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First Order Perturbation Theory - QM

  1. Sep 27, 2016 #1
    1. The problem statement, all variables and given/known data
    The ground state energy of the 1D harmonic oscillator with angular frequency ##\omega## is ##E_0 = \frac{\hbar \omega}{2}##. The angular frequency is perturbed by a small amount ##\delta \omega##. Use first order perturbation theory to estimate the ground state energy of the perturbed system.

    2. Relevant equations


    3. The attempt at a solution

    I have calculated the energy shift, but I don't think my units make sense.

    My perturbation: ##\Delta V(x) = V(x) - V_0(x)##, where ##V_0(x)## is my unperturbed potential. ##V(x) = \frac{m (\omega + \delta \omega)^2 x^2}{2} = \frac{mx^2}{2} (\omega^2 + \delta \omega^2 + 2 \omega \delta \omega)##.

    ##\Delta V(x) = \frac{m x^2}{2} (\delta \omega^2 + 2 \omega \delta \omega)##

    From first order perturbation theory, ##\Delta E_n^{(1)} = \langle \Delta V \rangle ^{(0)}_n## i.e the first order correction to the energy is the expectation value of the perturbation in the unperturbed state.

    Using the ground state wave function for the harmonic oscillator ##\psi_0 = (\frac{m \omega}{\pi \hbar})^{1/4} e^{-\frac{m \omega x^2}{\hbar}}## I calculate the expectation value of the perturbation.

    ##\langle \Delta V \rangle = \int^{\infty}_{-\infty} \psi* \Delta V \psi##

    I'm skipping the details of this integral, but I find that it is equal to ##\langle \Delta V \rangle = \frac{m}{2} (\delta \omega^2 + 2 \omega \delta \omega)##.

    This doesn't seem right to me as it does not have units of energy. Have I made a mistake in my reasoning, or does the error likely lie in how I have evaluated that integral? If it's the latter, I can try to find the error myself.

    Thanks for reading!
     
  2. jcsd
  3. Sep 27, 2016 #2

    DrClaude

    User Avatar

    Staff: Mentor

    You made a mistake in the integration. Be careful that you don't forget the normalization constant.
     
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