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WKB and perturbation theory.

  1. Apr 24, 2007 #1


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    for a Hamiltonian [tex] H=H_0 + \epsilon V(x) [/tex]

    my question is (for small epsilon) can WKB and perturbative approach give very different solutions ?? to energies eigenvalues and so on the index '0' means that is the Hamiltonian of a free particle.

    problem arises perhaps in calculation of:

    [tex] \int \mathcal D [x] exp(iS[x]/\hbar)[/tex]

    with the action [tex] S[x]=S_0 [x] +\epsilon V[x] [/tex]

    here the main problem is that in perturbation theory the functional integral may be divergent (due to IR and UV divergences) but in WKB (semiclassical approach) the integral can be 'calculated' (given finite meaning),

    hence i'd like to know if at least for perturbative case you can use WKB approach (with some re-scaled constant) to deal with perturbation theory..note that for the 'free particle' no interaction WKB gives exact methods..thankx.
  2. jcsd
  3. Apr 24, 2007 #2
    Technically, WKB and "perturbation theory" are both perturbative expansions in what are presumed to be "small quantities". In the case of your "perturbation theory", the small parameter is the [tex]\epsilon[/tex], whereas in the WKB approximation, you have expanded the propagator to first order in [tex]\hbar[/tex], so evidently the total action of the trajectories is large compared to Planck's constant, which sets the scale for action in quantum mechanics.

    Thus, you can get two very different answers depending on what exactly is "small".
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