Past exam of the advanced quantum physics course of the University of Cambridge / 2013 / Paper 1 / 1

Put $\mathbf{J}={\mathbf{S}}_1+{\mathbf{S}}_2$. Angular momentum addition gives $j=0,1,2,3$, and the spin-dot-product eigenvalue isThus the spin-three-halves pair spectrum isFor identical fermions, exchange parity of two coupled identical spins makes $j=0,2$ spin-antisymmetric and $j=1,3$ spin-symmetric. Corresponding spatial exchange symmetries make the full state antisymmetric; without a specified spatial state, none of these total-spin sectors is excluded.

Adjacent nodes of a wavefunction are separated by roughly $a/n$, so a locally sinusoidal wavefunction has typical wavevector $k\sim \pi n/a$. The kinetic-energy estimate from wavefunction nodes is thereforeEquivalently, $⟨K⟩={\hslash }^2\int |{\psi }^{\prime }{|}^{2}dx/(2m\int |\psi {|}^{2}dx)$. The robust scaling is ${\hslash }^2{n}^2/(m{a}^2)$; the precise coefficient depends on how the extent is defined and how the oscillations vary near classical turning points. For few nodes, use an order-one offset to $n$: the nodeless ground state still has confinement energy bound from position support of order ${\hslash }^2/(m{a}^2)$, rather than zero.

Centre the infinite square well at zero. Its ground state wavefunction is $\psi (x)=\sqrt{2/L}\cos (\pi x/L)$ inside the well and zero outside. The momentum representation in quantum mechanics gives the wavevector amplitudeConsequently the unnormalised wavevector distribution of an infinite-well ground state isThe apparent singularities at $k=\pm \pi /L$ are removable. The normalized probability density is $4\pi L$ times the boxed expression and has limit $L/(4\pi )$ there. Translating the well changes only the Fourier transform's phase. The wavevector is continuously distributed; a box energy eigenstate is not a free momentum eigenstate.