How Do You Compute the 1st Order Wave Function Correction in Quantum Mechanics?

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Homework Help Overview

The discussion revolves around the computation of first-order wave function corrections in quantum mechanics, specifically using perturbation theory in the context of an infinite potential well. The original poster expresses uncertainty about evaluating the sum for the first-order correction due to the presence of infinite terms.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster questions how to compute the matrix elements necessary for evaluating the sum in the first-order correction formula. Some participants suggest considering the form of the perturbation H' to identify non-zero terms in the summation.

Discussion Status

The discussion has progressed with participants exploring the implications of different forms of the perturbation. A suggestion has been made that if H' is a constant, it results in no first-order correction, as it does not alter the eigenstates of the Hamiltonian.

Contextual Notes

The original poster's question indicates a lack of clarity regarding the perturbation's structure and its impact on the wave function corrections. There is an implicit assumption that the perturbation can vary, which is being examined in the discussion.

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Homework Statement



A have a bit of a general question regarding 1st order wave function corrections using perturbation theory.

In a problem like the infinite potential well where you have states numbered like n = 1, 2, 3, ..., how do you compute the sum for the 1st order correction when you have infinite terms?:

[tex]\psi_n^{(1)} = \Sigma_{l \ne n} \frac{<\psi_n^{(0)}|H'|\psi_l^{(0)}>}{E_n^{(0)} - E_l^{(0)}} \psi_l^{(0)}[/tex]

I guess I don't know how to get <n|H'|l> so I can evaluate the sum
 
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Do you know what your perturbation H' looks like? Sometimes the non-zero terms in the summation result in something that can be summed analytically.
 
This was the thinking I was missing!

So for H' = constant there is no first-order correction because [itex]l \ne n[/itex], yes?
 
Correct. If you add a constant to your Hamiltonian, you shift the zero of energy but you do not change its eigenstates.
 

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