Angular Momentum and the Hydrogen Atom

$\hat{L}_x = -i\hbar\left(y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y}\right), \quad \hat{L}_y = -i\hbar\left(z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z}\right), \quad \hat{L}_z = -i\hbar\left(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}\right)$

Commutation relations:

$[\hat{L}_x, \hat{L}_y] = i\hbar\hat{L}_z, \quad [\hat{L}_y, \hat{L}_z] = i\hbar\hat{L}_x, \quad [\hat{L}_z, \hat{L}_x] = i\hbar\hat{L}_y$

$[\hat{L}^2, \hat{L}_i] = 0 \quad \mathrm{for\ all\ } i$

Simultaneous eigenstates: $|l, m\rangle$ with

$\hat{L}^2|l,m\rangle = \hbar^2 l(l+1)|l,m\rangle, \quad \hat{L}_z|l,m\rangle = \hbar m|l,m\rangle$

Where $l = 0, 1, 2, \ldots$ and $m = -l, -l+1, \ldots, l-1, l$.

Define the ladder operators:

$\hat{L}_{\pm} = \hat{L}_x \pm i\hat{L}_y$

Key commutation relations:

$[\hat{L}_z, \hat{L}_{\pm}] = \pm\hbar\hat{L}_{\pm}, \quad [\hat{L}^2, \hat{L}_{\pm}] = 0$

Proof. $[\hat{L}_z, \hat{L}_+] = [\hat{L}_z, \hat{L}_x] + i[\hat{L}_z, \hat{L}_y] = i\hbar\hat{L}_y + i(i\hbar\hat{L}_x) = \hbar(\hat{L}_y + i\hat{L}_x)\cdot(-1)$

Wait, let us redo this carefully:

$[\hat{L}_z, \hat{L}_+] = [\hat{L}_z, \hat{L}_x + i\hat{L}_y] = [\hat{L}_z, \hat{L}_x] + i[\hat{L}_z, \hat{L}_y] = i\hbar\hat{L}_y + i(-i\hbar\hat{L}_x) = i\hbar\hat{L}_y + \hbar\hat{L}_x = \hbar(\hat{L}_x + i\hat{L}_y) = \hbar\hat{L}_+$

Similarly, $[\hat{L}_z, \hat{L}_-] = -\hbar\hat{L}_-$. And:

$[\hat{L}^2, \hat{L}_+] = [\hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2, \hat{L}_+] = 0$

Since $\hat{L}^2$ commutes with each component. $\blacksquare$

Action on eigenstates. Since $[\hat{L}_z, \hat{L}_+] = \hbar\hat{L}_+$:

$\hat{L}_z(\hat{L}_+|l,m\rangle) = (\hat{L}_+\hat{L}_z + \hbar\hat{L}_+)|l,m\rangle = \hbar(m+1)(\hat{L}_+|l,m\rangle)$

So $\hat{L}_+|l,m\rangle$ is an eigenstate of $\hat{L}_z$ with eigenvalue $\hbar(m+1)$: it raises $m$ by 1. Similarly, $\hat{L}_-$ lowers $m$ by 1. Both preserve the $l$ value since $[\hat{L}^2, \hat{L}_{\pm}] = 0$.

Normalisation. Write $\hat{L}_+|l,m\rangle = C_+(l,m)|l,m+1\rangle$. Then:

$|C_+(l,m)|^2 = \langle l,m|\hat{L}_-\hat{L}_+|l,m\rangle$

Using $\hat{L}_-\hat{L}_+ = \hat{L}^2 - \hat{L}_z^2 - \hbar\hat{L}_z$:

$|C_+(l,m)|^2 = \hbar^2 l(l+1) - \hbar^2 m^2 - \hbar^2 m = \hbar^2[l(l+1) - m(m+1)]$

Therefore:

$\hat{L}_+|l,m\rangle = \hbar\sqrt{l(l+1) - m(m+1)}\,|l,m+1\rangle$

$\hat{L}_-|l,m\rangle = \hbar\sqrt{l(l+1) - m(m-1)}\,|l,m-1\rangle$

Theorem 6.1. The quantum numbers $l$ and $m$ satisfy:

• $l = 0, 1/2, 1, 3/2, 2, \ldots$ (integer or half-integer)
• For a given $l$: $m = -l, -l+1, \ldots, l-1, l$ (there are $2l+1$ values)
• For orbital angular momentum, $l$ is restricted to non-negative integers.

Proof. Starting from a state $|l,m\rangle$Repeatedly applying $\hat{L}_+$ raises $m$ by 1 each time. The norm of the resulting state is:

$\|\hat{L}_+|l,m\rangle\|^2 = \hbar^2[l(l+1) - m(m+1)]$

This must remain non-negative, so $m(m+1) \leq l(l+1)$Giving $m \leq l$. The raising process must Terminate at some maximum $m_{\max}$ where $\hat{L}_+|l, m_{\max}\rangle = 0$:

$l(l+1) - m_{\max}(m_{\max} + 1) = 0$

Similarly, the lowering process terminates at $m_{\min}$ where $\hat{L}_-|l, m_{\min}\rangle = 0$:

$l(l+1) - m_{\min}(m_{\min} - 1) = 0$

Subtracting: $m_{\max}(m_{\max}+1) - m_{\min}(m_{\min}-1) = 0$. Since we reach $m_{\max}$ from $m_{\min}$ in $N$ steps: $m_{\max} = m_{\min} + N$. Solving gives $m_{\max} = l$ and $m_{\min} = -l$ So $N = 2l$Meaning $2l$ must be a non-negative integer. Therefore $l = 0, 1/2, 1, 3/2, \ldots$ and $m$ takes $2l+1$ values from $-l$ to $l$. $\blacksquare$

For orbital angular momentum (defined as $\hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}}$), The wave function must be single-valued under a full rotation $\phi \to \phi + 2\pi$. This requires $e^{im\phi} = e^{im(\phi+2\pi)}$ So $m$ must be an integer, which restricts $l$ to integers.

The simultaneous eigenfunctions of $\hat{L}^2$ and $\hat{L}_z$ are the spherical harmonics $Y_l^m(\theta, \phi)$:

$Y_l^m(\theta, \phi) = (-1)^m\sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}\,P_l^m(\cos\theta)\,e^{im\phi}$

Where $P_l^m$ are the associated Legendre functions.

Properties:

• Orthonormality: \int Y_l^m^*\, Y_{l"}^{m'}\,d\Omega = \delta_{ll'}\delta_{mm'}
• Completeness: \sum_{l=0}^{\infty}\sum_{m=-l}^{l} Y_l^m(\theta,\phi)\,Y_l^m^*(\theta',\phi') = \delta(\cos\theta - \cos\theta')\delta(\phi - \phi')
• Parity: $Y_l^m(\pi-\theta, \phi+\pi) = (-1)^l\,Y_l^m(\theta,\phi)$

First few spherical harmonics:

The Hamiltonian for hydrogen (electron of mass $m_e$ and charge $-e$Proton of charge $+e$):

$\hat{H} = -\frac{\hbar^2}{2m_e}\nabla^2 - \frac{e^2}{4\pi\varepsilon_0 r}$

In spherical coordinates, the Laplacian separates, and we write $\psi(r,\theta,\phi) = R(r)\,Y_l^m(\theta,\phi)$. The radial equation is:

$-\frac{\hbar^2}{2m_e}\frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right) + \left[-\frac{e^2}{4\pi\varepsilon_0 r} + \frac{\hbar^2 l(l+1)}{2m_e r^2}\right]R = ER$

The term $\hbar^2 l(l+1)/(2m_e r^2$ acts as an effective centrifugal barrier.

Substitute $u(r) = rR(r)$ and define the Bohr radius $a_0 = 4\pi\varepsilon_0\hbar^2/(m_e e^2)$ and the Rydberg energy $E_R = e^2/(8\pi\varepsilon_0 a_0) = m_e e^4/(8\varepsilon_0^2 h^2)$. With the substitution $\rho = 2r/(na_0)$The radial equation becomes:

$\frac{d^2u}{d\rho^2} = \left[\frac{l(l+1)}{\rho^2} - \frac{1}{\rho} + \frac{n}{4}\left(\frac{1}{n^2} - \frac{E}{E_R}\right)\right]u$

For the solution to be well-behaved at both $\rho = 0$ and $\rho \to \infty$We require:

$E = -\frac{E_R}{n^2} = -\frac{m_e e^4}{2(4\pi\varepsilon_0)^2\hbar^2}\cdot\frac{1}{n^2}$

With $n = 1, 2, 3, \ldots$ and $l = 0, 1, \ldots, n-1$.

The radial wave functions are:

$R_{nl}(r) = \sqrt{{\left(\frac{2}{na_0}\right)}^3\frac{(n-l-1)!}{2n[(n+l)!]^3}}\,e^{-r/(na_0)}\!\left(\frac{2r}{na_0}\right)^l L_{n-l-1}^{2l+1}\!\left(\frac{2r}{na_0}\right)$

Where $L_q^p$ are the associated Laguerre polynomials.

Energy eigenvalues:

E_n = -\frac{m_e e^4}{2(4\pi\varepsilon_0)^2 \hbar^2} \cdot \frac{1}{n^2} = -\frac{13.6\,\mathrm{eV}{n^2}, \quad n = 1, 2, 3, \ldots}

Degeneracy: Each energy level $E_n$ has degeneracy $n^2$ (ignoring spin). The quantum numbers are:

• Principal: $n = 1, 2, 3, \ldots$
• Orbital angular momentum: $l = 0, 1, \ldots, n - 1$
• Magnetic: $m_l = -l, \ldots, l$

The ground state wave function ($n = 1, l = 0, m_l = 0$):

$\psi_{100}(r, \theta, \phi) = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0}$

Where $a_0 = \frac{4\pi\varepsilon_0 \hbar^2}{m_e e^2} \approx 0.529\,\mathrm{\AA}$ is the Bohr radius.

Example 6.1. Calculate $\langle r \rangle$, $\langle r^2 \rangle$ And $\langle 1/r \rangle$ for the Hydrogen ground state.

Electric dipole transitions between hydrogen states are governed by selection rules derived from the Wigner-Eckart theorem. For a transition $|n,l,m\rangle \to |n',l',m'\rangle$ induced by the electric Dipole operator $\hat{\mathbf{r}}$:

$\Delta l = l' - l = \pm 1, \quad \Delta m = m' - m = 0, \pm 1$

$\Delta n$ is unrestricted (energy conservation determines which transitions are allowed).

Proof sketch. The matrix element $\langle n'l'm'|\hat{z}|nlm\rangle$ involves the integral $\int Y_{l'}^{m'*}(\theta,\phi)\cos\theta\,Y_l^m(\theta,\phi)\,d\Omega$. Using the addition theorem For spherical harmonics, $\cos\theta = \sqrt{4\pi/3}\,Y_1^0$The integral becomes a product of Clebsch-Gordan coefficients that vanishes unless $l' = l \pm 1$ and $m' = m$. $\blacksquare$

The three quantum numbers characterise hydrogen atom eigenstates:

• $n$ (principal): Determines the energy and overall size. The mean radius scales as $\langle r \rangle \propto n^2 a_0$.
• $l$ (orbital angular momentum): Determines the shape. The spectroscopic notation is $l = 0$ (s), $l = 1$ (p), $l = 2$ (d), $l = 3$ (f), etc.
• $m_l$ (magnetic): Determines the spatial orientation. The angular dependence is $Y_l^{m_l}(\theta, \phi)$.

Radial probability distribution. The probability of finding the electron between $r$ and $r+dr$ is $P(r)\,dr = |R_{nl}(r)|^2 r^2\,dr$. For the $1s$ state, the maximum is at $r = a_0$ (the Bohr radius). For $2s$There is a node at $r = 2a_0$. For $2p$The distribution peaks closer to the nucleus.

Angular distributions. The $s$ orbitals ($l = 0$) are spherically symmetric. The $p$ orbitals ($l = 1$) have dumbbell shapes aligned along the $x$-, $y$-, or $z$-axis depending on $m_l$. The $d$ Orbitals ($l = 2$) have more complex cloverleaf patterns.

Radial nodes. The radial wave function $R_{nl}(r)$ has $n - l - 1$ nodes (zeros excluding $r = 0$ And $r = \infty$). The total number of nodes in the full wave function is $n - 1$Consistent with The general property that the $n$-th energy eigenstate has $n - 1$ nodes.

Fine structure. The non-relativistic Schrodinger equation gives energy levels depending only on $n$. Relativistic corrections (spin-orbit coupling, Darwin term, kinetic energy correction) split these Into fine structure multiplets, removing the $l$-degeneracy. The fine structure shift is of order $\alpha^2 E_n$ where $\alpha \approx 1/137$ is the fine structure constant.

flowchart TD
    A[6_Angular Momentum And The Hydrogen Atom] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Angular momentum in quantum mechanics is like a spinning top that can only point in certain discrete directions. The ladder operators are like climbing stairs: you can raise or lower the magnetic quantum number one step at a time, but you cannot stop between steps. The hydrogen atom is nature’s simplest atom and the hydrogen of quantum mechanics: the electron orbits the nucleus not in fixed paths but in probability clouds shaped like dumbbells, clovers, and spheres. The quantum numbers are like an address: n tells you the floor (energy), l tells you the room shape (angular momentum), and m tells you the orientation. The energy depends only on n, which is a hidden symmetry of the Coulomb potential. The selection rules are like dance rules: the electron can only jump between certain states, which is why hydrogen emits only specific colors of light.

• Operators and Observables: Operators represent physical observables and their commutation relations determine the structure of angular momentum.

• Spin: Spin is an intrinsic angular momentum that extends the orbital angular momentum formalism.

• Scattering Theory: Scattering theory applies angular momentum techniques to analyze particle collisions.

• Calculus

• Linear Algebra

• Vector Calculus

• Quantum Computing

Mistake 1: Confusing the orbital quantum number $l$ with the magnetic quantum number $m$ For a given $l$, the magnetic quantum number $m$ ranges from $-l$ to $l$ in integer steps, giving $2l + 1$ values. Students sometimes confuse these and assign $m$ values outside this range. The quantum number $l$ determines the total angular momentum magnitude $\hbar\sqrt{l(l+1)}$, while $m$ determines the $z$-component $\hbar m$.

Mistake 2: Assuming $L_z$ can equal $L$ The $z$-component of angular momentum satisfies $|L_z| \leq \hbar l < \hbar\sqrt{l(l+1)} = |\mathbf{L}|$. The magnitude of $\mathbf{L}$ is always greater than the magnitude of any single component. This is a consequence of the uncertainty relations between $L_x$, $L_y$, and $L_z$: they cannot all be simultaneously determined.

Mistake 3: Forgetting that $l$ is restricted to integers for orbital angular momentum For orbital angular momentum (defined as $\hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}}$), $l$ must be a non-negative integer because the wave function must be single-valued under a full rotation. Half-integer $l$ values are allowed for spin angular momentum but not for orbital angular momentum. Never assign $l = 1/2$ to an orbital state.