Deriving the ODE and solution with an arbitrary factor ordering

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Nakayama's Derivation

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Derivation of the ODE

Starting from

Taking the energy $E=0$, promoting $\Pi_l \to -i \hbar \frac{d}{d l}$, taking an arbitrary factor ordering, and multiplying by a $-1$ we have

(1)

\begin{align} \hbar^{2} l^{i} \frac{d}{d l} l^{j} \frac{d}{d l} i^{k} \Psi - [\left( m+ \frac{1}{2} \right)^2 l^{-1} + \lambda l] \Psi = 0 \end{align}

where $\Psi \equiv \Psi \left( l \right)$. The momentum term is also self-adjoint in $L^2 \left( \mathbb{R}^+, l^{k-i} dl \right)$ as it will be obvious later. Defining $V \equiv - [\left( m+ \frac{1}{2} \right)^2 l^{-1} + \lambda l]$ with $i+j+k=1$, $c \equiv i-k$, $i-c-k=0$, and $j=1-c-2 k$ this is:

(2)

\begin{align} 0 & = \hbar^{2} l^{i} \frac{d}{d l} l^{j} \frac{d}{d l} l^{k} \Psi + V \Psi \\ & = \hbar^{2} l^{i} \frac{d}{d l} l^{j} \left( k l^{k-1} \Psi + l^{k} \Psi^{\prime} \right) + V \Psi \\ & = \hbar^{2} l^{i} \frac{d}{d l} \left( k l^{j+k-1} \Psi + l^{j+k} \Psi^{\prime} \right) + V \Psi \\ & = \hbar^{2} l^{i} \frac{d}{d l} \left( k l^{1-c-k-1} \Psi + l^{1-c-k} \Psi^{\prime} \right) + V \Psi\\ & = \hbar^{2} l^{i} \frac{d}{d l} \left( k l^{-c-k} \Psi + l^{1-c-k} \Psi^{\prime} \right) + V \Psi \\ & = \hbar^{2} \left( k \left( -c-k \right) l^{i-c-k-1} \Psi + \left( k+1-c-k \right) l^{i-c-k} \Psi^{\prime} +l^{1+i-c-k} \Psi^{\prime \prime} \right) + V \Psi \\ & = \hbar^{2} \left( k \left( -c-k \right) l^{-1} \Psi + \left( 1-c \right) \Psi^{\prime} + l \Psi^{\prime \prime} \right) + V \Psi \end{align}

But we want 2 to be self-adjoint in $L^2 \left( \mathbb{R}^+, l^{-1/2} dl \right)$ of the form $l^{m} \left( \hbar^{2} l^{1/2} \frac{d}{d l} l^{1/2} \frac{d}{d l} l^{-m} \Psi +V_{QP} l^{-m} \Psi \right)$ where the conjugated operators $l^{m}$ and $l^{-m}$ go from the metric that is imposed upon 2 to $L^2 \left( \mathbb{R}^+, l^{-1/2} dl \right)$ . So, if we ignore the potential V for now, we assume the operator is of this form:

(3)

\begin{align} & \ \ \ \ \hbar^{2} l^{ \left( - \frac{1}{4} + \frac{c}{2} \right) } \left( l^{1/2} \frac{d}{d l} l^{1/2} \frac{d}{d l} l^{ \left( \frac{1}{4} - \frac{c}{2} \right) } \Psi \right) \\ & = \hbar^{2} l^{ \left( - \frac{1}{4} + \frac{c}{2} \right) + 1/2 } \left( \frac{d}{d l} l^{1/2} \frac{d}{d l} l^{ \left( \frac{1}{4} - \frac{c}{2} \right) } \Psi \right) \\ & = \hbar^{2} l^{ \left( - \frac{1}{4} + \frac{c}{2} \right) + 1/2 } \left( \frac{d}{d l} l^{1/2} \left( \left( \frac{1}{4} - \frac{c}{2} \right) l^{ \left( \frac{1}{4} - \frac{c}{2} \right) -1 } \Psi + l^{ \left( \frac{1}{4} - \frac{c}{2} \right) } \Psi^{\prime} \right) \right) \\ & = \hbar^{2} l^{ \left( - \frac{1}{4} + \frac{c}{2} \right) + 1/2 } \left( \frac{d}{d l} \left( \left( \frac{1}{4} - \frac{c}{2} \right) l^{ \left( \frac{1}{4} - \frac{c}{2} \right) -1 + 1/2 } \Psi + l^{ \left( \frac{1}{4} - \frac{c}{2} \right) +1/2 } \Psi^{\prime} \right) \right) \\ & = \hbar^{2} l^{ \left( - \frac{1}{4} + \frac{c}{2} \right) + 1/2 } \left( \left( \frac{1}{4} - \frac{c}{2} \right) \left( \left( \frac{1}{4} - \frac{c}{2} \right) -1 + 1/2 \right) l^{ \left( \frac{1}{4} - \frac{c}{2} \right) -2 + 1/2 } \Psi + \left( \frac{1}{4} - \frac{c}{2} \right) l^{ \left( \frac{1}{4} - \frac{c}{2} \right) -1 + 1/2 } \Psi^{ \prime } \\ & \ \ \ \ \ \ \ \ + \left( \frac{1}{4} - \frac{c}{2} + 1/2 \right) l^{ \left( \frac{1}{4} - \frac{c}{2} \right) -1 + 1/2 } \Psi^{\prime} + l^{ \left( \frac{1}{4} - \frac{c}{2} \right) +1/2 } \Psi^{\prime \prime} \right) \\ & = \hbar^{2} \left( \left( \frac{1}{4} - \frac{c}{2} \right) \left( \frac{-1}{4} - \frac{c}{2} \right) l^{-1} \Psi + \left( 1-c \right) \Psi^{\prime} + l \Psi^{\prime \prime} \right) \end{align}

Knowing that we want this form and seeing the first two derivatives' coefficients are the same, we add and subtract $\hbar^{2} \left[ \left( \frac{1}{4} - \frac{c}{2} \right) \left( \frac{-1}{4} - \frac{c}{2} \right) l^{-1} \right]$ onto the result of Eq.(2). We obtain:

(4)

\begin{align} \begin{align} 0 & = \hbar^{2} \left( k \left( -c-k \right) l^{-1} \Psi + \left( 1-c \right) \Psi^{\prime} + l \Psi^{\prime \prime} \right) + V \Psi + \hbar^{2} \left[ \left( \frac{1}{4} - \frac{c}{2} \right) \left( \frac{-1}{4} - \frac{c}{2} \right) l^{-1} \right] \Psi - \hbar^{2} \left[ \left( \frac{1}{4} - \frac{c}{2} \right) \left( \frac{-1}{4} - \frac{c}{2} \right) l^{-1} \right] \Psi \\ & = \hbar^{2} \left( \left[ \left( \frac{1}{4} - \frac{c}{2} \right) \left( \frac{-1}{4} - \frac{c}{2} \right) l^{-1} \right] \Psi + \left( 1-c \right) \Psi^{\prime} + l \Psi^{\prime \prime} \right) + V \Psi + \hbar^{2} k \left( -c-k \right) l^{-1} \Psi - \hbar^{2} \left[ \left( \frac{1}{4} - \frac{c}{2} \right) \left( \frac{-1}{4} - \frac{c}{2} \right) l^{-1} \right] \Psi \\ & = \hbar^{2} l^{ \left( - \frac{1}{4} + \frac{c}{2} \right) } \left( l^{1/2} \frac{d}{d l} l^{1/2} \frac{d}{d l} l^{ \left( \frac{1}{4} - \frac{c}{2} \right) } \Psi \right) + V \Psi + \hbar^{2} k \left( -c-k \right) l^{-1} \Psi - \hbar^{2} \left[ \left( \frac{1}{4} - \frac{c}{2} \right) \left( \frac{-1}{4} - \frac{c}{2} \right) l^{-1} \right] \Psi \end{align}

We pull out $l^{ \left( \frac{1}{4} - \frac{c}{2} \right) }$ to obtain the differential operator on $\Psi$

(5)

\begin{align} 0 = l^{ \left( - \frac{1}{4} + \frac{c}{2} \right) } \left[ \hbar^{2} l^{1/2} \frac{d}{d l} l^{1/2} \frac{d}{d l} + V \Psi + \hbar^{2} k \left( -c-k \right) l^{-1} - \hbar^{2} \left( \frac{1}{4} - \frac{c}{2} \right) \left( \frac{-1}{4} - \frac{c}{2} \right) l^{-1} \right] l^{ \left( \frac{1}{4} - \frac{c}{2} \right) } \Psi. \end{align} \end{align}

To be clear about what is going on here. If we were to multiply back in the $l^{ \left( \frac{1}{4} - \frac{c}{2} \right) }$ on the right, it would be operated on by the differential operators in the first term. We could in fact pull out the $\Psi$ out of the equation under the same assumption. Also, since $l^{ \left( - \frac{1}{4} + \frac{c}{2} \right) }$ and $l^{ \left( \frac{1}{4} - \frac{c}{2} \right) }$ are conjugations by a multiplication operator that goes between $L^{2} \left( \mathbb{R}^{+}, l^{k-i} dl \right)$ and $L^{2} \left( \mathbb{R}^{+}, l^{-1/2} dl \right)$, they can be dropped with the understanding that we are working in their metric. This finally yields our differential equation

(6)

\begin{align} \hbar^{2} l^{1/2} \frac{d}{d l} \left( l^{1/2} \frac{d}{d l} \Psi \right) - \left( \left( m+ \frac{1}{2} \right)^2 l^{-1} + \lambda l - \hbar^{2} \left( \frac{1}{4} - \frac{c}{2} \right) \left( \frac{-1}{4} - \frac{c}{2} \right) l^{-1} + \hbar^{2} k \left( -c-k \right) l^{-1} \right) \Psi = 0. \end{align}

Now to solve it.

Solving and a solution

First, we take and group the terms:

(7)

\begin{align} \hbar^{2} l^{1/2} \frac{d}{d l} \left( l^{1/2} \frac{d}{d l} \Psi \right) - \left( \left[ \left( m+ \frac{1}{2} \right)^2 - \hbar^{2} \left( \frac{1}{4} - \frac{c}{2} \right) \left( \frac{-1}{4} - \frac{c}{2} \right) + \hbar^{2} k \left( -c-k \right) \right] l^{-1} + \lambda l \right) \Psi = 0. \end{align}

Next, define $a = \left[ \left( m+ \frac{1}{2} \right)^2 - \hbar^{2} \left( \frac{1}{4} - \frac{c}{2} \right) \left( \frac{-1}{4} - \frac{c}{2} \right) + \hbar^{2} k \left( -c-k \right) \right]$ so this becomes

(8)

\begin{align} l \Psi^{\prime \prime} + \frac{1}{2} \Psi^{\prime} - \hbar^{-2} \left[ l^{-1} a + \lambda l \right] \Psi = 0. \end{align}

We now rescale $z = \alpha 2 \sqrt{\lambda} l$ and $\Phi (z) = e^{-\beta z} z^{-\gamma} \Psi (l)$ where we solve for $\alpha$, $\beta$, and $\gamma$ to yield a similar result to Nakayama's. With these substitutions, 8 becomes

(9)

\begin{align} z \Phi^{\prime \prime} + \Phi^{\prime} \left[ 2 \gamma + \frac{1}{2} +2 \beta z \right] + \Phi \left[ z^{-1} \left[ \gamma \left( \gamma -1 \right) + \frac{\gamma}{2} - \frac{a}{\hbar^{2}} \right] + z \left[ \beta^2 - \frac{1}{ \alpha^2 \hbar^2} \right] + 2 \gamma \beta + \frac{\beta}{2} \right] =0. \end{align}

To solve for $\beta$, we notice in Nakayama's solution and in the Associated Laguerre ODE, that in $\Phi^{\prime}$ the $z$ must be multiplied by a $-1$. So then setting $2 \beta = -1$ gets us $\beta = -\frac{1}{2}$. For $\alpha$, we set $\beta^2 - \frac{1}{ \alpha^2 \hbar^2} = \frac{1}{4} - \frac{1}{ \alpha^2 \hbar^2}=0$ to get us $\alpha = \frac{2}{\hbar}$. So our ODE is now

(10)

\begin{align} z \Phi^{\prime \prime} + \Phi^{\prime} \left[ 2 \gamma + \frac{1}{2} - z \right] + \Phi \left[ z^{-1} \left[ \gamma \left( \gamma -1 \right) + \frac{\gamma}{2} - \frac{a}{\hbar^{2}} \right] - \gamma - \frac{1}{4} \right] =0. \end{align}

Now solving for $\gamma$ we notice further that there is no $z^{-1}$ term, so setting that to 0 yields $\gamma \left( \gamma -1 \right) + \frac{\gamma}{2} - \frac{a}{\hbar^{2}}=0$ which when solving for $\gamma$ gets us $\gamma = \frac{1}{4} \pm \frac{1}{2} \sqrt{ \frac{1}{4} + 4 \frac{a}{\hbar^2}}$. This then gives us our final ODE

(11)

\begin{align} z \Phi^{\prime \prime} + \Phi^{\prime} \left[ \pm \sqrt{ \frac{1}{4} + 4 \frac{a}{\hbar^2}} + 1 - z \right] + \Phi \left[ \mp \frac{1}{2} \sqrt{ \frac{1}{4} + 4 \frac{a}{\hbar^2}} - \frac{1}{2} \right] =0. \end{align}

This is again similar to the Associated Laguerre ODE so in a manner similar to Nakayama's solution we get the solution

(12)

\begin{align} \Phi (z) = \sqrt{\frac{n!}{\left( n+k \right)!}} L_n^k (z). \end{align}

where $n = \left[ \mp \frac{1}{2} \sqrt{ \frac{1}{4} + 4 \frac{a}{\hbar^2}} - \frac{1}{2} \right]$ and $k = \pm \sqrt{ \frac{1}{4} + 4 \frac{a}{\hbar^2}}$. This gives us then the full solution of

(13)

\begin{align} \Phi (z) = \sqrt{\frac{\Gamma \left(n -1 \left) }{\Gamma \left( n+k - 1} \right)}} e^{-\frac{\sqrt{\lambda}}{\hbar} l} \left( \frac{2\sqrt{\lambda}}{\hbar} l \right)^{ \frac{1}{4} \pm \frac{1}{2} \sqrt{ \frac{1}{4} + 4 \frac{a}{\hbar^2}}} L_{n}^{k} (\frac{2 \sqrt{\lambda}}{\hbar} l) \end{align}

with $\Gamma (n)$ is the Gamma Function and $L_n^k (z)$ are the generalized Laguerre polynomials.

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