Past exam of the quantum mechanics quantum theory and quantum computers course of the University of Oxford / 2008 / B

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.

Put $K=m{\omega }^2/2$ and let $g(x)=(2\pi {\sigma }^2{)}^{-1/2}{e}^{-x^2/(2{\sigma }^2)}$ denote the ground-state density. Then $H^{\prime }=K(a^2-x^2)1_{[-a,a]}$ and $p={\int }_{-a}^{a}g(x)dx$. Because $g^{\prime }=-xg/{\sigma }^2$, integration by parts yieldsThus the two first-order shifts areThe first is the supplied ground-state expression; the second follows because the excited-state density is $x^{2}g/{\sigma }^2$, so $E_1^{\prime }=K(a^2{I}_2-I_4)/{\sigma }^2$.

Subtracting, and using $K{\sigma }^2=\hslash \omega /4$, givesThe bracket is zero at $a=0$, tends to one at infinity, and has derivative $2a^{2}g(a)/{\sigma }^2\ge 0$. Therefore the first-order predicted level gap isFor every $\lambda >2$ it becomes negative for all sufficiently large $a$, crossing when $p-2ag=2/\lambda$. It is still positive at $a=0$: the stated conclusion must be read with the large-$a$ regime, rather than for every interval width.

For $\lambda >1$ the inner potential curves downward, placing a central barrier between two symmetric minima at $\pm a$. The even unperturbed ground state has its greatest density at the central barrier, while the odd first excited state has a node there and is less penalized. This explains the relative first-order shifts. A negative predicted gap signals failure of that perturbation approximation, not inversion of the exact lowest two states: the true one-dimensional ground state remains nodeless and even, with the odd state above it. At a high barrier the actual low levels instead form a close double-well pair. The apparent level crossing from a finite-interval oscillator perturbation is useful precisely as a diagnostic of the approximation's range.

A density operator is a positive self-adjoint trace-class operator with $\boxed{\mathrm{Tr}\rho =1}$. A finite expectation is $\boxed{⟨A{⟩}_{\rho }=\mathrm{Tr}(\rho A)}$. It represents a pure state exactly when $\rho =|v⟩⟨v|$ for a unit vector, equivalently ${\rho }^2=\rho$ or $\mathrm{Tr}{\rho }^2=1$.

For a finite-dimensional spectral decomposition $\rho ={\sum }_j{p}_j|v_j⟩⟨v_j|$, introduce an auxiliary space with orthonormal vectors $f_j$ and use the enlarged space $\mathcal{K}=\mathcal{H}\otimes \mathcal{E}$. DefineThen $\Vert {\Omega }_{\rho }{\Vert }^2=\sum p_j=1$ and $⟨{\Omega }_{\rho },\tilde{A}{\Omega }_{\rho }⟩=\sum p_j⟨v_j,A{v}_j⟩=\mathrm{Tr}(\rho A)$. This purification shows that subsystem mixedness can arise from entanglement with an unobserved environment even when the total state is pure. Subsystem measurements alone cannot distinguish these two descriptions.

A nonzero pure bipartite vector is entangled when it cannot be written as a tensor product of two vectors; equivalently its Schmidt rank exceeds one. For mixed bipartite states, entanglement means absence of a convex decomposition into product density operators.

For the nonorthogonal expansion in the question, the unnormalized reduced operator isIndeed $⟨\Psi ,(A\otimes I)\Psi ⟩={\sum }_{j,k}⟨{\xi }_k,A{\xi }_j⟩⟨{\eta }_k,{\eta }_j⟩=\mathrm{Tr}(A{S}_1)$. Positivity follows because $⟨v,S_{1}v⟩=\Vert {\sum }_j⟨v,{\xi }_j⟩{\eta }_j{\Vert }^2\ge 0$, and $\mathrm{Tr}{S}_1=\Vert \Psi {\Vert }^2=:Z$. Thus $\boxed{{\rho }_1=S_1/Z}$ is a density operator; if $\Psi$ is already normalized, $Z=1$. The Gram factors cannot be dropped merely because the vectors are called bases. For unnormalized $\Psi$, physical expectations likewise divide its raw quadratic form by $Z$.

For the Schmidt decomposition diagonalize $S_1$ in an orthonormal basis ${\xi }_j$ with its positive eigenvalues $c_j^2$. Expand $\Psi ={\sum }_j{\xi }_j\otimes w_j$ with $w_j=(⟨{\xi }_j|\otimes I)\Psi$. The reduced matrix gives $⟨w_k,w_j⟩=c_j^2{\delta }_{jk}$. Therefore ${\eta }_j=w_j/c_j$ are orthonormal andZero-eigenvalue components vanish because their squared norms are zero. This proves the decomposition for an arbitrary nonzero vector; the zero vector uses the empty sum. For a unit vector $\sum c_j^2=1$.

In the numerical example the coefficient matrix in the two product bases is $C=\left(\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\end{array}\right)$. Taking the first-subsystem trace yieldsThe vector $(1,-2,1{)}^{\mathsf{T}}$ is in its nullspace, so rank is at most two; its leading $2\times 2$ minor has determinant one, so $\boxed{\mathrm{rank}{S}_2=2}$. This is the rank of $C$ and the Schmidt rank, proving entanglement. The displayed source ket has norm squared $19$, and the displayed reduced matrix has trace $19$. The physical density operator is $\boxed{{\rho }_2=S_2/19}$; its nonzero eigenvalues are $(19\pm \sqrt{337})/38$, since $C{C}^{\mathsf{T}}=\left(\begin{array}{cc}14& 8\\ 8& 5\end{array}\right)$. The raw matrix is correct for the unnormalized quadratic form actually printed.

Put $d=2^N$ and $\lambda =e^{2\pi i/d}$, with nonnegative integer $N$ so the register exists. The geometric sum satisfiesConsequently the matrix $F_{+}$ with entries $d^{-1/2}{\lambda }^{jk}$ has $F_{+}^{\ast }{F}_{+}=I$. This proves unitarity andFor a binary output string $j={\sum }_{r=1}^N{j}_r{2}^{N-r}$, expansion of the printed tensor product gives coefficientThus the printed product is exactly $F_{+}|k⟩$. It is a correct product decomposition of a Fourier phase state, but the printed recovery identity has inconsistent signs:by the square of the quantum Fourier transform. To recover $f(j)$ instead, use $F_{-}=F_{+}^{\ast }$, replacing every product phase by its complex conjugate. Equivalently one may reverse the sign in the definition of $\hat{f}$. The discrepancy is visible for $d\ge 4$ by taking $f$ concentrated at $1$: the printed expression returns the basis vector at $d-1$. The recovery display also uses $k$ as the summation variable on its right while its summand uses $j$; that dummy variable should be $j$.

For factorization first check $\gcd (3,91)=1$. Successive residues areThe first return is $\boxed{r=6}$. Its half power is $3^3=27$, which is neither $1$ nor $-1$ modulo $91$. Therefore

In Shor's algorithm prepare a uniform exponent register of size $Q=2^n$, conveniently $91^2\le Q<2\cdot 91^2$, and reversibly compute $3^x\mathrm{mod}91$ into a second register. Measuring that second register, or tracing it out, leaves periodic exponent correlations of period $r$. Conditional on one measured residue, the first register is a superposition over an arithmetic progression of exponents separated by $r$. The quantum Fourier transform converts this periodic structure into peaks near $k/Q=s/r$. Continued fractions recover $s/r$; when $s$ is coprime to $r$, its denominator is $r$. Verify by modular exponentiation and repeat when the reduced denominator is insufficient. For this example $Q=16384$ is suitable, and useful peaks include those near $Q/6$ and $5Q/6$. Either transform sign gives the same period information, so the source's inverse-sign error does not invalidate the order-finding mechanism.