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

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

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

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

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

$$L_+|\ell\ell\rangle = 0$$ 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.

\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 $$r=0$$.

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

Created: 2023-06-25 Sun 02:30

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