As we all know, the Schrödinger equation is $$\fr\d{\d t}\ket{\fc\psi t}=-\i\fc Ht\ket{\fc\psi t},$$ where $\ket\psi$ is the state vector and $H$ is the Hamiltonian operator (may be time-dependent). This is an ordinary differential equation (ODE), so we can express its solution as a sum $$\ket{\fc\psi t}=\sum_{n=0}^\infty\ket{\fc{\psi^{\p n}}t},$$ where each term in the sum is defined iteratively by (see also my past article) $$\ket{\fc{\psi^{\p0}}t}\ceq\ket{\fc\psi0},\quad \ket{\fc{\psi^{\p{n+1}}}t}\ceq-\i\int_0^t\d t'\,\fc H{t'}\ket{\fc{\psi^{\p n}}{t'}}.$$ This iteration is called the Picard iteration, which is most known as a method to prove the Picard–Lindelöf theorem.

Let us actually write out the general term in this sum as $$\ket{\fc{\psi^{\p n}}t}=\fc{K^{\p n}}t\ket{\fc\psi0},$$ where $$\fc{K^{\p n}}t=\p{-\i}^n\int_0^t\d t_n\,\fc H{t_n}\int_0^{t_n}\d t_{n-1}\,\fc H{t_{n-1}}\cdots\int_0^{t_2}\d t_1\,\fc H{t_1}.$$ Now with the trick of time-ordering, we can rewrite this as $$\begin{align*} \fc{K^{\p n}}t&=\fr{\p{-\i}^n}{n!}\int_0^t\d t_n\int_0^t\d t_{n-1}\cdots\int_0^t\d t_1\,\bfc{\mcal T}{\fc H{t_n}\fc H{t_{n-1}}\cdots\fc H{t_1}}\\ &=\fr{\p{-\i}^n}{n!}\mcal T\p{\int_0^t\d t'\,\fc H{t'}}^n, \end{align*}$$ where $\bfc{\mcal T}\cdots$ means to order the operators inside according to their time arguments. The factor of $1/n!$ appears because there are $n!$ ways to order $n$ time variables, but another way to see this is to note that the domain of integration is an $n$-simplex, whose volume is $1/n!$ of the corresponding $n$-parallelotope. When we then sum over all $n$, we get the time-ordered exponential $$\ket{\fc\psi t}=\fc Kt\ket{\fc\psi0},\quad \fc Kt=\sum_n\fc{K^{\p n}}t=\mcal T\fc\exp{-\i\int_0^t\d t'\,\fc H{t'}}.$$ $$(1)$$ The operator $\fc Kt$ has a bunch of equivalent names, such as the time evolution operator, the propagator, the Green’s function, the Dyson operator, and the S-matrix (well, they are not entirely equivalent because they are used under different contexts).

The interesting part comes when we consider the matrix elements of $\fc Kt$ and how they relate to the matrix elements of $\fc Ht$. We choose an orthonormal basis $\B{\ket x}$, and then insert a $\sum_x\ket x\bra x$ between each pair of Hamiltonian operators in the expression of $\fc{K^{\p n}}t$: $$\bra{x_n}\fc{K^{\p n}}t\ket{x_0} =\fr{\p{-\i}^n}{n!}\int_0^t\d^nt\sum_{x_1,\ldots,x_{n-1}} \fc{h_{x_nx_{n-1}}}{t_n}\fc{h_{x_{n-1}x_{n-2}}}{t_{n-1}}\cdots\fc{h_{x_1x_0}}{t_1},$$ where we have abbreviated $\fc{h_{xy}}t\ceq\bra x\fc Ht\ket y$, and $t_1\le\cdots\le t_n$ are a specific ordering of the integrated time variables. We can pull out the sum over intermediate basis states and call the summand a contribution from a walk ¹ from $x_0$ to $x_n$. In other words, $$\bra y\fc{K^{\p n}}t\ket x =\sum_{\B{x_i}}^{\substack{x_n=y\\x_0=x}}\fc K{\B{x_i},t},$$ where the sum is over all walks of length $n$ from $x$ to $y$, and $$\fc K{\B{x_i},t}\ceq\fr{\p{-\i}^n}{n!}\int_0^t\d^nt\, \prod_{i=1}^n\fc{h_{x_ix_{i-1}}}{t_i}.$$ Now, the matrix elements of the full propagator is then the sum over contributions from all walks: $$\bra y\fc Kt\ket x=\sum_{\text{walks }x\to y}\fc K{\text{walk},t}.$$ We can imagine a “Hamiltonian graph” formed by taking the basis states as vertices and the Hamiltonian matrix elements as edge weights (the weight of the directed edge from $x$ to $y$ is $-\i\bra y\fc Ht\ket x$). Note that an edge can be a self-loop. Then, the propagator contribution from a walk is given by integrating the product of all the edge weights along the walk (for a trivial walk, which has zero length, the contribution is simply $1$). This formulation may be called the discrete path integral. There is a paper on arXiv that delivers the idea of the Hamiltonian graph. Its difference from the current article is that it only focuses on time-independent Hamiltonians and that it treats self-loops separately instead of just like normal edges. The following two sections (excluding self-loops and the Feynman path integral) largely follow from the ideas from this paper.

Excluding self-loops

In some cases, it may be hard to consider self-loops on the Hamiltonian graph. We may then benefit from counting only walks without self-loops. However, the contributions from each walk will now be different because we have to account for the same walk with self-loops inserted at various positions. In other words, for a walk $\B{x_i}$ without self-loops, instead of contributing $\fc K{\B{x_i},t}$, we now want to find the contribution $\fc L{\B{x_i},t}$ that sums over all ways to insert self-loops into $\B{x_i}$: $$\fc L{\B{x_i},t}=\sum_{m_0,\ldots,m_n=0}^\infty \fc K{\B{x_i^{m_i}},t},$$ where $m_i$ is the number of self-loops inserted at the vertex $x_i$, and $\B{x_i^{m_i}}$ is an abbreviation of this walk: $$\underbrace{x_0,\ldots,x_0}_{m_0}, \underbrace{x_1,\ldots,x_1}_{m_1},\ldots, \underbrace{x_n,\ldots,x_n}_{m_n}.$$

I will show that we can find an expression for $\fc L{\B{x_i},t}$ for the case when the Hamiltonians at different times commute with each other. In this case, we have $$\fc K{\B{x_i},t}=\fr{1}{n!}\prod_{i=1}^n \p{-\i\int_0^t\d t\,\fc{h_{x_ix_{i-1}}}t}.$$ Therefore, by the definition of $\fc L{\B{x_i},t}$, we have $$\fc L{\B{x_i},t}=\fc K{\B{x_i},t}\sum_{m_0,\ldots,m_n} \fr{n!}{\p{n+\sum_i m_i}!} \prod_i\p{-\i\int_0^t\d t\,\fc{h_{x_ix_i}}t}^{m_i}.$$ For abbreviation, for the rest of this section, we denote $S_i\ceq-\i\int_0^t\d t\,\fc{h_{x_ix_i}}t$.

Now, we use a trick to replace the factorial in the denominator with an contour integral: $$\fr1{N!}=\fr1{2\pi\i}\oint\d z\fr{\e^z}{z^{N+1}},$$ where the contour is a counterclockwise simple closed curve around the origin in the complex plane. We then have $$\fr{\fc L{\B{x_i},t}}{n!\,\fc K{\B{x_i},t}} =\fr1{2\pi\i}\sum_{m_0,\ldots,m_n} \oint\d z\fr{\e^z}{z^{n+\sum_im_i+1}} \prod_i S_i^{m_i} =\fr1{2\pi\i}\oint\d z\fr{\e^z}{z^{n+1}} \prod_i\sum_{m_i}\fr{S_i^{m_i}}{z^{m_i}}.$$ We have thus separated the sums over different $m_i$. Each sum is a geometric series that converges when we choose the contour large enough, so $$\fr{\fc L{\B{x_i},t}}{n!\,\fc K{\B{x_i},t}} =\fr1{2\pi\i}\oint\d z\fr{\e^z}{z^{n+1}} \prod_i\fr1{1-S_i/z} =\sum_i\fr{\e^{S_i}}{\prod_{j\ne i}\p{S_i-S_j}},$$ where the last step used the residue theorem for each pole at $z=S_i$. This expression is exactly the expanded form of the divided difference of $\B{\e^{S_i}}$, often denoted $\e^{\b{S_0,\ldots,S_n}}$. Therefore, $$\fc L{\B{x_i},t}=n!\,\fc K{\B{x_i},t}\e^{\b{S_0,\ldots,S_n}}.$$ $$(2)$$

The discrete path integral can then be rewritten in a form that only involves collecting contributions from walks without self-loops: $$\bra y\fc Kt\ket x=\sum_{\text{walks }x\to y}^{\text{no self-loops}}\fc L{\text{walk},t}.$$

Feynman path integral

This discrete path integral formulation of the propagator already looks similar to the Feynman path integral, but we have to go a step further to take the continuum limit to actually get there. For simplicity, I will only consider a particle with unvarying mass $m$ moving in a time-independent potential in one dimension. Its Hamiltonian is $H=p^2/2m+\fc Vx$, and the orthonormal basis is chosen to be the position basis, also denoted as $\B{\ket x}$.

The more standard way to derive the Feynman path integral is to slice the time integral in Equation 1, to express the total exponentiation as a product as many small exponentiations, and then to insert $\int\d x\ket x\bra x$ between each pair of exponentiations (see, e.g., chapter 6 of Quantum Field Theory by Mark Srednicki). However, this approach does not make its connection to the discrete path integral clear. Instead, we will discretize the position space into a lattice with spacing $a$, use the discrete path integral formulation on this lattice, and then take the continuum limit $a\to0$ at the end. Now, instead of $x\in\bR$, we have $x\in a\bZ$. Each basis vector $\ket x$ now has two nearest neighbors $\ket{x-a}$ and $\ket{x+a}$.

In the position basis, the kinetic part of the Hamiltonian $p^2/2m$ is a second derivative operator. From numerical differentiation, we can approximate it on the lattice as $$\fr{p^2}{2m}=\fr1{2ma^2}\p{2\ket x-\ket{x+a}-\ket{x-a}}\bra x.$$ Therefore, the discretized Hamiltonian is $$H=\sum_x\p{\fr1{2ma^2}\p{2\ket x-\ket{x+a}-\ket{x-a}}+\fc Vx\ket x}\bra x.$$ Its matrix elements are then $$h_{yx}=\fr1{2ma^2}\p{2\dlt_{y,x}-\dlt_{y,x+a}-\dlt_{y,x-a}}+\fc Vx\dlt_{y,x}.$$ Conceptually, it consists of on-site energy $1/ma^2+\fc Vx$ and nearest-neighbor hops. The on-site energy looks bothersome, but we can remove it if we only consider walks without self-loops. Equation 2 becomes $$\bra y\fc Kt\ket x =\sum_{n=0}^\infty\p{\fr\beta{ma^2}}^n \sum_{\B{x_i}}^{\substack{x_n=y\\x_0=x}}\e^{\b{S_0,\ldots,S_n}} \prod_{i=1}^n\p{\fr12\dlt_{x_i,x_{i-1}+a}+\fr12\dlt_{x_i,x_{i-1}-a}},$$ $$(3)$$ where $\beta=\i t$ is the imaginary time, and $S_i\ceq-\beta\p{1/ma^2+\fc V{x_i}}$ is defined for the same abbreviation reason as the previous section. The terms proportional to $\dlt_{x_i,x_{i-1}}$ in the multiplicant are omitted because we only consider walks without self-loops. The rest of this section is done under a Wick rotation so that $\beta$ is assumed to be a positive real parameter.

First, let us tackle the divided difference $\e^{\b{S_0,\ldots,S_n}}$. Define $\Dlt S_i\ceq S_i-\bar S$, where $\bar S\ceq\sum_i S/\p{n+1}$ is the mean of $\B{S_i}$. Then, $\Dlt S_i$ is of order unity (while $S_i$ is of order $\beta/ma^2$, which is much larger than unity for small $a$). Then, from the expanded form of the divided difference, we can easily get $$\e^{\b{S_0,\ldots,S_n}}=\e^{\bar S}\e^{\b{\Dlt S_0,\ldots,\Dlt S_n}}.$$ Recalling how we initially derived the divided difference, we have $$\e^{\b{\Dlt S_0,\ldots,\Dlt S_n}}=\sum_{m_0,\ldots,m_n}\fr1{\p{n+\sum_im_i}!}\prod_i\Dlt S_i^{m_i}.$$ When $n$ is large (the reason of which will be explained in a minute), we have $n!\ll\p{n+1}!\ll\p{n+2}!$ etc., while $Q_i$ is of the order of unity, so we only need to consider the terms with the lowest $\sum_im_i$. The leading term is the term with $\sum_im_i=0$, which is trivially $1/n!$, so we have $$\e^{\b{S_0,\ldots,S_n}}=\fr1{n!}\e^{\i\bar S} =\fr1{n!}\fc\exp{-\fr{\beta}{ma^2}-\fr{\beta}{n+1}\sum_i\fc V{x_i}}.$$ $$(4)$$ This contributes to the potential part of the action.

Substitute Equation 4 into Equation 3, and we have $$\bra y\fc Kt\ket x =\sum_{n=0}^\infty\fr{\e^{-\lmd}\lmd^n}{n!} \sum_{\B{x_i}}^{\substack{x_n=y\\x_0=x}} \fc\exp{-\fr{\beta}{n+1}\sum_i\fc V{x_i}}\prod_{i=1}^n\p{\fr12\dlt_\cdots+\fr12\dlt_\cdots},$$ where $\lmd\ceq\beta/ma^2$ is a large positive number when $a$ is small. Observe that the factor $\e^{-\lmd}\lmd^n/n!$ is the probability mass function of the Poisson distribution with mean $\lmd$ evaluated at $n$. When $\lmd$ is very large, the Poisson distribution can be approximated by a delta distribution because the standard deviation $\sqrt\lmd$ is much smaller than $\lmd$. In other words, $\e^{-\lmd}\lmd^n/n!\approx\dlt_{n,\lmd}$. Therefore, $$\bra y\fc Kt\ket x =\sum_{\B{x_i}}^{\substack{x_\lmd=y\\x_0=x}} \fc\exp{-\fr{\beta}{n+1}\sum_i\fc V{x_i}} \prod_{i=1}^\lmd\p{\fr12\dlt_{x_i,x_{i-1}+a}+\fr12\dlt_{x_i,x_{i-1}-a}}.$$

If the factor involving $V$ were not there in the summand, the sum of products is exactly the probability that a random walk starting at $x$ ends at $y$ after $\lmd$ steps, where at each step the walk moves to the left or right nearest neighbor with equal probability $1/2$. Instead of considering one random walk with $\lmd$ steps, we can consider $N$ random walks with $l\ceq\lmd/N$ steps each, where both $l$ and $N$ are large. Because $l$ is large, we can approximate the distribution of the position at the end of the $j$th random walk as a normally distributed random variable $q_j$ with variance $la^2$. Note that even though $l$ is large, $la^2$ is still very small, so the majority of contribution in the sum only comes from those paths where $x_i$ does not differ too much from the $q_j$ of its corresponding part of random walk. Therefore, if we fix a set of $\B{q_j}$, the factor involving $V$ in the summand can be approximated by replacing $\fc V{x_i}$ with $\fc V{q_j}$ for all $x_i$ in the $j$th random walk segment. We can then pull this factor out of the sum over $\B{x_i}$ (but still inside the integral over $\B{q_j}$). Therefore, we get $$\begin{align*} \bra y\fc Kt\ket x &=\int_{\B{q_j}}^{\substack{q_N=y\\q_0=x}}\d q_1\cdots\d q_{N-1} \sum_{\B{x_i}}^{x_{jl}=q_j}\fc\exp{\tfr{-\beta}{n+1}\sum_i\fc V{x_i}} \prod_{i=1}^\lmd\p{\tfr12\dlt_{x_i,x_{i-1}+a}+\tfr12\dlt_{x_i,x_{i-1}-a}}\\ &=\int_{\B{q_j}}^{\substack{q_N=y\\q_0=x}}\d q_1\cdots\d q_{N-1} \fc\exp{\tfr{-\beta}{N+1}\sum_j\fc V{q_j}} \prod_{j=1}^N\p{a\tfr1{\sqrt{2\pi la^2}}\fc\exp{-\tfr{\p{q_j-q_{j-1}}^2}{2la^2}}}, \end{align*}$$ where the extra factor of $a$ in the multiplicant comes from converting a probability density to a probability (since the probability that the position ends up at the lattice site $y$ is $a$ times the probability density at $y$).

Combining the product of exponentiations into the exponentiation of a sum, we have $$\prod_{j=1}^N\p{a\fr1{\sqrt{2\pi la^2}}\fc\exp{-\fr{\p{q_j-q_{j-1}}^2}{2la^2}}} =\fr1{\sqrt{2\pi l}^N} \fc\exp{-\sum_{j=1}^N\fr{\p{q_j-q_{j-1}}^2}{2la^2}}.$$ If we introduce the time step $\Dlt t\ceq\beta/N=mla^2$, for large $N$, we have $$\sum_{j=1}^N\fr{\p{q_j-q_{j-1}}^2}{2la^2} =\sum_{j=1}^N\Dlt t\,\fr12m\p{\fr{q_j-q_{j-1}}{\Dlt t}}^2 =\int_0^\beta\d t'\,\fr12m\fc{\dot q}{t'}^2.$$ Similarly, for the potential part we introduce the time step $\Dlt t=\beta/\p{N+1}$ ², $$\fr\beta{N+1}\sum_{j=0}^N\fc V{q_j} =\sum_{j=0}^N\Dlt t\,\fc V{q_j} =\int_0^\beta\d t'\,\fc V{\fc q{t'}}.$$ Here, $\fc q{t'}$ and $\fc{\dot q}{t'}$ are the position and its time at time $t'$ for a particle undergoing these random walks. Therefore, we have $$\bra y\fc Kt\ket x =\sqrt{\fr{ma^2}{2\pi\Dlt t}}^N \int_{\B{q_j}}^{\substack{q_N=y\\q_0=x}}\d q_1\cdots\d q_{N-1} \fc\exp{-\int_0^\beta\d t'\p{\fr12m\fc{\dot q}{t'}^2+\fc V{\fc q{t'}}}}.$$

Finally, simply revert the Wick rotation by substituting $\beta=\i t$ and rewrite the integral over $\B{q_j}$ as a path integral over all paths $\fc q{t'}$. Then, we get the Feynman path integral expression of the propagator: $$\bra y\fc Kt\ket x =\int^{\substack{\fc qt=y\\\fc q0=0}} \mcal Dq\fc\exp{\i\int_0^t\d t'\p{\fr12m\fc{\dot q}{t'}^2-\fc V{\fc q{t'}}}}.$$ This completes the derivation.

Suppose there is an annulus defined by the region $\set{\p{\rho,\vphi}}{R_\mrm{in}<\rho<R_\mrm{out}}$. What are the functions $\Phi$ defined on this region that satisfy $$\lmd\Phi=\nabla^2\Phi=\fr1\rho\partial_\rho\p{\rho\partial_\rho\Phi}+\fr1{\rho^2}\partial_\vphi^2\Phi$$ and the boundary conditions $$\partial_\rho\fc\Phi{\rho=R_\mrm{in},\vphi}=\partial_\rho\fc\Phi{\rho=R_\mrm{out},\vphi}=0,$$ i.e., eigenfunctions of the Laplacian on the annulus with homogeneous Neumann boundary conditions?

For any boundary condition with azimuthal symmetry, an easy separation of variable gives you solutions of the general form $$\fc\Phi{\rho,\vphi}=\p{\Xi^{\p J}\fc{J_m}{z\rho/R_\mrm{out}}+\Xi^{\p Y}\fc{Y_m}{z\rho/R_\mrm{out}}}\e^{\i m\vphi},$$ where $\Xi^{\p J}$, $\Xi^{\p Y}$, and $z$ are constants fixed by the boundary conditions, and $m$ is an integer and the azimuthal quantum number. The functions $J_m$ and $Y_m$ are Bessel functions of the first and second kind, respectively, which are two linearly independent solutions of the Bessel equation $$z^2\fc{y''}z+z\fc{y'}z+\p{z^2-m^2}\fc{y}z=0$$ $$(1)$$ with some particular normalization. Plugging the general form into the boundary condition, and we have $$\Xi^{\p J}\fc{J_m'}{z}+\Xi^{\p Y}\fc{Y_m'}{z}=0,\quad \Xi^{\p J}\fc{J_m'}{rz}+\Xi^{\p Y}\fc{Y_m'}{rz}=0,$$ where $r\ceq R_\mrm{in}/R_\mrm{out}$. This is regarded as a homogeneous linear system of equations for $\Xi^{\p J}$ and $\Xi^{\p Y}$. In order for it to have nontrivial solutions, the determinant of the coefficient matrix must vanish, which gives $$\fc gz\ceq\fc{J_m'}{rz}\fc{Y_m'}z-\fc{Y_m'}{rz}\fc{J_m'}z=0.$$ $$(2)$$ We can then assign a radial quantum number $n$ to the eigenfunctions by making the value of $z$ the $n$th root of $g$, which we may casually call $z_{mn}$.

We then have the final solution (up to normalization) $$\fc{\Phi_{mn}}{\rho,\vphi}=\p{\fc{Y_m'}{z_{mn}}\fc{J_m}{z_{mn}\rho/R_\mrm{out}} -\fc{J_m'}{z_{mn}}\fc{Y_m}{z_{mn}\rho/R_\mrm{out}}}\e^{\i m\vphi},$$ where $m$ is the azimuthal quantum number, and $n$ is the radial quantum number, and $z_{mn}$ is defined as the $n$th root of the function $g$. The eigenvalue is $\lmd_{mn}=-z_{mn}^2/R_\mrm{out}^2$. In order for the eigenfunctions to be complete, we need to allow $m$ to be any integer, but we will restrict ourselves to non-negative $m$ in the rest of the article because the negative ones are practically the same as the positive ones.

Distribution of eigenvalues

From the well-known asymptotic form of the Bessel functions (source) $$\fc{J_m}{z\to\infty}=\sqrt{\fr2{\pi z}}\fc\cos{z-\fr{m\pi}2-\fr\pi4},\quad \fc{Y_m}{z\to\infty}=\sqrt{\fr2{\pi z}}\fc\sin{z-\fr{m\pi}2-\fr\pi4},$$ we can easily conclude that the asymptotic form of $g$ is $$\fc g{z\to\infty}=\fr{2\fc\sin{\p{1-r}z}}{\pi\sqrt r\,z}.$$ This inspires us to define a function in companion with $g$ as $$\fc fz\ceq\fc{J_m'}{rz}\fc{J_m'}z+\fc{Y_m'}{rz}\fc{Y_m'}z,$$ $$(3)$$ which will have the asymptotic form $$\fc f{z\to\infty}=\fr{2\fc\cos{\p{1-r}z}}{\pi\sqrt r\,z}.$$ Therefore, $f\sim\cos\fc\tht z$ and $g\sim\sin\fc\tht z$ are like the cosine and sine pair of oscillatory functions, with a phase angle $\fc\tht{z\to\infty}=\p{1-r}z$ asymptotically directly proportional to $z$.

Here is a plot that shows how $\tht$ behaves:

The blue line is $\fc\arctan{\fc gz/\fc fz}$, and the orange line is $\arctan\fc\tan{\p{1-r}z}$. we can see that the two functions are asymptotically equal.

A good thing about this description is that we can now see that $z_{mn}$ as a root of $g$ is just the solution to the equation $\fc\tht z=n\pi$. In other words, $z_{mn}$ are just the points at which the blue line crosses the horizontal axis in the plot above. Therefore, if we can find the inverse function $\fc z\tht$ (maybe in a series expansion), we can directly get $z_{mn}=\fc z{n\pi}$. We can see from the plot that $\fc z\tht=\tht/\p{1-r}$ is probably not a good enough approximation. Therefore, we would like to seek higher order terms.

Kummer’s equation

For this part, we employ a method similar to a paper by Bremer.

First, for a general homogeneous second-order linear ODE in the form $y''+2\chi y'+\psi y=0$, where $\chi$ and $\psi$ are known functions of $z$, we can define $$y_\mrm n\ceq py,\quad p\ceq C\exp\int\chi\,\d z, \quad q\ceq p\psi-p'',$$ and the ODE will become a normal form $y_\mrm n''+qy_\mrm n=0$. Therefore, for any second-order linear ODE, we only need to consider those without the first-derivative term.

Then, one can prove that, for the ODE $y''+qy=0$, its two linearly independent solutions can be expressed as $$u=\fr{\cos\alp}{\sqrt{\alp'}},\quad v=\fr{\sin\alp}{\sqrt{\alp'}},$$ $$(4)$$ where $\alp$ satisfy Kummer’s equation $$\alp^{\prime2}+\fr{\alp'''}{2\alp'}-\fr{3\alp^{\prime\prime2}}{4\alp^{\prime2}}=q.$$ $$(5)$$ This may seem like converting a problem to a more complicated one, but the advantage is that $\alp$ is not oscillatory while $u$ and $v$ are oscillatory, which means that it would be easier to expand $\alp$ as a series for large $z$. Notice that $\alp'=1/\p{u^2+v^2}$, which provides a means of finding $\alp$ if the solutions of the original ODE are known.

Now, we can try to apply this to $J_m'$ and $Y_m'$. First, we need to find the second-order ODE that those two functions satisfy. We already know that $J_m$ and $Y_m$ are solutions to Equation 1, so it would be straightforward to derive that $J_m'$ and $Y_m'$ satisfy $$\fc{y''}z+\fr1z\fr{z^2-3m^2}{z^2-m^2}\fc{y'}z +\p{1-\fr{m^2+1}{z^2}-\fr{2m^2}{z^2-m^2}}\fc yz=0.$$ To reduce this into the normal form without the first-derivative term, define $$\fc pz\ceq\sqrt{\fr\pi2}\fr{z^{3/2}}{\sqrt{z^2-m^2}},\quad \fc qz\ceq1-\fr{4m^2-1}{4z^2}-\fr{2m^2+z^2}{\p{m^2-z^2}^2},$$ $$(6)$$ and we have $\fc pz\fc{J_m'}z$ and $\fc pz\fc {Y_m'}z$ satisfy $\fc{y''}z+\fc qz\fc yz=0$. Equation 4 then gives $$\fc{Y_m'}z=\fr{\cos\fc\alp z}{\fc pz\sqrt{\fc{\alp'}z}},\quad -\fc{J_m'}z=\fr{\sin\fc\alp z}{\fc pz\sqrt{\fc{\alp'}z}}$$ $$(7)$$ (the normalization of $p$ and the initial value of $\alp$ are chosen to make sure they are correct). You may worry that $\fc p{z<m}$ is imaginary, but $\fc{\alp'}z$ will be negative in that region so that everything works out and is real in the end. Substitute Equation 7 into Equation 3 and 2, and we have $$\fc fz=\fr{\cos\fc\tht z}{\fc p{rz}\fc pz\sqrt{\fc{\alp'}{rz}\fc{\alp'}z}},\quad \fc gz=\fr{\sin\fc\tht z}{\fc p{rz}\fc pz\sqrt{\fc{\alp'}{rz}\fc{\alp'}z}},$$ where $\fc\tht z\ceq\fc\alp z-\fc\alp{rz}$. By this, we have a workable expression for $\fc\tht z$ so that we now just need to turn our attention to $\fc\alp z$.

We can substitute $\fc qz$ in Equation 6 into Equation 5 and expand on both sides as a series of $z$ at infinity and solve for the series coefficients to get $$\fc{\alp'}z=1+\fr{-4m^2-3}{8z^2}+\fr{-16m^4-184m^2+63}{128z^4}+\cdots,$$ $$(8)$$ which has a certain convergence radius which is not important at this stage.

Asymptotic expansion

Here we will work out another way of expanding $\fc\alp z$ as a series based on properties of the Bessel functions. From Equation 7, we have $$\fr1{\fc{\alp'}z}=\fc pz^2\p{\fc{J_m'}z^2+\fc{Y_m'}z^2}.$$ If we were working with $\fc{J_m}z^2+\fc{Y_m}z^2$ instead, we would be able to employ the handy Nicholson’s integral, but the same method cannot be applied here.

We can, however, use the asymptotic expansion of the Bessel functions directly (source): $$\begin{align*} \fc{Y_m'}z&=\sqrt{\fr2{\pi z}}\p{\fc\cos{z-\fr{m\pi}2-\fr\pi4} \sum_{k=0}^\infty\fr{\p{-1}^kc_{2k}}{z^{2k}} -\fc\sin{z-\fr{m\pi}2-\fr\pi4}\sum_{k=0}^\infty\fr{\p{-1}^kc_{2k+1}}{z^{2k+1}}},\\ -\fc{J_m'}z&=\sqrt{\fr2{\pi z}}\p{\fc\cos{z-\fr{m\pi}2-\fr\pi4} \sum_{k=0}^\infty\fr{\p{-1}^kc_{2k+1}}{z^{2k+1}} +\fc\sin{z-\fr{m\pi}2-\fr\pi4}\sum_{k=0}^\infty\fr{\p{-1}^kc_{2k}}{z^{2k}}}, \end{align*}$$ where $$c_k\ceq a_k+\p{k-\fr12}a_{k-1},\quad a_k\ceq\p{-1}^k\fr{\p{1/2-m}^{\overline k}\p{1/2+m}^{\overline k}}{2^kk!},$$ where the raising factorial notation is used. By combining them, we get $$\fc{J_m'}z^2+\fc{Y_m'}z^2=\fr2\pi\sum_{k=0}^\infty\fr{r_k}{z^{2k+1}},\quad r_k\ceq\p{-1}^k\sum_{l=0}^{2k}\p{-1}^lc_lc_{2k-l}.$$