Past exam of the quantum theory and quantum computers course of the University of Oxford / 2009 (C7.1b, DMAC 2A78)

Original solutions to the paper released at the source request. The complete editable transcription is retained in the private examination archive.

Let $U(t)=e^{-iHt/\hslash }$ and take an observable with no explicit time dependence. In the Schrödinger picture, ${\psi }_S(t)=U(t){\psi }_0$ and $A_S=A$. In the Heisenberg picture, ${\psi }_H={\psi }_0$ and $A_H(t)=U(t{)}^{†}AU(t)$. For the interaction picture choose a time-independent splitting $H=H_0+V$, put $U_0=e^{-i{H}_{0}t/\hslash }$, and defineThen $i\hslash {\dot{\psi }}_I=V_I{\psi }_I$ and ${\dot{A}}_I=(i/\hslash )[H_0,A_I]$. Unitarity gives the identical expectation in all three descriptions:Different choices of the split interpolate between pictures; $H_0=H$ makes the state constant, while $H_0=0$ gives the Schrödinger description.

Differentiate $U^{†}AU$ to obtain the Heisenberg equation of motionFor an explicitly time-dependent observable there is additionally $U^{†}({\partial }_{t}A)U$. Thus $H_H=H$ is constant. In a normalized energy eigenstate, $⟨[H,A_H]⟩=E⟨A_H⟩-E⟨A_H⟩=0$, so the expectation of its Heisenberg derivative vanishes whenever the products are defined.

For the harmonic oscillator, the canonical commutator givesThe last identity follows from the operator product rule without commuting $X$ and $P$. Although $XP$ is not itself self-adjoint, the same calculation can be applied to the self-adjoint dilation observable $(XP+PX)/2$ and gives the same derivative.

Stationarity of the three expectations gives $⟨P⟩=⟨X⟩=0$ and $⟨P^2⟩=m^2{\omega }^2⟨X^2⟩$. Substitution into $E=⟨P^2⟩/(2m)+m{\omega }^2⟨X^2⟩/2$ yields the quantum virial theoremThe vanishing means make the variances equal to the second moments, soThis satisfies the position-momentum Heisenberg uncertainty relation, with equality only for the ground state.

For a nonzero trial vector use the Rayleigh quotient $\mathcal{E}(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$. Its first variation in any allowed direction $\eta$ isVanishing for all complex directions implies $(H-\mathcal{E})\psi =0$; an eigenvector conversely makes it vanish. The ground energy is the infimum of this quotient for a lower-bounded Hamiltonian, so minimizing over a trial family provides an upper bound to it, exact when that family contains a ground eigenvector.

Set $a=⟨\varphi ,H\varphi ⟩$, $b=⟨\varphi ,I\varphi ⟩$ and $c=⟨\varphi ,HI\varphi ⟩$. All three are real: $H,I$ are self-adjoint and commute. Also $I^{†}I=I^2=1$. Expansion of the numerator and norm givesDifferentiate the quotient, retaining only nonzero trial vectors:Thus the stationarity equation is $\boxed{(t^2-1)(c-ab)=0}$. If $c=ab$, the quotient is identically $a$ wherever defined. Otherwise $t=\pm 1$ are the two extrema, withIn this nonconstant case $|b|<1$, so both denominators are positive. If $c-ab>0$, the plus value is the maximum and the minus value the minimum; the assignments reverse when $c-ab<0$. One can also see global extremality by resolving the trial vector into the orthogonal $I=\pm 1$ sectors: its energy is a weighted mean of their two fixed projected energies.

The null-vector exception matters. If $|b|=1$, then $I\varphi =\pm \varphi$, the covariance $c-ab$ vanishes and one of $\varphi \pm I\varphi$ is zero. That value has no Rayleigh quotient and is not an extra stationary physical state.

For particle exchange, change of integration variables shows that $I\varphi (x_1,x_2)=\varphi (x_2,x_1)$ is unitary and $I^2=1$, hence self-adjoint. Exchanging coordinates interchanges the two kinetic derivatives and leaves the symmetric potential unchanged, so $[I,H]=0$. The vectors $\varphi \pm I\varphi$ lie in its symmetric and antisymmetric sectors. Their normalized nonzero versions are respectively bosonic and fermionic spatial trial wavefunctions.

Because the energy of a general trial state is the weighted average of its sector energies, at least one nonzero symmetry projection has energy no greater than the original trial. Thus minimizing can be performed in a definite exchange-symmetry sector. This is a variational approximation within the trial subspace; it does not assert that projecting an arbitrary trial vector produces an exact eigenvector of $H$. For particles with a prescribed statistics sector one minimizes within that physical sector.

For classical events, $A\setminus C\subset (A\setminus B)\cup (B\setminus C)$: an outcome in $A$ but not in $C$ either fails $B$ or belongs to $B\setminus C$. Monotonicity and the union bound prove the Fréchet-Bell inequalityMalus' law gives transmission ${\cos }^2\gamma$ for a photon polarized by the first filter, so its conditional rejection probability at the second is $\boxed{{\sin }^2\gamma }$. Conditional and joint probabilities must be distinguished: $P(A\setminus B)=P(A)P(B^c\mid A)$.

For equal unbiased marginals $P(A)=P(B)=1/2$, the quantum two-setting rejection probabilities are $\frac{1}{2}{\sin }^2({\theta }_A-{\theta }_B)$. A classical common assignment of all three outcomes would therefore require ${\sin }^2({\theta }_A-{\theta }_B)+{\sin }^2({\theta }_B-{\theta }_C)\ge {\sin }^2({\theta }_A-{\theta }_C)$. Choose directions $0,\gamma ,2\gamma$. ThenFor example $\gamma =\pi /6$ gives $1/2<3/4$. More generally ${\sin }^{2}a+{\sin }^{2}b-{\sin }^2(a+b)=-2\sin a\sin b\cos (a+b)$. In a triangle whose side directions are these filters, the internal angle supplementary to $a+b$ is obtuse exactly in this violation range. This specifies the angle convention behind the triangle description: with polarization axes regarded as unoriented lines, a claim that every sufficiently large physical axis separation violates the inequality would not be correct. The contradiction concerns a single classical joint model for all settings, not the validity of ordinary probability or a comparison of arbitrarily renormalized sequential experiments.

Use $|0⟩=|{\alpha }_0⟩$ and $|1⟩=|{\alpha }_1⟩$. Projecting the first photon with $|0⟩⟨0|$ gives the unnormalized product vector $c_0|00⟩$, with success probability $|c_0{|}^2$. Projection of the second onto $|\theta ⟩=\cos \theta |0⟩+\sin \theta |1⟩$ givesConditioned on success, normalize these vectors; a zero amplitude means that outcome never occurs. For the maximally entangled equal-amplitude pair, rotation of both real polarization bases gives the same correlations for arbitrary settings. Comparing the appropriately normalized coincidence/rejection frequencies at three setting pairs tests the inequality above. A general unequal-amplitude state need not reproduce the unbiased rotationally invariant probabilities, so the maximally entangled choice supplies the clean explicit example.

For swapping, label the four systems in order ${\xi }_1,{\eta }_1,{\xi }_2,{\eta }_2$ and let $|\Phi {⟩}_{\eta }=2^{-1/2}{\sum }_{j=0}^1|{\eta }_j{\eta }_j⟩$. Its overlap with $|{\eta }_r{\eta }_s⟩$ is ${\delta }_{rs}/\sqrt{2}$. Applying its rank-one projector to Alice's two systems givesAfter regrouping the remote and local systems, this factorizes into $[{\sum }_r{b}_r^2|{\xi }_r{\xi }_r⟩/\sqrt{2}]\otimes |\Phi {⟩}_{\eta }$. The selected outcome has probability $p=\frac{1}{2}{\sum }_r|b_r{|}^4$. Therefore the normalized conditional remote state isIt has Schmidt rank at least two whenever the original pair was entangled, since squaring a nonzero coefficient does not make it zero. Thus the two photons that never interacted become entangled conditional on Alice's joint measurement outcome. This is entanglement swapping. Alice is projecting onto a specified Bell state, not performing a universal test of whether an arbitrary unknown state is entangled. Averaging all her outcomes leaves the remote reduced density operator unchanged, so no faster-than-light signal is available without her classical outcome label.