Past exam of the physics A course of the University of Cambridge / 2010 / Paper 2 / Section B

The parity operator acts as $P\psi (x)=\psi (-x)$, with $P^2=I$. For an even potential energy $[H,P]=0$, so energy eigenstates can be chosen even or odd, with parity eigenvalues $+1$ and $-1$. One-dimensional bound states are nondegenerate; the ground state is nodeless and even.

For the finite square well, square integrability excludes the growing exterior exponential, while even parity excludes the interior sine. Thus the interior solution is proportional to $\cos kx$, with matching decaying tails andContinuity at $a$ gives $B\cos ka=A{e}^{-qa}$ and $-Bk\sin ka=-qA{e}^{-qa}$. Their ratio yieldsThe conditions at $-a$ follow from the same parity.

For a narrow weak well, $ka\ll 1$, hence $q\simeq k^{2}a\simeq 2m{V}_{0}a/{\hslash }^2$ andThis is self-consistent because $|E_0|/V_0\simeq 2m{V}_0{a}^2/{\hslash }^2\ll 1$. It is the shallow finite-square-well energy, equivalently the attractive delta potential limit with integrated strength $2a{V}_0$. An arbitrarily weak attractive well of this form still has its even bound ground state.

The first excited state must be odd. Replace the interior cosine by $\sin kx$ and choose exterior tails with opposite signs. The matching condition becomes $q=-k\cot ka$. Its first root requires $\pi /2<ka<\pi$. At the binding threshold $q\to 0$, sois necessary and sufficient for the first odd bound state. Equality gives a zero-energy nonnormalizable threshold solution, not an additional bound state.

Writing $a=cx+idp$, where $c=\sqrt{m\omega /(2\hslash )}$ and $d=1/\sqrt{2m\omega \hslash }$, the canonical commutation relation givesConsequently $[a^{†}a,a]=-a$ and $\boxed{[H,a]=-\hslash \omega a}$. If $a|{\varphi }_0⟩$ were nonzero, this commutator would make it an energy eigenstate with energy $E_0-\hslash \omega$, contradicting the lowest-energy property. Therefore $a|{\varphi }_0⟩=0$.

The commutator product rule and induction giveActing on the vacuum and shifting the sum then shows that the unnormalized bosonic coherent state $|\tilde{\psi }⟩=e^{\alpha a^{†}}|0⟩$ satisfies $a|\tilde{\psi }⟩=\alpha |\tilde{\psi }⟩$.

The displayed series lacks its wavefunction normalization factor. Since $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$, its squared norm is ${\sum }_n|\alpha {|}^{2n}/n!=e^{|\alpha {|}^2}$. The normalized coherent state needed for the stated expectation values isUse $⟨a⟩=\alpha$, $⟨a^{†}⟩={\alpha }^{\ast }$, $⟨a^{†}a⟩=|\alpha {|}^2$ and $⟨a{a}^{†}⟩=|\alpha {|}^2+1$. For $x=x_0(a+a^{†})$, $x_0=\sqrt{\hslash /(2m\omega )}$,The displacement changes the mean position but not its ground-state uncertainty. Unnormalized bras and kets give the same physical expectation values only after division by their squared norm.

The angular-momentum commutators giveFor example $[L_x+i{L}_y,L_z]=-i\hslash L_y+i(i\hslash L_x)=-\hslash L_{+}$. In the displayed basis $|1,1⟩,|1,0⟩,|1,-1⟩$, the only nonzero upper entries of $L_{+}$ connect adjacent magnetic quantum numbers. Subtracting their column and row $L_z$ eigenvalues multiplies each by $-\hslash$, proving the first matrix commutator; the lower matrix similarly gives the second. Direct multiplication giveswhose difference is $2\hslash L_z$. Also $L^2=L_z^2+(L_{+}{L}_{-}+L_{-}{L}_{+})/2=2{\hslash }^{2}I$, confirming $\ell =1$.

A rotation changes the quantization axis without changing the spectrum, so $L_{\mathbf{n}}$ has eigenvalues $+\hslash ,0,-\hslash$. Its dimensionless matrix isSolving $(L_{\mathbf{n}}-\hslash I)v=0$ and normalizing gives the spin-one maximum-projection eigenvectorFor instance the first equation is $(\cos \theta -1){\cos }^2(\theta /2)+{\sin }^2\theta /2=0$, and the squared components sum to one. An overall phase is immaterial.

At $\theta =\pi /3$, $v_{+}=(3/4,\sqrt{6}/4,1/4{)}^{\mathsf{T}}$. The Born rule therefore predicts, in the requested order,This treats the Stern-Gerlach experiment as the stipulated ideal measurement of orbital projection. In an actual atom, other magnetic moments and spin-orbit coupling can change which angular momentum the apparatus resolves.

With angular-coordinate measure $d\varphi$, the normalized one-particle orbitals are $u_l(\varphi )=e^{il\varphi }/\sqrt{2\pi }$. Single-valuedness requires integer $l$. Their products obeyIf arc-length measure is used instead, each one-particle wavefunction normalization includes $R^{-1/2}$; the physical energy is unchanged. For the following parts impose antisymmetry on the full spatial-and-spin state of the two fermions.

Both particles can occupy $l=0$ with opposite spins, giving a unique singlet state at $E=0$, quantum degeneracy 1. At the next distinct energy ${ϵ}_R$, one orbital is $l=0$ and the other is $l=+1$ or $-1$. For each orbital choice there is one symmetric spatial state paired with the singlet state, and one antisymmetric spatial state paired with three triplet state spin projections. HenceSwapping particle labels adds no independent state after the required exchange symmetry is imposed.

The lowest singlet state isFor the lowest triplet states take $\lambda =\pm 1$ and $m_s=-1,0,1$:orthogonality of $u_0$ and $u_{\lambda }$ verifies wavefunction normalization of the antisymmetric spatial factor. At this same energy there are also two excited singlet states, obtained by replacing its minus sign with a plus sign and the spin factor by $|0,0⟩$. Together they reproduce the eight states counted in part a.

At coincidence ${\varphi }_1={\varphi }_2$, the triplet state spatial factor vanishes, while the lowest singlet state has finite, uniform pair probability. Consequently a weak short-range repulsion raises the singlet state energy more strongly: antisymmetry suppresses close encounters in the triplet state. In the contact-interaction limit the triplet state contribution is exactly zero, whereas the singlet state shift is positive. A finite-range interaction gives a nonzero but suppressed triplet state shift. This explains a relative energy increase; it does not by itself establish a crossing or justify using unchanged noninteracting wavefunctions for arbitrarily strong repulsion.