Past exam of the quantum theory and quantum computers course of the University of Oxford / 2010 (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.

A permutation $\pi$ of identical particles multiplies a fermionic state by $\mathrm{sgn}(\pi )$; a bosonic state is invariant. Particle exchange keeps $y_{+}$ fixed and sends $y_{-}$ to $-y_{-}$. Consequently the spatial wave function is even in $y_{-}$ for the stated bosonic model and odd for the stated fermionic model. This assumes no additional spin variables in the supplied model; exchange of full particle states is the underlying rule.

The coordinate change is orthogonal, with unit Jacobian determinant, so the kinetic energy becomeson the square $|y_{+}|,|y_{-}|<a$ with zero boundary values. Product states ${\psi }_r(y_{+}){\psi }_s(y_{-})$ have energyThe one-dimensional eigenfunction is even for odd $s$ and odd for even $s$. Bosons therefore allow odd $s$, whereas fermions require even $s$. The lowest bosonic pair is $(r,s)=(1,1)$; its first excited pair is $(2,1)$. The lowest fermionic pair is $(1,2)$. ThusThe printed bosonic ground-energy claim omits the factor ${\pi }^2/4$; the value above follows directly from the supplied one-dimensional spectrum and kinetic energy.

The separation is $|x_1-x_2|=\sqrt{2}|y_{-}|$. The normalized $y_{+}$ factor integrates out. In the fermionic ground state, the relative-coordinate density is $a^{-1}{\sin }^2(\pi y_{-}/a)$, givingFor the bosonic ground state it is $a^{-1}{\cos }^2(\pi y_{-}/(2a))$, givingBoth integrals follow by replacing squared trigonometric functions by their constant and cosine parts and integrating once by parts. Henceand the difference is $\boxed{2\sqrt{2}a/{\pi }^2}$. The odd relative state vanishes at coincidence, which increases its expected separation.

Let $E=f_H(\psi )$. For any admissible variation $\eta$, differentiation with respect to a real parameter givesStationarity for all variations, including their imaginary multiples, forces $(H-E)\psi =0$; conversely this eigenvector equation makes the derivative vanish. Thus stationary Rayleigh quotients are exactly eigenvectors, with their eigenvalues as the quotient values.

Put $\kappa =e^2/(4\pi {ϵ}_0)$ and assume $\alpha >0$. The squared norm of the radial trial function is $4\pi {\int }_0^{\infty }{r}^2{e}^{-2\alpha r}dr=\pi /{\alpha }^3$. Its kinetic expectation, computed through $\int |\nabla \psi {|}^2$, is ${\hslash }^2{\alpha }^2/(2m)=\kappa a{\alpha }^2/2$. Splitting the potential integral at $R$ givesFor example, the first integral is $[1-e^{-2s}(1+2s+2s^2)]/(4{\alpha }^3)$ and the second is $e^{-2s}(R/(2\alpha )+1/(4{\alpha }^2))$, where $s=\alpha R$. ThereforeIts derivative is $\kappa [a\alpha -(1+2\alpha R)e^{-2\alpha R}]$. The actual stationary condition is consequentlyThe extra Coulomb factor in the printed stationary equation is an error: the common prefactor cancels on differentiation. The corrected equation is dimensionless on both sides. Its left side increases from zero, while its right side decreases from one, so it has one positive solution and gives the minimum in this trial family.

Let $\epsilon =R/a$ and $y=a\alpha$. Expansion of $(1+2s)e^{-2s}$ gives $1-2s^2+O(s^3)$. ThusThere is no linear correction in $R/a$. By the quantum variational principle, any normalized trial state gives an upper bound on the true ground energy. Choosing $\alpha =1/a$ already gives the exact boundIts small-radius expansion isUsing the optimized $\alpha$ improves this bound starting at order ${\epsilon }^4$, and hence gives the same leading finite-size correction. The true energy is also at least the point-nucleus value $-\kappa /(2a)$, since the finite-radius potential is nowhere more attractive.

A density operator is a positive self-adjoint trace-class operator with trace one. In finite dimensions this is a positive semidefinite Hermitian matrix with $\mathrm{tr}\rho =1$. Its expectation of an observable is $\boxed{⟨A⟩=\mathrm{tr}(A\rho )}$ whenever the expectation is defined.

In the source's square-root notation, take the unique positive square root $W=\sqrt{\rho }$. It has Hilbert-Schmidt norm one and satisfies $W{W}^{\ast }=\rho$, soThe standard vector purification is the associated vector $|W⟩⟩={\sum }_{r,s}{W}_{rs}|{\xi }_r⟩|{\eta }_s⟩$ on a system and a chosen copy. Tracing out the copy gives $\rho$; measuring $A\otimes I$ gives the same trace formula. Thus the operator square root and the purified vector are two related objects, distinguished by this vectorization.

For a general coefficient matrix $C$, expansion of the expectation givesThus $\boxed{{\rho }_1=C{C}^{\ast }}$. Positivity is immediate from $v^{\ast }C{C}^{\ast }v=\Vert C^{\ast }v{\Vert }^2$, and its trace is one by normalization of $\Psi$.

If $C$ is nonsingular, it is square and ${\rho }_1$ is positive definite. DefineThen $U{U}^{\ast }=C^{-1}{\rho }_1(C^{\ast }{)}^{-1}=I$, which for a square matrix also implies $U^{\ast }U=I$. Thus $U$ is unitary, $\sqrt{{\rho }_1}=CU$, and $C=\sqrt{{\rho }_1}{U}^{\ast }$.

In an eigenbasis of ${\rho }_1$, its square root is $\mathrm{diag}(\sqrt{p_r})$. Hence $c_{rs}=\sqrt{p_r}\overline{U_{sr}}$. DefineUnitarity gives $⟨{\zeta }_r|{\zeta }_t⟩={\delta }_{rt}$, and substitution gives the Schmidt decompositionFor singular or rectangular $C$, the same conclusion follows on its support by a singular value decomposition; the stated nonsingular case needs only the unitary calculation above.

For the supplied Bell state, the requested matrices areThe Schmidt vectors can be chosen as ${\zeta }_1={\eta }_2$, ${\zeta }_2=-{\eta }_1$, and the two coefficients are $1/\sqrt{2}$. The square root's canonical vector purification is $(|{\xi }_1{\eta }_1⟩+|{\xi }_2{\eta }_2⟩)/\sqrt{2}$; the original Bell vector is a different purification of the same reduced state, related by a unitary on the second factor. This illustrates the unitary freedom of purification.