I am dealing with scalar field theory in $0+1$ dimension with the following free theory Hamiltonian, $$ \mathcal{H}_0 =\frac{1}{2}\big[\pi^2+m^2\phi^2 \big]\tag{1} $$ with a quartic interaction of the form, $\frac{\lambda}{4!}\phi^4$.

I calculated the first-order energy correction to the $n$th state for the given interaction potential using time-independent perturbation theory as, $$ E_n^{^{(1)}} = \frac{\lambda}{4!}\left(\frac{\hbar}{2m}\right)^2 (6n^2+6n+3).\tag{2} $$

For the ground state $(n=0)$ I see that the first-order energy correction can be calculated using the vaccum bubble of the theory for the given order. My doubt is how does those two results actually matches? What is the reasoning behind it? Also how can I calculate the corrections to the energy using Feynman diagrams for excited states $(n>0)$?


1 Answer 1

  1. Yes, there is total agreement between the first-order correction (2) in Rayleigh-Schrödinger (RS) perturbation theory to the anharmonic oscillator with a quartic perturbation, and the vacuum bubble diagram $O\!O$ from the corresponding Feynman path integral.

  2. This correspondence is explained and worked out in detail for a cubic perturbation in my Phys.SE answer here. The quartic calculation works in a similar way.

  • $\begingroup$ How can I understand this without using path integrals? Thanks. $\endgroup$
    – Fermion
    Jul 23 at 19:39
  • $\begingroup$ Hi @Fermion: That leaves you with the operator formalism. The operator formalism is naturally organized in terms of the RS perturbation theory. $\endgroup$
    – Qmechanic
    Jul 24 at 7:15

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