Past exam of the physics A course of the University of Cambridge / 2014 / Paper 1 / Section C

Each electromagnetic normal mode is a quantum harmonic oscillator with $E_n=(n+1/2)h\nu$. For $x=h\nu /(k_{B}T)$ its canonical partition function and occupation probabilities areThe geometric distribution is normalized and gives thermal mode energy $h\nu /(e^x-1)$. The zero-point term $h\nu /2$ is not part of the thermally emitted black-body intensity. The equilibrium photon chemical potential is zero because photon number is not conserved.

Counting two transverse polarizations gives $8\pi {\nu }^2/c^3$ modes per volume per frequency interval. Thus the spectral energy density is $u_{\nu }=8\pi h{\nu }^3/[c^3(e^x-1)]$. In an isotropic radiation field, spectral radiance is $B_{\nu }=c{u}_{\nu }/(4\pi )$. Transforming the spectral measure, $B_{\lambda }=B_{\nu }|d\nu /d\lambda |$, givesHere intensity means radiance per wavelength and solid angle; the hemispheric spectral exitance of a black surface is $M_{\lambda }=\pi B_{\lambda }$. The extra Jacobian is essential: the wavelength peak is not found by simply converting the frequency peak using $\lambda =c/\nu$.

At short wavelength the exponential suppresses radiation. At long wavelength the Rayleigh-Jeans limit is $B_{\lambda }\simeq 2c{k}_{B}T/{\lambda }^4$; using classical equipartition at every frequency instead would produce the ultraviolet catastrophe. The wavelength maximum solves $5(1-e^{-x})=x$, with nonzero root $x=4.9651$, yielding Wien's displacement law. For the specified temperature,Integrating the hemispheric flux gives the Stefan-Boltzmann law, $M=\sigma T^4\simeq 3.01\times 10^{-6}\mathrm{W}{\mathrm{m}}^{-2}$. The plot uses the specified ideal temperature, rather than substituting a later measured value. The physical black-body description of the cosmic microwave background is tested by comparing the sky spectrum with an absolute reference blackbody and separating instrumental and Galactic contributions.^[ref]

For a normalized top-hat wavefunction of width $L$ centred at zero, $\psi (x)=L^{-1/2}$ inside the interval and zero outside. Its position distribution has $⟨x⟩=0$ and $\Delta x=L/\sqrt{12}$. With Fourier convention $(2\pi \hslash {)}^{-1/2}$, the momentum amplitude isThe momentum density has long oscillatory $1/p^2$ tails. Consequently $\int p^2|\tilde{\psi }{|}^{2}dp$ diverges: the ideal discontinuous top hat has infinite r.m.s. momentum spread. Its symmetric momentum density has formal principal-value mean zero, but no finite absolute first moment either. The finite-variance form $\Delta x\Delta p\ge \hslash /2$ is therefore not saturated and cannot be evaluated using the width of the central diffraction lobe as $\Delta p$.

The first momentum zeros occur at $p=\pm 2\pi \hslash /L$, showing the inverse-width scale. Smoothing the edges sufficiently gives a finite momentum variance and restores the ordinary finite-variance uncertainty comparison; sharper edges require larger high-momentum components. It is the ideal jump, not a violation of the uncertainty principle, that causes the divergence.

In a classically forbidden region the local wavevector becomes imaginary. For a slowly varying barrier, the WKB transmission probability has leading exponentialwhere $r_1,r_2$ are turning points and $\mu$ is the reduced mass. This is a probability exponent; the wave amplitude has only half that exponent. Near a turning point one uses matching formulas rather than the unmodified local WKB expression.

For alpha decay, an attractive nuclear interior supports a quasibound state behind the repulsive Coulomb barrier of the daughter nucleus. For zero orbital angular momentum, write the exterior potential as $C/r$, $C=2Z_d{e}^2/(4\pi {ϵ}_0)$, with nuclear matching radius $R$ and outer turning point $b=C/Q>R$. The exterior action can be integrated:The decay rate is approximately $\lambda =\nu P_{\mathrm{pre}}{e}^{-2G}$ and the half-life is $\ln 2/\lambda$. Here $\nu$ is the interior assault-frequency scale and $P_{\mathrm{pre}}$ is the alpha preformation probability. They are model-dependent; barrier penetration alone does not determine them. Nonzero orbital angular momentum adds a centrifugal barrier.

For $R/b\ll 1$, the leading dependence is $G\propto Z_d/\sqrt{Q}$. Thus a modest increase in decay energy can cause an enormous increase in decay rate, giving the Geiger-Nuttall law relating logarithmic half-life to inverse square-root energy along a suitable isotope sequence. Tunnelling conserves the total energy: the escaping alpha particle does not acquire energy from the barrier.

electron spin is intrinsic angular momentum, independent of an orbital wavefunction's $\mathbf{r}\times \mathbf{p}$. It has $s=1/2$, soA spatial orbital therefore carries a two-component spinor degree of freedom. Noncommuting components do not describe an ordinary vector with simultaneously definite projections; measurements along one axis give only the two allowed values. Even a spherically symmetric $l=0$ orbital retains electron spin.

The associated magnetic moment is $\mathbf{\mu }=-g_{e}e\mathbf{S}/(2m_e)$ with $e>0$ and $g_e\simeq 2$. A spinor changes sign under a $2\pi$ rotation and returns under $4\pi$; that sign is observable only relative to another coherent amplitude, not as an isolated global phase. A literal classically rotating charged sphere does not account for this fixed half-integer representation, component algebra and spinor rotation law. Classical angular momentum provides useful analogies for torque and precession, but no ordinary classical orientation reproduces the full intrinsic quantum spin structure.