Anharmonic oscillator first-order correction to energy

  • #1
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Homework Statement


I have ##H'=ax^3+bx^4##, and wish to find the general perturbed wave-functions.

Homework Equations


First-order correction to the wave-function is given by, $$\psi_n^{(1)}=\Sigma_{m\neq n}\frac{\langle\psi_m^{(0)}|H'|\psi_n^{(0)}\rangle}{n-m}|\psi_m^{(0)}\rangle.$$

The Attempt at a Solution


The formula becomes (focusing exclusively on the cubic term), $$\psi_n^{(1)}=\Sigma_{m\neq n}\frac{\langle\psi_m^{(0)}|ax^3|\psi_n^{(0)}\rangle}{n-m}|\psi_m^{(0)}\rangle.$$

(I shall denote, ##|\psi_{n,m}^{(0)}\rangle## as ##|n,m\rangle## to avoid typesetting overly complicated formulas. Using the ladder operator definition of ##x## then, I have, $$|n^{(1)}\rangle=\frac{a(\frac{\hbar}{2m\omega}^{3/2})}{\hbar\omega}\Sigma_{m\neq n}\frac{\langle m|(a+a^+)^3|n\rangle}{n-m}|m\rangle.$$

Now the trouble is, according to my textbook, the cubic term contributes a first-order correction to the wavefunction, but I'm not seeing it. There would never be an equal number of raising and lowering operators, such that by orthogonality all the coefficients would become ##0##.
 

Answers and Replies

  • #2
blue_leaf77
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There would never be an equal number of raising and lowering operators, such that by orthogonality all the coefficients would become 00.
You seem to forget that ##|n\rangle## and ##|m\rangle## are different states. For example Something like ##\langle n+3|(a^+)^3| n \rangle \neq 0##.
 
  • #3
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You seem to forget that ##|n\rangle## and ##|m\rangle## are different states. For example Something like ##\langle n+3|(a^+)^3| n \rangle \neq 0##.
Oh, okay, I see. So first, I'd expand the operator as, $$(a+a^+)=aa+aa^++a^+a+a^+a^+(a+a^+)^3=aaa+aaa^++aa^+a+aa^+a^++a^+aa+a^+aa^++a^+a^+a+a^+a^+a^+.$$
Then say for the first term, ##m=n-3## in order for them to be non-orthogonal, such that the coeffiecient is $$\frac{\sqrt{n}\sqrt{n-1}\sqrt{n-2}}{3}.$$

Is that right?
 
  • #4
blue_leaf77
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