Copyright © University of Cambridge. All rights reserved.

This problem is really pushing well into
university concepts, but the solution given here is well worth a
read by keen school students who wish to further their
understanding of differential equations.

Task 0: The lowest energy set of equation corresponds to $ n = 0$. This implies that $l =0$ and $m_l = 0$ also.

Thus:

a) $\frac{1}{R} \frac{d}{dr}\left[r^2 \frac{dR}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$

b) $\frac{sin\theta}{P} \frac{d}{d\theta} \left[sin\theta \frac{dP}{d\theta}\right] = 0$

c) $\frac{1}{F} \frac{d^2F}{d\phi^2} = 0$

Task 1: If $R_1$ is a solution, then for $aR_1$:

$\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$

Since $a$ is a constant, it can be removed outside of the differential:

$\frac{a}{aR_1} \frac{d}{dr}\left[r^2 \frac{dR_1}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$

$\frac{1}{R_1} \frac{d}{dr}\left[r^2 \frac{dR_1}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$

Therefore, since $R_1$ is a solution, so is $aR_1$ for any non-zero constant $a$.

If $P_1$ is a solution, then for $bP_1$:

$\frac{sin\theta}{bP_1} \frac{d}{d\theta} \left[sin\theta \frac{d(bP_1)}{d\theta}\right] = 0$

Since $b$ is a constant, it can be removed outside of the differential:

$\frac{b\ sin\theta}{bP_1} \frac{d}{d\theta} \left[sin\theta \frac{dP_1}{d\theta}\right] = 0$

$\frac{sin\theta}{P_1} \frac{d}{d\theta} \left[sin\theta \frac{dP_1}{d\theta}\right] = 0$

Therefore, since $P_1$ is a solution, so is $bP_1$ for any non-zero constant $b$.

If $F_1$ is a solution, then for $cF_1$:

$\frac{1}{cF_1} \frac{d^2(cF_1)}{d\phi^2} = 0$

Since $c$ is a constant, it can be removed outside of the differential:

$\frac{c}{cF_1} \frac{d^2F_1}{d\phi^2} = 0$

$\frac{1}{F_1} \frac{d^2F_1}{d\phi^2} = 0$

Therefore, since $F_1$ is a solution, so is $cF_1$ for any non-zero constant $c$.

Task 2: If R is a constant, then $\frac{dR}{dr} = 0$

$\therefore \frac{1}{R} \frac{d}{dr}\left[0\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = l(l+1)$

$ -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = l(l+1)$

Since $r > 0$, and $l> 0$, this equation is not satisfied by a constant value of $R$.

If P is a constant, then $\frac{dP}{d\theta} = m_l^2$

$\therefore \frac{sin\theta}{P} \frac{d}{d\theta} \left[0\right] + l(l+1)sin^2\theta= m_l^2$

$l(l+1)sin^2\theta = m_l^2$

The equation is not satisfied since there is still a dependence on $\theta$, which means that the wavefunction would only be permitted for certain values of $\theta$ Therefore a constant value of $P$ is not permissible.

If F is a constant, then:

$m_l^2 = 0$

$m_l = 0$

Task 3:

a) Let $P = sin\theta$

$\therefore \frac{dP}{d\theta} = cos\theta$

$\rightarrow \frac{sin\theta}{sin\theta} \frac{d}{d\theta} \left[sin\theta cos\theta \right] + l(l+1)sin^2 \theta= m_l^2$

$cos(2\theta) + l(l+1)sin^2 \theta = m_l^2$

Expanding $cos (2\theta)$ as $cos^2\theta - sin^2\theta$, and simplifying gives:

$cos^2\theta + (l^2 +l -1)sin^2\theta = m_l^2$

To eliminate the trigonometric terms, we wish to use the identity $cos^2\theta + sin^2\theta = 1$

$\therefore l^2 + l -2 = 0$

$(l+2)(l-1) = 1$

$\mathbf{ l =1}$ since $l > 0$

$\mathbf{m_l = \pm 1}$

b) Now let $P = cos\theta$

$\therefore \frac{dP}{d\theta} = -sin\theta$

$\frac{sin\theta}{cos\theta}\frac{d}{d\theta}\left[-sin^2\theta\right] + l(l+1)sin^2\theta = m_l^2$

Differentiating, simplifying and collecting terms yields:

$(l^2 + l -2) sin^2\theta = m_l^2$

For this to be valid, $\mathbf{m_l^2 = 0}$, which means that either $sin^2 \theta$ or $(l^2 + l -2)$ must be equal to zero.

Clearly, for the wavefunction to exist:

$l^2+ 1 -2 = 0$

$\mathbf{l=1}$ since $l > 0$

c) Substituting in $P = tan\theta$ yields an equation still containing trigonometry. There is no way for this to be removed, and so the original trial solution is invalid.

If you haven't tried any other trigonometric functions yet, why not try $cos^2\theta$, $sin(2\theta)$ and $sin^2\theta$...?

Task 4:

$\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = l(l+1)$

Given that $a_0 = \frac{\pi e^2 \mu}{h^2 \epsilon_0}$, $\alpha = \frac{e^2}{4\pi \epsilon_0}$, $E = -\frac{\mu e^4}{8n^2h^2\epsilon_0^2}$:

$\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[-\frac{\mu e^4 r^2}{8n^2h^2\epsilon_0^2} +\frac{e^2r}{4\pi \epsilon_0}] = l(l+1)$

$\frac{1}{R}\frac{d}{dr} \left[r^2\frac{dR}{dr}\right] + (\frac{a_0r}{n})^2 -2a_0r = l(l+1)$

Task 0: The lowest energy set of equation corresponds to $ n = 0$. This implies that $l =0$ and $m_l = 0$ also.

Thus:

a) $\frac{1}{R} \frac{d}{dr}\left[r^2 \frac{dR}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$

b) $\frac{sin\theta}{P} \frac{d}{d\theta} \left[sin\theta \frac{dP}{d\theta}\right] = 0$

c) $\frac{1}{F} \frac{d^2F}{d\phi^2} = 0$

Task 1: If $R_1$ is a solution, then for $aR_1$:

$\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$

Since $a$ is a constant, it can be removed outside of the differential:

$\frac{a}{aR_1} \frac{d}{dr}\left[r^2 \frac{dR_1}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$

$\frac{1}{R_1} \frac{d}{dr}\left[r^2 \frac{dR_1}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$

Therefore, since $R_1$ is a solution, so is $aR_1$ for any non-zero constant $a$.

If $P_1$ is a solution, then for $bP_1$:

$\frac{sin\theta}{bP_1} \frac{d}{d\theta} \left[sin\theta \frac{d(bP_1)}{d\theta}\right] = 0$

Since $b$ is a constant, it can be removed outside of the differential:

$\frac{b\ sin\theta}{bP_1} \frac{d}{d\theta} \left[sin\theta \frac{dP_1}{d\theta}\right] = 0$

$\frac{sin\theta}{P_1} \frac{d}{d\theta} \left[sin\theta \frac{dP_1}{d\theta}\right] = 0$

Therefore, since $P_1$ is a solution, so is $bP_1$ for any non-zero constant $b$.

If $F_1$ is a solution, then for $cF_1$:

$\frac{1}{cF_1} \frac{d^2(cF_1)}{d\phi^2} = 0$

Since $c$ is a constant, it can be removed outside of the differential:

$\frac{c}{cF_1} \frac{d^2F_1}{d\phi^2} = 0$

$\frac{1}{F_1} \frac{d^2F_1}{d\phi^2} = 0$

Therefore, since $F_1$ is a solution, so is $cF_1$ for any non-zero constant $c$.

Task 2: If R is a constant, then $\frac{dR}{dr} = 0$

$\therefore \frac{1}{R} \frac{d}{dr}\left[0\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = l(l+1)$

$ -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = l(l+1)$

Since $r > 0$, and $l> 0$, this equation is not satisfied by a constant value of $R$.

If P is a constant, then $\frac{dP}{d\theta} = m_l^2$

$\therefore \frac{sin\theta}{P} \frac{d}{d\theta} \left[0\right] + l(l+1)sin^2\theta= m_l^2$

$l(l+1)sin^2\theta = m_l^2$

The equation is not satisfied since there is still a dependence on $\theta$, which means that the wavefunction would only be permitted for certain values of $\theta$ Therefore a constant value of $P$ is not permissible.

If F is a constant, then:

$m_l^2 = 0$

$m_l = 0$

Task 3:

a) Let $P = sin\theta$

$\therefore \frac{dP}{d\theta} = cos\theta$

$\rightarrow \frac{sin\theta}{sin\theta} \frac{d}{d\theta} \left[sin\theta cos\theta \right] + l(l+1)sin^2 \theta= m_l^2$

$cos(2\theta) + l(l+1)sin^2 \theta = m_l^2$

Expanding $cos (2\theta)$ as $cos^2\theta - sin^2\theta$, and simplifying gives:

$cos^2\theta + (l^2 +l -1)sin^2\theta = m_l^2$

To eliminate the trigonometric terms, we wish to use the identity $cos^2\theta + sin^2\theta = 1$

$\therefore l^2 + l -2 = 0$

$(l+2)(l-1) = 1$

$\mathbf{ l =1}$ since $l > 0$

$\mathbf{m_l = \pm 1}$

b) Now let $P = cos\theta$

$\therefore \frac{dP}{d\theta} = -sin\theta$

$\frac{sin\theta}{cos\theta}\frac{d}{d\theta}\left[-sin^2\theta\right] + l(l+1)sin^2\theta = m_l^2$

Differentiating, simplifying and collecting terms yields:

$(l^2 + l -2) sin^2\theta = m_l^2$

For this to be valid, $\mathbf{m_l^2 = 0}$, which means that either $sin^2 \theta$ or $(l^2 + l -2)$ must be equal to zero.

Clearly, for the wavefunction to exist:

$l^2+ 1 -2 = 0$

$\mathbf{l=1}$ since $l > 0$

c) Substituting in $P = tan\theta$ yields an equation still containing trigonometry. There is no way for this to be removed, and so the original trial solution is invalid.

If you haven't tried any other trigonometric functions yet, why not try $cos^2\theta$, $sin(2\theta)$ and $sin^2\theta$...?

Task 4:

$\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = l(l+1)$

Given that $a_0 = \frac{\pi e^2 \mu}{h^2 \epsilon_0}$, $\alpha = \frac{e^2}{4\pi \epsilon_0}$, $E = -\frac{\mu e^4}{8n^2h^2\epsilon_0^2}$:

$\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[-\frac{\mu e^4 r^2}{8n^2h^2\epsilon_0^2} +\frac{e^2r}{4\pi \epsilon_0}] = l(l+1)$

$\frac{1}{R}\frac{d}{dr} \left[r^2\frac{dR}{dr}\right] + (\frac{a_0r}{n})^2 -2a_0r = l(l+1)$