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 $V ({\vec{r}}_{1}, {\vec{r}}_{2}) = V ({\vec{r}}_{1} - {\vec{r}}_{2})$ . $L = T - V = \frac{m_{1}}{2} {| {\dot{\vec{r}}}_{1} |}^{2} + \frac{m_{1}}{2} {| {\dot{\vec{r}}}_{1} |}^{2} - V ({\vec{r}}_{1}, {\vec{r}}_{2}) = \frac{m_{1}}{2} {| {\dot{\vec{r}}}_{1} |}^{2} + \frac{m_{1}}{2} {| {\dot{\vec{r}}}_{1} |}^{2} - V (| {\vec{r}}_{1} - {\vec{r}}_{2} |)$ . Let $\vec{r} = {\vec{r}}_{1} - {\vec{r}}_{2}$ and ${\vec{r}}_{C M} = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2}}{m_{1} + m_{2}}$ . Hence, $L = T - V = \frac{m_{1} + m_{2}}{2} {| {\dot{\vec{r}}}_{C M} |}^{2} \frac{m_{1} m_{2}}{2 (m_{1} + m_{2})} {| \dot{\vec{r}} |}^{2} - V (\vec{r})$ . Let $M = m_{1} + m_{2}$ and $μ = \frac{m_{1} m_{2}}{M}$ . So,

\begin{array}{r} L = T - V = \frac{M}{2} {| {\dot{\vec{r}}}_{C M} |}^{2} \frac{μ}{2} {| \dot{\vec{r}} |}^{2} - V (\vec{r}) . \end{array}

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{array}{r} \frac{d}{d t} \frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q} = 0 \end{array}

C.f. $\frac{d p}{d t} - F = 0$ .

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.

${\vec{p}}_{C M} = M {\dot{\vec{r}}}_{C M}$ and $\vec{p} = μ \dot{\vec{r}}$ . Hence,

\begin{array}{r} H = \frac{{\vec{p}}_{C M}}{2 M} + \frac{{\vec{p}}^{2}}{2 μ} + V (| \vec{r} |) = H_{C M} + H_{r} . \end{array}

$[H_{C M}, H_{r}] = 0$ .

$H_{C M} | ψ ⟩ = E_{C M} | ψ ⟩$ and $H_{r} | ψ ⟩ = E_{r} | ψ ⟩$ so $H | ψ ⟩ = E | ψ ⟩$ .

$- \frac{ℏ^{2}}{2} [\frac{1}{M} {\vec{\nabla}}_{C M}^{2} + \frac{1}{μ} {\vec{\nabla}}_{r}^{2} + V (r) - E] Ψ_{C M} (\vec{r}) Ψ_{C M} ({\vec{r}}_{C M})$ .

Dividing out the product of the wavefunctions,

\begin{array}{r} - \frac{ℏ^{2}}{2 M} \frac{{\vec{\nabla}}_{C M}^{2} ψ_{C M}}{ψ_{C M}} - \frac{ℏ^{2}}{2 μ} \frac{{\vec{\nabla}}_{r}^{2} ψ_{r}}{ψ_{r}} + V (r) = E_{C M} + E_{r} = E . \end{array}

This gives us two separate equations,

\begin{array}{r} - \frac{ℏ^{2}}{2 M} \frac{{\vec{\nabla}}_{C M}^{2} ψ_{C M}}{ψ_{C M}} = E_{C M}, \end{array}

\begin{array}{r} - \frac{ℏ^{2}}{2 μ} \frac{{\vec{\nabla}}_{r}^{2} ψ_{r}}{ψ_{r}} + V (r) = E_{r} . \end{array}

For the center of mass, it is a free particle - $ψ_{C M} = \frac{1}{(2 π ℏ)^{\frac{3}{2}}} \exp \frac{i}{ℏ} {\vec{p}}_{C M} \cdot {\vec{r}}_{C M}$ .

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

\begin{array}{r} x = r \sin θ \cos φ, y = r \sin θ \sin φ, z = r \cos θ, {\vec{\nabla}}^{2} = \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial}{\partial r}) + \frac{1}{r^{2}} [\frac{1}{\sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial}{\partial θ}) + \frac{1}{\sin^{2} θ} \frac{\partial^{2}}{\partial φ^{2}}] . \end{array}

Using spherical coordinates, $ψ_{r} = R (r) Y (θ, φ)$ .

\begin{array}{r} \frac{1}{r^{2}} \frac{d}{d r} (r^{2} \frac{d R}{d r}) Y + \frac{R}{r^{2}} [\frac{1}{\sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial Y}{\partial θ} + \frac{1}{\sin^{2} θ} \frac{\partial^{2} Y}{\partial φ^{2}})] + \frac{2 μ}{ℏ^{2}} [E - V (r)] R Y . \end{array}

Dividing out by $R Y / r^{2}$ ,

\begin{array}{r} \frac{d}{d r} (r^{2} \frac{d R}{d r}) \frac{1}{R} + \frac{1}{Y} [\frac{1}{\sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial Y}{\partial θ} + \frac{1}{\sin^{2} θ} \frac{\partial^{2} Y}{\partial φ^{2}})] + \frac{2 μ r^{2}}{ℏ^{2}} [E - V (r)] . \end{array}

Then, we have,

\begin{array}{r} \frac{d}{d r} (r^{2} \frac{d R}{d r}) + \frac{2 μ r^{2}}{ℏ^{2}} [E - V (r)] R = λ R, \end{array}

\begin{array}{r} \frac{1}{\sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial Y}{\partial θ}) + \frac{1}{\sin^{2} θ} \frac{\partial^{2} Y}{\partial φ^{2}} = - λ Y . \end{array}

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

Orbital Angular Momentum

$L_{x}$	$L_{y}$	$L_{z}$
$y p_{z} - z p_{y}$	$z p_{x} - x p_{z}$	$x p_{y} - y p_{x}$

Consider the angular momentum: $\vec{L} = \vec{r} \times \vec{p} = \hat{i} (y p_{z} - z p_{y}) - \hat{j} (x p_{z} - z p_{x}) + \hat{k} (x p_{y} - y p_{z})$ . Then, ${\hat{L}}_{x} = \hat{y} {\hat{p}}_{z} - \hat{z} \hat{p_{y}}$ .

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

$Y_{l}^{m} (θ, φ)$ .

$L_{x} = i ℏ (\sin φ \frac{\partial}{\partial θ} + \frac{\cos φ}{\tan θ} \frac{\partial}{\partial φ})$ $L_{y} = i ℏ (- \cos φ \frac{\partial}{\partial θ} + \frac{\sin φ}{\tan θ} \frac{\partial}{\partial φ})$ $L_{z} = - i ℏ \frac{\partial}{\partial φ}$

This gives us, $L^{2} Y = ℏ^{2} λ Y$ . Also, $L_{z} Y = ℏ m Y$ .

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

Commutators

$[L^{2}, L_{i}] = 0$
$[L_{i}, L_{j}] = i ℏ L_{k} ε_{i j k}$

Restrictions on Quantum Numbers

The Magnetic Quantum Number

From $L_{z} Y = ℏ m Y$ we see $L_{z} Y = - i ℏ \frac{\partial}{\partial φ} Y = ℏ m Y$ so $Y = f_{l} (θ) f_{z} (φ)$ . Then, $- i ℏ \frac{d}{d φ} f_{z} (φ) = ℏ m f_{z} (φ)$ gives $f_{z} (φ) = C \exp (i m φ)$ . Thus, it is periodic hence $m \in Z$ .

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

$⟨ L^{2} ⟩ = ⟨ L_{x}^{2} ⟩ + ⟨ L_{y}^{2} ⟩ + ⟨ L_{z}^{2} ⟩ = ⟨ L_{x} ψ | L_{x} ⟩ + ⟨ L_{y} ψ | L_{y} ⟩ + ⟨ L_{z} ψ | L_{z} ⟩ = | | L_{x} | ψ ⟩ | |^{2} + | | L_{y} | ψ ⟩ | |^{2} + | | L_{z} | ψ ⟩ | |^{2} \geq 0$ . Thus, let $λ = ℓ (ℓ + 1)$ and so $ℓ (ℓ + 1) \geq 0$ . Thus, $ℓ \geq 0$ or $ℓ \leq - 1$ . By convention, we choose $ℓ$ non-negative.

More Notes on Angular States

$⟨ ℓ^{'} m^{'} | ℓ m ⟩ = δ_{ℓ^{'} ℓ} δ_{m^{'} m}$ .

So, $\int Y_{ℓ^{'}}^{m^{'} *} Y_{ℓ}^{m} d Ω = δ_{ℓ^{'} ℓ} δ_{m^{'} m}$ .

$d Ω = \sin θ d θ d φ$ .

Ladder Operators

Let $L_{+} = L_{x} + i L_{y}$ and $L_{-} = L_{x} - i L_{y}$ .

Eq 1: $⟨ L^{2} - L_{z}^{2} \mp ℏ L_{z} ⟩ = ℏ^{2} ℓ (ℓ + 1) - ℏ^{2} m^{2} \mp ℏ^{2} m = ℏ^{2} (ℓ \mp m) (ℓ \pm m + 1) \geq 0$ .

Note, $L_{+}^{†} = L_{-}$ .

Then, $⟨ ℓ m | L \mp L_{\pm} | ℓ m ⟩ = ⟨ ℓ m | L_{\pm}^{†} L_{\pm} | ℓ m ⟩ = | | L_{\pm} | ℓ m ⟩ | |^{2} \geq 0$ . Thus, this with Eq 1 gives $| m | \leq ℓ$ .

$L_{+} | ℓ ℓ ⟩ = 0$ since the ladder operator is at the limit. Similarly, $L_{-} | ℓ - ℓ ⟩ = 0$ .

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

Then, $L^{2} L_{\pm} | ℓ m ⟩ = L_{\pm} L^{2} | ℓ m ⟩ = ℏ^{2} ℓ (ℓ + 1) L_{\pm} | ℓ m ⟩$ . Thus, $L_{\pm} | ℓ m ⟩$ does not change $ℓ$ . So, $L_{\pm} | ℓ_{⟩} = | ℓ_{⟩}$ .

Note: $[L_{z}, L_{\pm}] = [L_{z}, L_{x}] \pm [L_{z}, L_{y}] = \pm ℏ L_{\pm}$ . Then, $L_{z} L_{\pm} | ℓ m ⟩ = L_{\pm} L_{z} | ℓ m ⟩ \pm L_{\pm} | ℓ m ⟩ = L_{\pm} [ℏ (m \pm 1)] | ℓ m ⟩$ . Thus, $L_{\pm} | ℓ m ⟩ = c_{\pm} | ℓ m \pm 1 ⟩$ .

$| | L_{\pm} | ℓ m ⟩ | |^{2} = ℏ^{2} (ℓ \mp m) (ℓ \pm m + 1) = | | c_{\pm} | ℓ m \pm 1 | |^{2} = | c_{\pm} |^{2}$ .

$L_{+} = - i ℏ \exp (i φ) [i \frac{\partial}{\partial θ} - \cot θ \frac{\partial}{\partial φ}]$

$L_{+} | ℓ ℓ ⟩ = 0$ so $L_{+} Y_{ℓ}^{ℓ} = 0$ . This gives us a differential equation to solve to find $Y_{ℓ}^{ℓ}$ .

This gives us, $Y_{ℓ}^{ℓ} c_{ℓ} \exp (i ℓ φ) \sin^{ℓ} θ$ . Use the lowering operators to find the other $Y_{ℓ}^{m}$ .

This gives us, $Y_{ℓ}^{m} = \frac{(- 1)^{ℓ}}{2^{ℓ} ℓ!} \sqrt{\dots} \exp (i m φ) \frac{1}{\sin^{m} θ} \frac{d^{ℓ - m}}{d (\cos θ)^{ℓ - m}} (\sin θ)^{2 ℓ}$ .

Note: $L_{\pm} | ℓ m ⟩ = ℏ \sqrt{(ℓ \mp) (ℓ \pm m + 1)} | ℓ m \pm 1 ⟩$ .

Semiclassical Approach

Why do we have $ℓ (ℓ + 1)$ well our maximum $m$ is $ℓ$ and we note that $| L_{z} < | L |$ (due to the uncertainty principle) so $ℏ ℓ \leq ℏ \sqrt{ℓ (ℓ + 1)}$

Assume $L_{z}$ is exactly equal to $| L |$ then $L_{x} = L_{y} = 0$ fully specifying the state but the commutator $[L_{i}, L_{j}] = i ℏ L_{k} ε_{i j k}$ 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: $I = m_{1} r_{1}^{2} + m_{2} r_{2}^{2} = μ r_{e}^{2}$ . $| \vec{L} | = I ω_{R} = μ r_{e}^{2} ω_{R}$ . The Hamiltonian is $\frac{L^{2}}{2 I} = \frac{I ω_{R}}{2} = \frac{L^{2}}{2 μ r_{e}^{2}}$ . Then, $H | ℓ m ⟩ = E_{ℓ} | ℓ m ⟩ = \frac{ℏ^{2} ℓ (ℓ + 1)}{2 μ r_{e}^{2}} | ℓ m ⟩ = B ℓ (ℓ + 1) | ℓ m ⟩$

$E_{ℓ} - E_{ℓ - 1}$ energy level difference: $B ℓ (ℓ + 1) - B (ℓ - 1) ℓ = 2 B ℓ$ . 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 $H_{i n t e r a c t i o n} = - {\vec{d}}_{0} \cdot {\vec{E}}_{l i g h t} = - e Z E_{l i g h t}$ $H_{i n t e r a c t i o n} = - e r_{0} \cos θ E_{l i g h t}$ .

$- e r_{0} E_{l i g h t} \int Y_{ℓ^{'}}^{m^{'} *} \cos θ Y_{ℓ}^{m} d Ω$ Orthogonality: $\cos θ Y_{ℓ}^{m} = c_{1} Y_{ℓ - 1}^{m} + c_{2} Y_{ℓ + 1}^{m}$

$Δ ℓ = ℓ^{'} - ℓ = \pm 1$ , $Δ m = 0, \pm 1$ are selection rules. $Δ m = 0$ for z-polarized and $\pm 1$ for x,y polarized.

$⟨ ℓ^{'} m^{'} | H_{i n t} | ℓ m ⟩ \neq 0$ implies a valid transition arising from the interaction.

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

$P \sim | ⟨ ℓ^{'} m^{'} | H_{i n t} | ℓ m ⟩ |^{2}$

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: $ℏ Ω_{i n}$ goes in and you measure $ℏ Ω_{s c}^{'}$ . $Ω_{s c}^{'} = Ω_{i n} \pm ω_{R}, ω_{R} = \frac{E_{ℓ^{'}} - E_{ℓ}}{ℏ}$ .

The interaction goes as $H_{i n t} \sim Z^{2}$ instead of $Z$ , and so we get $\cos^{2}$ . Thus, the selection rules go as $Δ ℓ = \pm 2, 0$ .

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

Radial Degree of Freedom

\begin{array}{r} \frac{d}{d r} (r^{2} \frac{d R}{d r}) + \frac{2 μ r^{2}}{ℏ^{2}} [E - V (r)] R = λ R, \end{array}

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

We can rearrange to obtain: $\frac{- ℏ^{2}}{2 μ r^{2}} (2 r \frac{d R}{d r} + r^{2} \frac{d^{2} R}{d r^{2}}) + (V (r) + \frac{ℓ (ℓ + 1) ℏ^{2}}{2 μ r^{2}}) R (r) = \frac{- ℏ^{2}}{2 μ r^{2}} (2 r \frac{d R}{d r} + r^{2} \frac{d^{2} R}{d r^{2}}) + V_{e f f} (r) R (r) = E R (r)$ Let $R (r) = \frac{u (r)}{r} = - \frac{ℏ^{2}}{2 μ} \frac{d^{2} u}{d r^{2}} + U_{e f f} (r) u (r) = E u (r)$ . This is like a 1D Schrödinger equation. Some constraints we have: $r \geq 0$ and $u (0) = 0$ . Let $u = u_{k}^{ℓ}$ .

Hydrogen Atom

$V (r) = - \frac{q^{2}}{4 π ε_{0} r} = - \frac{e^{2}}{r}$ Then, $V_{e f f} (r) = - \frac{e^{2}}{r} + \frac{ℓ (ℓ + 1) ℏ^{2}}{2 μ r^{2}}$ . Bohr radius: $a_{0} = \frac{ℏ^{2}}{μ e^{2}}$ , then $ρ = \frac{r}{a_{0}}$ . Also, $E \to E / E_{I}, E_{I} = \frac{μ e^{4}}{2 ℏ^{2}}$ . Let $λ_{k ℓ} = \sqrt{- E_{k, ℓ} / E_{I}}$ , to find the bound state energies. Hence, $V_{e f f} (r) = - \frac{e^{2}}{a_{0} ρ} + \frac{e^{4} ℓ (ℓ + 1)}{4 E_{I} a_{0}^{2} ρ^{2}}$ .

With this, we get $[\frac{d^{2}}{d ρ^{2}} - \frac{ℓ (ℓ + 1)}{ρ} + \frac{2}{ρ} - λ_{k ℓ}^{2}] u_{k}^{ℓ} (ρ) = 0$ .

For $ρ \to \infty$ , we have $[\frac{d^{2}}{d ρ^{2}} - λ_{k ℓ}^{2}] u_{k}^{ℓ} (ρ) = 0$ . $u_{k}^{ℓ} (ρ) = C_{1} \exp (ρ λ_{k ℓ}) + C_{2} \exp (- ρ λ_{k ℓ})$ . Since $C_{1} \neq 0$ has a solution that blows up, we find that at infinity we get $u_{k, ℓ} (ρ) = C \exp (- ρ λ_{k, ℓ})$ . Then, $u_{k}^{ℓ} (ρ) = \exp (- ρ λ_{k ℓ}) y_{k}^{ℓ} (ρ)$ .

From this, our original differential equation becomes: $[\frac{d^{2}}{d ρ^{2}} - 2 λ_{k, ℓ} \frac{d}{d ρ} + \frac{2}{ρ} - \frac{ℓ (ℓ + 1)}{ρ^{2}}] y_{k}^{ℓ} (ρ) = 0$ . We use a power series to solve: $y_{k}^{ℓ} (ρ) = ρ^{s} \sum_{q = 0}^{\infty} c_{q} ρ^{q}$ , Frobenius series.

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

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

If we check convergence: $\frac{c_{q}}{c_{q - 1}} \to \frac{2 λ_{k ℓ}}{q}$ . 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 $λ_{k ℓ} (k + ℓ) = 1$ . Then, $λ_{k ℓ} = \frac{1}{k + ℓ}$ . Note that $λ$ was the energy so we get quantization of energy. Hence, $E_{k, ℓ} = - \frac{E_{I}}{k + ℓ}$ with $k = 1, 2, 3 \dots$ .

Typically, we have the principle quantum number $n = k + ℓ$ then $E_{n} = - \frac{E_{I}}{n^{2}}$ with $n = 1, 2, \dots$ . Then in this case we get $ℓ = n - k \Rightarrow n - 1, n - 2, \dots, 0$ .

Therefore, $R_{n, ℓ} (r) = - {(\frac{2}{n a_{0}})}^{\frac{3}{2}} \sqrt{\frac{(n - ℓ - 1)!}{2 n (n - ℓ)!^{3}}} {(\frac{2 r}{n a_{0}})}^{ℓ} \exp (\frac{- r}{n a_{0}}) L_{n - ℓ - 1}^{2 ℓ + 1} (\frac{2 r}{n a_{0}})$ .

Hydrogen-Like Atoms

$Ψ_{n ℓ m} (r, θ, φ) = R_{n}^{ℓ} (r) Y_{ℓ}^{m} (θ, φ)$ . Note, that $ℓ = 0$ is special since you will have a non-zero probability for $r \to 0$ . All other $ℓ$ ’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? $r = 0$ @ $100 = n ℓ m$ . Well, if you see in the probability density in the integral, we get a radial integration of $| R_{n}^{ℓ} (r) |^{2} r^{2}$ which gives a probability density of $| R_{n}^{ℓ} (r) r |^{2}$ 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. $⟨ r ⟩_{100} = \int_{0}^{\infty} | R_{10} |^{2} r \cdot r^{2} d r = \int_{0}^{\infty} | R_{10} |^{2} r^{3} d r$ . Use a gamma function! $\int_{0}^{\infty} x^{n} \exp (- x) d x = n! = Γ (n + 1)$ . Thus, we get $\int d r = 4 a_{0}^{3} {(\frac{a_{0}}{2})}^{4} 3! = a_{0} \frac{3}{2}$ .

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 $| R |^{2} r^{2}$ , 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. $g_{n} = \sum_{ℓ = 0}^{n - 1} (2 ℓ + 1) = n (n - 1) + n = n^{2}$ . With spin, we get $g_{n} = 2 n^{2}$ .

Shells

Shells: $n = 1234 \dots = K L M N \dots$ . Subshells: $ℓ = 01234 \dots = s p d f g \dots$ .

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 $- Z e^{2} / r$ .

We expect $E_{n} = - Z^{2} \frac{E_{I}}{n^{2}}$ . Recall $E_{I} = \frac{μ e^{4}}{2 ℏ^{2}}$ , 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 $Z e$ 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. $a_{0} (Z) = \frac{a_{0}^{H}}{Z}$ . $a_{0}^{H} = \frac{ℏ^{2}}{μ e^{2}}$ .

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 $V (r < R) = 0$ and $V = \infty$ otherwise. Let $Ψ (r, θ, φ) = R_{? ℓ} (r) Y_{ℓ}^{m} (θ, φ)$ . We now just need to solve for the $R (r)$ equation. Note, $R (r) = u (r) / r$ gives the SE: $- \frac{ℏ^{2}}{2 m} \frac{d^{2} u}{d r^{2}} + V_{e f f} (r) u (r) = E u (r)$ with $V_{e f f} = V (r) + \frac{ℏ^{2} ℓ (ℓ + 1)}{2 μ r^{2}}$ . So, $V_{e f f} (r < R) = \frac{ℏ^{2} ℓ (ℓ + 1)}{2 μ r^{2}}$ and $V_{e f f} (r > R) = \infty$ . The outer limit means the wave function must be zero at the boundary. Hence, $\frac{d^{2} u_{ℓ}}{d r^{2}} + [\frac{2 μ E}{ℏ^{2}} - \frac{ℓ (ℓ + 1)}{r^{2}}] (r) u_{ℓ} (r) = 0$ .

Ansatz: $R_{n}^{ℓ} (r) = A j_{n}^{ℓ} (r)$ .

By letting $u_{ℓ} (r) = \sqrt{z} η_{ℓ} (z)$ we get $\frac{d^{2} η}{d z^{2}} + \frac{1}{z} \frac{d η_{ℓ}}{d z} + (1 - \frac{(ℓ + 1 / 2)^{2}}{z^{2}}) η_{ℓ} (z) = 0$ . $η_{ℓ} (z) = C_{1} J_{ℓ + 1 / 2} (z) + C_{2} J_{- (ℓ + 1 / 2)} (z)$ .

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

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

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

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

Thus, $u_{ℓ} (r) = k r (C_{1} j_{ℓ} (k r) + C_{2} n_{ℓ} (k r))$ .

So, for our problem, $u_{ℓ} (r) = k r C_{1} j_{ℓ} (k r)$ from $r \to 0$ implies $u (r) \to 0$ . Then, we find for $r = R$ , $j_{ℓ} (k R) = 0$ . So, we get $u_{n}^{ℓ} (r) = k r C j_{ℓ} (k_{n}^{ℓ} r)$ such that $j_{ℓ} (k_{n}^{ℓ} R) = 0$ .

Note: $j_{0} (z) = \frac{\sin z}{z}$ , $j_{1} (z) = \frac{\sin z}{z^{2}} - \frac{\cos z}{z}$ and $j_{2} (z) = (\frac{3}{z^{3}} - \frac{1}{z}) \sin (z) - \frac{3}{z} \cos z$ . So, $R_{n}^{ℓ} (r) = C j_{ℓ} (k_{n}^{ℓ} r)$ .

Note: $E = \frac{ℏ^{2} k_{n}^{ℓ 2}}{2 μ}$ .