I don't seem to understand what perturbation theory really is, and what it is needed for. Can someone please provide an explanation for what it is, and why it is needed in QM?

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    $\begingroup$ Perturbation theory allows you to calculate the response of a system to a small perturbation. For example, it can tell you, how energy levels shift under the influence of the perturbation, how energy level degeneracies are lifted, and it even allows you to calculate linear response functions. You should be able to find more detailed answers to your question in standard QM books and on the web. $\endgroup$ – flaudemus Feb 22 '19 at 11:24
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    $\begingroup$ This seems to be a side effect of an incomplete physics education. Perturbation theory is an incredibly broad set of techniques useful in all subfields of physics and much of applied mathematics at large. But often people are not introduced to it until a quantum mechanics course, probably because there's not enough time to look at nontrivial (i.e. no simple closed form solution) problems earlier in the curriculum, which would motivate introducing perturbation theory. $\endgroup$ – knzhou Feb 22 '19 at 16:19
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    $\begingroup$ As such this question ends up sounding like... "what is calculus and why is it introduced in quantum field theory class?", or "what is the quadratic formula and why is it useful for statistical mechanics?" $\endgroup$ – knzhou Feb 22 '19 at 16:21
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    $\begingroup$ @knzhou I don't see where is the OP is implying that perturbation theory is useful only for QM. Also, the question "why is perturbation theory needed in QM?" is very on point, IMO. $\endgroup$ – pppqqq Feb 22 '19 at 19:14
  • $\begingroup$ wikipedia has a fairly good explanation of this, have you read it? $\endgroup$ – niels nielsen Feb 22 '19 at 23:53

There are two main reasons.

The first is practical, and it is that QM is hard. Very few systems are solvable, particularly in the presence of interactions. There's a wide variety of situations where perturbation theory is the only available way to make any headway at all.

The second one is conceptual and it is basically that it allows us to establish a hierarchy in the strengths of the different components of a system, and it allows us to have a clearer picture of how this strengths scale with the parameters of the system. Thus, even if you can solve the full system, if one component is weak, it often makes sense to treat it explicitly as a perturbation, so that you can keep a clearer track of how the different aspects of the solution (energies, eigenfunctions, dipole moments, what have you) scale.

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    $\begingroup$ Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks. $\endgroup$ – Dvij Mankad Feb 22 '19 at 12:16
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    $\begingroup$ @DvijMankad I have no idea what scope you're thinking of. Ask separately. $\endgroup$ – Emilio Pisanty Feb 22 '19 at 12:50
  • $\begingroup$ I was thinking schemes where we have a small parameter but we are not doing perturbation theory, for example, the adiabatic approximation. I will try to formulate a separate question. $\endgroup$ – Dvij Mankad Feb 22 '19 at 13:10

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