Semiclassical Regime: What Is It & When to Use It

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Many papers say "we are treating a semiclassical regime".
I don't know what is a semiclassical regime and when a system can be treated as a semiclassical regime?
Thank you!
 
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This term is usually used quite loosely.
The idea is that you look at what the system is doing classically, and then treat quantum fluctuations around the classical trajectory as being small corrections.

I would suggest you take a look at the WKB approximation (in say, Griffiths, or some other QM book; Wikipedia's explanation is terrible)

Ok, so here's the money:

In quantum mechanics we're often interested in the path integral, which is of course a sum over paths weighted by their classical action:

Z = \int\!\mathcal{D}q\, e^{i S[q]}

Let's Wick rotate to make things convergent

Z = \int\!\mathcal{D}q\, e^{-S[q]}

The assumption is that S is quadratic around it's minimum, and that it is very steep (compared to \hbar which I have set to 1), meaning that the paths that veer slightly from the classical minimum get heavily suppressed by the exponential. Thus, we may use a saddle-point approximation:

S[q] \approx S[q_0] + \frac{1}{2!}(q-q0) S''[q_0] (q-q0)

where q_0 is the classical solution, and q-q0 is a small fluctuation.

Z \approx e^{-S[q_0]} \int\!\mathcal{D}q\, e^{-(q-q0)S''[q](q-q0)}

or

Z \approx e^{-S[q_0]} \sqrt{ \det S''[q_0] }

Even more interesting than the path integral or partition function is the logarithm of it, or the free energy, which generates connected Feynman diagrams. Thus

W = \ln Z = -S[q_0] - \ln \det S''[q_0]

Why would we want to do such a thing? Well, sometimes you can treat one part of the system semiclassically and leave the other part fully quantum mechanically. Then you'd still have to integrate over some other dynamical degree of freedom, say, x, but you have an effective action description of q. An example is to have electrons in a "background" or "classical" electromagnetic field where you treat the field's quantum corrections on a first order basis.
 
many thanks.
You have said about this problem so detailedly.
 
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