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You can consider degenerate perturbation theory as the more general case. So that you can think of the non-degenerate case, as a degenerate Hilbert subspace of dimension one. In this case there is no fundamental distinction, and the methods do not depend on whether other subspaces are degenerate or not. Possibly what the statement in books mean is to ...

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The ambiguity that people normally refer to is due to the lack of Borel summability of the perturbation series. Consider a series of the form $$A(g) = \sum_{n=1}^{\infty}{(-1)^n g^n (n!)}$$ If the coefficients $b_n$ are of order $1$, this series is obviously divergent. But we can compute its Borel sum. First compute the Borel transform: ...

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Maybe this would be better as a comment, since it is not a full answer, but I don't have enough reputation for that. The most important ambiguity is that there is an infinite number of functions that have the same asymptotic expansion. As an example, if $f(g)$ has some asymptotic expansion in $g$ as $g \to 0$ than $f(g) + e^{-1/g^2}$ has exactly the same ...

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As I cannot comment (reputation>50 needed), I put my comment here, as this might be considered an answer, too. The answer and picture of DanielSank is very nice, so just to add what I think is a satisfactory answer as well, just in words. On page 139 of my favourite textbook:-), the explanation is right in the beginning of the "Secular equation" paragraph. ...

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You can also have a look in Landau and Lifshitz (Quantum Mechanics - Non-relativistic Theory, where in §39. The secular equation, degenerate perturbation theory is treated, then there is specifically to your question Problem 2.: "Derive the formulae for the correction to the eigenfunctions in the first approximation and to the eigenvalues in the second ...

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This is the total transition rate for a transition to occur. For instance, in a scattering experiment, there is a probability that an incoming particle will or will not be scattered by the target. If the quantity in which one is interested is the probability that a particle is scattered, then the quantity you have given above would give you the answer. This ...

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QTF is pretty messed up, although many physicists won't probably agree with this. The current methods are good enough to predict outcomes of experiments, but they are quite dubious from a mathematical point of view. Consider Dirac's interaction picture for instance, which is usually invoked by many physicists around the world to predict the outcome of an ...

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First: Scientific theories are never proven, only not falsified. Repeat that until it sinks in. Now, for the actual content of the question: That we only have perturbative ways to compute the S-matrix/scattering amplitudes for the Standard Model is not a reason to doubt its validity. Almost no physical system, apart from toy models, can be solved exactly, ...

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