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I am currently studying the WKB approximation, and certain parts of the argument (mostly when dealing with turning points and patching wavefunctions) rely on the fact that the WKB approximation is a semi-classical approximation, and in the semi-classical regime, $\hbar \ \rightarrow \ 0$.

I understand how certain aspect of classical mechanics can be recovered as Planck's constant gets smaller, but my question is: why are we allowed to do this? After all, we are using the WKB approximation in the context of regular quantum mechanics, where $\hbar$ is just a fixed number. It doesn't really make sense to me why we are able to make this assumption.

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We never "let $\hbar \to 0$". As you said, that doesn't make sense because $\hbar$ is dimensionful, and also fixed in our universe.

What we mean by $\hbar \to 0$ is that we're considering only physical situations where the action $S$ is large, and thus taking the limit $\hbar/S \to 0$. That's what the semiclassical regime means.

It's like how the "nonrelativistic regime" means considering only objects with speeds $v$ small compared to the speed of light, $v/c \to 0$. Sometimes people sloppily write that as $c \to \infty$.

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    $\begingroup$ I don't agree with this answer. Even if $c$ and $\hbar$ are dimensionfull, they can be very large (or very small) in any given unit system. An infinite or zero value remains infinite or zero in any finite (i.e. usefull) units. $\endgroup$
    – Cham
    Commented Jan 25, 2020 at 19:22
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    $\begingroup$ @Cham Any mention of a unit system is a distraction. You can't change whether a given physical system is accurately described by, e.g. classical mechanics by just changing the units. What matters is what the system is actually doing. $\endgroup$
    – knzhou
    Commented Jan 25, 2020 at 19:23
  • $\begingroup$ 0 can have dimensions. I can say a distance is 0 meters for example. However, this answer does address that we generally speak in terms of a ratio because this is a more consistent and precise way $\endgroup$
    – Liam Clink
    Commented Jan 25, 2020 at 19:24
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    $\begingroup$ @LiamClink The point is that when you take a dimensionful limit like $\hbar \to 0$, it's ambiguous what it means. You can make the numeric value of $\hbar$ small by changing the unit system, for example, but we know that doesn't actually change anything -- that's precisely the issue the OP is pointing out. Writing $\hbar/S \to 0$ makes more sense because it disentangles dependence on an arbitrary unit system. If $\hbar/S$ is small for a particular system, then it is in any unit system. $\endgroup$
    – knzhou
    Commented Jan 25, 2020 at 19:27
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    $\begingroup$ @LiamClink Another issue that $\hbar/S \to 0$ fixes is that it makes clear that taking the limit is possible at all. "$\hbar \to 0$" is confusing because you can't actually change $\hbar$ in real life, and if you actually did, god knows what would happen? "$\hbar/S \to 0$" tells us the limit is taken by fixing $\hbar$, as it is fixed in real life, and considering systems with larger and larger $S$. $\endgroup$
    – knzhou
    Commented Jan 25, 2020 at 19:30
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What is happening is that the solution to the Schroedinger Equation is being Taylor series expanded in $\hbar$, and the semiclassical approximation is good when the first and higher order terms are relatively small, and so you can drop them (which is the same as setting $\hbar$ to zero) without affecting the form of the solution much. We are "allowed" to do this only when the higher order coefficients are near zero, and when they are that is the reason.

The physical reason is that when you get to large enough scales, $\hbar$ may as well be zero because of how little difference it makes. It's not exact, but this is an approximation so you already aren't exact. You're just making another assumption which restricts the regime of validity, and you just need to keep that in mind.

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    $\begingroup$ Thanks for the answer! When you say "large enough scales", what exactly are you speaking in terms of? $\endgroup$
    – Jack Ceroni
    Commented Jan 25, 2020 at 18:33
  • $\begingroup$ It depends on the precise details, but ultimately "large enough" is when the error introduced by this approximation is acceptable, which you need to specify. Typically one is interested in a small relative (fractional) error, and how small it needs to be is up to you. It's an approximation, so you must've decided that some level of inaccuracy is acceptable, so that is what determines it. Generally we say an approximation is valid if the relative error is much smaller than 1. That's not an exact statement though. $\endgroup$
    – Liam Clink
    Commented Jan 25, 2020 at 18:45
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    $\begingroup$ Ok, that makes sense, thank you! $\endgroup$
    – Jack Ceroni
    Commented Jan 25, 2020 at 18:55

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