Perturbed in the harmonic oscillator

  • Thread starter Fatimah od
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  • #1
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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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Answers and Replies

  • #2
pasmith
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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)} =
\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.
 
  • #3
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ok , I do it thanks

any help for second part please
 
  • #4
pasmith
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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)\\
\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 Schrodinger 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]).
 
  • #5
vela
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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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