Past exam of the physics A course of the University of Cambridge / 2016 / Paper 1 / Section B

For the oppositely travelling components, write $u=c_1{e}^{ikx}$ and $v=c_2{e}^{-ikx}$ after removing the common time phase. Then ${\partial }_x\psi =ik(u-v)e^{-i\omega t}$. In $\mathrm{Im}({\psi }^{\ast }{\partial }_x\psi )$ the interference terms contribute no imaginary part, soA standing-wave interference pattern can therefore change density without adding a cross-term to the net current.

For barrier-top scattering, the stationary equation inside the barrier is $-{\hslash }^2{\psi }_2^{\prime \prime }/(2m)=0$. Thus its spatial part is linear: ${\psi }_2=(bx+c)e^{-i\omega t}$. Take incident amplitude one, $k=\sqrt{2mE}/\hslash$ and $\omega =E/\hslash$. Continuity of the spatial wave function and its derivative at the two finite steps givesSolving these four equations yieldsThe equal exterior velocities mean flux ratios are squared amplitude ratios:This is the threshold limit of finite-barrier scattering, not exponential decay with a nonzero interior decay constant.

The interior constants areSince $\mathrm{Im}[(bx+c{)}^{\ast }b]=\mathrm{Im}(c^{\ast }b)=k/(1+k^2{a}^2)$,It is positive and independent of position, matching the transmitted flux. Although the classical kinetic energy inside is zero, the complex linear quantum wave function carries conserved probability current; its phase varies with position.

Differentiating the expectation value and using ${\partial }_t|\psi ⟩=-i\hat{H}|\psi ⟩/\hslash$ and ${\partial }_t⟨\psi |=i⟨\psi |\hat{H}/\hslash$ givesThe second term means explicit operator time dependence, distinct from the state's evolution.

The commutator product rule follows by adding and subtracting $\hat{B}\hat{A}\hat{C}$:For the mathematical induction step, let $\hat{D}=[\hat{A},\hat{B}]$ and use $[\hat{A},\hat{D}]=0$. The case $n=1$ is immediate. If $[{\hat{A}}^m,\hat{B}]=m\hat{D}{\hat{A}}^{m-1}$, thenThus the claimed identity holds for every positive integer. Commutation of $\hat{B}$ with $\hat{D}$ is not needed for this particular step. This is the central-commutator differentiation rule. Termwise application to a convergent power series, on a common domain where the operations are justified, gives

For position and momentum, $[\hat{x},\hat{p}]=i\hslash \hat{I}$. Hence $[{\hat{p}}^2,\hat{x}]=-2i\hslash \hat{p}$, $[V(\hat{x}),\hat{x}]=0$ and $[V(\hat{x}),\hat{p}]=i\hslash V^{\prime }(\hat{x})$. Substitution gives Ehrenfest theorem:The mean momentum changes by the mean force. It does not follow for a general nonlinear Potential energy that $⟨V^{\prime }(\hat{x})⟩=V^{\prime }(⟨\hat{x}⟩)$: a closed classical trajectory for the mean position requires an additional condition, such as a quadratic Potential energy or a sufficiently localized wave packet.

Expanding with the angular momentum commutators givesFor ${\hat{L}}_z|\psi ⟩=\alpha \hslash |\psi ⟩$, commuting ${\hat{L}}_z$ past a ladder operator givesThus the raised or lowered vector, if nonzero, is an eigenvector with the shifted eigenvalue; an angular momentum ladder endpoint can instead be annihilated. Also $[{\hat{L}}^2,{\hat{L}}_{\pm }]=0$, so ${\hat{L}}^2{\hat{L}}_{\pm }|\psi ⟩=\Lambda {\hslash }^2{\hat{L}}_{\pm }|\psi ⟩$ for every nonzero ladder result.

For orbital angular momentum, ${\hat{L}}_z=-i\hslash {\partial }_{\varphi }$ gives an azimuthal factor $e^{i\alpha \varphi }$. Single-valuedness under $\varphi \mapsto \varphi +2\pi$ requires integer $\alpha$. Positivity of ladder norms givesfor normalized states. A fixed finite $\Lambda$ therefore has upper and lower endpoints. Writing the upper endpoint as $m=\ell$, its vanishing raising norm gives $\Lambda =\ell (\ell +1)$; the lower endpoint is $m=-\ell$. Consequentlywith $\ell =0,1,2,\dots$. This integer conclusion is for orbital angular momentum; spin representations can also have half-integer labels.

For the specified Gaussian-radial state, use a Cartesian p-orbital decomposition: write $x+y+z=r(\sin \theta \cos \varphi +\sin \theta \sin \varphi +\cos \theta )$. Its angular factor isOnly $\ell =1$ occurs. The squared angular coefficients are $2,2,2$ before common normalization, so each allowed $m$ has probability one third. Therefore the simultaneous outcome isThe common normalized radial factor cancels from this ratio. The standard symmetric $m$ sequence used above resolves the duplicated central positive entry in the printed list.

Direct matrix multiplication givesThe normalized $S_x$ eigenvectors are $(|\uparrow ⟩\pm |\downarrow ⟩)/\sqrt{2}$, and the $S_y$ eigenvectors are $(|\uparrow ⟩\pm i|\downarrow ⟩)/\sqrt{2}$, with eigenvalues $\pm \hslash /2$ in each case.

For $N=|a{|}^2+|b{|}^2>0$, allowing for a possibly unnormalized input pair of coefficients,For a normalized state $N=1$. The second result follows because each squared Spin operator matrix equals ${\hslash }^2\hat{I}/4$.

The tilted component is ${\hat{S}}_{z^{\prime }}=(\hslash /2)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{array}\right)$. Its eigenvalues remain $\pm \hslash /2$, with one real phase convention for the normalized eigenvectorsProjecting the initial $|\uparrow ⟩$ onto them gives $P_{+}={\cos }^2(\theta /2)$ and $P_{-}={\sin }^2(\theta /2)$. ThusA smaller expectation does not imply a smaller individual spin eigenvalue; it describes the probabilities of the same two measurement outcomes.