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Is there any source that explicitly writes down the WKB "function" (to be defined soon) in orders of time derivative of the frequency over the frequency? Of course only to some finite order.

That is: $$\ddot{f}(t)-\omega(t)^2f(t)=0$$ is solved by the WKB ansatz $$f(t)=\frac{1}{\sqrt{2W(t)}}e^{i\int^t W(t')}.$$

I need the equation for $W(t)$ dependent on only $\omega(t)$. (The wikipedia article doesn't really help since it's a recursive formula, and it makes a different ansatz, i.e. one where $W$ is written as a sum. Although the solution will also be a sum of increasing order in $\omega^{(k)}/\omega^{k+1}$ or powers of lower derivatives.) The differential equation $W$ needs to solve is:

$$W^2=\omega^2-\frac{1}{2}\frac{\ddot{W}}{W}+\frac{3}{4}\frac{\dot{W}^2}{W^2}$$ (exact equation)

So yeah: Any source that gives $W$ to some order explicitly?

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Consider the equation \begin{equation} -\hbar^2\frac{d^2 f(t)}{dt^2} + \omega(t)f(t) = 0\,. \end{equation} I added an $\hbar$ as a power counting parameter (just set $\hbar = 1$ at the end of the calculations to recover your equation).

The WKB ansatz is of the form \begin{equation} f(t) = \exp\left(\frac{i}{\hbar}\int^t \mathcal{P}(t^\prime) dt^\prime\right),. \end{equation} Here $\mathcal{P}(t)$ is sometimes called the quantum momentum in the context of quantum mechanics.

Substituting the ansatz in the ODE you get the Riccati equation \begin{equation}\label{eq:exact wkb riccati} \mathcal{P}^2(t) - i \hbar \frac{d\mathcal{P}(t)}{dt} = -\omega(t)\,. \end{equation}

Now you expand $\mathcal{P}(t)$ in a formal series in $\hbar$, this series is in general not convergent, but asymptotic \begin{equation} \mathcal{P}(t) = \sum_{n=0}^\infty p_n(t) \hbar^n\,. \end{equation} Then you put this into the Riccati equation and equate each order of the resulting series to 0. You get the initial condition \begin{equation} p_0(t) = \sqrt{-\omega(t)}\,, \end{equation} notice the two branches of the square root correspond to two different independent solutions of the original second order equation. Also you obtain the recursion relation \begin{equation} p_{n}(t) = \frac{1}{2 p_0(t)}\left(i\frac{d p_{n-1}(t)}{dt} - \sum_{m=1}^{n-1} p_{m}(t)p_{n-m}(t)\right)\,. \end{equation}

In particular one can divide the formal series in the even and odd powers, call $W(t)$ the even series i.e. $W(t) = \sum_n p_{2n}(t)\hbar^{2n}$ and $W_{\text{odd}}(t)$ the odd series. It turns out that \begin{equation} W_{\text{odd}}(t) = \frac{i\hbar}{2}\frac{d}{dt}\log W(t)\,. \end{equation} Thus we can express $f$ in terms only of the even powers of $\hbar$ \begin{equation} f(t) = \frac{1}{\sqrt{W(t)}}\exp\left(\frac{i}{\hbar}\int^t W(t^\prime) dt^\prime\right)\,. \end{equation}

I do not see the problem in the recursive formula, you can use it in principle to get any to any finite order, it is pretty pointless to be lazy, the formula is pretty easy and the first few orders can be easily done by hand. Also even if you found it on a resource double-checking papers result is basically a fundamental step in reading most papers since you may find typos often (and I invite you to double check my formulas too, the ideas are right, the calculations hopefully are too).

There are many references on what I explained here, a good one which goes way deeper than what I wrote is https://arxiv.org/abs/1811.04812 and the references therein. Here also the points I made about the series being asymptotic and how to make sense of it via Borel resummations is explained as well as the framework of resugent quantum mechanics. The parts on the Thermodynamic Bethe Ansatz are interesting but most likely not too relevant

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  • $\begingroup$ Thanks! I didn't actually know the WKB ansatz had much of a derivation up to the physical intuition. Could you also do this ansatz instead? $\frac{1}{W}e^{...}$ rather than the square root? When I tried today, I got a weird imaginary term which was odd in $\bar{h}$ $\endgroup$ Commented Apr 7 at 17:39
  • $\begingroup$ This 1/2, which leads to the square root seems to be kind of fixed and not convention... $\endgroup$ Commented Apr 7 at 17:44
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    $\begingroup$ The recursion relation seems wrong, I get $p_n=\frac{1}{2p_0}(i\dot{p}_{n-1}-\sum_{k=1}^{n-1}p_kp_{n-k})$. I.e. there's a shift in the sum, and an $i$ is missing in front of the time derivative. $\endgroup$ Commented Apr 7 at 19:49
  • $\begingroup$ And I really don't see how the relation between odd and even series is derived. Any hints? $\endgroup$ Commented Apr 7 at 20:06
  • $\begingroup$ The recursion relation indeed is wrong, I had copied from notes and modified the equation because they had some factor of $i$ difference. You should be able to derive the relation between the odd end even power series by first looking at $p_0$ and $p_1$ for which it holds and then proceed by induction if iirc $\endgroup$
    – lucabtz
    Commented Apr 8 at 1:40
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Thank god I finally found the source...

It only goes to the third order (the one I refuse to calculate myself). If anyone finds higher orders, I would also be interested in those.

https://arxiv.org/pdf/gr-qc/0510001.pdf

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