Motion in a Central Potential

Two-Body Problem

Two particles $(m_1;\vec{r}_1, m_2;\vec{r}_2)$ in 3D space with potential $V\left(\vec{r}_1,\vec{r}_2\right) = V\left(\vec{r}_1-\vec{r}_2\right)$. $\mathcal{L} = T-V = \frac{m_1}{2}\left|\dot{\vec{r}}_1\right|^2 + \frac{m_1}{2}\left|\dot{\vec{r}}_1\right|^2 - V\left(\vec{r}_1, \vec{r}_2\right) = \frac{m_1}{2}\left|\dot{\vec{r}}_1\right|^2 + \frac{m_1}{2}\left|\dot{\vec{r}}_1\right|^2 - V\left(\left|\vec{r}_1 - \vec{r}_2\right|\right)$. Let $\vec{r}=\vec{r}_1-\vec{r}_2$ and $\vec{r}_{CM}=\frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}$. Hence, $\mathcal{L} = T-V = \frac{m_1+m_2}{2}\left|\dot{\vec{r}}_{CM}\right|^2 \frac{m_1m_2}{2(m_1+m_2)}\left|\dot{\vec{r}}\right|^2 - V(\vec{r})$. Let $M=m_1+m_2$ and $\mu=\frac{m_1m_2}{M}$. So,

\begin{align*} \mathcal{L} = T-V = \frac{M}{2}\left|\dot{\vec{r}}_{CM}\right|^2 \frac{\mu}{2}\left|\dot{\vec{r}}\right|^2 - V(\vec{r}). \end{align*}

This simplifies the problem to only consider this reduced mass in a potential and a free center of mass particle.

The equation of motion is then,

\begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q} = 0 \end{align*}

C.f. $\frac{dp}{dt}-F=0$.

If $\frac{\partial L}{\partial q} = 0$ then $q$ is a cyclic variable.

Then, in our case the center of mass is a cyclic variable. So, the momentum is constant for the center of mass.

$\vec{p}_{CM}=M\dot{\vec{r}}_{CM}$ and $\vec{p}=\mu\dot{\vec{r}}$. Hence,

\begin{align*} H=\frac{\vec{p}_{CM}}{2M} + \frac{\vec{p}^2}{2\mu} + V(|\vec{r}|) = H_{CM} + H_{r}. \end{align*}

$[H_{CM},H_r]=0$.

$H_{CM}|\psi\rangle = E_{CM}|\psi\rangle$ and $H_r|\psi\rangle=E_r|\psi\rangle$ so $H|\psi\rangle = E|\psi\rangle$.

$-\frac{\hbar^2}{2}\left[\frac{1}{M}\vec{\nabla}^2_{CM}+\frac{1}{\mu}\vec{\nabla}^2_r + V(r) - E\right]\Psi_{CM}(\vec{r})\Psi_{CM}(\vec{r}_{CM})$.

Dividing out the product of the wavefunctions,

\begin{align*} -\frac{\hbar^2}{2M}\frac{\vec{\nabla}^2_{CM}\psi_{CM}}{\psi_{CM}}-\frac{\hbar^2}{2\mu}\frac{\vec{\nabla}^2_r\psi_r}{\psi_r} + V(r) = E_{CM} + E_r = E. \end{align*}

This gives us two separate equations,

\begin{align*} -\frac{\hbar^2}{2M}\frac{\vec{\nabla}^2_{CM}\psi_{CM}}{\psi_{CM}} = E_{CM}, \end{align*} \begin{align*} -\frac{\hbar^2}{2\mu}\frac{\vec{\nabla}^2_r\psi_r}{\psi_r} + V(r) = E_r. \end{align*}

For the center of mass, it is a free particle - $\psi_{CM}=\frac{1}{(2\pi\hbar)^{\frac{3}{2}}}\exp\frac{i}{\hbar}\vec{p}_{CM}\cdot\vec{r}_{CM}$.

For the reduced mass system, we can transform this into a radial and angular variables to separate out the potential.

\begin{align*} x=r\sin\theta\cos\varphi,y=r\sin\theta\sin\varphi,z=r\cos\theta,\vec{\nabla}^2=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}\right]. \end{align*}

Using spherical coordinates, $\psi_r = R(r)Y(\theta,\varphi)$.

\begin{align*} \frac{1}{r^2}\frac{d}{dr}(r^2\frac{dR}{dr})Y+\frac{R}{r^2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial Y}{\partial\theta}+\frac{1}{\sin^2\theta}\frac{\partial^2Y}{\partial\varphi^2}\right)\right] + \frac{2\mu}{\hbar^2}[E-V(r)]RY. \end{align*}

Dividing out by $RY/r^2$,

\begin{align*} \frac{d}{dr}(r^2\frac{dR}{dr})\frac{1}{R}+\frac{1}{Y}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial Y}{\partial\theta}+\frac{1}{\sin^2\theta}\frac{\partial^2Y}{\partial\varphi^2}\right)\right] + \frac{2\mu r^2}{\hbar^2}[E-V(r)]. \end{align*}

Then, we have,

\begin{align*} \frac{d}{dr}(r^2\frac{dR}{dr}) + \frac{2\mu r^2}{\hbar^2}[E-V(r)]R = \lambda R, \end{align*} \begin{align*} \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial Y}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2Y}{\partial\varphi^2} = -\lambda Y. \end{align*}

The second part has the Spherical Harmonic solutions from the Angular Motion.

Orbital Angular Momentum

$L_x$	$L_y$	$L_z$
$yp_z-zp_y$	$zp_x-xp_z$	$xp_y-yp_x$

Consider the angular momentum: $\vec{L}=\vec{r}\times\vec{p} = \hat{i}(yp_z-zp_y) - \hat{j}(xp_z-zp_x) + \hat{k}(xp_y-yp_z)$. Then, $\hat{L}_x = \hat{y}\hat{p}_z - \hat{z}\hat{p_y}$.

We can then simplify the second part to $L^2Y=\cdots Y$.

$Y_l^m(\theta,\varphi)$.

$L_x = i\hbar\left(\sin\varphi\frac{\partial}{\partial\theta} + \frac{\cos\varphi}{\tan\theta}\frac{\partial}{\partial\varphi}\right)$ $L_y = i\hbar\left(-\cos\varphi\frac{\partial}{\partial\theta} + \frac{\sin\varphi}{\tan\theta}\frac{\partial}{\partial\varphi}\right)$ $L_z=-i\hbar\frac{\partial}{\partial\varphi}$

This gives us, $L^2Y = \hbar^2\lambda Y$. Also, $L_zY=\hbar mY$.

$\{H,L^2,L_z\}$ is a C.S.C.O. and covers all of the degrees of freedom.

Commutators

$[L^2,L_i] = 0$
$[L_i,L_j] = i\hbar L_k \varepsilon_{ijk}$

Restrictions on Quantum Numbers

The Magnetic Quantum Number

From $L_zY=\hbar mY$ we see $L_zY=-i\hbar\frac{\partial}{\partial \varphi}Y=\hbar mY$ so $Y=f_l(\theta)f_z(\varphi)$. Then, $-i\hbar\frac{d}{d\varphi}f_z(\varphi)=\hbar mf_z(\varphi)$ gives $f_z(\varphi)=C\exp(im\varphi)$. Thus, it is periodic hence $m\in\mathbb{Z}$.

$m$ is called the magnetic quantum number since applying the magnetic field splits the energies for given $m$ states.
Oribtal Quantum Number

$\langle L^2\rangle = \langle L_x^2\rangle + \langle L_y^2\rangle + \langle L_z^2\rangle = \langle L_x\psi|L_x\rangle + \langle L_y\psi|L_y\rangle + \langle L_z\psi|L_z\rangle = ||L_x|\psi\rangle||^2 + ||L_y|\psi\rangle||^2 + ||L_z|\psi\rangle||^2 \geq 0$. Thus, let $\lambda = \ell(\ell+1)$ and so $\ell(\ell+1)\geq 0$. Thus, $\ell\geq 0$ or $\ell\leq -1$. By convention, we choose $\ell$ non-negative.

More Notes on Angular States

$\langle \ell'm'|\ell m\rangle = \delta_{\ell'\ell}\delta_{m'm}$.

So, $\int Y_{\ell'}^{m'*}Y_\ell^m d\Omega = \delta_{\ell'\ell}\delta_{m'm}$.

$d\Omega = \sin\theta d\theta d\varphi$.

Ladder Operators

Let $L_+=L_x+iL_y$ and $L_-=L_x-iL_y$.

Eq 1: $\langle L^2-L_z^2\mp\hbar L_z\rangle = \hbar^2\ell(\ell+1)-\hbar^2m^2\mp\hbar^2m=\hbar^2(\ell\mp m)(\ell\pm m+1)\geq 0$.

Note, $L_+^\dagger = L_-$.

Then, $\langle\ell m|L\mp L_\pm|\ell m\rangle = \langle \ell m|L_\pm^\dagger L_\pm|\ell m\rangle = ||L_\pm|\ell m\rangle||^2\geq 0$. Thus, this with Eq 1 gives $|m|\leq \ell$.

$L_+|\ell\ell\rangle = 0$ since the ladder operator is at the limit. Similarly, $L_-|\ell-\ell\rangle = 0$.

Note: $[L^2,L_\pm] = 0$.

Note: $[L_z,L_\pm] = [L_z,L_x] \pm [L_z,L_y] = \pm\hbar L_\pm$. Then, $L_zL_{\pm}|\ell m\rangle = L_\pm L_z|\ell m\rangle \pm L_\pm|\ell m\rangle = L_\pm\left[\hbar(m\pm1)\right]|\ell m\rangle$. Thus, $L_\pm|\ell m\rangle = c_{\pm}|\ell m\pm 1\rangle$.

$||L_\pm|\ell m\rangle||^2 = \hbar^2(\ell\mp m)(\ell \pm m+1) = ||c_\pm|\ell m\pm 1||^2 = |c_\pm|^2$.

$L_+=-i\hbar\exp(i\varphi)\left[i\frac{\partial}{\partial\theta}-\cot\theta\frac{\partial}{\partial\varphi}\right]$

$L_+|\ell\ell\rangle = 0$ so $L_+Y_\ell^\ell=0$. This gives us a differential equation to solve to find $Y_\ell^\ell$.

This gives us, $Y_\ell^\ell c_\ell \exp(i\ell\varphi)\sin^\ell\theta$. Use the lowering operators to find the other $Y_\ell^m$.

This gives us, $Y_\ell^m=\frac{(-1)^\ell}{2^\ell\ell!}\sqrt{\cdots}\exp(im\varphi)\frac{1}{\sin^m\theta}\frac{d^{\ell-m}}{d(\cos\theta)^{\ell-m}}(\sin\theta)^{2\ell}$.

Note: $L_\pm|\ell m\rangle = \hbar\sqrt{(\ell\mp)(\ell\pm m+1)}|\ell m\pm1\rangle$.

Semiclassical Approach

Why do we have $\ell(\ell+1)$ well our maximum $m$ is $\ell$ and we note that $|L_z\lt|L|$ (due to the uncertainty principle) so $\hbar\ell\leq\hbar\sqrt{\ell(\ell+1)}$

Assume $L_z$ is exactly equal to $|L|$ then $L_x=L_y=0$ fully specifying the state but the commutator $[L_i,L_j]=i\hbar L_k\varepsilon_{ijk}$ gives a contradition.

Diatomic Molecules

Rigid Rotors

We have a fixed distance between the two particles, $r_e=|\vec{r}_1-\vec{r}_2|$ fixed. Moment of inertia: $I = m_1r_1^2 + m_2r_2^2 = \mu r_e^2$. $|\vec{L}|=I\omega_R=\mu r_e^2\omega_R$. The Hamiltonian is $\frac{L^2}{2I} = \frac{I\omega_R}{2} = \frac{L^2}{2\mu r_e^2}$. Then, $H|\ell m\rangle = E_\ell|\ell m\rangle = \frac{\hbar^2\ell(\ell+1)}{2\mu r_e^2}|\ell m\rangle=B\ell(\ell+1)|\ell m\rangle$

$E_\ell-E_{\ell-1}$ energy level difference: $B\ell(\ell+1) - B(\ell-1)\ell = 2B\ell$. B is called the rotational constant.

Transitions between levels are mediated by absorption and emission.

Heteropolar Molecules

$\vec{d}_0 = e\vec{r}_0$, permanent dipole moment, excite molecule with $\vec{E}$ (i.e. light) implies $H_{interaction}=-\vec{d}_0\cdot\vec{E}_{light}=-eZE_{light}$ $H_{interaction}=-er_0\cos\theta E_{light}$.

$-er_0E_{light}\int Y_{\ell'}^{m'*}\cos\theta Y_\ell^m\:d\Omega$ Orthogonality: $\cos\theta Y_{\ell}^m=c_1Y_{\ell-1}^m+c_2Y_{\ell+1}^m$

$\Delta\ell=\ell'-\ell=\pm 1$, $\Delta m=0,\pm 1$ are selection rules. $\Delta m=0$ for z-polarized and $\pm1$ for x,y polarized.

$\langle \ell'm'|H_{int}|\ell m\rangle\neq 0$ implies a valid transition arising from the interaction.

$\nu_{\ell,\ell-1} = \frac{E_{\ell}-E_{\ell-1}}{h} = \frac{2B}{h}\ell$. Spacing between frequencies: $\nu_{\ell+1,\ell}-\nu_{\ell,\ell-1}=\frac{2B}{h}$, equally spaced.

$\mathcal{P}\sim|\langle\ell'm'|H_{int}|\ell m\rangle|^2$

Any deviation might be due to vibrational modes unaccounted for in rigid rotor approximation. We also may have a magnetic interaction unaccounted for that splits the lines, we then get more transition lines.

Electronic degrees of freedom ~1 eV, vibrational is 0.1 eV, and rotational is around 0.01 eV.

Homopolar Molecules

Two identical atoms rather than distinct. Use Raman-spectroscopy.

Interaction: $\hbar\Omega_{in}$ goes in and you measure $\hbar\Omega_{sc}'$. $\Omega_{sc}'=\Omega_{in}\pm\omega_R,\omega_R=\frac{E_{\ell'}-E_\ell}{\hbar}$.

The interaction goes as $H_{int}\sim Z^2$ instead of $Z$, and so we get $\cos^2$. Thus, the selection rules go as $\Delta\ell=\pm 2,0$.

Raman Spectrum: Rayeleigh: symmetric graph about $\Delta\ell=0$ with spacing around the center to be $6B/h$ and the rest to be $4B/h$. Frequencies that are larger are called anti-Stokes lines and those that are lower are called Stokes lines.

Radial Degree of Freedom

\begin{align*} \frac{d}{dr}\left(r^2\frac{dR}{dr}\right) + \frac{2\mu r^2}{\hbar^2}[E-V(r)]R = \lambda R, \end{align*}

From our Orbital solution, we see that $R=R_{?}^\ell(r)$.

We can rearrange to obtain: $\frac{-\hbar^2}{2\mu r^2}\left(2r\frac{dR}{dr}+r^2\frac{d^2R}{dr^2}\right) + \left(V(r)+\frac{\ell(\ell+1)\hbar^2}{2\mu r^2}\right)R(r)=\frac{-\hbar^2}{2\mu r^2}\left(2r\frac{dR}{dr}+r^2\frac{d^2R}{dr^2}\right) + V_{eff}(r)R(r)=ER(r)$ Let $R(r)=\frac{u(r)}{r}=-\frac{\hbar^2}{2\mu}\frac{d^2u}{dr^2}+U_{eff}(r)u(r)=Eu(r)$. This is like a 1D Schrödinger equation. Some constraints we have: $r\geq0$ and $u(0)=0$. Let $u=u_k^\ell$.

Hydrogen Atom

$V(r)=-\frac{q^2}{4\pi\varepsilon_0r}=-\frac{e^2}{r}$ Then, $V_{eff}(r) = -\frac{e^2}{r}+\frac{\ell(\ell+1)\hbar^2}{2\mu r^2}$. Bohr radius: $a_0=\frac{\hbar^2}{\mu e^2}$, then $\rho=\frac{r}{a_0}$. Also, $E\to E/E_{I}, E_{I}=\frac{\mu e^4}{2\hbar^2}$. Let $\lambda_{k\ell}=\sqrt{-E_{k,\ell}/E_I}$, to find the bound state energies. Hence, $V_{eff}(r)=-\frac{e^2}{a_0\rho}+\frac{e^4\ell(\ell+1)}{4E_I a_0^2\rho^2}$.

With this, we get $\left[\frac{d^2}{d\rho^2}-\frac{\ell(\ell+1)}{\rho}+\frac{2}{\rho}-\lambda_{k\ell}^2\right]u_k^\ell(\rho)=0$.

For $\rho\to\infty$, we have $\left[\frac{d^2}{d\rho^2}-\lambda_{k\ell}^2\right]u_k^\ell(\rho)=0$. $u_k^\ell(\rho)=C_1\exp(\rho\lambda_{k\ell})+C_2\exp(-\rho\lambda_{k\ell})$. Since $C_1\neq0$ has a solution that blows up, we find that at infinity we get $u_{k,\ell}(\rho)=C\exp(-\rho\lambda_{k,\ell})$. Then, $u_{k}^\ell(\rho)=\exp(-\rho\lambda_{k\ell})y_k^\ell(\rho)$.

From this, our original differential equation becomes: $\left[\frac{d^2}{d\rho^2}-2\lambda_{k,\ell}\frac{d}{d\rho}+\frac{2}{\rho}-\frac{\ell(\ell+1)}{\rho^2}\right]y_k^\ell(\rho)=0$. We use a power series to solve: $y_k^\ell(\rho)=\rho^s\sum_{q=0}^\infty c_q\rho^q$, Frobenius series.

This gives us: $\sum_{q=0}^\infty(q+s)(q+s-1)c_q\rho^{q+s-2}-2\lambda_{k\ell}\sum_{q=0}^\infty(q+s)c_q\rho^{q+s-1}+2\sum_{q=0}^\infty c_q\rho^{q+s-1}-\ell(\ell+1)\sum_{q=0}^\infty c_q\rho^{q+s-2}=0$. Labeling each term sequentially from 1 to 4. Note: the lowest power is $s-2$, which is 1 and 4. Then, $(s(s-1)-\ell(\ell+1))c_0\rho^{s-2}=0$ implies that either $c_0=0$ or $s(s-1)=\ell(\ell+1)$. We assume $c_0\neq0$ so $s(s-1)=\ell(\ell+1)$, giving us a quadratic equation. Thus, $s=-\ell$ or $s=\ell+1$. If we have $s=-\ell$, then for $\ell=0$ we have singularities at zero. Thus, we have that $s=\ell+1$.

Let us shift the indices of 2 and 3. We get This gives us: $\sum_{q=0}^\infty(q+s)(q+s-1)c_q\rho^{q+s-2}-2\lambda_{k\ell}\sum_{q=1}^\infty(q+s-1)c_{q-1}\rho^{q+s-2}+2\sum_{q=1}^\infty c_{q-1}\rho^{q+s-2}-\ell(\ell+1)\sum_{q=0}^\infty c_q\rho^{q+s-2}=0$. Hence, for a power of $\rho$ we have a coefficient $[(q+s)(q+s-1)-\ell(\ell+1)]c_q+c_{q-1}(2-2\lambda(q+s-1))$. Replacing $s$, we get $[(q+\ell+1)(q+\ell)-\ell(\ell+1)]c_q+c_{q-1}(2-2\lambda(q+\ell))$. Since this coefficient must be zero: $[(q+\ell+1)(q+\ell)-\ell(\ell+1)]c_q+c_{q-1}(2-2\lambda(q+\ell))=0$. Then, $c_q=\frac{2(\lambda_{k\ell}(q+\ell)-1)}{q(q+2\ell+1)}c_{q-1}$.

If we check convergence: $\frac{c_q}{c_{q-1}}\to\frac{2\lambda_{k\ell}}{q}$. Taylor expanding $exp(2ρλ_kℓ) gives this ratio at the limit, which we cannot allow to happen. Thus, we need a cutoff. Let $k$ be the cutoff such that $c_k=0$. Then let $q$ run from $0$ to $k-1$. So, we need $\lambda_{k\ell}(k+\ell)=1$. Then, $\lambda_{k\ell}=\frac{1}{k+\ell}$. Note that $\lambda$ was the energy so we get quantization of energy. Hence, $E_{k,\ell}=-\frac{E_I}{k+\ell}$ with $k=1,2,3\cdots$.

Typically, we have the principle quantum number $n=k+\ell$ then $E_n=-\frac{E_I}{n^2}$ with $n=1,2,\cdots$. Then in this case we get $\ell=n-k\Rightarrow n-1,n-2,\cdots,0$.

Therefore, $R_{n,\ell}(r)=-\left(\frac{2}{na_0}\right)^{\frac{3}{2}}\sqrt{\frac{(n-\ell-1)!}{2n(n-\ell)!^3}}\left(\frac{2r}{na_0}\right)^\ell\exp\left(\frac{-r}{na_0}\right)L_{n-\ell-1}^{2\ell+1}\left(\frac{2r}{na_0}\right)$.

Hydrogen-Like Atoms

$\Psi_{n\ell m}(r,\theta,\varphi)=R_n^\ell(r)Y_\ell^m(\theta,\varphi)$. Note, that $\ell=0$ is special since you will have a non-zero probability for $r\to0$. All other $\ell$’s have a zero probability of the particle to be at $r=0$.

Questions

Electron is in the ground state

At what $r$ is the probability density maximal? $r=0$ @ $100=n\ell m$. Well, if you see in the probability density in the integral, we get a radial integration of $|R_n^\ell(r)|^2r^2$ which gives a probability density of $|R_n^\ell(r)r|^2$ with the Jacobian. Then, we see that the probability density is actually maximized at $r_1=a_0$ (and $0$ but we know that is when the function is zero, not maximized). This gives the physical meaning to the Bohr radius.

Note, the expectation value of position is a bit larger than $a_0$ since the expectation value is an average. $\langle r\rangle_{100} = \int_0^\infty |R_{10}|^2r\cdot r^2\:dr = \int_0^\infty |R_{10}|^2r^3 dr$. Use a gamma function! $\int_0^\infty x^n\exp(-x)dx=n!=\Gamma(n+1)$. Thus, we get $\int\:dr = 4a_0^3\left(\frac{a_0}{2}\right)^4 3! = a_0\frac{3}{2}$.

Probability Density versus Wavefunction Squared

Example 1

Lets say we have a detector that has one shot to detect an electron. Where should we place the detector? Place it at the center due to the spherical symmetry.
Example 2

Lets say we have many detectors on a ring or sphere around it? Place it at the Bohr radii. If we can take advantage of the surface area, use the $|R|^2r^2$, but if we cannot take advantage (directional and one-shot) of it then $|R|^2$ is better.

Energy Levels

Degeneracies increase with the principle quantum number. $g_n=\sum_{\ell=0}^{n-1}(2\ell+1) = n(n-1) + n = n^2$. With spin, we get $g_n=2n^2$.

Shells

Shells: $n=1234\cdots=KLMN\cdots$. Subshells: $\ell=01234\cdots=spdfg\cdots$.

Like Atoms

Originally, we had $V=-e^2/r$.

For Deuterium, we have 1 proton, neutron, and electron. For Tritium, we have 1 proton, 2 neutrons, and 1 electron. For He$^+$, we have 2 proton, 2 neutrons, and 1 electron. For Li$^{2+}$, we have 3 proton, 2 neutrons, and 1 electron.

For more protons, we get $-Ze^2/r$.

We expect $E_n=-Z^2\frac{E_I}{n^2}$. Recall $E_I=\frac{\mu e^4}{2\hbar^2}$, so the reduced mass also changes (also note that the $Z^2$ came since you can idealize one of the $e$ in the original equations as $Ze$ and so each $e$ gets squared and so we get a $Z^2$ factor). Additionally, we expect that the electrons will tend to hang out closer to the nucleus for the higher charged nuclei. $a_0(Z)=\frac{a_0^H}{Z}$. $a_0^H=\frac{\hbar^2}{\mu e^2}$.

Positron: $e^-,e^+$. Equal mass, so the reduced mass is exactly half of the original mass. So, the ionization energy is roughly half of the Hydrogen atom.

Particle in Spherically Symmetric Potential

Example: Spherical Box (Quantum Dot)

Let $V(r\lt R)=0$ and $V=\infty$ otherwise. Let $\Psi(r,\theta,\varphi) = R_{?\ell}(r)Y_\ell^m(\theta,\varphi)$. We now just need to solve for the $R(r)$ equation. Note, $R(r)=u(r)/r$ gives the SE: $-\frac{\hbar^2}{2m}\frac{d^2 u}{dr^2}+V_{eff}(r)u(r)=Eu(r)$ with $V_{eff}=V(r)+\frac{\hbar^2\ell(\ell+1)}{2\mu r^2}$. So, $V_{eff}(r\lt R)=\frac{\hbar^2\ell(\ell+1)}{2\mu r^2}$ and $V_{eff}(r\gt R)=\infty$. The outer limit means the wave function must be zero at the boundary. Hence, $\frac{d^2 u_\ell}{dr^2}+\left[\frac{2\mu E}{\hbar^2}-\frac{\ell(\ell+1)}{r^2}\right](r)u_\ell(r)=0$.

Ansatz: $R_n^\ell(r) = A j_n^\ell(r)$.

By letting $u_\ell(r)=\sqrt{z}\eta_\ell(z)$ we get $\frac{d^2\eta}{dz^2}+\frac{1}{z}\frac{d\eta_\ell}{dz} + \left(1-\frac{(\ell+1/2)^2}{z^2}\right)\eta_\ell(z)=0$. $\eta_\ell(z)=C_1J_{\ell+1/2}(z)+C_2J_{-(\ell+1/2)}(z)$.

$j_\ell(z)=\sqrt{\frac{\pi}{2}}\frac{J_{\ell+1/2}(z)}{\sqrt{z}}$, $n_\ell(z)=\sqrt{\frac{\pi}{2}}(-1)^{\ell+1}\frac{J_{-(\ell+1/2)}(z)}{\sqrt{z}}$.

As $z\to 0$, $j_\ell$ behave like a power law $\left(\frac{z^\ell}{(2\ell+1)!!}\right)$ and $n_\ell(z)$ behaves inversely $-\frac{(2\ell-1)!!}{z^{\ell+1}}$.

$j_\ell(z) \to \frac{1}{z}\cos[z-(\pi(\ell+1))/2]$ $n_\ell(z) \to \frac{1}{z}\sin[z-(\pi(\ell+1))/2]$ (Are these flipped?)

$j_0(0) = \mathbb{C}$ but all other give $j_\ell(0)=0$. $n_\ell(0) \to -\infty$.

Thus, $u_\ell(r)=kr(C_1j_\ell(kr)+C_2n_\ell(kr))$.

So, for our problem, $u_\ell(r)=krC_1j_\ell(kr)$ from $r\to 0$ implies $u(r)\to 0$. Then, we find for $r=R$, $j_\ell(kR)=0$. So, we get $u_n^\ell(r) = krCj_\ell(k_n^\ell r)$ such that $j_\ell(k_n^\ell R) = 0$.

Note: $j_0(z)=\frac{\sin z}{z}$, $j_1(z) = \frac{\sin z}{z^2}-\frac{\cos z}{z}$ and $j_2(z)=\left(\frac{3}{z^3}-\frac{1}{z}\right)\sin(z) - \frac{3}{z}\cos z$. So, $R_n^\ell(r)=Cj_\ell(k_n^\ell r)$.

Note: $E=\frac{\hbar^2 k_n^{\ell 2}}{2\mu}$.