Past exam of the advanced quantum physics course of the University of Cambridge / 2014 / Paper 1 (Advanced Quantum Physics, Physics (3))

The normalized quantum harmonic oscillator ground state has $|A{|}^2=\sqrt{m\omega /(\pi \hslash )}$ and energy $E_0=\hslash \omega /2$. Its classical turning points are $x=\pm \sqrt{\hslash /(m\omega )}$. With $y=x\sqrt{m\omega /\hslash }$, the ground-state oscillator probability beyond the turning points isThe stated rounded error function gives $0.16$; the more precise Gaussian integral is $0.15730$.

An impulse integrated across its full support produces the impulsive quantum kick $K=e^{-i\hat{U}/\hslash }$. Thus the leading transition amplitude is $-i{U}_{21}/\hslash$ andHere $U_{21}$ has units of action. The off-diagonal two-state impulse is exactly $P={\sin }^2(|U_{21}|/\hslash )$ for a Hermitian two-state impulse with zero diagonal. No energy denominator occurs: the impulse is instantaneous.

For two equivalent $p$ electrons, antisymmetry permitsTheir $9+5+1=15$ states exhaust $\left(\genfrac{}{}{0ex}{}{6}{2}\right)$. Hund's rules first maximize spin, then orbital angular momentum; the less-than-half-filled $p$ shell puts the smallest $J$ lowest. Hence the ground level is ${}^3{P}_0$. This is the allowed LS terms of a p-squared configuration.^[ref]

The postulates of quantum mechanics assign a pure state to a normalized ray in a complex Hilbert space, and an observable to a self-adjoint operator. A spectral outcome projector $P_a$ has Born rule probability $p_a=⟨\psi |P_a|\psi ⟩$. An ideal projective measurement updates the state to $P_a|\psi ⟩/\sqrt{p_a}$ when $p_a>0$; degeneracy requires the full outcome projector.

Between measurements, an isolated system obeys $i\hslash {\partial }_t|\psi ⟩=H|\psi ⟩$, giving unitary evolution. Composite systems use tensor products; identical particles additionally obey the appropriate exchange symmetry. Mixed preparations use a density matrix, with $p_a=\mathrm{tr}(\rho P_a)$. These postulates require a specified Hamiltonian and measurement model; they do not assign those automatically.^[ref]

The central field approximation replaces the anisotropic many-electron environment by an average spherical potential, giving one-electron orbitals labelled by $n,l,m$. Screening changes the effective radial attraction and removes the pure Coulomb degeneracy between different $l$.

The Hartree method improves the potential self-consistently. For normalized occupied orbitals,Guess orbitals, construct their mean fields, solve the Hartree equations, and iterate to convergence. The $j\ne i$ sum excludes self-interaction in this product-state model. A spherical average can enforce the central-field specialization. A Hartree product state omits fermionic exchange and dynamical correlation; Hartree-Fock method adds antisymmetry and exchange through a Slater determinant. Converged orbital eigenvalues must not simply be added as the total energy, because that double counts pair interactions.^[ref]

A laser combines pumping, an amplifying medium and optical feedback. Stimulated emission adds photons to an occupied optical mode; net amplification requires population inversion between the relevant levels after accounting for their degeneracies. Pumping through additional levels and rapid lower-level relaxation can sustain inversion; a closed two-level system under steady incoherent resonant pumping cannot sustain it.

An optical resonator selects modes and feeds light back through the gain medium. For intensity gain coefficient $g$, internal intensity loss ${\alpha }_i$, length $L$ and mirror intensity reflectivities $R_1,R_2$, the round-trip laser threshold condition isAbove threshold, saturation limits gain and a partially transmitting mirror provides output. Typical properties are strong directionality, substantial coherence and narrow bandwidth; multiple modes and pulsed operation are possible, so perfect monochromaticity is not universal.^[ref]

For spin one-half, $s_{\pm }=s_x\pm i{s}_y=(\hslash /2)({\sigma }_x\pm i{\sigma }_y)$, henceThey raise or lower the spin projection by one unit of $\hslash$ and annihilate the corresponding endpoint. The exchange identity ${\mathbf{\sigma }}_1\cdot {\mathbf{\sigma }}_2={\sigma }_{1z}{\sigma }_{2z}+2({\sigma }_{1+}{\sigma }_{2-}+{\sigma }_{1-}{\sigma }_{2+})$ preserves total spin projection.

Write $\Delta =\Delta E$, $u={\mu }_{0}B(g_1+g_2)$ and $w={\mu }_{0}B(g_1-g_2)$. In the listed order $(\uparrow \uparrow ,\downarrow \downarrow ,\uparrow \downarrow ,\downarrow \uparrow )$, direct action givesThe printed matrix instead uses $(\uparrow \uparrow ,\uparrow \downarrow ,\downarrow \uparrow ,\downarrow \downarrow )$. Permuting the listed indices to $(1,3,4,2)$ recovers its central block and diagonal entries. This is a basis-order mismatch in the source, not a different Hamiltonian.

The two-spin exchange spectrum with unequal signed magnetic moments isUse the products $g_j{\mu }_0$ as the signed magnetic coefficients in the stated Hamiltonian. Opposite physical charges must be represented in those signs; they do not by themselves imply equal moment magnitudes. Here ${\mu }_0$ denotes the source's magneton, not vacuum permeability.

At zero field, the spin-one-half triplet state has energy $\Delta /4$ and degeneracy three; the spin-one-half singlet state has energy $-3\Delta /4$ and degeneracy one:For $\Delta >0$ the singlet is lower; for $\Delta <0$ the triplet is lower. If $\Delta =0$, all four states coincide.

For $|w|\gg |\Delta |$, the mixed states approach opposite-spin product states andThe omitted mixed-branch correction is of order ${\Delta }^2/|w|$. The four levels are generically nondegenerate, apart from special moment choices or crossings.

For equal-and-opposite signed moments, $g_2=-g_1$, $u=0$ and the parallel-spin level remains exactly twofold degenerate at $\Delta /4$; the other two approach $-\Delta /4\pm |g_1{\mu }_{0}B|$. Equal moment magnitudes are an extra assumption, not implied by opposite charges alone. If $g_1=g_2$, the differential Zeeman term vanishes and the singlet/triplet-zero block stays unmixed by the field. This is the strong-field limit of a two-spin exchange spectrum.

The square-root branches avoid one another when $\Delta \ne 0$; crossings with a parallel-spin branch are allowed because total spin projections differ. At zero field the positive-$\Delta$ sketch has one lower singlet and a triplet above it. The source supplies no numerical moment ratio, so the two panels show explicit illustrative signed choices, including the exactly compensated case.

The quoted expansion is nondegenerate perturbation theory for a known isolated eigenstate of a time-independent Hamiltonian. A useful small-mixing condition is $|⟨k|H_1|i⟩|\ll |E_i-E_k|$ for coupled states. First order is a diagonal expectation value; second order describes virtual mixing. A degenerate or nearly degenerate coupled subspace must first be treated by diagonalization there. Symmetry can eliminate terms, and convergence still needs checking. Ground-state second-order shifts are nonpositive because all denominators are negative.^[ref]

For the rotational Stark effect, $H_1=-pE\cos \theta$, where $\theta$ is the polar angle to the field. The source calls it azimuthal, which is inconsistent with this coupling. Rotor states have parity $(-1{)}^J$, while $\cos \theta$ is odd, so every first-order matrix element within one fixed-$J$ manifold vanishes. The $m$ degeneracy therefore causes no first-order splitting.

Only $J\pm 1$ with the same $m$ contribute in second order. Using their energy differences $2BJ$ and $-2B(J+1)$,For $J\ge 1$ this simplifies to the quadratic Stark shift of a polar linear rotor,For $J=0$, only $J^{\prime }=1,m=0$ contributes, with squared angular matrix element $1/3$ and denominator $-2B$:Here $B$ is the rotational energy constant, unlike the magnetic field in Question3.

The unperturbed line gives $B/h=6.08\mathrm{GHz}$. The $J=1$ shifts are $(pE{)}^2/(10B)$ for $m=0$ and $-(pE{)}^2/(20B)$ for $m=\pm 1$. Subtracting the common lower-state shift gives transition shifts $4(pE{)}^2/(15Bh)$ and $7(pE{)}^2/(60Bh)$. Their separation determines the dipole moment from a rotational Stark doublet:The $\mathrm{D}$ unit is the debye. Using external $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{D}=3.33564\times 10^{-30}\mathrm{C}\mathrm{m}$ gives $pE/B\simeq 0.0593$, consistent with weak-field perturbation theory. The two shifted frequencies are about $12.165707$ and $12.162497\mathrm{GHz}$; polarization selects $\Delta m=0$ or $\pm 1$. The inferred value is a magnitude; these shifts do not determine the dipole's sign.^[ref][ref]