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Error in perturbation series question

  1. Jul 6, 2013 #1
    Suppose we know the perturbation series

    E = E(\epsilon) = E_0 + \epsilon E_1 + \epsilon^2 E_2 + \ldots

    converges, where [itex]E_0[/itex] is a discrete eigenvalue of [itex]H_0[/itex] and we are considering a Hamiltonian [itex]H = H_0 + \epsilon H_1[/itex]. Does this mean that we know

    E - E_0 = O(\epsilon)

    as [itex]\epsilon \to 0[/itex] in the precise sense that we know there exists a [itex]\delta > 0[/itex] and a [itex]C > 0[/itex] such that if [itex]|\epsilon| < \delta[/itex], then [itex]|E - E_0| \leq C|\epsilon|[/itex]?
  2. jcsd
  3. Jul 6, 2013 #2


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    If the perturbation series converges, this C has to exist - that is just mathematics (taylor series).
  4. Jul 6, 2013 #3
    If you really have a convergent series, then I think your statement holds just as a general property of power series, and you can pick any ##C > |E_1|##. As I understand it, however, perturbation series typically do *not* converge. They are usually asymptotic series, meaning that if you truncate them after order ##\epsilon^N##, the error between the truncated series and the actual eigenvalue vanishes like ##\epsilon^{N+1}## as ##\epsilon \to 0## (which statement you can formalize if you like in the same epsilon-delta style).
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