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When a condition of some physical quantity $a$ and $b$ is presented, say $a \gg b$, how much bigger do we say $a$ has to be than $b$? $10$ times? $1000$ times? $10^6$ times?

Some context:

I am trying to use adiabatic theory to derive the energy spectrum for a particle confined in a ring of radius $b$ in the x-y plane, when a Dirac monopole is slowly moved at a constant velocity $v$ from $z = -\infty$ to $z = \infty$. The condition for applicability of the adiabatic theorem is that the time $T$ that it takes for a varying parameter to change appreciably is much larger than $\frac{\hbar}{\Delta E}$, where $\Delta E$ denotes some typical energy level difference between eigenstates of the Hamiltonian. My problem is, I can't directly quantify an appreciable change, say by $10 \%$ in $z$, since we can write $z = z_0 + vt$ but $z_0$ is $-\infty$. So, I was thinking of taking $z$ to be essentially infinitely far away from the ring when $|z| \gg b$, the radius of the ring. This would give me a value for $z_0$. But I am not sure how much bigger to make $z$ than $b$, hence my question.

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    $\begingroup$ Sorry, the answer to the first question differs wildly between fields. In thermodynamics, it may be $10^{23}$. In mechanical engineering it may be $10^3$. In the soft sciences it may be $10$. In QCD it may be $3$. In the epsilon expansion it is literally $1$. Usually the error would be "on the order of $b/a$" because that's the next term in the Taylor series. $\endgroup$ – knzhou May 17 '18 at 9:05
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    $\begingroup$ Your second question seems to be totally different, it's about bounding the error of the adiabatic approximation. I don't know the answer, it looks pretty tricky! Definitely not just a straightforward Taylor series. $\endgroup$ – knzhou May 17 '18 at 9:07
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    $\begingroup$ The varying parameter for your applicability condition would be the term directly entering Hamiltonian of a particle and not $z$ itself. I would guess it would be something like circulation of vector potential around the ring. $\endgroup$ – A.V.S. May 17 '18 at 17:14
  • $\begingroup$ @A.V.S. The magnetic vector potential $A$ enters the Hamiltonian, which depends on the z-coordinate of the monopole. Are you suggesting examining $A$? I will try that now. $\endgroup$ – user154080 May 18 '18 at 0:38
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The usual treatment for the approximation is to consider the ratio $1\gg \frac{b}{a}$ and proceed to Taylor expand the function of your interest around some point, such that the displacements (perturbations) are written in terms of this ratio. Then you can decide where to truncate your series expansion and estimate what is the error you are making by bounding the truncated terms.

My suggestion for your case would be then to try to write exact expressions first, if possible. Then do as I say above. It is possible that you may need to translate the adiabatic condition into some other variables that appear directly in your formulas.

EDIT: A common example would be the expansion of a term of the form: $$\left(1-\frac{b}{a}\right)^{-n} =\sum_{k=0}^\infty \binom{n+k-1}{k}x^k = 1 + n\frac{b}{a} + \mathcal{o}\left(\left(\frac{b}{a}\right)^2\right) \approx 1 + n\frac{b}{a}$$

(You can read about the little-o notation in Wikipedia which has the precise definition). In the last step, the series was truncated to first order (all higher order terms are ignored). However, you can prove that the error you are commiting can be bounded by the next derivative (after truncation), so in our case by the second derivative, and by the displacement. This series is centered at 0, so the displacement would be just |x|. For the details, see Taylor's Theorem. I hope the example works to understand the principle.

In the end is up to you to say how much precision you need, to state at which point your conclusions are valid.

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  • $\begingroup$ This is the answer I was coming to write. An example is "small angle approximation" $\sin x \approx x$, where the next term in the series goes like $x^3$. So for angles smaller than 0.1 radian, the small angle approximation gives a fractional error better than 1%. $\endgroup$ – rob May 17 '18 at 12:34
  • $\begingroup$ Could you please elaborate further on truncating the series expansion and bounding the truncated terms? $\endgroup$ – user154080 May 18 '18 at 23:25

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