quantum mechanics ii

The Stark effect is the splitting of atomic energy levels by an external electric field $\mathcal{E}$. For hydrogen, the perturbation is $H" = e\mathcal{E}z$ (taking the field along $z$).

Linear Stark effect (degenerate case). For hydrogen, states with the same $n$ but different $l$ are degenerate. Consider the $n = 2$ manifold $\{|200\rangle, |210\rangle, |211\rangle, |21{-1}\rangle\}$. The perturbation matrix within the degenerate subspace:

$\langle n'l'm'|z|nlm\rangle \propto \delta_{l',l\pm 1}\,\delta_{m',m}$

Only $\langle 200|z|210\rangle$ and $\langle 210|z|200\rangle$ are nonzero. Defining the “parabolic basis” states:

$|\psi_1\rangle = \frac{1}{\sqrt{2}}(|200\rangle + |210\rangle), \quad |\psi_2\rangle = \frac{1}{\sqrt{2}}(|200\rangle - |210\rangle)$

The first-order energy shifts are:

$E^{(1)} = \pm 3ea_0\mathcal{E}, \quad 0, \quad 0$

The quadratic Stark effect arises in non-hydrogenic atoms (where $l$-degeneracy is lifted). The second-order energy shift is proportional to $\mathcal{E}^2$:

$E_n^{(2)} = -\frac{1}{2}\alpha_n\mathcal{E}^2$

where $\alpha_n$ is the electric polarizability. For the ground state of hydrogen, $\alpha_1 = \frac{9}{2}a_0^3$.

An external magnetic field $\mathbf{B} = B\hat{z}$ splits atomic levels. The perturbation $H' = -\boldsymbol{\mu}_L \cdot \mathbf{B} - \boldsymbol{\mu}_S \cdot \mathbf{B}$:

$H' = \frac{eB}{2m_e}(L_z + 2S_z) = \frac{e\hbar B}{2m_e}(J_z + S_z)$

Normal Zeeman effect (spin neglected): $H' = \omega_L L_z$ where $\omega_L = eB/(2m_e)$ is the Larmor frequency. Each level $l$ splits into $2l + 1$ equally spaced sublevels separated by $\hbar\omega_L$.

Anomalous Zeeman effect (weak field, $B \ll B_{\text{SO}}$): Use $|n, l, s, j, m_j\rangle$ as the unperturbed basis. The first-order shift:

$E_Z = g_J\mu_B B\,m_j$

where $\mu_B = e\hbar/(2m_e)$ is the Bohr magneton and the Landé $g$-factor is:

$g_J = 1 + \frac{j(j+1) + s(s+1) - l(l+1)}{2j(j+1)}$

For a pure orbital state ($s = 0$): $g_J = 1$. For a pure spin state ($l = 0$): $g_J = 2$.

Paschen–Back effect (strong field, $B \gg B_{\text{SO}}$): $L_z$ and $S_z$ decouple, and the spin-orbit interaction is treated as the perturbation. The energy shifts approach the normal Zeeman pattern.

Using the effective nuclear charge trial function $\psi_{\text{trial}} = (Z_{\text{eff}}^3/\pi a_0^3)\exp[-Z_{\text{eff}}(r_1 + r_2)/a_0]$:

$E(Z_{\text{eff}}) = \left[Z_{\text{eff}}^2 - 4Z_{\text{eff}} + \frac{5}{4}Z_{\text{eff}}\right]\text{Ry} = \left[Z_{\text{eff}}^2 - \frac{11}{4}Z_{\text{eff}}\right]\text{Ry}$

Minimising: $dE/dZ_{\text{eff}} = 0$ gives $Z_{\text{eff}} = 27/16 = 1.6875$:

$E_{\min} = -\left(\frac{27}{16}\right)^2 \times 13.6\,\text{eV} = -77.5\,\text{eV}$

The experimental value is $-79.0$ eV. The variational result is within 2% and captures the screening effect: each electron partially shields the nucleus, reducing the effective charge below $Z = 2$.

Given two angular momenta $\mathbf{J}_1$ and $\mathbf{J}_2$, the total $\mathbf{J} = \mathbf{J}_1 + \mathbf{J}_2$. The possible values of $j$ are:

$j = |j_1 - j_2|, \; |j_1 - j_2| + 1, \; \ldots, \; j_1 + j_2$

The coupled basis $|j_1 j_2; jm\rangle$ is related to the uncoupled basis $|j_1 m_1\rangle|j_2 m_2\rangle$ by the Clebsch–Gordan (CG) decomposition:

$|j_1 j_2; jm\rangle = \sum_{m_1, m_2} \langle j_1 m_1\, j_2 m_2 | jm\rangle \; |j_1 m_1\rangle |j_2 m_2\rangle$

where the sum runs over $m_1 + m_2 = m$ and the coefficients $\langle j_1 m_1\, j_2 m_2 | jm\rangle$ are CG coefficients.

Properties of CG coefficients:

• Orthogonality:

$\sum_{m_1, m_2} \langle j_1 m_1\, j_2 m_2 | jm\rangle\langle j_1 m_1\, j_2 m_2 | j'm'\rangle = \delta_{jj'}\delta_{mm'}$

$\sum_{j, m} \langle j_1 m_1\, j_2 m_2 | jm\rangle\langle j_1 m_1'\, j_2 m_2' | jm\rangle = \delta_{m_1 m_1'}\delta_{m_2 m_2'}$

• Symmetry: $\langle j_1 m_1\, j_2 m_2 | jm\rangle = (-1)^{j_1 + j_2 - j}\langle j_2 m_2\, j_1 m_1 | jm\rangle$

• Special values: $\langle j_1 j_1\, j_2 (m - j_1) | jm\rangle$ and $\langle j_1 (-j_1)\, j_2 (j) | j_0\rangle$ are positive.

Example: Two spin-1/2 particles. The addition $s_1 = s_2 = 1/2$ gives $S = 0, 1$:

$|1, 1\rangle = |{+}\rangle|{+}\rangle$ $|1, 0\rangle = \frac{1}{\sqrt{2}}\bigl(|{+}\rangle|{-}\rangle + |{-}\rangle|{+}\rangle\bigr)$ $|1, {-1}\rangle = |{-}\rangle|{-}\rangle$ $|0, 0\rangle = \frac{1}{\sqrt{2}}\bigl(|{+}\rangle|{-}\rangle - |{-}\rangle|{+}\rangle\bigr)$

The spin-orbit interaction in hydrogen (from the electron’s perspective in the nuclear rest frame):

$H_{\text{SO}} = \frac{1}{2m_e^2 c^2}\frac{1}{r}\frac{dV}{dr}\,\mathbf{L}\cdot\mathbf{S}$

For the Coulomb potential $V = -e^2/(4\pi\varepsilon_0 r)$:

$H_{\text{SO}} = \frac{e^2}{8\pi\varepsilon_0 m_e^2 c^2}\frac{\mathbf{L}\cdot\mathbf{S}}{r^3}$

Using $\mathbf{L}\cdot\mathbf{S} = \frac{1}{2}[\mathbf{J}^2 - \mathbf{L}^2 - \mathbf{S}^2]$:

$\langle H_{\text{SO}}\rangle = \frac{\hbar^2 e^2}{16\pi\varepsilon_0 m_e^2 c^2}\frac{j(j+1) - l(l+1) - s(s+1)}{a_0^3 n^3 l(l+1/2)(l+1)}$

This vanishes for $l = 0$ (the Darwin term takes over for $s$-states).

LS (Russel–Saunders) coupling. Applicable when the residual electrostatic interaction between electrons dominates over spin-orbit coupling (light atoms, $Z \lesssim 30$):

1. Couple orbital momenta: $\mathbf{L} = \sum_i \mathbf{l}_i$, giving total $L$
2. Couple spins: $\mathbf{S} = \sum_i \mathbf{s}_i$, giving total $S$
3. Couple to total: $\mathbf{J} = \mathbf{L} + \mathbf{S}$, with $J = |L - S|, \ldots, L + S$

States are labelled $^{2S+1}L_J$ (spectroscopic notation). For carbon ($1s^2 2s^2 2p^2$): the valence configuration gives $L = 0, 1, 2$ and $S = 0, 1$. The allowed terms are $^1S_0$, $^3P_{0,1,2}$, $^1D_2$.

jj coupling. Dominates when spin-orbit coupling exceeds the electrostatic interaction (heavy atoms, $Z \gtrsim 80$):

1. Couple each electron: $\mathbf{j}_i = \mathbf{l}_i + \mathbf{s}_i$
2. Couple to total: $\mathbf{J} = \sum_i \mathbf{j}_i$

The matrix elements of a spherical tensor operator $T_q^{(k)}$ (rank $k$, component $q$) between angular momentum eigenstates are:

$\langle j'm' | T_q^{(k)} | jm \rangle = \langle jm\, kq | j'm' \rangle \;\langle j' || T^{(k)} || j \rangle$

The first factor is a CG coefficient (containing all angular dependence) and the second is the reduced matrix element (independent of $m, m', q$).

Consequences:

• Selection rules follow directly: $|j - j'| \leq k \leq j + j'$, $m' = m + q$
• All matrix elements with the same $j, j', k$ are determined by a single reduced matrix element
• Parity: $T_q^{(k)}$ has parity $(-1)^k$, so $\langle j'm'|T_q^{(k)}|jm\rangle \neq 0$ only if $(-1)^{l' + k + l} = +1$

For identical particles, the field operators create and annihilate particles at points in space:

$\hat{\psi}(\mathbf{r}) = \sum_i a_i \phi_i(\mathbf{r}), \quad \hat{\psi}^\dagger(\mathbf{r}) = \sum_i a_i^\dagger \phi_i^*(\mathbf{r})$

Creation/annihilation operator commutation relations:

Bosons: $[a_i, a_j^\dagger] = \delta_{ij}$, $[a_i, a_j] = 0$, $[a_i^\dagger, a_j^\dagger] = 0$

Fermions: $\{a_i, a_j^\dagger\} = \delta_{ij}$, $\{a_i, a_j\} = 0$, $\{a_i^\dagger, a_j^\dagger\} = 0$

The many-body Hamiltonian for interacting fermions:

$\hat{H} = \sum_{ij} \langle i | h | j \rangle a_i^\dagger a_j + \frac{1}{2}\sum_{ijkl} \langle ij | V | kl \rangle a_i^\dagger a_j^\dagger a_l a_k$

where $h$ is the single-particle Hamiltonian and $V$ is the two-body interaction.

The Heitler–London model treats the hydrogen molecule $H_2$ using valence bond theory. The spatial wavefunction for two electrons near protons $A$ and $B$:

$\Psi_S = \frac{1}{\sqrt{2(1 + S^2)}}\bigl[\psi_A(1)\psi_B(2) + \psi_B(1)\psi_A(2)\bigr] \quad \text{(singlet, bonding)}$

$\Psi_T = \frac{1}{\sqrt{2(1 - S^2)}}\bigl[\psi_A(1)\psi_B(2) - \psi_B(1)\psi_A(2)\bigr] \quad \text{(triplet, antibonding)}$

where $S = \langle\psi_A|\psi_B\rangle$ is the overlap integral. The energies:

$E_{\text{singlet}}(R) = \frac{Q + A}{1 + S^2}, \qquad E_{\text{triplet}}(R) = \frac{Q - A}{1 - S^2}$

where $Q$ is the Coulomb integral and $A$ is the exchange integral (positive). The exchange integral $A$ is responsible for covalent bonding — it has no classical analogue and is purely quantum-mechanical. The singlet state has a minimum at $R_e \approx 1.64\,a_0$ with binding energy $\sim 3.15$ eV (experiment: 4.75 eV).

The Hartree–Fock method finds the best single Slater determinant by solving self-consistent equations. For orbital $\phi_i$:

$\hat{F}\,\phi_i(\mathbf{r}) = \varepsilon_i\,\phi_i(\mathbf{r})$

where the Fock operator is:

$\hat{F} = h + \sum_{j=1}^{N}\bigl(\hat{J}_j - \hat{K}_j\bigr)$

• Coulomb operator: $\hat{J}_j\phi_i(\mathbf{r}) = \left[\int \frac{|\phi_j(\mathbf{r}')|^2}{|\mathbf{r} - \mathbf{r}'|}\,d^3r'\right]\phi_i(\mathbf{r})$
• Exchange operator: $\hat{K}_j\phi_i(\mathbf{r}) = \left[\int \frac{\phi_j^*(\mathbf{r}')\phi_i(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\,d^3r'\right]\phi_j(\mathbf{r})$

The total Hartree–Fock energy:

$E_{\text{HF}} = \sum_{i=1}^{N}\langle\phi_i|h|\phi_i\rangle + \frac{1}{2}\sum_{i,j=1}^{N}\bigl(\langle ij|ij\rangle - \langle ij|ji\rangle\bigr)$

where $\langle ij|ij\rangle = J$ is the direct integral and $\langle ij|ji\rangle = K$ is the exchange integral.

Koopmans’ theorem: The Hartree–Fock orbital energy $\varepsilon_i$ approximates the ionisation energy of electron $i$ (and the electron affinity for unoccupied orbitals). This provides a theoretical justification for photoelectron spectroscopy interpretation.

The scattering state $|\psi^{(+)}\rangle$ satisfies the Lippmann–Schwinger equation:

$|\psi^{(+)}\rangle = |\phi\rangle + \hat{G}_0^{(+)}V|\psi^{(+)}\rangle$

where $|\phi\rangle$ is a free plane wave, $V$ is the scattering potential, and $\hat{G}_0^{(+)}$ is the free retarded Green’s function:

$\hat{G}_0^{(+)}(E) = \frac{1}{E - \hat{H}_0 + i\epsilon}$

Iterating gives the Born series:

$|\psi^{(+)}\rangle = |\phi\rangle + \hat{G}_0 V|\phi\rangle + \hat{G}_0 V\hat{G}_0 V|\phi\rangle + \cdots$

The first iteration reproduces the Born approximation; higher iterations include multiple scattering events.

The transition matrix (T-matrix) encapsulates all scattering information:

$\hat{T}(E) = V + V\hat{G}_0^{(+)}(E)\hat{T}(E)$

The scattering amplitude is:

$f(\mathbf{k}', \mathbf{k}) = -\frac{m}{2\pi\hbar^2}\langle\mathbf{k}'|\hat{T}(E)|\mathbf{k}\rangle$

The Born approximation is $T \approx V$, and the full solution sums all repeated scatterings.

At low energies, electron scattering off noble gas atoms exhibits a pronounced minimum in the total cross section. For electron–argon scattering, $\sigma$ drops to near zero at $E \approx 0.7$ eV.

Explanation via partial wave analysis. The $l = 0$ phase shift passes through zero ($\delta_0 \approx n\pi$) at a specific energy. Since $\sigma \approx 4\pi a_s^2$ at low energy and the scattering length $a_s = -\tan(\delta_0)/k \approx 0$, the cross section nearly vanishes. This is a quantum-mechanical transparency caused by destructive interference between the incoming and scattered waves.

For low-energy $s$-wave scattering, the phase shift is parameterised by:

$k\cot\delta_0 = -\frac{1}{a_s} + \frac{1}{2}r_e\,k^2 + O(k^4)$

where $a_s$ is the scattering length and $r_e$ is the effective range. The cross section:

$\sigma = \frac{4\pi}{k^2 + (k\cot\delta_0)^2} \;\xrightarrow{k \to 0}\; \frac{4\pi a_s^2}{1 + k^2 a_s^2}$

A large positive scattering length ($|a_s| \gg r_e$) signals a near-threshold bound state (as in the deuteron, $a_s \approx 5.4$ fm).

For a spin-0 particle of mass $m$, imposing $E^2 = p^2c^2 + m^2c^4$ as an operator equation gives:

$\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2 + \frac{m^2c^2}{\hbar^2}\right)\psi = 0$

Problems with the Klein–Gordon equation:

• Second-order in time (requires initial conditions on $\psi$ and $\partial\psi/\partial t$)
• Negative energy solutions: $E = \pm\sqrt{p^2c^2 + m^2c^4}$
• Probability density $\rho = (i\hbar/2mc^2)(\psi^*\dot\psi - \psi\dot\psi^*)$ is not positive-definite
• The conserved current is $j^\mu = (i\hbar/2m)(\phi^*\partial^\mu\phi - \phi\partial^\mu\phi^*)$ but $\rho$ can be negative

These issues are resolved by interpreting negative-energy states as antiparticles (charge conjugation).

Dirac sought a first-order equation linear in both $\partial/\partial t$ and $\nabla$:

$i\hbar\frac{\partial\psi}{\partial t} = \left(c\,\boldsymbol{\alpha}\cdot\hat{\mathbf{p}} + \beta mc^2\right)\psi$

where $\boldsymbol{\alpha} = (\alpha_1, \alpha_2, \alpha_3)$ and $\beta$ are $4 \times 4$ matrices satisfying:

$\alpha_i\alpha_j + \alpha_j\alpha_i = 2\delta_{ij}\mathbf{1}, \quad \alpha_i\beta + \beta\alpha_i = 0, \quad \beta^2 = \mathbf{1}$

In the Dirac representation:

$\beta = \begin{pmatrix} \mathbf{1} & 0 \\ 0 & -\mathbf{1} \end{pmatrix}, \quad \alpha_i = \begin{pmatrix} 0 & \sigma_i \\ \sigma_i & 0 \end{pmatrix}$

where $\sigma_i$ are the Pauli matrices. The four-component wavefunction is called a bispinor.

Free-particle solutions. Plane wave $\psi = u(p)\,e^{-i p \cdot x/\hbar}$ with $u(p)$ satisfying:

$(\gamma^\mu p_\mu - mc)\,u(p) = 0$

where $\gamma^0 = \beta$, $\gamma^i = \beta\alpha_i$, and $\{ \gamma^\mu, \gamma^\nu \} = 2g^{\mu\nu}$.

There are two positive-energy spinors ($E = +\sqrt{p^2c^2 + m^2c^4}$, spin up/down) and two negative-energy spinors ($E = -\sqrt{p^2c^2 + m^2c^4}$). The negative-energy solutions are reinterpreted via the Dirac sea: all negative energy states are filled; a hole is an antiparticle (positron).

Writing the bispinor as $\psi = \begin{pmatrix} \phi \\ \chi \end{pmatrix}$ where $\phi$ and $\chi$ are two-component spinors:

• $\phi$ is the “large” component (dominates at low energy)
• $\chi$ is the “small” component ($|\chi|/|\phi| \sim p/(mc) \sim v/c$)

In the non-relativistic limit ($v \ll c$), the upper component $\phi$ satisfies the Pauli equation:

$i\hbar\frac{\partial\phi}{\partial t} = \left[\frac{\hat{\mathbf{p}}^2}{2m} + V - \frac{\hat{\mathbf{p}}^4}{8m^3c^2} + \frac{e\hbar}{4m^2c^2}\boldsymbol{\sigma}\cdot(\mathbf{E} \times \hat{\mathbf{p}}) + \frac{e\hbar}{8m^2c^2}(\nabla\cdot\mathbf{E})\right]\phi$

The extra terms are: relativistic kinetic correction, spin-orbit coupling, and the Darwin term.

The Dirac equation predicts antimatter (confirmed experimentally by Anderson’s discovery of the positron, 1932). For hydrogen, the exact Dirac energy levels are:

$E_{n,j} = mc^2\left[1 + \left(\frac{Z\alpha}{n - (j+1/2) + \sqrt{(j+1/2)^2 - Z^2\alpha^2}}\right)^2\right]^{-1/2}$

Expanding to order $\alpha^4$:

$E_{n,j} \approx mc^2\left[1 - \frac{Z^2\alpha^2}{2n^2} - \frac{Z^4\alpha^4}{2n^4}\left(\frac{n}{j+1/2} - \frac{3}{4}\right)\right]$

The $\alpha^4$ term gives the fine structure. Crucially, the Dirac equation predicts that $2S_{1/2}$ and $2P_{1/2}$ are exactly degenerate — a result confirmed experimentally and explained by QFT as due to the Lamb shift ($\sim 1$ GHz, arising from vacuum fluctuations).

QFT treats particles as excitations of underlying fields. For a scalar field $\hat{\phi}(\mathbf{x})$:

$\hat{\phi}(\mathbf{x}) = \sum_{\mathbf{k}}\frac{1}{\sqrt{2\omega_k V}}\left(a_{\mathbf{k}}\,e^{i\mathbf{k}\cdot\mathbf{x}} + a_{\mathbf{k}}^\dagger\,e^{-i\mathbf{k}\cdot\mathbf{x}}\right)$

where $\omega_k = \sqrt{k^2 + m^2}$ and $a_{\mathbf{k}}^\dagger$, $a_{\mathbf{k}}$ create and annihilate particles with momentum $\mathbf{k}$.

The Fock space is the direct sum of $N$-particle Hilbert spaces:

$\mathcal{F} = |0\rangle \oplus \bigoplus_{N=1}^{\infty}\mathcal{H}^{\otimes N}_\text{sym/antisym}$

• $a_{\mathbf{k}}^\dagger|0\rangle = |1_{\mathbf{k}}\rangle$ (one-particle state)
• $a_{\mathbf{k}_1}^\dagger a_{\mathbf{k}_2}^\dagger|0\rangle = |1_{\mathbf{k}_1}, 1_{\mathbf{k}_2}\rangle$ (two-particle state)
• For bosons: $(a_{\mathbf{k}}^\dagger)^n|0\rangle \propto |n_{\mathbf{k}}\rangle$ (Bose–Einstein condensation)
• For fermions: $(a_{\mathbf{k}}^\dagger)^2|0\rangle = 0$ (Pauli exclusion)

The number operator: $\hat{N}_{\mathbf{k}} = a_{\mathbf{k}}^\dagger a_{\mathbf{k}}$, with $\hat{N}_{\mathbf{k}}|n_{\mathbf{k}}\rangle = n_{\mathbf{k}}|n_{\mathbf{k}}\rangle$.

The vector potential for the free electromagnetic field:

$\hat{\mathbf{A}}(\mathbf{x}) = \sum_{\mathbf{k},\lambda}\sqrt{\frac{\hbar}{2\omega_k \varepsilon_0 V}}\left[a_{\mathbf{k},\lambda}\,\boldsymbol{\epsilon}_{\mathbf{k},\lambda}\,e^{i\mathbf{k}\cdot\mathbf{x}} + a_{\mathbf{k},\lambda}^\dagger\,\boldsymbol{\epsilon}_{\mathbf{k},\lambda}^*\,e^{-i\mathbf{k}\cdot\mathbf{x}}\right]$

where $\lambda = 1, 2$ labels the two transverse polarisations, $\boldsymbol{\epsilon}_{\mathbf{k},\lambda}$ are polarisation vectors, and $\omega_k = c|\mathbf{k}|$.

The Hamiltonian is $\hat{H} = \sum_{\mathbf{k},\lambda}\hbar\omega_k\left(a_{\mathbf{k},\lambda}^\dagger a_{\mathbf{k},\lambda} + \tfrac{1}{2}\right)$.

The zero-point energy $\sum_{\mathbf{k},\lambda}\hbar\omega_k/2$ diverges — this is the origin of the Casimir effect and vacuum energy in cosmology.

Two parallel perfectly conducting plates separated by distance $d$ modify the allowed electromagnetic modes between them. The vacuum energy per unit area between the plates:

$\mathcal{E}(d) = \frac{\hbar c \pi^2}{720\,d^3}$

The Casimir force per unit area (attractive):

$F/A = -\frac{\partial \mathcal{E}}{\partial d} = -\frac{\hbar c \pi^2}{240\,d^4}$

For $d = 1\,\mu$m: $F/A \approx 1.3 \times 10^{-3}$ Pa. This force has been measured experimentally (Lamoreaux, 1997; Mohideen & Roy, 1998) and confirms the reality of vacuum fluctuations.

The four maximally entangled two-qubit states (Bell basis):

$|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle), \quad |\Phi^-\rangle = \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle)$ $|\Psi^+\rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle), \quad |\Psi^-\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$

These states cannot be written as a product $|\psi_A\rangle \otimes |\psi_B\rangle$. Measuring one qubit instantly determines the state of the other, regardless of spatial separation.

Einstein, Podolsky, and Rosen (1935) argued that QM is incomplete. Their argument:

1. For the entangled state $|\Psi^-\rangle = (|01\rangle - |10\rangle)/\sqrt{2}$, measuring particle 1 in the $z$-basis gives $+1$ or $-1$. If particle 1 yields $+1$, particle 2 must be in $|-1\rangle$.
2. We could equally choose to measure particle 1 in the $x$-basis. The result determines particle 2’s $x$-spin.
3. Since particle 2 was not disturbed by the measurement on particle 1 (locality), particle 2 must have had definite values of both $S_z$ and $S_x$ simultaneously — contradicting the uncertainty principle.

Resolution (Bell’s theorem): No local hidden variable theory can reproduce all QM predictions.

Consider measurements on two spin-1/2 particles in directions $\mathbf{a}$ and $\mathbf{b}$. For a local hidden variable theory with hidden variable $\lambda$:

$P(A = a, B = b | \lambda) = P(A = a | \lambda)\,P(B = b | \lambda)$

CHSH inequality. For four measurement settings $\mathbf{a}, \mathbf{a}', \mathbf{b}, \mathbf{b}'$:

$|S| \leq 2$

where $S = E(\mathbf{a}, \mathbf{b}) + E(\mathbf{a}, \mathbf{b}') + E(\mathbf{a}', \mathbf{b}) - E(\mathbf{a}', \mathbf{b}')$ and $E(\mathbf{a}, \mathbf{b}) = \langle\sigma_{\mathbf{a}} \otimes \sigma_{\mathbf{b}}\rangle$.

QM prediction for the Bell state $|\Phi^-\rangle$ with optimally chosen angles:

$S_{\text{QM}} = 2\sqrt{2} \approx 2.83 > 2$

This violation has been confirmed experimentally (Aspect 1982; Zeilinger 1998; Hensen 2015; loophole-free experiments 2015–2023), ruling out local hidden variables.

Quantum teleportation transmits an unknown quantum state $|\chi\rangle = \alpha|0\rangle + \beta|1\rangle$ from Alice to Bob using shared entanglement and classical communication.

Protocol:

1. Alice and Bob share the Bell pair $|\Phi^+\rangle_{AB}$.
2. Alice performs a Bell measurement on her particle and the unknown state $|\chi\rangle$.
3. Alice sends the 2-bit measurement outcome to Bob classically.
4. Bob applies a Pauli correction ($I$, $X$, $Z$, or $ZX$) based on Alice’s message.

The teleported state is $|\chi\rangle$ with fidelity 1. No faster-than-light communication occurs because the quantum information is unusable without the classical bits.

A qubit is a two-level quantum system: $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ with $|\alpha|^2 + |\beta|^2 = 1$. Unlike a classical bit (0 or 1), a qubit can be in a superposition of both states simultaneously. On the Bloch sphere:

$|\psi\rangle = \cos\frac{\theta}{2}|0\rangle + e^{i\phi}\sin\frac{\theta}{2}|1\rangle$

Quantum gates are unitary operations on qubits:

Single-qubit gates:

$H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \quad X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

$R_\phi = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\phi} \end{pmatrix}, \quad T = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/4} \end{pmatrix}$

Hadamard $H$ creates superpositions: $H|0\rangle = (|0\rangle + |1\rangle)/\sqrt{2}$.

Multi-qubit gate — CNOT:

$\text{CNOT} = |0\rangle\langle 0| \otimes I + |1\rangle\langle 1| \otimes X = \begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0 \end{pmatrix}$

CNOT flips the target qubit if and only if the control qubit is $|1\rangle$. Combined with Hadamard, it creates entanglement: CNOT$(H \otimes I)|00\rangle = |\Phi^+\rangle$.

A quantum circuit is a sequence of gates applied to qubits. Measurements at the end collapse superpositions into classical bitstrings. The circuit model is universal: any unitary operation on $n$ qubits can be decomposed into single-qubit gates and CNOT gates.

Key fact: Quantum circuits are reversible (all gates are unitary). Classical circuits need irreversible gates (AND, OR) — quantum computation gains power from superposition and entanglement, not from irreversible logic.

Problem: Given a function $f: \{0,1\}^n \to \{0,1\}$ promised to be either constant (all 0s or all 1s) or balanced (exactly half 0s, half 1s), determine which.

Classical: Requires $2^{n-1} + 1$ queries in the worst case.

Quantum: One query suffices.

Circuit:

1. Initialise $n$ input qubits and 1 ancilla to $|0\rangle^{\otimes n}|1\rangle$.
2. Apply $H^{\otimes n} \otimes H$ to all qubits.
3. Apply the oracle $U_f: |x\rangle|y\rangle \mapsto |x\rangle|y \oplus f(x)\rangle$.
4. Apply $H^{\otimes n}$ to input qubits.
5. Measure all input qubits.

Result: All zeros $\Rightarrow$ constant. Any other outcome $\Rightarrow$ balanced. This demonstrates an exponential quantum speedup, though for this artificial problem.

Problem: Search an unstructured database of $N$ items for a marked item.

Classical: $O(N)$ queries.

Grover’s algorithm: $O(\sqrt{N})$ queries — a quadratic speedup.

The Grover iterate $G = (2|\psi\rangle\langle\psi| - I)\,O$ where:

• $O$ is the oracle (flips the phase of the marked state)
• $2|\psi\rangle\langle\psi| - I$ is the diffusion operator (inversion about the mean)

After $\lfloor(\pi/4)\sqrt{N}\rfloor$ iterations, the amplitude of the marked state is near 1. Measurement then finds the target with high probability ($> 1 - 1/N$).

Grover’s algorithm is provably optimal for unstructured search (Bennett et al., 1997). It has found applications in optimisation, cryptanalysis, and amplitude amplification.

1. Confusing entanglement with superposition. Superposition is a property of a single quantum system; entanglement is a property of a composite system that cannot be factorised. A Bell state is entangled; $H|0\rangle$ is a superposition but not entangled.

2. Misapplying the variational principle to excited states. The variational principle gives an upper bound on the ground state only. For excited states, use the Hylleraas–Undheim theorem: the $k$-th variational eigenvalue is an upper bound on the $(k+1)$-th true eigenvalue.

3. Ignoring the off-diagonal elements in degenerate perturbation theory. When degeneracy is present, diagonalising the perturbation matrix in the degenerate subspace is mandatory. Skipping this step and applying non-degenerate formulas gives undefined (division by zero) or incorrect results.

4. Assuming all cross sections decrease at high energy. While the Born approximation predicts $\sigma \to 0$ for finite-range potentials, the total cross section for some interactions (e.g., photon–atom) approaches a constant (Thomson cross section $\sigma_T = 8\pi r_e^2/3$).

5. Treating the Dirac equation as a single-particle wave equation. The Dirac equation inherently describes a many-body system (particles and antiparticles). Single-particle interpretations are only approximate, valid when pair production is negligible ($E \ll 2m_ec^2$).

6. Believing Bell inequality violations imply faster-than-light communication. The correlations are nonlocal but cannot transmit information faster than $c$. The no-communication theorem guarantees that local operations and classical communication cannot transmit quantum information without the classical channel.

7. Overestimating quantum speedups. Quantum algorithms provide speedups for specific problem structures (period-finding, search, simulation). Generic computation is not exponentially faster on a quantum computer. The class BQP is believed to be a strict subset of EXP but a strict superset of P.

• Perturbation theory extends systematically via degenerate diagonalisation, selection rules ($\Delta l = \pm 1$ for E1), and specific atomic effects (Stark, Zeeman, Paschen–Back).
• Angular momentum coupling through CG coefficients, LS/jj schemes, and the Wigner–Eckart theorem provides the framework for atomic and nuclear structure.
• Identical particles lead to exchange interactions (Heitler–London bonding), Hartree–Fock self-consistent fields, and second-quantised many-body formalism.
• Scattering theory generalises through the Lippmann–Schwinger equation, T-matrix, and effective range expansion, with quantum phenomena like the Ramsauer–Townsend effect.
• Relativistic QM (Klein–Gordon, Dirac) reveals spin as a relativistic phenomenon, predicts antimatter, and yields the hydrogen fine structure — with discrepancies (Lamb shift) pointing to QFT.
• QFT concepts (Fock space, quantised fields, Casimir effect) show that particles are field excitations and vacuum fluctuations have measurable physical consequences.
• Quantum entanglement (Bell states, Bell/CHSH inequalities, teleportation) demonstrates that quantum correlations exceed any classical description.
• Quantum computing (qubits, gates, Deutsch–Jozsa, Grover) harnesses superposition and entanglement for computational advantages over classical algorithms.

Problem: Calculate the most probable radius for the electron in the ground state of hydrogen (1s orbital). Solution: The radial probability density P(r) = 4r^2 |R_10(r)|^2 = (4/a_0^3) r^2 exp(-2r/a_0). Set dP/dr = 0: d/dr [r^2 exp(-2r/a_0)] = 0. r(2 - 2r/a_0) exp(-2r/a_0) = 0. Solutions: r = 0, r = a_0. The most probable radius is r = a_0 = 0.529 Angstrom, which is the Bohr radius.

Problem: Calculate the spin-orbit coupling energy for a single valence electron in the 2p state of hydrogen-like sodium (Zeff = 11). Solution: The spin-orbit coupling energy is Delta E = (Z_eff^4 * alpha^2 * E_n) / (n * l * (l + 1/2) * (l + 1)), where alpha = 1/137. For n=2, l=1: the 2p level splits into 2p{3/2} and 2p_{1/2}. The splitting is Delta E proportional to Z_eff^4 _ alpha^2 _ E_n / n^3, which gives the D-line splitting observed in the sodium spectrum (589.0 nm and 589.6 nm).