Past exam of the quantum mechanics quantum theory and quantum computers course of the University of Oxford

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

In the position representation the Time-dependent Schrödinger equation isFor a normalized state define probability density and probability current byMultiplying the equation by $\overline{\psi }$ and its complex conjugate by $\psi$, then subtracting, cancels the real potential. The resulting identity isThus $\boxed{{\partial }_t\rho +{\partial }_{x}j=0}$. Integrating over an interval gives $d{\int }_b^c\rho dx/dt=j(b)-j(c)$: probability changes only through flux across the endpoints. This is local conservation of quantum probability.

Continuity with the zero exterior imposes zero endpoint values on the infinite square well. Solving the stationary equation with these boundary conditions giveswith zero extension outside. The integral of ${\sin }^2(n\pi x/a)$ is $a/2$, which fixes the normalization. The initial coefficients are $3/5$ and $4/5$, whose squared moduli sum to one. Each energy component acquires its stationary phase:At the midpoint, ${\Psi }_1=\sqrt{2/a}$, ${\Psi }_1^{\prime }=0$, ${\Psi }_2=0$, ${\Psi }_2^{\prime }=-(2\pi /a)\sqrt{2/a}$. HenceTaking its imaginary part proves

For bound hydrogen atom states, the radial wave function is regular and finite at the origin, and decays at infinity so that $4\pi {\int }_0^{\infty }{r}^2|\Psi (r){|}^{2}dr<\infty$. Equivalently the reduced radial function $w(r)=r\Psi (r)$ has $w(0)=0$ and is square integrable on the half-line. A singular $1/r$ branch is not an admissible regular Coulomb eigenfunction; its derivative energy diverges and it introduces an origin singularity absent from the Hamiltonian domain.

Write $\kappa =e^2/(4\pi {ϵ}_0)$, so $a={\hslash }^2/(m\kappa )$. Multiplication by $r$ turns the radial equation intoSince $w=a{e}^{-\lambda R}f(R)$ and $r=aR$, substitution givesChoose $\boxed{\lambda =\mu a>0}$ to remove the constant coefficient and factor out the decaying exponential. For $f={\sum }_{k\ge 0}{a}_k{R}^{k+c}$, the lowest power has coefficient $c(c-1)a_0$, so $\boxed{c(c-1)=0}$. The $c=0$ behavior would give $\Psi \sim a_0/R$, violating regularity; choose $c=1$.

Matching successive powers then yieldsA nonterminating tail eventually has one sign, and its coefficient ratio is asymptotic to $2\lambda /k$. More precisely, for any $0<\delta <\lambda$, the ratio is eventually at least $(2\lambda -\delta )/(k+1)$. Its same-sign tail consequently dominates a positive constant times the exponential series of rate $2\lambda -\delta$, up to finitely many terms. The factor $e^{-\lambda R}$ cannot then make $\Psi$ decay. Thus square integrability forces termination.

If the last nonzero term is $a_{n-1}{R}^n$, termination requires $\lambda n=1$. ThereforeFor $n=1$, $\lambda =1$, $f$ is proportional to $R$, andThe radial volume element is $4\pi r^{2}dr$, so the mean radius is the ratio of the radius-weighted norm integral to the norm integral. With $r=aR$, the common angular and normalization factors cancel:In the ground state $\boxed{⟨r⟩=3a/2}$.

For a normalized state and a self-adjoint operator, the expectation value and dispersion areUse the physics convention that the inner product is conjugate-linear in its first argument, and assume the displayed products and second moments are defined. Self-adjointness gives $\overline{⟨\psi |A\psi ⟩}=⟨A\psi |\psi ⟩=⟨\psi |A\psi ⟩$, so the expectation is real. MoreoverIt vanishes exactly when $\psi$ is an eigenstate, with eigenvalue its expectation.

The product rule gives $XPf=-i\hslash x{f}^{\prime }$ and $PXf=-i\hslash (f+x{f}^{\prime })$, hence $\boxed{[X,P]f=i\hslash f}$ on their common differentiable domain. For general operators, put $z=⟨A\psi |B\psi ⟩$. ThenThe Cauchy-Schwarz inequality yieldsApply this to centered position and momentum operators, whose commutator is unchanged. Their squared norms are the variances, givingEquality requires both proportionality in Cauchy-Schwarz and a purely imaginary inner product. Thus for real means $x_0,p_0$,with real $\gamma$. Solving this first-order equation givesNormalizability on the real line forces $\gamma >0$. Writing ${\sigma }^2=\hslash /(2\gamma )$, all normalized equality states are thereforeThey have $\Delta X=\sigma$ and $\Delta P=\hslash /(2\sigma )$. A nonconstant quadratic phase would increase the product, so the full family here has a real positive Gaussian width and only a linear spatial phase.

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.

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

Inside the infinite well, the time-independent Schrödinger equation is $-{\hslash }^2{\Psi }^{\prime \prime }/(2m)=E\Psi$ with Dirichlet endpoints. Negative or zero energy gives no nonzero solution satisfying both endpoints. For $E>0$, the sine solution requires $ka=n\pi$. ThusThe normalization follows from ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$. These functions form an orthonormal basis. Substitute the expansion into $i\hslash {\partial }_t\psi =H\psi$ and take the inner product with each basis vector: orthonormality gives $i\hslash {\dot{c}}_n=E_n{c}_n$, equivalentlyFor a general infinite expansion this is interpreted in the usual Hamiltonian domain or its unitary evolution; the particular state below has only two terms.

Put $\theta =\pi x/a$. The identity $\sin \theta (16{\cos }^2\theta -1)=4\sin 3\theta +3\sin \theta$ gives $c_1=3/5$, $c_3=4/5$, all others zero. Their squared magnitudes sum to one. HenceThe evolved state is $(3e^{-i{E}_{1}t/\hslash }{\Psi }_1+4e^{-i{E}_{3}t/\hslash }{\Psi }_3)/5$. Its position probability includes an interference term with frequency $\Omega =(E_3-E_1)/\hslash =4{\pi }^2\hslash /(m{a}^2)$. Specifically,The three required integrals are $\pi /8-1/4$, $\pi /8+1/12$ and $1/4$, respectively. ThereforeThe energy expectation is constant while the spatial probability oscillates, because relative energy phases affect interference but not the energy probabilities.

Let $u=rF$. Multiplication of the radial equation by $-2mr/{\hslash }^2$ gives $u^{\prime \prime }+[A/r-\lambda (\lambda +1)/r^2-{\alpha }^2]u=0$. With $u=e^{-\alpha r}f$ the constant terms cancel, leavingThe indicial equation is $c(c-1)=\lambda (\lambda +1)$, with roots $c=\lambda +1$ and $c=-\lambda$. The usual physical Coulomb radial origin condition selects $c=\lambda +1$: it imposes the regular reduced radial boundary condition $u(0)=0$. Normalizability alone at the origin is insufficient to make this selection when an irregular branch is locally square-integrable, notably at $\lambda =0$.

Write $f=r^{\lambda +1}{\sum }_{j\ge 0}{a}_j{r}^j$. Coefficient comparison givesA nonterminating regular series has the exponentially growing asymptotic branch, suggested by $a_{j+1}/a_j\sim 2\alpha /j$: it supplies an $e^{2\alpha r}$ component, which the prefactor $e^{-\alpha r}$ cannot normalize. Equivalently the regular confluent-hypergeometric solution grows unless its first parameter is a nonpositive integer. Thus a bound state requires the numerator to vanish at some $j=k\ge 0$, givingThis is series-termination quantization of the Coulomb radial equation. It requires $A>0$ for the assumed $\alpha >0$; at $A=0$ there are no such negative-energy Coulomb bound states.

For $k=0$, $F=N{r}^{\lambda }{e}^{-\alpha r}$. The angular integral cancels when computing $⟨1/r⟩$, while the radial volume element contributes $r^{2}dr$. ConsequentlyThe ratio follows by one integration by parts and remains valid for real $\lambda \ge 0$; no integer-only factorial identity is needed. Since now $\alpha =A/[2(\lambda +1)]$,For a classical particle with that negative total energy, nonnegative kinetic energy requires $E-V={\hslash }^2(A/r-{\alpha }^2)/(2m)\ge 0$, hence $\boxed{r\le A/{\alpha }^2}$. This is a necessary radial bound; extra angular momentum can impose a stronger restriction.

At $k=\lambda =0$, $\alpha =A/2$ and the classical boundary is $r=2/\alpha$. The normalized radial probability density is $4{\alpha }^3{r}^2{e}^{-2\alpha r}$. With $z=2\alpha r$, the exterior quantum probability isThe final integration gives $(4^2+2\cdot 4+2)e^{-4}/2$. This is the nonzero quantum tail beyond the classical turning region.

With the specified ladder-operator normalization, $A_{+}^{†}=A_{-}$ andTherefore $\Vert A_{-}\psi {\Vert }^2=2m(E-\hslash \omega /2)\Vert \psi {\Vert }^2\ge 0$. This proves $\boxed{E\ge \hslash \omega /2}$, with equality exactly when $A_{-}\psi =0$. In position representation this equation is $\hslash {\psi }^{\prime }+m\omega x\psi =0$, whose general solution isFor a unit vector $|C|=(m\omega /\pi \hslash {)}^{1/4}$; its phase is arbitrary. The commutators $[H,A_{\pm }]=\pm \hslash \omega A_{\pm }$ make $A_{+}$ raise and $A_{-}$ lower energies by $\hslash \omega$. Repeated raising of the Gaussian produces all levels $(n+1/2)\hslash \omega$. Conversely lowering any nonzero energy eigenvector must eventually terminate, since otherwise it would produce an energy below the lower bound. A terminating vector is annihilated by $A_{-}$ and hence has energy $\hslash \omega /2$. Thus no other eigenvalues occur.

Bilinearity of the two-dimensional commutators giveswhile all momentum-momentum and position-position commutators remain zero. Equal angles give a canonical position-momentum pair. Perpendicular angles give commuting components, which belong to independent coordinate directions.

The orthogonal coordinates $X_{+}=(X_1+X_2)/\sqrt{2}$, $X_{-}=(X_1-X_2)/\sqrt{2}$ preserve the kinetic sum, andHence $\boxed{{\omega }_{+}=\omega ,{\omega }_{-}=3\omega }$ and $H=H_{+}+H_{-}$. Every cross-mode position/momentum pair commutes, so $[H_{+},H_{-}]=0$. On the tensor-product oscillator basis,For each $N\ge 0$, allowed pairs have $n_{-}=0,1,\dots ,\lfloor N/3\rfloor$ and $n_{+}=N-3n_{-}$. ThusEach pair contributes one product eigenvector; counting them accounts for the energy-level degeneracy.

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.