/L} is linearly dependent you can write the wave-function as a Fourier series with these functions as a basis. If L tends to infinity your sum of plane wave bases becomes a Fourier transform of the continuous variable p, rather than the discrete sum of p's (or n's). This is just the same procedure one goes through in generalising the Fourier series of functions with finite period to those with infinite period.