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Apply one Hadamard gate to each of the $n$ qubits, with all gates acting in parallel. This quantum circuit implements the tensor product $H^{\otimes n}$; it needs $n$ single-qubit gates and one layer. For one qubit,Taking the tensor product of these expressions multiplies the probability amplitudes and adds their exponents, givingHere the dot product is evaluated using modular arithmetic modulo two.

In particular, an initially zero quantum register becomes the uniform quantum superpositionThis is useful at the start of a quantum algorithm because a subsequent quantum oracle acts on all these components coherently. Later quantum interference can reveal a global property of the function. Measuring immediately would merely produce a uniformly random binary string; the useful resource is the preserved quantum coherence, rather than the ability to read every function value in one run.

Write $\oplus$ for bitwise exclusive or. Both quantum registers begin in $|0^n⟩$. After the first Hadamard transform their joint quantum state isThe quantum oracle $U_f$ maps each component $|x⟩|0^n⟩$ to $|x⟩|f(x)⟩$. By linearity, the required quantum state is thereforeThe different first-register computational-basis states are orthogonal, so there are $2^n$ orthogonal joint components of squared probability amplitude $2^{-n}$, even though some second-register values coincide. Thus the displayed quantum state is normalized.

Apply the Hadamard transform to the conditional quantum state from c. Since $(x_k\oplus s)\cdot z=x_k\cdot z+s\cdot z$ modulo two, its output isIf $s\cdot z=1$, the two probability amplitudes cancel by destructive quantum interference. If $s\cdot z=0$, they add, givingThus every observed string obeys $\boxed{z\cdot s=0}$. This Simon Fourier sampling is uniform on the $(n-1)$-dimensional vector subspace $s^{\perp }$ over the two-element field. Indeed, the nonzero linear functional $z\mapsto z\cdot s$ has a kernel of dimension $n-1$ by the rank-nullity theorem. The distribution does not depend on $k$, so measuring or discarding the second quantum register gives the same statistics for the first.

Represent the single-photon state by its two path probability amplitudes. The first Hadamard gate sends $|0⟩$ to $(|0⟩+|1⟩)/\sqrt{2}$. With $P(\varphi )=\mathrm{diag}(1,e^{i\varphi })$ for the lower-arm phase shifter, the complete Mach-Zehnder interferometer givesThe global phase has no effect on detection. By the Born rule,They sum to one. At $\varphi =0$, port 1 is a dark port; at $\varphi =\pi$, the roles reverse. The result uses coherent addition of the path amplitudes, rather than addition of separate path probabilities.

The beta barium borate crystal supplies spontaneous parametric down-conversion: a weak ultraviolet pump can produce a pair of lower-frequency photons, with their frequencies and wavevectors selected by conservation of energy and phase matching. A suitably aligned and compensated source makes the two polarization alternatives coherent, giving the stated Bell state. The crystal alone does not guarantee that every collected pair has that state.

The two polarizing beam splitters route horizontal polarization and vertical polarization to different ports. Take $H$ from arm $a$ to the trigger $T$ and $V$ to arm $c$. The ordinary beam splitter in arm $b$ is a lossless, polarization-independent 50:50 splitter, with the unused input in the vacuum state. It splits probability amplitudes, rather than measuring which output a photon takes. The half-wave plate changes the relative phase of its principal polarizations by $\pi$; orient it so that a $V$ photon in $c$ becomes $(H+V)/\sqrt{2}$ in $c^{\prime }$, up to a common phase. Finally, photodetectors absorb arriving photons and supply electrical clicks. They need not resolve photon number for the ideal four-photon, fourfold-coincidence calculation: a click at every detector already requires one photon at each.

Choose the relative wave phases so that the final polarizing beam splitter sends $H_{c^{\prime }}$ to $H_2$, $V_{c^{\prime }}$ to $V_1$, $H_d$ to $H_1$ and $V_d$ to $V_2$. Such phase conventions must be used consistently: an uncompensated relative phase would give a phase-shifted GHZ state.

Use the resonant interaction picture and define $g>0$ by the one-atom Jaynes-Cummings model in part (a). Equal coupling to both atoms gives the Tavis-Cummings modelThe initial state has two excitations and is symmetric under atom exchange. Excitation conservation and this symmetry confine the evolution toThe antisymmetric atomic spin singlet state is uncoupled and unpopulated. Applying the annihilation operator and creation operator givesThe first off-diagonal entry combines two indistinguishable atomic emission paths; the second also includes the $\sqrt{2}$ enhancement for creating the second photon. Its eigenvalues are $0,\pm \sqrt{6}g$. Put $\Omega =\sqrt{6}g$ and $K=H_I/\hslash$, so $K^3={\Omega }^{2}K$. Its exponential isApplying this to $|A⟩$ yieldsAll three states have the same bare energy on resonance, so restoring the free Schrödinger-picture evolution adds only a common global phase. The expression starts at $|ee;0⟩$, has initial derivative $-i\sqrt{2}g|B⟩$, and its three squared probability amplitudes sum to one.

Suppose the conditional atomic pure state were a product stateComparing its four probability amplitudes with $(|eg⟩+|ge⟩)/\sqrt{2}$ givesThe last two equations make all four factors nonzero, contradicting either of the first two. Equivalently, the coefficient matrix has determinant $-1/2$, while a product state has a rank-one coefficient matrix and zero determinant. Thus the atoms are entangled; indeed each reduced density matrix is $I/2$, making the conditional state maximally entangled.

Set $A=\alpha a^{†}$ and $B=-{\alpha }^{\ast }a$. Their commutator isIt commutes with both operators, so the stated Baker--Campbell--Hausdorff formula terminates:In particular $e^{-{\alpha }^{\ast }a}|0⟩=|0⟩$. Expanding the remaining exponential and using $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$ gives the number state expansionThe orthonormality of the number states givesThis is a Poisson distribution with parameter $\mu =|\alpha {|}^2$. Its probabilities sum to $e^{-\mu }{e}^{\mu }=1$. The phase of $\alpha$ affects the coherent state's relative probability amplitudes but not its photon number probabilities. At $\alpha =0$ the distribution is concentrated at $n=0$.

Let $x\in \{00,01,10,11\}$ and let $f(x)$ be one only at the marked input. A three-qubit implementation of Grover's algorithm starts with $|00⟩|1⟩$, applies a Hadamard gate to each qubit, queries the reversible quantum oracle $U_f:|x,y⟩\mapsto |x,y\oplus f(x)⟩$, applies the Grover diffusion operator $D=2|s⟩⟨s|-I$ to the first two qubits, and measures those two qubits in the computational basis. Here $|s⟩=\frac{1}{2}{\sum }_x|x⟩$. The full three-wire quantum circuit is the composition

Here $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Since $X|-⟩=-|-⟩$, phase kickback gives $U_f(|x⟩|-⟩)=(-1{)}^{f(x)}|x⟩|-⟩$. The third qubit remains separable and unchanged. A direct marked-state phase oracle $O_f|x⟩=(-1{)}^{f(x)}|x⟩$ therefore replaces the three-qubit query using only two qubits:For the explicit choice $f(x)=1$ only at $x=11$, $O_f$ is a Controlled-Z gate. Immediately after it,The mean of these four probability amplitudes is $\overline{a}=1/4$. The Grover diffusion operator maps each probability amplitude $a_x$ to $2\overline{a}-a_x$. The three unmarked probability amplitudes become zero and the marked one becomes one. A single query consequently finds $11$ with probability one.

A gate-level implementation of $D$ is $H^{\otimes 2}(2|00⟩⟨00|-I)H^{\otimes 2}$. Up to an irrelevant global phase, it is $H^{\otimes 2}{X}^{\otimes 2}\mathrm{CZ}{X}^{\otimes 2}{H}^{\otimes 2}$. For any other chosen marked word, conjugate the Controlled-Z gate by Pauli X gates on precisely the positions where the marked word has a zero.

Starting at port 1, the first beam splitter encounter sends amplitude $c$ straight out through port 0 and stores amplitude $-c$ towards M2. A return from M2 enters port $1^{\prime }$; it sends the stored amplitude times $-c$ out through port 1 and times $c$ towards M1. A return from M1 enters port $0^{\prime }$; it sends amplitude times $c$ out through port 0 and times $c$ towards M2. Both internal continuations carry $+c$, so two additional encounters multiply the surviving amplitude by $+c^2$.

With the fixed mirror and propagation phases absorbed as in a, the nonzero coefficients of the outgoing single-photon state areand $c_{0,2j}=c_{1,2j+1}=0$. The first nonzero terms are $c_{0,1}=1/\sqrt{2}$, $c_{1,2}=1/2$, $c_{0,3}=-1/(2\sqrt{2})$ and $c_{1,4}=1/4$. If fixed reflection amplitudes $r_1,r_2$ are retained instead, multiply $c_{0,2j+1}$ by $(r_1{r}_2{)}^j$ and $c_{1,2j}$ by $r_2^j{r}_1^{j-1}$, as well as the appropriate delay propagation phases. These unit-modulus factors do not alter the following probabilities.

Distinct exit events are orthogonal because they differ in output port or arrival-time bin. This requires the incident wavepacket to resolve the delay. Summing the squared moduli as geometric series givesTheir sum is one. After $n$ encounters the probability of still being inside is $2^{-n}$, which tends to zero, so no trapped component is omitted.

The source drawing places the delay line in the M2 arm. A visit to M2 crosses that line in both directions. If $\tau$ is its one-way delay, then an exit after $n$ encounters has crossed it $2\lfloor n/2\rfloor$ times:If the phrase “delay time $\tau$” instead denotes the complete M2-arm round-trip delay, replace $2\tau$ by $\tau$ in the floor formula. The wording does not explicitly distinguish these conventions. Consecutive encounters can produce the same exit time in different ports: for example $n=2,3$ both follow one M2 visit. If the incident wavepacket were much longer than the arm delay, different histories in the same port would overlap and their probability amplitudes would have to be added before squaring; the asserted time-bin expansion would then not be an orthonormal event basis.

Let $S$ denote immediate transmission through the BS1 subsystem between Bob's lower-right port and Alice's upper-left port; let $L$ denote transmission after one visit to each internal mirror. From the same beam splitter rules as in b, the corresponding amplitudes in either direction areup to a common fixed two-mirror propagation phase. The minus sign in $L$ comes from the $1^{\prime }\to 1$ or reverse $1\to 1^{\prime }$ scattering. The two paths selected by D0's time gate are $L$ on the outward journey followed by $S$ on the return, and $S$ outward followed by $L$ on the return. They have the same total delay and fixed propagation phase, so their probability amplitudes interfere.

For the first history, Alice applies $e^{i{\varphi }_A}$ to the late component, while Bob's return-time modulation is not encountered in the return delay arm. For the second, Alice's early component is unmodulated and Bob applies $e^{i{\varphi }_B}$ in the return delay arm. Bob must time his modulation to the return passage, leaving the outward reference preparation unchanged; this is implicit in the stated encoding. Each enabled modulator is traversed twice, so its $\pi /2$ single-pass phase becomes the stipulated $\pi$.

Include the amplitude $c$ to enter BS1 through BS2, and the amplitude of modulus $c$ to reach D0 through BS2 on return. Both histories therefore contain six 50:50 beam splitter factors. Ignoring one common unobservable global phase,Hence the unconditional probability per emitted photon, with no renormalization after discarding paths, isFor the allowed values $0,\pi$, this is $1/16$ when ${\varphi }_A={\varphi }_B$ and zero otherwise. The ignored trajectories exit other ports or other time bins; excluding them from D0's acceptance does not increase the success probability. Equal mirror reflection phases, channel phases and Alice's mirror phase multiply both accepted histories equally and cancel from the interference condition. A residual relative phase would need calibration.

The two given energy eigenstates are orthonormal. Resolve the initial state as $\psi (0)=\cos \theta {\psi }_0-\sin \theta {\psi }_1$. The Schrödinger equation therefore givesThe second component, which is the probability amplitude for the requested final state, is $\sin \theta \cos \theta (e^{-i{E}_{0}t/\hslash }-e^{-i{E}_{1}t/\hslash })$. By the Born rule,This includes $P(t)=0$ when the energy eigenvalues coincide or when the initial state is itself an energy eigenstate.

Inside the infinite square well, the time-independent Schrödinger equation is $-{\hslash }^2{u}^{\prime \prime }/(2m)=Eu$ with $u(-a)=u(a)=0$. For $E<0$ the exponential solutions cannot satisfy both conditions except trivially, and for $E=0$ neither can a linear function. For $E>0$ put $k=\sqrt{2mE}/\hslash$. The first boundary condition selects $u=C\sin (k(x+a))$, and the second gives $\sin (2ak)=0$. Thuson $[-a,a]$, with $u_n=0$ outside. Their squared integrals are one, and the sine orthogonality identity gives an orthonormal set. The sine Fourier basis on the interval is complete: odd extension to an interval of length $4a$ reduces this to completeness of the Fourier series basis. Hence these are a complete set of normalized energy eigenstates.

Choosing the harmless sign of the second energy eigenstate so that $v_1=a^{-1/2}\cos (\pi x/(2a))$ and $v_2=a^{-1/2}\sin (\pi x/a)$, the specified wave function is $v_1/\sqrt{5}+2v_2/\sqrt{5}$. The Born rule therefore giveswith no other possible measured energies. The two energy eigenvalues are $E_1={\hslash }^2{\pi }^2/(8m{a}^2)$ and $E_2=4E_1$.

After the first measurement gives the ground state, the old state is $v_1$. Doubling the width puts the walls at $\pm 2a$, and the new normalized ground state is $w_1=(2a{)}^{-1/2}\cos (\pi x/(4a))$. The stipulated unchanged state is zero outside the original interval, so the probability amplitude isSquaring givesThe ground-state overlap after sudden expansion of a square well stays unchanged during subsequent evolution in the fixed new well, because the energy-basis coefficients acquire only phases. The answer is conditional on the already observed old ground state; its earlier probability $1/5$ is not another factor in this conditional question.

Use the symmetric lossless beam splitter convention with transmission amplitude $t=1/\sqrt{2}$ and reflection amplitude $r=i/\sqrt{2}$. The two mirrors contribute the same fixed quantum phase, which can be suppressed. At the rightward port $D_1$, the upper-arm route has one reflection at BS1 and one transmission at BS2, while the lower-arm route has one transmission at BS1 and one reflection at BS2. Their probability amplitudes therefore add asBy the Born rule,For the upward port, the two amplitudes are $r^2{e}^{i\theta }$ and $t^2$, giving $a_0=(1-e^{i\theta })/2$ and $P(D_0)={\sin }^2(\theta /2)$. They sum to one, as required for an ideal lossless Mach-Zehnder interferometer. Thus $D_1$ is bright at zero relative quantum phase and dark at quantum phase $\pi$. A fixed unequal reflection quantum phase or a swapped port labeling would shift the fringe; the stated convention fixes the port and quantum phase calibration used below.

Write $\varphi =2\pi n/N+ϵ$, with $|ϵ|\le \delta \ll 2\pi /N$. Decode the measured bits as above. The probability amplitude of label $m$ isUsing the finite geometric series,Interpret removable singularities by continuity. In particular, near-grid quantum phase estimation returns the nearby exact label $n\mathrm{mod}N$ with probabilityThis is near one under the stated small-offset assumption, but not exactly one when $ϵ\ne 0$.

There is also a rigorous error bound. Taking the squared modulus of the sum giveswhere $1-\cos x\le x^2/2$ and the elementary sums of $k$ and $k^2$ evaluate the last expression. HenceOn that event the circular quantum phase error is at most $\delta$. For other labels, the exact probabilities areTheir total is $1-P(n)=O(N^2{\delta }^2)$, describing the leakage from the perfect solution. The physical output bit string remains reversed in the printed circuit; this changes its decoding, not these probabilities.

Expand the initial wavefunction in the complete orthonormal energy eigenstates, with $c_n=⟨{\psi }_n,\Psi (0)⟩$. The unitary time evolution isOn the two-dimensional subspace spanned by ${\psi }_1,{\psi }_2$, the observable has matrix $\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$. Its normalized eigenvectors are the symmetric and antisymmetric combinations. A complete orthonormal eigenbasis is thereforeThe eigenvalues are $1,3,0$, with $0$ infinitely degenerate. The first two are nondegenerate, regardless of the distinct energies $E_n$.

The first measurement in quantum mechanics, with outcome $3$, leaves the state ${\varphi }_2$ up to an irrelevant phase. At time $t$ it has becomeIts probability amplitude for outcome $1$ isThe Born rule gives the two-level observable transition probability

Set $L=2^N$. Applying the inverse quantum Fourier transform to the register in part (a) gives probability amplitude at label $m$ equal toChoose $m$ nearest to $\eta =L{\varphi }_j/(2\pi )$, with $m$ reduced modulo $L$ at the phase boundary. Then the difference can be represented with $|\delta |\le \pi /L$. For $\delta \ne 0$, sum the finite geometric series to obtainLet $z=L|\delta |/2\le \pi /2$. On $[0,\pi /2]$, concavity of sine gives $\sin z\ge 2z/\pi$, the line through its endpoint values. Also $|\sin (\delta /2)|\le |\delta |/2$. Thereforeand the Born rule proves the nearest-integer success bound for quantum phase estimation:If $\delta =0$, every summand is one and the probability is exactly one. If the scaled phase is halfway between two integers, either nearest label separately satisfies the displayed bound. Interpreting the measured integer as $m/L$ gives the closest available $N$-bit phase estimate modulo one.

Choose an orthonormal eigenbasis of $V$, with eigenphases ${\varphi }_0,{\varphi }_1$, counting multiplicity. The maximally mixed state is basis independent, soLinearity means that the experiment is the equal mixture of the two eigenstate experiments. After the inverse quantum Fourier transform, the joint state is $\frac{1}{2}{\sum }_{j=0}^1|{\chi }_j⟩⟨{\chi }_j|\otimes |v_j⟩⟨v_j|$, where $|{\chi }_j⟩={\sum }_m{a}_m({\varphi }_j)|m⟩$ andThus the exact output distribution for quantum phase estimation with a maximally mixed target isFor a nonzero phase mismatch, the squared probability amplitude is ${\sin }^2[L(\varphi -2\pi m/L)/2]/\{L^2{\sin }^2[(\varphi -2\pi m/L)/2]\}$, with value one at an exact match. If both eigenphases are exactly representable on the $L$-point grid, their two labels occur with probability $1/2$ each; coincident labels have their probabilities added. For general eigenphases there can be nonzero probabilities at other labels, so it would not be exact to list only the two best approximations with probability $1/2$ each. Each eigenphase is sampled with spectral weight $1/2$, and its nearest-label event has conditional probability at least $4/{\pi }^2$. In particular a nearest label for one eigenphase has unconditional probability at least $2/{\pi }^2$, with additional contributions possible from the other phase.

Use the bit-query Boolean quantum oracle $O_x|i,b,z⟩=|i,b\oplus x_i,z⟩$, where $z$ denotes workspace. Initially every probability amplitude is independent of $x$, hence is a constant polynomial. An input-independent unitary gate only forms linear combinations of probability amplitudes, so it does not increase their polynomial degree.

If $a_{i,b,z}(x)$ is an amplitude before a query, its new value isOne query increases polynomial degree by at most one. Inductively, after $T$ queries each amplitude has polynomial degree at most $T$. The acceptance probability is a sum of squared absolute values of the accepting amplitudes, so it is a real polynomial of degree at most $2T$. Intermediate quantum measurements and classical adaptation can be retained coherently, or handled by summing unnormalized branch probabilities; either approach yields the same degree bound.

An exact algorithm has acceptance probability precisely $f(x)$ at every Boolean input. The multilinear reduction on the Boolean cube replaces positive powers of each $x_i$ by $x_i$, preserving these values without increasing polynomial degree. Therefore the polynomial method for quantum query lower bounds givesHere $\mathrm{deg}(f)$ is the degree of the unique multilinear polynomial representing the Boolean function, and $Q_E$ is its exact quantum query complexity. Uniqueness follows, for example, by evaluating successively on the indicator vectors of subsets: the value on a subset determines its coefficient once all smaller-subset coefficients are known.

Choose the prepared wavefunction real and positive on its support. Its normalization isIn the ramp-state expansion in a symmetric infinite well, write $\psi ={\sum }_{n\ge 1}{c}_n{\psi }_n$. Orthogonality givesFor $\kappa =n\pi /(2a)$, a primitive is $-x\cos (\kappa (x+a))/\kappa +\sin (\kappa (x+a))/{\kappa }^2$. Evaluating both endpoints givesAn equivalent parity-separated form isThese formulas with ${\sum }_n{c}_n{\psi }_n$ are the requested explicit expansion. The convergence is in the square-integrable norm; at the idealized discontinuity at the right wall, a pointwise boundary value does not change the quantum state or its probability amplitudes.