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*}


\(H_{CM}|\psi\rangle = E_{CM}|\psi\rangle\) 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 \(L^2Y=\cdots Y\).


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

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.


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

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: \([L^2,L_\pm] = 0\).

Then, \(L^2L_\pm|\ell m\rangle = L_\pm L^2|\ell m\rangle = \hbar^2\ell(\ell+1)L_\pm|\ell m\rangle\). Thus, \(L_\pm|\ell m\rangle\) does not change \(\ell\). So, \(L_\pm|\ell _\rangle = |\ell _\rangle\).

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, \(L_\pm|\ell m\rangle = c_{\pm}|\ell m\pm 1\rangle\).

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


\(L_+|\ell\ell\rangle = 0\) 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(2ρλkℓ) 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 \(r=0\).


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: \(n=1234\cdots=KLMN\cdots\). 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: \(E=\frac{\hbar^2 k_n^{\ell 2}}{2\mu}\).

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:30