Perturbed in the harmonic oscillator

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Homework Help Overview

The discussion revolves around a problem related to perturbation theory in quantum mechanics, specifically focusing on the harmonic oscillator. Participants are exploring how to express the perturbed Hamiltonian and its implications on energy levels and wave functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to express the perturbed Hamiltonian in power series form and are questioning how to properly set up the integration for calculating energy corrections. There is also discussion about the definitions of terms like H', E_n^1, and ψ_n^0 as they relate to the perturbation method.

Discussion Status

Some participants have made progress in understanding the first part of the problem, while others are seeking further assistance with the second part. There are indications of differing interpretations regarding the correct form of the perturbed Hamiltonian and the evaluation of integrals, with no clear consensus reached yet.

Contextual Notes

Participants are working under the constraints of their textbook definitions and are addressing potential divergences in integrals, which may affect their calculations. There is an emphasis on the need for clarity in definitions and expressions related to the problem.

Fatimah od
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Homework Statement



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Homework Equations



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The Attempt at a Solution



for part a I do not know how to write it in power series form ?

for part b :
I chose the perturbed H' is v(x)= (1+ε )K x^2 /2
then I started integrate E_1 = ∫ H' ψ^2 dx

the problem was , the result equals to ∞ !

shall I integrate in the interval [ 0,∞ ] !
 
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Fatimah od said:
for part a I do not know how to write it in power series form ?

You are given [itex]E_n = (n + \frac12)\hbar\sqrt{k/m}[/itex]. [itex]k[/itex] is now [itex]k(1+\epsilon)[/itex], so [itex]E_n[/itex] is now
[tex] E_n = \left(n + \frac12\right)\hbar\sqrt{\frac km(1 + \epsilon)} = <br /> \left(n + \frac12\right)\hbar\sqrt{\frac km}(1 + \epsilon)^{1/2}[/tex]
Since [itex]|\epsilon| < 1[/itex], [itex](1 + \epsilon)^{1/2}[/itex] can be expanded in a binomial series.
 
ok , I do it thanks

any help for second part please
 
Fatimah od said:
ok , I do it thanks

any help for second part please

I can't help you with that unless you tell me how your textbook defines [itex]H'[/itex], [itex]E_n^1[/itex] and [itex]\psi_n^0[/itex]. I am prepared to guess that you have
[tex]E_n = E_n^0 + \epsilon E_n^1 + O(\epsilon^2)\\<br /> \psi_n = \psi_n^0 +\epsilon\psi_n^1 + O(\epsilon^2)[/tex]
but I'm not prepared to guess what [itex]H'[/itex] might be. However I suspect the method is to substitute the above into the Schrödinger equation to get
[tex] (E_n^0 + \epsilon E_n^1)(\psi_n^0 + \epsilon \psi_n^1) = -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2} (\psi_n^0 + \epsilon\psi_n^1) + \frac{kx^2}{2}(1 + \epsilon)(\psi_n^0 +\epsilon\psi_n^1)[/tex]
and then require that the coefficients of [itex]\epsilon^0[/itex] and [itex]\epsilon^1[/itex] should vanish. At some stage you may want to take an inner product with [itex]\langle\psi_n^0|[/itex], and recall that
[tex]H_0 = -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2} + \frac{kx^2}{2}[/tex]
is self-adjoint ([itex]\langle f | H_0 | g \rangle = \langle g | H_0 | f \rangle[/itex] for all [itex]f[/itex] and [itex]g[/itex]).
 
Your H' is wrong, or you're not expressing yourself clearly. H' is the difference between the complete Hamiltonian and the unperturbed Hamiltonian.

Show us how you're getting that the integral diverges. The problem seems to lie in your evaluation of the integral.
 

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