Past exam of the physics A course of the University of Cambridge / 2017

The original diagram places hard walls at $0$ and $3L$, and the central plateau on $L<x<2L$ is exactly at the indicated ground-state energy. This equality determines the middle of the requested sketch. In the Time-independent Schrödinger equation,the central interval has $V=E$, hence ${\psi }^{\prime \prime }=0$. The ground state can be chosen real and positive inside the walls by the nodeless theorem for a one-dimensional ground state. Reflection symmetry makes it even about $x=3L/2$, so its linear central segment must be constant.

The hard-wall boundary conditions are $\psi (0)=\psi (3L)=0$. On both outer intervals $V=0<E$, so the positive wavefunction has negative curvature and is sinusoidal. Wavefunction matching at a finite potential step requires both $\psi$ and ${\psi }^{\prime }$ to remain continuous at $L,2L$; the derivative therefore vanishes where each sinusoidal flank meets the flat middle. A normalized representative isEach side contributes $L/2$ and the plateau contributes $L$ to the squared norm before multiplying by $1/(2L)$. The corresponding consistency condition is $E={\pi }^2{\hslash }^2/(8m{L}^2)$; a numerical energy is not required for the sketch.

Thus the ground state at the top of a central rectangular barrier has no internal nodes, continuous value and slope, and a flat middle, rather than an exponential dip through a classically forbidden region. Multiplication by an overall constant phase changes none of these physical features.

Use the canonical commutation relation $[\hat{x},\hat{p}]=i\hslash I$ on a common invariant domain of the operator products. Write $A=\hat{p}\hat{x}$, so $\hat{x}\hat{p}=A+i\hslash I$. Associativity then giveswhereasThereforeAs a direct position-representation check, the momentum operator is $\hat{p}=-i\hslash d/dx$. Acting on a smooth function $f$ givesThis is the equality of two quadratic position-momentum orderings; it does not make arbitrary rearrangements of noncommuting linear operators interchangeable.

For addition of angular momentum, the allowed total quantum number runs in unit steps from $|\ell -s|$ to $\ell +s$. With $\ell =2$ and $s=1$,Thus the complete list of multiplets isThe total angular momentum operator has squared magnitude eigenvalue ${\hslash }^{2}j(j+1)$ and chosen-axis component eigenvalue $\hslash m_j$. The number of states is $3+5+7=15$, matching the uncoupled dimension $(2\ell +1)(2s+1)=5\times 3$.

Use an inverting operational amplifier. Ground the non-inverting input, connect the source to the inverting input through $R_{\mathrm{in}}$, and connect $R_f$ from the output back to that same input.

In linear operation the ideal operational amplifier draws no input electric current, and feedback holds the inverting input at the same electric potential as the grounded non-inverting input. This node is a virtual ground. Kirchhoff's current law there givesThe source supplies current $I_i=V_i/R_{\mathrm{in}}$, so its input impedance is $R_{\mathrm{in}}$. The required design is thereforeOnly these two resistors and the single ideal operational amplifier are needed in the signal circuit; its power supplies are implicit. The result assumes an input small enough to keep the output within its linear operating range.

Use the sample mean to estimate each magnetic susceptibility, and the unbiased sample variance with denominator $n-1=4$ to estimate the repeatability of individual readings. In the numerical scale supplied by the paper,If the repeated measurements have independent random errors, their mean estimates have standard errorsThe difference and its estimated standard uncertainty are thereforeThe separation is about $2.53$ standard errors. Under an approximately normal distribution model with no experimenter-dependent bias, Welch's t-test gives about $7.34$ statistical degrees of freedom and a two-sided p-value of $0.038$. Thus there is modest evidence against equal underlying susceptibilities at a conventional $5\%$ significance level; overlap of the ranges of individual readings is not an adequate comparison of the means. Five readings per sample provide limited information about the error distribution.

Crucially, material and student exhibit complete confounding: each student measured only one sample. An experimenter-dependent systematic error could mimic the difference. Paramagnetism also depends on conditions such as temperature, so equal material alone does not guarantee equal readings without controlling those conditions. The defensible physical conclusion is the data suggest a difference, but do not establish different materials. Both students should measure both samples in randomized or alternating order, with calibration and temperature controlled, to separate a sample effect from an operator effect. No unit or scale factor absent from the original table has been invented.

Equation and boundary conditions. The delta potential has strength $V_0$ with units of energy times length, because the Dirac delta function has units of inverse length. The Time-independent Schrödinger equation isThe wavefunction is continuous at the origin. A finite jump in $\psi$ would produce a derivative of a Dirac delta function in ${\psi }^{\prime \prime }$, with no corresponding term in this potential. A jump in ${\psi }^{\prime }$ does produce a delta contribution and is allowed. Integrate through a symmetric interval:For continuous finite $\psi$, the right-hand side vanishes as $ϵ\to 0$. ConsequentlyThis derivative jump is the boundary condition associated with the attractive delta potential.

Unique bound state. For $E<0$, write $\kappa =\sqrt{-2mE}/\hslash >0$. Away from the origin, ${\psi }^{\prime \prime }={\kappa }^2\psi$. Normalizability excludes exponentials growing at either infinity, leavingContinuity gives $A=B=C$. The derivative jump then becomes $-2\kappa C=-2m{V}_{0}C/{\hslash }^2$. For a nonzero state,There is exactly one positive value of $\kappa$, so exactly one negative-energy bound state. An odd solution would have $\psi (0)=0$; the allowed decaying exponentials would then both vanish identically. Normalize the even solution usingChoosing the overall phase real and positive gives the bound state in a delta potential:It peaks at the origin, has a downward cusp there, and decays exponentially on both sides. Its value, but not its slope, is continuous.

Scattering amplitudes and fluxes. For $E>0$, put $k=\sqrt{2mE}/\hslash$ and retain $\kappa =m{V}_0/{\hslash }^2$ as a strength parameter. A unit-amplitude wave incident from the left hasContinuity gives $t=1+r$. The jump condition givesSolving,The probability current of a plane wave of amplitude $A$ is $\hslash k|A{|}^2/m$ in its propagation direction. Since the asymptotic wavenumber is the same on both sides, reflection probability and transmission probability are just $|r{|}^2$ and $|t{|}^2$:They satisfy $R+T=1$, expressing conserved probability current in this real potential.

The right-hand plot uses $\eta =\kappa /k=V_0\sqrt{m/(2E)}/\hslash$, proportional to $V_0$ at fixed incident energy. Then $R={\eta }^2/(1+{\eta }^2)$ rises monotonically from zero to one, while $T=1/(1+{\eta }^2)$ falls from one to zero; they are equal at $\eta =1$. Despite the absence of a classical forbidden barrier, an abrupt attractive interaction reflects a quantum wave. Increasing attraction strengthens the derivative mismatch and can approach complete reflection at fixed positive energy. At fixed $V_0$, increasing incident energy instead makes the well increasingly transparent. The negative-energy bound state and the positive-energy scattering states are different stationary solutions, not an absorptive loss channel.

Hamiltonian and commutators. Use the canonical commutation relation $[\hat{x},\hat{p}]=i\hslash I$ and define $c_0=\sqrt{m\omega /(2\hslash )}$. Expanding the product of the given creation and annihilation operators,Therefore the quantum harmonic oscillator Hamiltonian operator isSimilarly,Using $[AB,C]=A[B,C]+[A,C]B$ gives

Lowering a normalized energy eigenstate. The commutator impliesFor $n>0$, nondegeneracy of this one-dimensional energy level means $\hat{a}|{\varphi }_n⟩=c_n|{\varphi }_{n-1}⟩$. Its norm is fixed by the number operator:Choosing successive eigenstate phases in the conventional way makes $c_n=\sqrt{n}$, soFor $n=0$, the same norm is zero, hence $\hat{a}|{\varphi }_0⟩=0$. A nonzero lowered ground state would have energy below the lowest allowed level, and positivity of ${\hat{a}}^{†}\hat{a}$ forbids that possibility.

Coherent-state eigenvalue. Applying the annihilation operator term by term, with the $n=0$ contribution zero, givesThusThe coherent state is normalized because $e^{-|\alpha {|}^2}{\sum }_n|\alpha {|}^{2n}/n!=1$. Its annihilation operator is not Hermitian, so the eigenvalue $\alpha$ may be complex.

Time evolution. Under the time-independent Hamiltonian operator, each energy component acquires its own phase:Since $|\alpha e^{-i\omega t}|=|\alpha |$, the normalization prefactor is unchanged. HenceThe state remains a coherent state, with its complex parameter rotating at the oscillator's angular frequency; the displayed extra factor is a global phase from the zero-point energy.

Reduced radial equation. A spherical harmonic satisfies ${\hat{L}}^2{Y}_{\ell m}=\ell (\ell +1){\hslash }^2{Y}_{\ell m}$. Substitute $\psi =R{Y}_{\ell m}$ into the Time-independent Schrödinger equation, divide out the angular function, and put $R=U/r$. Direct differentiation givesMultiplication by $r$ then gives the Radial Schrodinger equationHere $\mu =m_1{m}_2/(m_1+m_2)$ is the reduced mass, and the extra positive term is the centrifugal potential.

Boundary conditions. At the origin this finite central potential has no point interaction. The physical radial solution is regular there, requiring $U(0)=0$; for $\ell =0$, a nonzero constant $U(0)$ would give $R\sim 1/r$ and a point-source singularity of the three-dimensional Laplacian. Mere square integrability alone would not remove that singular solution. For a bound state, the radial probability integral is ${\int }_0^{\infty }|U{|}^{2}dr$, so the exterior solution must decay and $U(r)\to 0$ as $r\to \infty$.

Solutions for zero orbital angular momentum. For $\ell =0$ and $-V_0<E<0$, defineThe interior equation is $U^{\prime \prime }+k^{2}U=0$; $U(0)=0$ excludes the cosine. The exterior equation is $U^{\prime \prime }-q^{2}U=0$; normalizability excludes the growing exponential. ThusThe constants cannot remain independent once the interface is matched. Since the potential step is finite, both $U$ and $U^{\prime }$ are continuous at $a$:Dividing the equations yields

Existence and number of s-wave bound states. Put $z=ka$, $y=qa$, and ${\rho }_0=a\sqrt{2\mu V_0}/\hslash$. ThenThe lowest branch lies on $\pi /2<z<\pi$. Along it, $y=-z\cot z>0$ and ${\rho }_0=z/\sin z$ increases continuously from $\pi /2$ to infinity. Therefore a first negative-energy solution exists precisely when ${\rho }_0>\pi /2$, orFurther branches begin at $z=3\pi /2,5\pi /2,\dots$ as $q\to 0^{+}$. Hence the bound-state thresholds of a spherical square well give exactly one negative-energy $\ell =0$ state whenThe upper equality is included: the second branch is then only at $E=0$, with constant exterior $U$, and is not normalizable. If threshold cases are excluded by convention, the commonly quoted open interval has a strict upper inequality; the physical negative-energy state count at equality is still one.

Deuteron estimate. Using $m_p{c}^2\simeq 938.27\mathrm{MeV}$, $m_n{c}^2\simeq 939.57\mathrm{MeV}$ and $\hslash c\simeq 197.327\mathrm{MeV}\mathrm{fm}$ gives $\mu c^2\simeq 469.46\mathrm{MeV}$. At $a=2.1\mathrm{fm}$,The threshold for a second $s$ state is $9V_{\ast }\simeq 209\mathrm{MeV}$, far above the stipulated depth $33.5\mathrm{MeV}$. Thus this model has no radially excited s-wave bound state.

Checking other orbital angular momentum sectors matters before ruling out every excitation. At the first $p$-wave threshold, the regular interior reduced radial function is $r{j}_1(kr)$ and the zero-energy exterior function is proportional to $r^{-1}$. Matching their logarithmic derivatives requires $j_0(ka)=0$, whose first root is $ka=\pi$. HenceLarger $\ell$ only increases the centrifugal potential. Therefore no bound orbital excited states are expected in the specified central square-well model. Its spin dependence and tensor nuclear forces have not been specified; the calculation does not resolve the real deuteron's internal spin structure.

Distinguishable noninteracting particles. Interpret $V$ as the same external one-particle potential for each particle; the total potential is $V({\mathbf{r}}_1)+V({\mathbf{r}}_2)$. The Time-dependent Schrödinger equation isLet $\hat{h}{\varphi }_a={ϵ}_a{\varphi }_a$ and $\hat{h}{\varphi }_b={ϵ}_b{\varphi }_b$, where $\hat{h}=-{\hslash }^2{\nabla }^2/(2m)+V$. Applying ${\hat{h}}_1+{\hat{h}}_2$ to the product multiplies it by ${ϵ}_a+{ϵ}_b$. The time derivative of $e^{-iEt/\hslash }$ multiplies it by $E$. Thus the proposed stationary product is a solution providedThe tensor product separates the two independent one-particle dynamics.

Identical spin-one-half particles. Identical fermions require antisymmetry under exchange of the complete position-and-spin labels. First take $a,b$ to be distinct orthonormal spatial orbitals, and defineThe symmetric spatial state must multiply the antisymmetric spin-one-half singlet state; the antisymmetric spatial state must multiply one of the three symmetric spin-one-half triplet states. The four normalized possibilities, each with the common time factor $e^{-iEt/\hslash }$, areThese are the required product states under fermionic exchange symmetry. If $a=b$, the antisymmetric spatial combination vanishes. Only the product ${\varphi }_a({\mathbf{r}}_1){\varphi }_a({\mathbf{r}}_2)$ times the singlet remains, as required by the Pauli exclusion principle. A nonorthogonal pair would require different normalization factors; the displayed $1/\sqrt{2}$ assumes an orthonormal stationary-state basis.

Weak repulsive Coulomb interaction. For fixed orbitals $a,b$, first-order perturbation theory gives spatial energy shifts $J\pm K$. Here $J$ is the direct interaction energy, while the exchange integral for repulsion strength $g>0$ isThe positivity follows from the positive Fourier transform of the Coulomb kernel. Consequently Coulomb exchange splitting of two orbitals givesAll three triplet spin components remain degenerate for a spin-independent Coulomb interaction. Their antisymmetric spatial factor vanishes at coincidence and reduces close encounters. The inequality is strict when $K>0$; for negligible orbital overlap, the splitting may vanish. This comparison fixes one particle in each of the specified orbitals and does not rank singlets with a different orbital occupation.

Spin-zero bosons. For one particle in each of the same distinct orthonormal orbitals, the spin factor is symmetric and bosonic exchange symmetry requires ${\Phi }_{+}$. At coincident positions,Thus the bosonic coincidence enhancement in two orthogonal orbitals isThese are joint probability densities, or small-detector-volume limits; an exactly specified point has zero probability in a continuous distribution. If both particles occupy the same orbital, the properly normalized product is already symmetric and the corresponding enhancement relative to distinguishable particles in that same orbital is one. The paper's usual factor-two answer presumes distinct orthogonal input orbitals.

Operators, quantization and rotations. Angular momentum is the generator of spatial rotations. In quantum mechanics, its components are Hermitian operators on their appropriate domains and obeyThese angular momentum commutation relations prevent all three components from having simultaneous sharp values. For example, the Robertson uncertainty principle gives $\Delta J_x\Delta J_y\ge \hslash |⟨J_z⟩|/2$. One conventionally diagonalizes ${\hat{J}}^2$ and a selected component ${\hat{J}}_z$:The allowed labels are $j=0,\frac{1}{2},1,\frac{3}{2},\dots$ and $m=-j,-j+1,\dots ,j$. An angular-momentum multiplet has $2j+1$ states. The magnitude $\hslash \sqrt{j(j+1)}$ differs from the largest possible component $\hslash j$; the remaining components fluctuate even in a state with maximal $m$.

The angular momentum lowering operator and its raising counterpart, ${\hat{J}}_{\pm }={\hat{J}}_x\pm i{\hat{J}}_y$, satisfyThe ladder terminates at $m=\pm j$. A rotation through angle $\theta$ about unit vector $\hat{\mathbf{n}}$ acts through the unitary operator $\exp [-i\theta \hat{\mathbf{n}}\cdot \hat{\mathbf{J}}/\hslash ]$. Thus quantized components and the continuous rotation of a state are compatible.

Orbital angular momentum. The orbital contribution is associated with motion in space:For a spinless spatial wavefunction, the simultaneous angular eigenfunctions are the spherical harmonics $Y_{\ell m}(\theta ,\varphi )$, with $\ell =0,1,2,\dots$ and $m=-\ell ,\dots ,\ell$. Single-valued ordinary scalar spatial wavefunctions give integer orbital labels. In a central potential energy, $[\hat{H},{\hat{L}}^2]=[\hat{H},{\hat{L}}_z]=0$, so stationary states may be labelled by these angular quantum numbers. The angular kinetic contribution appears in the radial equation as $\ell (\ell +1){\hslash }^2/(2m{r}^2)$, the centrifugal potential. Noncentral interactions need not conserve these orbital labels.

Intrinsic spin and addition. Spin angular momentum is an intrinsic internal degree of freedom with the same commutation algebra. It is not the angular momentum of a small classical charged sphere literally rotating about its centre. For an Electron, $s=1/2$ and $m_s=\pm 1/2$, so ${\hat{S}}_i=(\hslash /2){\sigma }_i$ in terms of the Pauli matrices. Its squared spin is $3{\hslash }^2/4$. Other particles can have integer or half-integer spin, including spin zero.

For one particle with orbital and spin degrees of freedom, $\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}}$; the two component sets commute with each other. Addition of angular momentum permits $j=|\ell -s|,|\ell -s|+1,\dots ,\ell +s$. The coupled states $|j,m_j⟩$ are linear combinations of the uncoupled $|\ell ,m_{\ell }⟩|s,m_s⟩$ with $m_j=m_{\ell }+m_s$, determined by Clebsch-Gordan coefficients. A rotationally invariant spin-orbit interaction can conserve $J^2,J_z$ while mixing the separate component labels.

Angular momentum also carries a magnetic moment. For an electron's orbital motion and intrinsic spin,The minus signs reflect negative electric charge. With the Bohr magneton ${\mu }_B=e\hslash /(2m_e)$, a spin projection $m_s$ has ${\mu }_z=-g_s{\mu }_B{m}_s$. A magnetic field can therefore distinguish angular-momentum states through their interaction energy $-\mathbf{\mu }\cdot \mathbf{B}$.

Stern–Gerlach measurement. In the Stern-Gerlach experiment, atoms emerge from a heated source, pass through collimating apertures, traverse a strongly inhomogeneous magnetic field, and strike a position-sensitive screen. The beam is neutral, so its separation is interpreted through the atomic magnetic moment rather than the ordinary charged-particle Lorentz force. Near a chosen quantization direction, the interaction givesA uniform field can exert a torque but would not separate the beam by this translational force; the gradient is essential. Different allowed magnetic-moment components become correlated with different trajectories. A classical ensemble with continuously varying orientations would yield a continuous spread, whereas discrete quantum components yield separated channels.

In the modern electronic description of ground-state silver, filled shells cancel their electronic angular momenta and the unpaired $5s$ electron has $L=0$, $S=1/2$, $J=1/2$. The effective electronic measurement therefore has two components, approximately ${\mu }_z=\pm {\mu }_B$, rather than an orbital $\ell =1/2$, which is not allowed. For positive field gradient the electronic $m_J=+1/2$ component has negative ${\mu }_z$ and deflects downward in this coordinate convention. Nuclear spin and hyperfine structure are neglected in this simplified two-channel account. Historically the original splitting preceded the electron-spin interpretation, so the modern explanation should not be confused with the original investigators' interpretation.

More generally a prepared spin state $a|+z⟩+b|-z⟩$ gives the two outcomes with probabilities $|a{|}^2,|b{|}^2$. Selecting one channel prepares a state for further measurements. An ideal second analyser along the same axis then gives a definite result, whereas one along a perpendicular axis gives two equally likely outcomes for a spin-one-half eigenstate. This connects the apparatus directly to incompatible spin components and the Born rule. Angular momentum is quantized through its operator algebra; the Stern–Gerlach gradient turns that discrete internal information into spatially distinguishable outcomes.

The Bohr model combines classical circular motion in the Coulomb potential with explicitly quantum postulates. A one-electron atom of nuclear charge $+Ze$ has electron–nucleus reduced mass $\mu$. With $C=Z{e}^2/(4\pi {\epsilon }_0)$, the relative circular orbit obeysBohr postulates permitted nonradiating stationary orbits with $L=n\hslash$, for $n=1,2,\dots$. These assumptions giveFor hydrogen with the nucleus taken infinitely heavy, the $n=1$ radius is the Bohr radius, $a_0\simeq 5.29\times 10^{-11}\mathrm{m}$. The finite reduced mass gives the small nuclear-motion correction, and increasing $Z$ contracts the one-electron orbit.

The circular-motion equation gives $T=C/(2r)$ while the potential energy is $-C/r$, so the total energy is $-C/(2r)$. Substituting the quantized radius yieldsFor the heavy-nucleus hydrogen approximation, $E_n\simeq -13.6\mathrm{eV}/n^2$. The ionization threshold is zero energy, so removing the ground-state electron requires about $13.6\mathrm{eV}$.

The model further postulates that radiation is emitted or absorbed only in transitions between stationary orbits, with photon energy $h\nu =|E_i-E_f|$. Emission from $n_i>n_f$ givesThus it reproduces the principal hydrogen spectral series, their limiting wavelengths, and the gross hydrogenic energy levels. For example, $n=3\to 2$ in hydrogen has $\Delta E\simeq 1.89\mathrm{eV}$ and $\lambda \simeq 656\mathrm{nm}$, a visible line. Absorption reverses the transition when the photon matches the energy gap.

The achievement is that quantized energies explain discrete spectra, but the model's stationary orbits are postulates rather than consequences of classical electrodynamics: an accelerating classical charge would radiate. It does not provide a general calculation of line intensities or transition selection rules and fails as a quantitative theory of many-electron atoms. Spin, fine structure and the full Zeeman effect require further physics. Modern quantum mechanics replaces definite electron orbits by wavefunctions and quantum probabilities. In particular, the true hydrogen ground state has $\ell =0$, so its orbital angular momentum vanishes; the model's rule $L=n\hslash$ should not be identified with the modern orbital eigenvalue rule $L^2=\ell (\ell +1){\hslash }^2$. Bohr's energy quantization remains a successful hydrogenic approximation, while its literal orbital picture does not.

Ehrenfest theorem relates the dynamics of quantum expectation values to operator commutators. For a normalized state obeying $i\hslash |\dot{\psi }⟩=\hat{H}|\psi ⟩$ and a possibly time-dependent observable $\hat{A}$, differentiate $⟨A⟩=⟨\psi |\hat{A}|\psi ⟩$. Using the adjoint Schrödinger equation for the bra givesThusThe identity assumes a common domain on which the required products and derivatives are defined.

For $H=p^2/(2m)+V(x)$, the canonical commutation relation gives $[x,H]=i\hslash p/m$ and $[p,H]=-i\hslash V^{\prime }(x)$. ThereforeThese resemble Hamilton's equations and Newton's second law, but generally $⟨V^{\prime }(x)⟩\ne V^{\prime }(⟨x⟩)$. If the state is localized about $\overline{x}=⟨x⟩$, a local expansion givesThe classical trajectory equation for the mean is accurate when the force changes little across the wave packet. For a quadratic harmonic potential, $V^{\prime }$ is linear and this closure is exact for every state with the relevant finite moments: $m{d}^2⟨x⟩/d{t}^2=-m{\omega }^2⟨x⟩$. This exact classical evolution of the centroid does not eliminate its quantum position uncertainty or make every state a point particle.