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Anharmonic oscillator first-order correction to energy

  1. Mar 4, 2017 #1
    1. The problem statement, all variables and given/known data
    I have ##H'=ax^3+bx^4##, and wish to find the general perturbed wave-functions.

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

    3. 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##.
     
  2. jcsd
  3. Mar 4, 2017 #2

    blue_leaf77

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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##.
     
  4. Mar 5, 2017 #3
    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?
     
  5. Mar 5, 2017 #4

    blue_leaf77

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    Yes.
     
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