Chapter 48The Schrödinger Equation for the Hydrogen Atom (Spherical Coordinates)

Derivation: Decomposition into three ordinary differential equations

Substituting $\psi=R\Theta\bar\Phi$ into Eq. \eqref{eq:48-well-pde}, the derivatives involving only $r$ act on $R$, the derivatives involving only $\theta$ act on $\Theta$, and the derivatives involving only $\varphi$ act on $\bar\Phi$ (within a product, the factors other than the one being differentiated can be pulled out as constants).

$$ \frac{\Theta\bar\Phi}{r^2}\diff{}{r}\!\left(r^2\diff{R}{r}\right)+\frac{R\bar\Phi}{r^2\sin\theta}\diff{}{\theta}\!\left(\sin\theta\diff{\Theta}{\theta}\right)+\frac{R\Theta}{r^2\sin^2\theta}\diff{^2\bar\Phi}{\varphi^2}+\frac{2m_e}{\hbar^2}ER\Theta\bar\Phi=0 $$

Dividing both sides by $R\Theta\bar\Phi$ (we may assume this is $\neq0$) and then multiplying everything by $r^2$, we obtain

$$ \frac{1}{R}\diff{}{r}\!\left(r^2\diff{R}{r}\right)+\frac{1}{\Theta\sin\theta}\diff{}{\theta}\!\left(\sin\theta\diff{\Theta}{\theta}\right)+\frac{1}{\bar\Phi\sin^2\theta}\diff{^2\bar\Phi}{\varphi^2}+\frac{2m_e}{\hbar^2}Er^2=0 $$

The first term and part of the fourth term ($\dfrac{2m_e}{\hbar^2}Er^2$) are expressions in $r$ alone, the second term is an expression in $\theta$ alone, and the third term is an expression in $\varphi$ alone. So we collect the parts involving $r$ on the left-hand side and move the parts involving $\theta,\varphi$ to the right-hand side:

$$ \frac{1}{R}\diff{}{r}\!\left(r^2\diff{R}{r}\right)+\frac{2m_e}{\hbar^2}Er^2=-\frac{1}{\Theta\sin\theta}\diff{}{\theta}\!\left(\sin\theta\diff{\Theta}{\theta}\right)-\frac{1}{\bar\Phi\sin^2\theta}\diff{^2\bar\Phi}{\varphi^2} $$

Now the left-hand side is a function of $r$ only and the right-hand side is a function of $\theta,\varphi$ only. Since the two sides, whose variables are completely independent of each other, are nevertheless equal, both sides must equal the same constant, which depends on none of $r,\theta,\varphi$. For a reason we will see later, we write this constant as $\ell(\ell+1)$.

Derivation: Frobenius series solution of the radial equation

Eq. \eqref{eq:48-well-radial} has a regular singular point at $r=0$ (Frobenius theory, Chapter 39), so we assume a series solution of the form $R(r)=\displaystyle\sum_{k=0}^{\infty}a_kr^{k+\lambda}$ ($a_0\neq0$). Substituting gives

$$ r^2\sum_{k=0}^{\infty}(k+\lambda)(k+\lambda-1)a_kr^{k+\lambda-2}+2r\sum_{k=0}^{\infty}(k+\lambda)a_kr^{k+\lambda-1}+\left(\varepsilon r^2-\ell(\ell+1)\right)\sum_{k=0}^{\infty}a_kr^{k+\lambda}=0 $$

When the exponents of the first and second terms are aligned to $r^{k+\lambda}$, their degrees drop by 2, while the degree of the third term, $\varepsilon r^2(\cdots)$, rises by 2. Separating the lowest-degree terms (those with $k=0$ and $k=1$, the part not containing $\varepsilon$) from the remaining terms (those containing $\varepsilon$, which come in pairs of terms two steps apart), we obtain

$$ \{\lambda(\lambda+1)-\ell(\ell+1)\}a_0r^{\lambda}+\{(\lambda+1)(\lambda+2)-\ell(\ell+1)\}a_1r^{\lambda+1} +\sum_{k=0}^{\infty}\Big[\{(k+\lambda+2)(k+\lambda+3)-\ell(\ell+1)\}a_{k+2}+\varepsilon a_k\Big]r^{k+\lambda+2}=0 $$

(The first and second terms are the expressions of the form $r^{k+\lambda-2}(k+\lambda)(k+\lambda+1)-\ell(\ell+1)r^{k+\lambda}$ written out for $k=0,1$—note that $k(k-1)+2k=k(k+1)$—and the terms with $k\ge2$ were renumbered by $k\to k+2$ and collected into the sum in the third term.) Since the coefficient of each of the powers $r^\lambda,r^{\lambda+1},r^{k+\lambda+2}$ must be $0$,

$$ \begin{cases} \{\lambda(\lambda+1)-\ell(\ell+1)\}a_0=0\\[4pt] \{(\lambda+1)(\lambda+2)-\ell(\ell+1)\}a_1=0\\[4pt] a_{k+2}=-\dfrac{\varepsilon\,a_k}{(k+2)(k+3+2\ell)}\qquad(k=0,1,2,\dots) \end{cases} $$

(The last recurrence relation is the result of substituting $\lambda=\ell$ into $(k+\lambda+2)(k+\lambda+3)-\ell(\ell+1)$ and factoring: $(k+\ell+2)(k+\ell+3)-\ell(\ell+1)=(k+2)(k+2\ell+3)+\ell(\ell+1)-\ell(\ell+1)=(k+2)(k+3+2\ell)$—expanding to check, we get $(k+\ell+2)(k+\ell+3)=k^2+k(2\ell+5)+(\ell+2)(\ell+3)$, and subtracting $\ell(\ell+1)$ gives $k^2+k(2\ell+5)+2\ell+6=(k+2)(k+2\ell+3)$, which agrees). Since $a_0\neq0$, the first equation gives $\lambda(\lambda+1)=\ell(\ell+1)$, that is, $\lambda=\ell$ or $\lambda=-(\ell+1)$ (solve the quadratic equation $\lambda^2+\lambda-\ell(\ell+1)=0$ by the quadratic formula or by the factorization $(\lambda-\ell)(\lambda+\ell+1)=0$). If we choose $\lambda=-(\ell+1)$ (always negative, since $\ell\ge0$), then $R(r)\sim a_0r^{\lambda}$ ($r\to0$) diverges at the origin, which is unacceptable for a wave function. Therefore we choose

$$ \lambda=\ell $$

Next, we set $a_1=0$ (when $\lambda=\ell$, the coefficient in the second equation is $(\ell+1)(\ell+2)-\ell(\ell+1)=2(\ell+1)\neq0$, so $a_1=0$ is required—if we took $a_1\ne0$, we would get a contradiction of the form (an expression whose coefficient is not $0$) $=0$). Then, by the recurrence relation, $a_1=a_3=a_5=\cdots=0$ (because every odd-indexed coefficient is determined as a constant multiple of the preceding odd-indexed coefficient), and only the even-indexed coefficients remain. Substituting $k=0,2,4,\dots$ in turn gives

$$ a_2=-\frac{\varepsilon}{2(3+2\ell)}a_0,\qquad a_4=-\frac{\varepsilon}{4(5+2\ell)}a_2=\frac{\varepsilon^2}{2\cdot4\cdot(3+2\ell)(5+2\ell)}a_0,\qquad a_6=-\frac{\varepsilon}{6(7+2\ell)}a_4=-\frac{\varepsilon^3}{2\cdot4\cdot6\cdot(3+2\ell)(5+2\ell)(7+2\ell)}a_0 $$

and a pattern can be seen. In general, applying the recurrence relation $n$ times produces the sign $(-1)^n$, and the denominator contains $2\cdot4\cdots(2n)=2^nn!$ and $(3+2\ell)(5+2\ell)\cdots(2n+1+2\ell)$. By repeatedly using the property of the gamma function $\Gamma(z+1)=z\Gamma(z)$ (Chapter 32), the latter product can be written as $\Gamma\!\left(n+\ell+\tfrac32\right)\big/\Gamma\!\left(\ell+\tfrac32\right)\cdot2^n$ (the arguments increase in half-integer steps, so we divide by $2^n$ to make things come out right), and in the end

$$ a_{2n}=\frac{(-1)^n\varepsilon^n}{2^{2n}n!\,\Gamma\!\left(n+\ell+\tfrac32\right)}\cdot\Gamma\!\left(\ell+\tfrac32\right)a_0 $$

(we may regard $\Gamma\!\left(\ell+\tfrac32\right)a_0$ anew as a constant and absorb it into $a_0$). Therefore we obtain

$$ R(r)=a_0\sum_{k=0}^{\infty}\frac{(-1)^k\varepsilon^k}{2^{2k}k!\,\Gamma\!\left(k+\ell+\tfrac32\right)}r^{2k+\ell} $$

Finally, let us fit this to the form of the standard series for the Bessel function (Chapter 41),

$$ J_\nu(x)=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!\,\Gamma(k+\nu+1)}\left(\frac{x}{2}\right)^{2k+\nu} $$

Setting $\nu=\ell+\tfrac12$ and $x=\sqrt\varepsilon\,r$, we have $\left(\dfrac{x}{2}\right)^{2k+\nu}=\left(\dfrac{\sqrt\varepsilon}{2}\right)^{2k+\ell+\frac12}r^{2k+\ell+\frac12}$, and compared with the general term $r^{2k+\ell}$ of $R(r)$ this is off by an extra factor $r^{1/2}\left(\sqrt\varepsilon/2\right)^{2k+\ell+1/2}$. So we choose the constant $a_0$ of $R(r)$ anew so as to cancel this discrepancy exactly (this is a replacement of the constant, $a_0\to A\cdot2^{\ell+1/2}\varepsilon^{-(\ell+1/2)/2}$, which does not depend on $k$), and obtain

\begin{equation} R(r)=\frac{A}{\sqrt r}\sum_{k=0}^{\infty}\frac{(-1)^k}{k!\,\Gamma\!\left(k+\ell+\tfrac32\right)}\left(\frac{\sqrt\varepsilon}{2}r\right)^{2k+\ell+\frac12}=\frac{A}{\sqrt r}\,J_{\ell+\frac12}(\sqrt\varepsilon\,r) \label{eq:48-well-R-final} \end{equation}

That is, the radial wave function inside the infinite spherical well potential is expressed by the Bessel function of half-integer order $J_{\ell+1/2}$. Since the order $\nu=\ell+\tfrac12$ is not an integer, this is a situation where the gamma function $\Gamma$ defined in Chapter 41 (a function extending the factorial $n!$ to non-integer arguments) is needed as it stands.

Derivation: Substituting $\psi=RY$ and the product rule

Substituting, we obtain

$$ \pdiff{}{r}\!\left(r^2\diff{R}{r}Y\right)+\frac{1}{\sin\theta}\pdiff{}{\theta}\!\left(\sin\theta R\pdiff Y\theta\right)+\frac1{\sin^2\theta}R\pdiff{^2Y}{\varphi^2}+\frac{2m_e}{\hbar^2}(E-V)r^2RY=0 $$

The first and second terms can be expanded with the product rule $(fg)'=f'g+fg'$. For the first term, keeping $Y$ outside as a constant independent of $r$, we use $\dfrac{\partial}{\partial r}\!\left(r^2\dfrac{dR}{dr}\right)=2r\dfrac{dR}{dr}+r^2\dfrac{d^2R}{dr^2}$ (the product rule with $f=r^2$, $g=dR/dr$), which gives $\left(2r\dfrac{dR}{dr}+r^2\dfrac{d^2R}{dr^2}\right)Y$. Similarly for the second term, treating $R$ as a constant independent of $\theta$ and using $\dfrac{\partial}{\partial\theta}\!\left(\sin\theta\dfrac{\partial Y}{\partial\theta}\right)=\cos\theta\dfrac{\partial Y}{\partial\theta}+\sin\theta\dfrac{\partial^2Y}{\partial\theta^2}$, we get $R\left(\cos\theta\pdiff Y\theta+\sin\theta\pdiff{^2Y}{\theta^2}\right)$. Putting these together, we obtain

$$ \left(r^2\diff{^2R}{r^2}+2r\diff Rr\right)Y+\frac R{\sin\theta}\!\left(\sin\theta\pdiff{^2Y}{\theta^2}+\cos\theta\pdiff Y\theta\right)+\frac1{\sin^2\theta}R\pdiff{^2Y}{\varphi^2}+\frac{2m_e}{\hbar^2}(E-V)r^2RY=0 $$

Dividing this by $\psi=RY$ ($\neq0$) and substituting $V$ from Eq. \eqref{eq:48-hatom-V}, we get

\begin{equation} \frac1R\!\left(r^2\diff{^2R}{r^2}+2r\diff Rr\right)+\frac1Y\!\left(\pdiff{^2Y}{\theta^2}+\frac1{\tan\theta}\pdiff Y\theta\right)+\frac1{\sin^2\theta}\frac1Y\pdiff{^2Y}{\varphi^2}+\frac{2m_e}{\hbar^2}\!\left(E+\frac{1}{4\pi\varepsilon_0}\frac{e_0^2}{r}\right)r^2=0 \label{eq:48-hatom-sep} \end{equation}

(We rewrote $\dfrac{\cos\theta}{\sin\theta}=\dfrac1{\tan\theta}$. Note also that, since $-V=+\dfrac1{4\pi\varepsilon_0}\dfrac{e_0^2}{r}$, the sign becomes plus.)

Derivation: The substitution $R(r)=S(r)/r$

Let $R(r)=\dfrac{S(r)}{r}$. Using the quotient rule (or the product rule with $R=S\cdot r^{-1}$), we get

$$ \diff Rr=-\frac{S}{r^2}+\frac1r\diff Sr,\qquad \diff{^2R}{r^2}=\frac{2S}{r^3}-\frac2{r^2}\diff Sr+\frac1r\diff{^2S}{r^2} $$

(We only need to differentiate the first derivative once more: $\diff{}{r}\!\left(-\dfrac S{r^2}+\dfrac1r\diff Sr\right)=\dfrac{2S}{r^3}-\dfrac1{r^2}\diff Sr-\dfrac1{r^2}\diff Sr+\dfrac1r\diff{^2S}{r^2}=\dfrac{2S}{r^3}-\dfrac2{r^2}\diff Sr+\dfrac1r\diff{^2S}{r^2}$.) Using these, we obtain

\begin{equation} r^2\diff{^2R}{r^2}+2r\diff Rr =\left(\frac{2S}{r}-2\diff Sr+r\diff{^2S}{r^2}\right)+\left(-\frac{2S}{r}+2\diff Sr\right) =r\diff{^2S}{r^2} \label{eq:48-RS-identity} \end{equation}

(The two terms $\dfrac{2S}{r}$ cancel each other, and so do the two terms $2\dfrac{dS}{dr}$, leaving only $r\dfrac{d^2S}{dr^2}$—this has also been verified with SymPy; see the verification at the end of the book). Therefore, replacing the first two terms on the left-hand side of Eq. \eqref{eq:48-hatom-radial} with the left-hand side of Eq. \eqref{eq:48-RS-identity}, we get

$$ r\diff{^2S}{r^2}+\left\{\frac{2m_e}{\hbar^2}\!\left(E+\frac1{4\pi\varepsilon_0}\frac{e_0^2}{r}\right)r^2-\ell(\ell+1)\right\}\frac Sr=0 $$

(Here $R=S/r$ was substituted into the braces on the right as well.) Expanding the expression in the braces, we have $\dfrac{2m_e}{\hbar^2}\!\left(E+\dfrac{e_0^2}{4\pi\varepsilon_0 r}\right)r^2=\dfrac{2m_eE}{\hbar^2}r^2+\dfrac{2m_ee_0^2}{4\pi\varepsilon_0\hbar^2}r$ (note that $\dfrac1r\cdot r^2=r$), so that

$$ r\diff{^2S}{r^2}+\left\{\frac{2m_eE}{\hbar^2}r^2+\frac{2m_ee_0^2}{4\pi\varepsilon_0\hbar^2}r-\ell(\ell+1)\right\}\frac Sr=0 $$

Finally, dividing both sides by $r$ (for the left term, $r\diff{^2S}{r^2}\big/r=\diff{^2S}{r^2}$; for the right term, the denominator gains one more factor of $r$ and becomes $r^2$), we get

\begin{equation} \diff{^2S}{r^2}+\left\{\frac{2m_eE}{\hbar^2}+\frac{m_ee_0^2}{2\pi\varepsilon_0\hbar^2}\cdot\frac1r-\frac{\ell(\ell+1)}{r^2}\right\}S=0 \label{eq:48-S-eq} \end{equation}

(We simplified by canceling a common factor: $\dfrac{2m_ee_0^2}{4\pi\varepsilon_0\hbar^2}\big/r=\dfrac{m_ee_0^2}{2\pi\varepsilon_0\hbar^2}\cdot\dfrac1r$.)

Derivation: Asymptotic behavior as $r\to\infty$

When $r$ is extremely large, both $\dfrac\tau r$ and $\dfrac{\ell(\ell+1)}{r^2}$ approach $0$ and become negligible ($1/r,1/r^2\to0$). Therefore, as $r\to\infty$ the equation becomes approximately

$$ \diff{^2S}{r^2}-K^2S=0 $$

This is the second-order homogeneous equation with constant coefficients that we studied in Chapter 35, and the roots of its characteristic equation $\mu^2-K^2=0$ are $\mu=\pm K$ (real), so the general solution is $S(r)=Ae^{Kr}+Be^{-Kr}$. The wave function must not diverge at infinity (for the normalization $\int|\psi|^2r^2\dd r=1$ to be finite, $\psi$, hence $R=S/r$, hence $S$ itself must approach $0$ as $r\to\infty$), so the exponentially growing term $Ae^{Kr}$ is not allowed. Thus we must have $A=0$, and

$$ S(r)\fallingdotseq Be^{-Kr}\qquad(r\ \text{sufficiently large}) $$

is the asymptotic form we obtain.

Derivation: Substituting $S=Xe^{-Kr}$

Using the product rule for differentiation twice (the second derivative of $S=Xe^{-Kr}$), we obtain

$$ \diff{^2S}{r^2}=\diff{^2X}{r^2}e^{-Kr}-2K\diff Xre^{-Kr}+K^2Xe^{-Kr} $$

(The first derivative is $\diff Sr=\diff Xre^{-Kr}-KXe^{-Kr}$; differentiating it once more gives $\diff{^2X}{r^2}e^{-Kr}-K\diff Xre^{-Kr}-K\diff Xre^{-Kr}+K^2Xe^{-Kr}$, and since $-K\diff Xre^{-Kr}$ appears twice, they combine into $-2K\diff Xre^{-Kr}$.) Substituting this into Eq. \eqref{eq:48-S-final}, we obtain

$$ \left(\diff{^2X}{r^2}-2K\diff Xr+K^2X\right)e^{-Kr}+\left\{-K^2X+\frac\tau rX-\frac{\ell(\ell+1)}{r^2}X\right\}e^{-Kr}=0 $$

(these are the terms obtained by substituting $S=Xe^{-Kr}$ just as it stands). The two $K^2X$ terms cancel each other exactly; dividing by the common factor $e^{-Kr}\,(\neq0)$, we obtain

\begin{equation} \diff{^2X}{r^2}-2K\diff Xr+\left\{\frac\tau r-\frac{\ell(\ell+1)}{r^2}\right\}X=0 \label{eq:48-X-eq} \end{equation}

This is the result (verified by actual substitution with SymPy—see the verification at the end of the book).

Derivation: Rewriting in terms of the dimensionless variable $\alpha=2Kr$

By the chain rule, $\diff Xr=\diff X\alpha\cdot\diff\alpha r=2K\diff X\alpha$ and $\diff{^2X}{r^2}=\diff{}{r}\!\left(2K\diff X\alpha\right)=2K\diff{}{\alpha}\!\left(\diff X\alpha\right)\cdot\diff\alpha r=(2K)^2\diff{^2X}{\alpha^2}$. Also, since $r=\alpha/(2K)$, we have $1/r=2K/\alpha$ and $1/r^2=4K^2/\alpha^2$. Substituting these into Eq. \eqref{eq:48-X-eq}, we obtain

$$ (2K)^2\diff{^2X}{\alpha^2}-2K\cdot2K\diff X\alpha+\left\{\tau\cdot\frac{2K}{\alpha}-\ell(\ell+1)\cdot\frac{4K^2}{\alpha^2}\right\}X=0 $$

Dividing both sides by $(2K)^2=4K^2$, we obtain

\begin{equation} \diff{^2X}{\alpha^2}-\diff X\alpha+\left\{\frac{\tau}{2K}\cdot\frac1\alpha-\frac{\ell(\ell+1)}{\alpha^2}\right\}X=0 \label{eq:48-X-alpha-eq} \end{equation}

This is the result (verified with SymPy).

Derivation: The Frobenius series method and the indicial equation

Eq. \eqref{eq:48-X-alpha-eq} has a regular singular point at $\alpha=0$, so we set $X(\alpha)=\displaystyle\sum_{k=0}^\infty a_k\alpha^{k+\lambda}$ ($a_0\neq0$) and substitute. Carrying this out in exactly the same way as the calculation in Section 48.1 (set the coefficient of the lowest-order term equal to $0$ to form the indicial equation, then shift the indices of the remaining terms to form the recurrence relation), we obtain

$$ \{\lambda(\lambda-1)-\ell(\ell+1)\}a_0\alpha^{\lambda-2} +\sum_{k=0}^\infty\left[(k+\lambda+\ell+1)(k+\lambda-\ell)a_{k+1}+\left\{\frac\tau{2K}-(k+\lambda)\right\}a_k\right]\alpha^{k+\lambda-1}=0 $$

and setting the coefficient of each power equal to $0$ gives

\begin{equation} \{\lambda(\lambda-1)-\ell(\ell+1)\}a_0=0,\qquad (k+\lambda+\ell+1)(k+\lambda-\ell)a_{k+1}+\left\{\frac\tau{2K}-(k+\lambda)\right\}a_k=0 \label{eq:48-X-frobenius} \end{equation}

Since $a_0\neq0$, we have $\lambda(\lambda-1)=\ell(\ell+1)$, that is, $\lambda=\ell+1$ or $\lambda=-\ell$ (from the factorization $\lambda^2-\lambda-\ell(\ell+1)=(\lambda-\ell-1)(\lambda+\ell)=0$). If we choose $\lambda=-\ell$, then $X(\alpha)\sim a_0\alpha^{-\ell}$ ($\alpha\to0$) diverges at the origin when $\ell\ge1$ (even for $\ell=0$, $R=S/r=Xe^{-Kr}/r$ will later acquire one more factor of $1/r$, so the behavior near the origin becomes even worse). Therefore, for the same reason that we chose $\lambda=\ell$ in Section 48.1, we choose

$$ \lambda=\ell+1 $$

Substituting this into the recurrence relation, we have $(k+\lambda+\ell+1)=(k+2\ell+2)$ and $(k+\lambda-\ell)=(k+1)$, so we obtain the recurrence relation

\begin{equation} a_{k+1}=\frac{(k+\ell+1)-\dfrac\tau{2K}}{(k+1)(k+2\ell+2)}\,a_k\qquad(k=0,1,2,\dots) \label{eq:48-X-recurrence} \end{equation}

(verified with SymPy).

Derivation: Correspondence with the associated Laguerre differential equation

Let us substitute $X=\alpha^{\ell+1}F(\alpha)$ (the form of the solution for $\lambda=\ell+1$) into Eq. \eqref{eq:48-X-alpha-eq}. By the product rule, $\diff X\alpha=(\ell+1)\alpha^\ell F+\alpha^{\ell+1}F'$ and $\diff{^2X}{\alpha^2}=\ell(\ell+1)\alpha^{\ell-1}F+2(\ell+1)\alpha^\ell F'+\alpha^{\ell+1}F''$. Substituting these into Eq. \eqref{eq:48-X-alpha-eq} and dividing the whole equation by $\alpha^\ell$ ($\neq0$ in the range $\alpha>0$), we obtain the equation

\begin{equation} \alpha F''+\{2(\ell+1)-\alpha\}F'+\left\{\frac\tau{2K}-(\ell+1)\right\}F=0 \label{eq:48-laguerre-ode-instance} \end{equation}

(verified by substitution with SymPy—see the verification at the end of the book). This is none other than an equation of the following form—the associated Laguerre differential equation. Here, using the same letter $m$ as the magnetic quantum number for the indices would be confusing, so in this book we use the letters $p,q$ (integers with $0\le p\le q$).

Derivation: The radial wave function and the energy levels

Connecting in order $X(\alpha)=\alpha^{\ell+1}F(\alpha)=a_0\alpha^{\ell+1}L_{n+\ell}^{2\ell+1}(\alpha)$, $\alpha=2Kr$, $S(r)=X(r)e^{-Kr}$, and $R(r)=S(r)/r$, we obtain

\begin{equation} R(r)=a_0\,\frac{e^{-Kr}}{r}\,(2Kr)^{\ell+1}L_{n+\ell}^{2\ell+1}(2Kr) \label{eq:48-Rnl-raw} \end{equation}

What remains is to determine $K$ explicitly. From Eq. \eqref{eq:48-Ktau-def} ($K=\sqrt{-2m_eE}/\hbar$, $\tau=m_ee_0^2/(2\pi\varepsilon_0\hbar^2)$) and Eq. \eqref{eq:48-quantum-numbers} ($n=\tau/2K$), we have

\begin{equation} K=\frac\tau{2n}=\frac{m_ee_0^2}{4\pi\varepsilon_0\hbar^2n} \label{eq:48-K-of-n} \end{equation}

Substituting this back into $K^2=-2m_eE/\hbar^2$ gives the energy.

$$ E=-\frac{\hbar^2K^2}{2m_e}=-\frac{\hbar^2}{2m_e}\cdot\frac{m_e^2e_0^4}{16\pi^2\varepsilon_0^2\hbar^4n^2}=-\frac{m_ee_0^4}{32\pi^2\varepsilon_0^2\hbar^2}\cdot\frac1{n^2} $$

Now, rewriting in terms of $h$ using the reduced Planck constant $\hbar=h/2\pi$ (hence $\hbar^2=h^2/4\pi^2$), we have $32\pi^2\hbar^2=32\pi^2\cdot h^2/4\pi^2=8h^2$, so

\begin{equation} E_n=-\frac{m_ee_0^4}{8\varepsilon_0^2h^2}\cdot\frac1{n^2}\qquad(n=1,2,3,\dots) \label{eq:48-energy} \end{equation}

This is the formula for the energy levels, determined solely by the principal quantum number $n$ (independent of $\ell,m$!) (everything, including the rewriting $\hbar\to h$, has been verified with SymPy—see the verification at the end of the book).

Derivation: The substitution $x=\cos\theta$ and the associated Legendre differential equation

Eq. \eqref{eq:48-theta-eq} is an equation in the variable $\theta$, but $\sin\theta,\cos\theta$ appear mixed together, which makes it awkward to handle. So we set $x:=\cos\theta$ ($x\in[-1,1]$ when $\theta\in[0,\pi]$) and rewrite the equation, regarding $\Theta$ as a function of $x$. We have $\sin^2\theta=1-\cos^2\theta=1-x^2$, and by the chain rule,

$$ \diff\Theta\theta=\diff\Theta x\cdot\diff x\theta=(-\sin\theta)\diff\Theta x $$

(where $\dfrac{dx}{d\theta}=\dfrac{d}{d\theta}\cos\theta=-\sin\theta$). Differentiating once more with respect to $\theta$, we obtain

$$ \diff{^2\Theta}{\theta^2}=\diff{}{\theta}\!\left(\diff\Theta\theta\right)=\diff{}{x}\!\left(\diff\Theta\theta\right)\cdot\diff x\theta=\diff{}{x}\!\left((-\sin\theta)\diff\Theta x\right)\cdot(-\sin\theta) =\left(\diff{^2\Theta}{x^2}\cdot(-\sin\theta)+\diff\Theta x\cdot(-\cos\theta)\right)\cdot(-\sin\theta) $$

(Since we also treat $-\sin\theta$ as a function of $x$, we have $\dfrac{d}{dx}(-\sin\theta)=\dfrac{d}{dx}\left(-\sqrt{1-x^2}\right)=\dfrac{x}{\sqrt{1-x^2}}=\dfrac{\cos\theta}{\sin\theta}\cdot\dfrac{\sin\theta}{1}$, and if we keep track of the signs it ends up as $-\cos\theta$—here, instead of differentiating $\sin\theta=\sqrt{1-x^2}$ with respect to $x$, it is clearer to compute $\dfrac{d}{d\theta}(-\sin\theta)=-\cos\theta$ first and then use $\dfrac{d}{dx}=\dfrac1{-\sin\theta}\dfrac d{d\theta}$ via the chain rule. If we use only the result, we have)

$$ \diff{^2\Theta}{\theta^2}=\sin^2\theta\diff{^2\Theta}{x^2}-\cos\theta\diff\Theta x=(1-x^2)\diff{^2\Theta}{x^2}-x\diff\Theta x $$

Substituting these into Eq. \eqref{eq:48-theta-eq}, and noting that $\dfrac{\cos\theta}{\sin\theta}\dfrac{d\Theta}{d\theta}=\dfrac x{\sin\theta}\cdot(-\sin\theta)\dfrac{d\Theta}{dx}=-x\dfrac{d\Theta}{dx}$ (to be used in the sense that $\sin\theta\cos\theta\dfrac{d\Theta}{d\theta}=\sin\theta\cdot x\cdot(-\sin\theta)\dfrac{d\Theta}{dx}=-x(1-x^2)\dfrac{d\Theta}{dx}$), we rearrange to get

$$ (1-x^2)\left\{(1-x^2)\diff{^2\Theta}{x^2}-x\diff\Theta x\right\}-x(1-x^2)\diff\Theta x+\left\{\ell(\ell+1)(1-x^2)-m^2\right\}\Theta=0 $$

and canceling one common factor $(1-x^2)$ (in the range where $0\lt1-x^2$), we obtain

\begin{equation} (1-x^2)\diff{^2\Theta}{x^2}-2x\diff\Theta x+\left\{\ell(\ell+1)-\frac{m^2}{1-x^2}\right\}\Theta=0 \label{eq:48-legendre-assoc-ode} \end{equation}

This is called the associated Legendre differential equation (it is a generalization of Legendre's differential equation that we studied in Chapter 40—the case $m=0$).

Derivation: The azimuthal solution and the periodic boundary condition

Eq. \eqref{eq:48-phi-eq} ($\bar\Phi''+m^2\bar\Phi=0$, in the case $m\neq0$) is the constant-coefficient second-order homogeneous equation of Chapter 35, and from the roots $\mu=\pm im$ (purely imaginary) of the characteristic equation $\mu^2+m^2=0$, the general solution is

$$ \bar\Phi(\varphi)=C_0e^{im\varphi}+d_0e^{-im\varphi} $$

Imposing the periodic boundary condition $\bar\Phi(0)=\bar\Phi(2\pi)$ ($\varphi$ is an angle, so after one full turn we return to the original position), we obtain $C_0+d_0=C_0e^{i2\pi m}+d_0e^{-i2\pi m}$. If $m$ is an integer, then $e^{\pm i2\pi m}=1$ (Euler's formula, Chapter 21), so this condition is satisfied automatically and cannot determine $C_0,d_0$. However, $e^{im\varphi}$ and $e^{-im\varphi}$ are independent solutions that are not constant multiples of each other (when $m\neq0$), so there is no need to keep both in the general solution—the term $d_0e^{-im\varphi}$ is contained in the term $C_0e^{im\varphi}$ with $m\to-m$ substituted (since we already allow $m$ to be any integer, positive or negative, we would be counting it twice). We may therefore take $d_0=0$.

The case $m=0$ must be checked separately. In this case Eq. \eqref{eq:48-phi-eq} becomes $\bar\Phi''=0$, and the general solution is not an exponential function but the linear function $\bar\Phi(\varphi)=A\varphi+B$. Substituting the periodic boundary condition $\bar\Phi(0)=\bar\Phi(2\pi)$ gives $B=2\pi A+B$, that is, $A=0$, so only $\bar\Phi(\varphi)=B$ (a constant) remains. This is the same conclusion as the one obtained by substituting $m=0$ directly into $\bar\Phi=C_0e^{im\varphi}$ ($e^{i0\cdot\varphi}=1$, a constant), and so we have confirmed that there is no need to treat the case $m=0$ specially.

Summarizing the above, we have

\begin{equation} \bar\Phi(\varphi)=C_0e^{im\varphi}\qquad(m=0,\pm1,\pm2,\dots) \label{eq:48-phi-final} \end{equation}

for the azimuthal solution.