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

This is the standalone Part C paper for Quantum Theory and Quantum Computers, held on Wednesday 15 June 2011, 9.30am–11.00am. Its three questions match questions 4–6 of section B in the combined Part B paper B7.1/C7.1, formal code DMAB 2686. The original solutions are reused here with the standalone question numbering, including the correction to the missing Fourier normalization factor.

Source release: Oxford examination archive release thread, Part C bundle pages 438–440.

Answer availability checked on 28 October 2026: bounded official-source searches by year, title, module and formal code found no exact full-paper or selected-part solution advertisement. The check was independent of any release policy for later years.

In the scalar two-particle model, the fermionic exchange symmetry and bosonic exchange symmetry conditions areFor particles with spin, exchange acts on all their labels; the stated coordinate-only model fixes the spatial symmetry sector. Define $S\Psi (x_1,x_2)=\Psi (x_2,x_1)$. Equal masses make the kinetic operator invariant under exchange, and the assumed potential is invariant too, so $HS=SH$. Therefore $H\Psi =E\Psi$ implies $H(S\Psi )=E(S\Psi )$. The componentsare respectively symmetric and antisymmetric eigenfunctions whenever nonzero. Their sum is the original nonzero quantum state, so at least one is nonzero. An eigenspace may contain both sectors if the energy is degenerate.

The normal modes of two harmonically coupled identical particles provide the separation. The coordinate transformation to $y_{+},y_{-}$ is orthogonal, with absolute Jacobian one. Exchanging the particles leaves $y_{+}$ fixed and changes $y_{-}$ to $-y_{-}$. Hence bosonic quantum states are even and fermionic quantum states odd in the relative coordinate. The chain rule givesAlso $x_1^2+x_2^2=y_{+}^2+y_{-}^2$ and $(x_1-x_2{)}^2=2y_{-}^2$. Combining the two trapping potentials with the interaction therefore gives the separated Hamiltonian operatorThese are independent oscillators of frequencies $\omega$ and $\omega \cos \alpha$. Both frequencies are positive in the stated range. Their product eigenstates have energiesThe parity of an oscillator level is $(-1{)}^n$. Consequently bosons allow even $n_{-}$ and fermions odd $n_{-}$, while $n_{+}$ is unrestricted.

The bosonic ground state uses $n_{+}=n_{-}=0$, and the fermionic ground state uses $n_{+}=0,n_{-}=1$. The oscillator length in the relative mode isUsing the normalized Gaussian for each mode givesIndeed ${\varphi }_{\tau }(y)=\sqrt{2}y{\psi }_{\tau }(y)/\tau$ by differentiating the Gaussian, so it is the normalized odd first excited state. The ground energies are $E_{\mathrm{B}}=\hslash \omega (1+\cos \alpha )/2$ and $E_{\mathrm{F}}=\hslash \omega (1+3\cos \alpha )/2$.

Choose normalized factors for a normalized product quantum state; this can always be done by reciprocal rescaling of its factors. Since the coordinate Jacobian has absolute value one,If the supplied factors are not individually normalized, the general expression has an additional factor $\Vert \psi {\Vert }^2$; choosing normalized factors is the convention in the displayed formula.

For the ground states, put $\rho (y)=|{\psi }_{\tau }(y){|}^2=e^{-y^2/{\tau }^2}/(\sqrt{\pi }\tau )$. Then $|{\varphi }_{\tau }(y){|}^2=2y^2\rho (y)/{\tau }^2$ and ${\rho }^{\prime }=-2y\rho /{\tau }^2$. Thus, for differentiable $f$ with suitable Gaussian integrability and vanishing boundary terms,This is integration by parts, and every polynomial satisfies the required endpoint conditions. The relative-coordinate moments in even and odd oscillator ground states follow by substituting $f(y)=y^n$:For odd $n$, both expectations vanish by parity, so this is an equality of zeros rather than an increase in signed mean separation. The even moments show the physical difference. In particular ${\mathbb{E}}_{\mathrm{B}}[y_{-}^2]={\tau }^2/2$ and ${\mathbb{E}}_{\mathrm{F}}[y_{-}^2]=3{\tau }^2/2$, so the fermionic root-mean-square particle separation is $\sqrt{3}$ times the bosonic value. The fermionic probability density vanishes at coincidence $y_{-}=0$, expressing Pauli exclusion principle in this spatial symmetry sector. This is an exchange constraint on the wave function, not an additional classical force. As $\alpha$ approaches $\pi /2$, the relative oscillator softens and its width diverges; the excluded endpoint would no longer have this normalizable oscillator ground state.

Use an expansion for the eigenstate continuing the nondegenerate level, with intermediate normalization $⟨{\psi }_0,\psi (ϵ)⟩=1$:Insert these expressions into the eigenvalue equation. At first order,Taking the inner product with ${\psi }_0$ eliminates the left side by self-adjointness. ThusFor $j\ne 0$, projection onto ${\psi }_j$ instead givesAt second order, projecting the eigenvalue equation onto ${\psi }_0$ gives $E^{(2)}=⟨{\psi }_0,H^{\prime }\chi ⟩$. Expand $\chi$ in the orthonormal energy basis and use self-adjointness of $H^{\prime }$:Nondegeneracy ensures the denominators for this level do not vanish; convergence and validity of the expansion are among the assumed perturbation-theory hypotheses.

For the cosine perturbation of the ground level in a one-dimensional box, write $u=\pi x/a$. The product-to-sum identity givesso, with the given common normalization of the sine eigenfunctions,The diagonal matrix element is $-ϵ/2$, and the only nonzero off-diagonal matrix element out of the ground state is $⟨{\varphi }_3,V{\varphi }_1⟩=ϵ/2$. Let $E_{\mathrm{box}}={\pi }^2{\hslash }^2/(2m{a}^2)$. Since ${\mathcal{E}}_3=9E_{\mathrm{box}}$, the ground energy isThe second-order term is negative, as expected for a nondegenerate ground level: every excited-quantum state denominator is negative. The first-order shift is also negative for positive $ϵ$, because the ground-quantum state density is largest near the center, where this perturbing potential is negative.

Use the Dirac notation convention that the inner product is linear in the ket. Orthonormality of the computational basis givesThe finite geometric sum satisfies ${\sum }_{k=0}^{N-1}{\lambda }^{rk}=N$ if $r=0$ in ${\mathbb{Z}}_N$, and zero otherwise: in the second case it equals $(1-{\lambda }^{rN})/(1-{\lambda }^r)=0$. ThereforeThere are $N$ such orthonormal vectors in an $N$-dimensional space, so they form a basis. The inverse expansion is verified by the same sum:This is the unitary discrete Fourier transform and its inverse.

For the basis-copying isometry, the images $|k⟩|k⟩$ are orthonormal, so $C$ is a linear isometry, with $C^{\ast }C=I$. Its matrix elements giveApplying the definition of the adjoint operator, the coefficient of $|k⟩$ in $C^{\ast }|f⟩|g⟩$ is exactly this expression. ThusEquivalently $C^{\ast }|k⟩|\ell ⟩={\delta }_{k\ell }|k⟩$. The adjoint of basis copying acts as pointwise multiplication of coefficient functions.

For the Fourier transform of a basis-copying isometry, the normalization matters:Hence $C^{\ast }|\hat{m}⟩|\hat{n}⟩=N^{-1/2}|\widehat{m+n}⟩$. Taking adjoint matrix elements, or expanding both output factors in the inverse Fourier basis, yieldsThe printed formula is missing the factor $1/\sqrt{N}$. There are $N$ mutually orthogonal terms in its sum, so without that factor its right side has norm $\sqrt{N}$, whereas its left side has norm one because $C$ is an isometry. The formula above is the normalized identity consistent with all the preceding definitions. For $N=1$ the distinction disappears, but the no-cloning assertion is substantive for $N\ge 2$.

The normalized output is not $|\hat{j}⟩|\hat{j}⟩$. Its coefficient matrix pairs every Fourier label $m$ with $j-m$; tracing out either factor gives $I/N$. Thus it is a maximally entangled state, while two clones of a pure Fourier quantum state would have pure reduced quantum states. A copier for the computational basis therefore does not copy the Fourier basis.

More generally, suppose a linear isometry with a fixed blank input could clone every normalized quantum state. Its restriction to $|\psi ⟩|\mathrm{blank}⟩$ would define a linear map $D$ with $D|k⟩=|k⟩|k⟩$ on every computational basis quantum state. Linearity forces $D=C$, which has just been shown to fail on Fourier quantum states. This proves the no-cloning theorem in the pure-quantum state isometric setting.

The inner-product proof makes the obstruction independent of the chosen basis. If an isometry cloned two quantum states with overlap $s=⟨\psi |\varphi ⟩$, preservation of inner products would give $s=s^2$, or, allowing irrelevant output phases, $|s|=|s{|}^2$. Distinct nonorthogonal rays have $0<|s|<1$, which is impossible. Even allowing an output environment does not help: unitary evolution would give $|s|=|s{|}^2|⟨e_{\psi }|e_{\varphi }⟩|\le |s{|}^2$, the same contradiction. Arbitrary unknown quantum states cannot be cloned, although any fixed orthonormal basis can be copied and a one-dimensional quantum state space is trivial.

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.

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.

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 write ${\psi }_S(t)=U(t){\psi }_0$ in the Schrödinger picture, with observables $A_S$ constant unless they have an explicit time dependence. In the Heisenberg picture the state is ${\psi }_H={\psi }_0$ andDifferentiate the two unitary factors to obtain the equation. Unitarity gives $⟨{\psi }_S,A_S{\psi }_S⟩=⟨{\psi }_0,A_H{\psi }_0⟩$.

For the interaction picture choose a reference split $H=H_0+V$, put $U_0=e^{-i{H}_{0}t/\hslash }$, and define ${\psi }_I=U_0^{\ast }{\psi }_S$, $A_I=U_0^{\ast }{A}_S{U}_0$. Then $i\hslash {\dot{\psi }}_I=V_I{\psi }_I$, $V_I=U_0^{\ast }V{U}_0$, while $A_I$ evolves with $H_0$ (and its explicit time dependence). AgainThese are different unitary allocations of the same dynamics between states and observables, so measured expectations agree.

The canonical commutation relation now givesMultiplication of the last equation by $e^{i\omega t}$ and integration givesSince $A_t^{\ast }-A_t=2im\omega X_t$ and $A_0{\psi }_0=0$, both initial expectations of $A_0,A_0^{\ast }$ vanish. ThereforeThe printed positive displacement endpoint has the wrong sign for the Hamiltonian's positive $\mathcal{E}X$ term. The two turning values are $0$ and $-2\mathcal{E}/(m{\omega }^2)$; their ordering depends on the sign of $\mathcal{E}$. The result oscillates about the displaced equilibrium $-\mathcal{E}/(m{\omega }^2)$, exactly as the force $-\mathcal{E}$ predicts.

Choose an orthonormal eigenbasis $H_0|n⟩=E_n|n⟩$ and the phase convention $⟨0|{\psi }^{\prime }⟩=0$. Comparing coefficients of the eigenvalue equation at first order yieldsProjecting onto $|0⟩$ and the other eigenvectors givesProjection of the next-order equation onto $|0⟩$ givesThese are coefficients of the printed series, so $E^{\prime \prime }$ denotes the quadratic coefficient rather than the second derivative itself. Without a phase convention an arbitrary purely imaginary multiple of ${\psi }_0$ can be added to the first vector correction. Continuous spectrum, if present, requires the corresponding spectral integral; the oscillator has a discrete complete eigenbasis.