Past exam of the introduction to quantum information course of the University of Oxford

Source: original released examination PDF, bundle pages 93–96; paper pages 1–4. The printed sitting is 7 June 2021 despite the bundle filename.

Solutions to all three questions. We use the usual Pauli matrices, column state vectors, and the density operator convention $\rho \mapsto U\rho U^{†}$ for a gate. The CHSH conclusions require independent, uniformly chosen settings and no communication between the devices during a run. In Question 3, a reliable unknown-basis transformation means one deterministic operation that works for every common input basis.

The Pauli multiplication identity gives $(\mathbf{s}\cdot \mathbf{\sigma }{)}^2=|\mathbf{s}{|}^{2}I$. Its eigenvalues are therefore $\pm |\mathbf{s}|$, and those of the density matrix areHermiticity and trace one are automatic; positive semidefiniteness holds exactly for $|\mathbf{s}|\le 1$. Thus physical qubit states form the Bloch ball, with pure states on the Bloch sphere and mixed states inside.

For the mixture of the two standard-basis states,These form the entire $z$-axis diameter, shown in the left panel of the diagram in part b. Generate the state by selecting $|0⟩$ with classical probability $p_0$ and $|1⟩$ with probability $p_1$, then discarding the classical selection record. A coherent superposition is generally a different state: the forgotten classical record removes off-diagonal coherence.

The six Pauli eigenstates have Bloch vectors $\pm {\mathbf{e}}_x$, $\pm {\mathbf{e}}_y$, $\pm {\mathbf{e}}_z$:Classical convex combinations of density operators produce the corresponding convex combinations of their Bloch vectors. Their convex hull is the stabilizer octahedron:The six vertices are pure Pauli states; the faces and interior represent their mixtures. To verify the reverse inclusion, give the vertex in the sign of each coordinate weight $|s_j|$ and use the leftover weight at the origin, itself the equal mixture of either opposite pair.

Conjugation by the Hadamard gate sends $(X,Y,Z)$ to $(Z,-Y,X)$. Therefore its Bloch vector action isIt is a rotation through $\pi$ about $({\mathbf{e}}_x+{\mathbf{e}}_z)/\sqrt{2}$: $H=(X+Z)/\sqrt{2}$ differs by a global phase from the corresponding spin rotation.

The S gate and T gate differ by global phases from $z$-axis rotations:Indeed $\mathrm{diag}(1,e^{i\theta })=e^{i\theta /2}{e}^{-i\theta Z/2}$, which sends $(s_x,s_y,s_z)$ to $(s_x\cos \theta -s_y\sin \theta ,s_x\sin \theta +s_y\cos \theta ,s_z)$.

Direct matrix multiplication givesThus each generating Clifford gate conjugates every Pauli operator to another Pauli operator up to sign. Repeated conjugation proves the same for every product of generators and their inverses, hence for the Clifford group. For Hermitian Pauli observables the possible phases here are signs; arbitrary Pauli-group elements also include their conventional phases.

If $P|\psi ⟩=\lambda |\psi ⟩$, then $(CP{C}^{†})C|\psi ⟩=\lambda C|\psi ⟩$. Therefore $C|\psi ⟩$ is another Pauli eigenstate. Every single-qubit Clifford circuit consequently permutes the six Pauli vertices of the stabilizer octahedron.

While the sequence is remembered, the conditional quantum state is the known pure endpoint $C|0⟩$, so its Bloch vector is one of the six Pauli vertices. Partial forgetting leaves a probability distribution $p_j$ over possible endpoints, givingHence the vector lies in the stabilizer octahedron. The mixing describes lost classical information, not a different unitary evolution of the physical qubit.

Under complete unbiased uncertainty over the six endpoints, $p_j=1/6$, the opposite vectors cancel andThis is the maximally mixed state at the center. Mere loss of the particular sequence does not mathematically force uniform probabilities: a known biased distribution of endpoint states would leave its corresponding noncentral mixture. The central conclusion uses the unbiased randomization implicit in complete forgetting in this setting.

Write $|\psi ⟩=a|0⟩+b|1⟩$ and $\omega =e^{i\pi /4}$. After the controlled-NOT gate, the joint state isAn ancilla measurement outcome zero leaves the unnormalized data state $(a|0⟩+\omega b|1⟩)/\sqrt{2}=T|\psi ⟩/\sqrt{2}$. Outcome one leaves $(\omega a|0⟩+b|1⟩)/\sqrt{2}=\omega T^{†}|\psi ⟩/\sqrt{2}$. Both outcomes have probability $1/2$. The correction for outcome one is $S$, and $S{T}^{†}=T$, so after correction the data is always $T|\psi ⟩$, up to an irrelevant global phase. This proves magic-state injection of a T gate.

The resource state's Bloch vector is $(1/\sqrt{2},1/\sqrt{2},0)$, outside the stabilizer octahedron. Clifford gates alone preserve stabilizer structure and, with standard-basis preparation and measurement, admit efficient classical simulation by the Gottesman--Knill theorem. Adding the non-Clifford T gate supplies a universal quantum gate set together with Clifford gates: arbitrary finite-qubit unitaries can be approximated to arbitrary precision. The assertion is approximate universality, not exact representation of every unitary by a finite circuit.

Each local reduced density operator of the Bell state is $I/2$. Alice's unrecorded local measurement leaves Bob's reduced state unchanged: he still sees a fair random bit. Only after learning her outcome through a classical channel can he identify the corresponding conditional state. Alice cannot choose the random outcome to encode a bit, so the correlations obey the no-signalling theorem.

If the shared state is the trusted pure Bell state, measuring both sides in the $Z$ basis gives identical, uniformly random raw bits. A pure joint state of Alice and Bob has no correlations with an external system, so these bits can form a shared secret. In an actual quantum key distribution protocol, the parties must verify the necessary state/device assumptions and authenticate their classical communication; correlation alone, without a security test, does not certify secrecy against an untrusted source.

Expand the Bell state in the standard basis:The transpose is not an adjoint: the formula is complex-linear in both operators. The equality follows since $(A^T{)}_{kj}=A_{jk}$.

The matrices $X,Z$ are real symmetric, $\mathrm{tr}(X^2)=\mathrm{tr}(Z^2)=2$, and $\mathrm{tr}(XZ)=0$. Part b therefore givesSince the local measurement outcomes are $\pm 1$, their product is one exactly on agreement. Thus $E[ab]=2\mathbb{P}(a=b)-1$, and

The first three angle differences have cosine $1/\sqrt{2}$, while the fourth has cosine $-1/\sqrt{2}$. The first three winning conditions require agreement and the fourth requires disagreement. Each consequently succeeds with the same probabilityUniform independent selection of the four setting pairs gives this same average. Equivalently the CHSH inequality expression is $E_{11}+E_{12}+E_{21}-E_{22}=2\sqrt{2}$, attaining the CHSH Tsirelson bound.

For predetermined outcomes, the four products multiply to one:The four desired signs are $(+1,+1,+1,-1)$, whose product is minus one. Thus at least one winning condition must fail for every deterministic assignment. Uniform independent settings give success at most $3/4$ per assignment, and averaging over hidden assignments preserves the bound.

An observed rate near $0.85$, with enough trials to exclude statistical fluctuation, therefore contradicts Eve's predetermined-local-outcome claim, provided the settings are independent of her devices and the devices cannot communicate during a run. It violates the local bound of the CHSH inequality while remaining consistent with the quantum optimum ${\cos }^2(\pi /8)\approx 0.853553$.

The observation is not by itself exact certification of a perfect two-qubit state. Maximal quantum CHSH correlations support CHSH self-testing only with the appropriate quantum/locality assumptions and up to local changes of representation and unused ancillary systems; a near-maximal finite data set requires a quantitative error analysis. An unrestricted communicating pair of black boxes can instead win every run, so the quantum bound is not a bound on arbitrary devices without these assumptions.

No. The independent-setting assumption can fail if Eve controls the random-number generators. Her hidden information can then include the settings that will be selected, and the outcome assignment can depend on that information. For each known setting pair she can prearrange its one tested winning relation, without having to satisfy all four counterfactual relations at once.

For example, arrange equal outputs for the first three setting pairs and opposite outputs for the fourth, conditioning the programmed outputs on the known pair. This can give success one with predetermined records. It does not contradict the $3/4$ proof, which assumes the same hidden-assignment distribution for every independently selected pair. This is the measurement-independence loophole, so the claimed security conclusion is not established by such a compromised test.

The partial trace removes the other orthonormal factor in the Schmidt decomposition:For instance, tracing $|a_k{b}_k⟩⟨a_{\ell }{b}_{\ell }|$ over $B$ gives $⟨b_{\ell }|b_k⟩|a_k⟩⟨a_{\ell }|$, which vanishes unless $k=\ell$. Consequently the nonzero eigenvalues of the two reduced density operators agree, with multiplicity. If the subsystem dimensions differ, the orthonormal Schmidt families can be completed to full bases and the extra eigenvalues are zero.

No: the advertised pure joint state is incompatible with these marginals. Their eigenvalue multisets are $\{0.9,0.1\}$ and $\{0.7,0.3\}$, which differ. Part a shows that a bipartite pure state must give the same nonzero Schmidt spectrum on both sides.

This excludes Eve's purity claim, not every possible joint state or every possibility of entanglement. A mixed state can have unequal marginal spectra; the product ${\rho }_A\otimes {\rho }_B$, for example, has exactly the stated marginals.

By linearity, the isotropic qubit map acts asIt contracts or expands every direction by the same factor and reverses the vector when $r<0$. More generally ${\mathcal{E}}_r(A)=rA+(1-r)\mathrm{tr}(A)I/2$, so the map preserves trace and Hermiticity for every real $r$.

The output of a state of Bloch radius $u$ has eigenvalues $(1\pm ru)/2$. These are nonnegative for every $0\le u\le 1$ exactly whenIf $|r|>1$, a pure input already gives a negative eigenvalue. Conversely, $|r|\le 1$ preserves the Bloch ball, hence positivity for states and, by positive scaling, for all positive operators. This is positivity of the map alone; physical implementation on part of an entangled system needs complete positivity.

Use the unnormalized Choi matrix convention $J_r={\sum }_{j,k=0}^1|j⟩⟨k|\otimes {\mathcal{E}}_r(|j⟩⟨k|)$, which has trace two. In the ordered basis $|00⟩,|01⟩,|10⟩,|11⟩$,Equivalently $J_r=2r|\Omega ⟩⟨\Omega |+(1-r)I_4/2$. The convention obtained by applying the channel to one half of the normalized Bell state gives $J_r/2$; its positivity criterion is the same.

The Bell state $|\Omega ⟩$ is an eigenvector of $J_r$ with eigenvalue $(1+3r)/2$. Its three-dimensional orthogonal complement has eigenvalue $(1-r)/2$. The Choi criterion for complete positivity therefore givesFor sufficiency one can also write a Kraus representation explicitly:In this range the four coefficients are nonnegative and sum to one, giving a physical mixture of Pauli conjugations. This is the complete positivity range of the depolarizing channel in the Bloch-scaling parameterization.

Write the unknown common eigenbasis as an unknown unit Bloch vector $\mathbf{n}$. Then ${\rho }_A$ has vector $0.8\mathbf{n}$, while ${\rho }_B$ has vector $-0.4\mathbf{n}$. A single deterministic channel working for every possible basis must therefore map $0.8\mathbf{n}$ to $-0.4\mathbf{n}$ for every unit $\mathbf{n}$.

By the basis-independent conversion of qubit Bloch radii, linearity forces that channel to be ${\mathcal{E}}_{-1/2}$. Explicitly, add the conditions for $\mathbf{n}$ and $-\mathbf{n}$ to get $\mathcal{E}(I)=I$, then use the three coordinate directions to fix its action on every Pauli matrix. But $-1/2<-1/3$, so part iv excludes complete positivity. Thus the forward conversion cannot be implemented reliably without knowing the basis.

The reverse conversion would require ${\mathcal{E}}_{-2}$, which is not even positive by part ii. Therefore the reverse conversion is also impossible. Unitary operations interspersed with physically allowed depolarizing maps cannot evade these restrictions: their composition remains a physical quantum channel. The question concerns one basis-independent operation on the supplied qubit, not an additional resource of many copies for learning its basis.

With the basis known, choose a unitary operator that interchanges its two basis states. On the Bloch sphere this is a rotation by $\pi$ about an axis perpendicular to $\mathbf{n}$, sending $\mathbf{n}$ to $-\mathbf{n}$. Follow it by ${\mathcal{E}}_{1/2}$, which is completely positive. ThenThus ${\rho }_A$ can be converted to ${\rho }_B$ using the allowed operations and a suitable physically available depolarizing parameter.

The reverse direction would have to increase Bloch radius from $0.4$ to $0.8$. Unitaries preserve radius, and every completely positive ${\mathcal{E}}_r$ has $|r|\le 1$, so any composition of the allowed operations cannot increase it. Hence ${\rho }_B$ still cannot be converted to ${\rho }_A$ using these resources. This conclusion uses the stated unitary/depolarizing resources; an unrestricted state-preparation channel would be a different allowance.

Source: released examination PDF, bundle pages 103–106.

Solutions to all surviving material, Question 3. TODO: Questions 1 and 2 cannot be answered from this release: printed pages 2–3 are blank. No missing question text or subpart structure is inferred.

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The Pauli matrices are Hermitian matrices with zero trace, so a real Bloch vector makes $\varrho$ Hermitian and gives $\mathrm{tr}\varrho =1$. Their Pauli multiplication identity implies $(\mathbf{s}\cdot \mathbf{\sigma }{)}^2=|\mathbf{s}{|}^{2}I$. Consequently the two eigenvalues of $\mathbf{s}\cdot \mathbf{\sigma }$ are $\pm |\mathbf{s}|$, including the zero-vector case, andA density matrix must also be positive semidefinite. The displayed eigenvalues are nonnegative exactly when $\boxed{|\mathbf{s}|\le 1}$. Thus real coordinates alone are insufficient: the allowed vectors fill the Bloch ball. Its boundary consists of pure states, with eigenvalues one and zero; its center is the maximally mixed state $I/2$.

For any column vector $v$ and positive semidefinite input $\varrho$, the Kraus representation givesThus the output is positive semidefinite. The cyclic property of the trace and completeness of the Kraus operators giveMore strongly, tensoring with an arbitrary ancilla replaces $A_k$ by $A_k\otimes I$, so the same argument proves complete positivity rather than only positivity on single-system density matrices.

If $\mathcal{F}$ has Kraus operators $B_j$, the composition of quantum channels hasThis is again a trace-preserving Kraus representation. The composition of completely positive maps is completely positive even without the trace-preserving assumption; in the general case, the product Kraus operators still establish complete positivity.

Conjugation by one Pauli matrix fixes that same matrix and reverses the other two. For example, $XXX=X$, $YXY=-X$, and $ZXZ=-X$. Applying this to each component of the Bloch vector givesHence the quantum depolarizing channel isThe Kraus operators $\sqrt{1-p}I$ and $\sqrt{p/3}{\sigma }_j$ establish complete positivity for $0\le p\le 1$. At $p=3/4$ every state becomes maximally mixed. For $p>3/4$ the factor is negative: it reverses the vector as well as reducing its length. Its most negative value is $-1/3$, consistent with the complete-positivity range of the isotropic qubit map.

The universal NOT gate is the isotropic qubit map with factor $-1$. It sendsFor $|\mathbf{s}|\le 1$, the output has the same two nonnegative eigenvalues as the input, exchanged, and still has trace one. Thus it maps every single-qubit density matrix to a density matrix. Its linear extension to all operators is $\mathcal{N}(A)=\mathrm{tr}(A)I-A$, not $I-A$ for arbitrary unnormalized $A$. Positivity on one qubit does not establish complete positivity when an ancilla is present.

The input is the Bell state $\rho =|{\Phi }^{+}⟩⟨{\Phi }^{+}|$, where $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. Acting on the first qubit reverses the signs of all three terms containing its Pauli matrices, so in computational basis order $00,01,10,11$ the output isIts eigenvalues are $\boxed{-1/2,1/2,1/2,1/2}$: in particular, its expectation value in $|{\Phi }^{+}⟩$ is $-1/2$. It therefore fails to be positive semidefinite, although its trace remains one. This single entangled state proves that the universal NOT gate is positive but not completely positive, so it cannot be a deterministic physical quantum channel on an arbitrary input.

Source: released examination PDF, bundle pages 110–113.

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.

An observed value $k$ must lie in the image of $f$. Choose either preimage $x_k$. The two-to-one function promise and $s\ne 0^n$ imply that its preimages are exactly $x_k$ and $x_k\oplus s$. A computational-basis measurement of the second quantum register projects the joint quantum state onto the unnormalized vectorBy the Born rule, $\mathrm{Pr}(k)=2^{1-n}$. Dividing by its square root gives the conditional first-register pure stateChoosing the other preimage merely interchanges the two summands. The result is a Simon coset state, retaining quantum coherence between the two inputs that share the measured function value. Outcomes outside the image have probability zero and hence no conditional state to normalize.

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.

Each run of Simon's algorithm uses one oracle call and returns one uniform sample from $s^{\perp }$. Collect samples as rows of a matrix $Z$ and solve $Zs=0$ over the two-element field using Gaussian elimination. Once the rows have matrix rank $n-1$, the kernel is $\{0,s\}$, so the promised nonzero vector is uniquely determined. Dependent samples, including $z=0$, must be allowed for; exactly $n-1$ runs need not suffice.

For a rough quantitative estimate, put $d=n-1$. When the existing samples span an $r$-dimensional vector subspace of $s^{\perp }$, the next sample increases the matrix rank with probability $1-2^{r-d}$. The random spanning of a binary vector space estimate givesAlternatively, $m=d+t$ samples fail to span with probability less than $2^{-t}$: failure means some nonzero linear functional on the $d$-dimensional space vanishes on every sample, and a union bound gives $(2^d-1)2^{-m}$. ConsequentlyThe Gaussian elimination requires only a polynomial amount of classical work, for example $O(n^3)$ bit operations for $O(n)$ samples. The degenerate case $n=1$ needs no calls because the promise already specifies $s=1$.

Classically, distinct inputs with equal function values reveal $s=x\oplus x^{\prime }$. Uniformly sampling distinct inputs finds such a function collision after roughly $\sqrt{2^n}$ calls, by the birthday square-root scale. For $q$ sampled inputs, the expected number of colliding pairs is $\left(\genfrac{}{}{0ex}{}{q}{2}\right)/(2^n-1)$. The usual randomized classical query complexity of Simon's problem is therefore exponential in $n$, on the scaleA straightforward exhaustive classical search instead uses $O(2^n)$ calls. The square-root collision estimate is the sharper classical comparison; the quantum advantage concerns access to the promised quantum oracle, rather than an explicit table of all function values.

For any normalized vector $|u⟩$, the Born rule assigns the probability $⟨u|\rho |u⟩$ to the corresponding rank-one orthogonal projection. This must be nonnegative for every $u$, which is exactly positive semidefiniteness. For any complete orthonormal basis $\{|j⟩\}$, the outcome probabilities must sum to one:Thus a Hermitian matrix represents a physical density matrix only ifThese conditions are also sufficient in finite dimension. The spectral theorem writes $\rho ={\sum }_j{p}_j|u_j⟩⟨u_j|$ with $p_j\ge 0$ and ${\sum }_j{p}_j=1$, describing a quantum state ensemble of pure states with valid mixing probabilities. A Hermitian matrix alone could have negative eigenvalues, hence negative measurement probabilities.

Expand the given pure state in the ordered computational basis $00,01,10,11$:The coefficients have squared magnitudes $1/3,1/6,1/3,1/6$, summing to one. The density matrix is the outer product $v{v}^{†}$, not the diagonal matrix of these four probabilities: its off-diagonal entries retain quantum coherence. ThereforeThe outer product is a Hermitian matrix and is positive semidefinite: $u^{†}\rho u=|v^{†}u{|}^2\ge 0$. Its trace is one, and ${\rho }^2=\rho$, as required for a pure state.

Let $a_{ij}$ be the coefficient of $|ij⟩$. The partial trace contracts the repeated index of the discarded qubit:For the first qubit, each diagonal entry is $1/3+1/6=1/2$ and the upper off-diagonal entry is $1/3-1/6=1/6$. For the second qubit, the diagonal entries are $2/3$ and $1/3$, while the upper off-diagonal entry is $-1/\sqrt{18}+1/\sqrt{18}=0$. Thus the two reduced density matrices, each in computational basis order $0,1$, areBoth have eigenvalues $2/3,1/3$ and trace one, so both are mixed states even though the joint quantum state is pure. Their different eigenvectors are visible in the useful rewriteAbsorbing the minus sign into one local basis vector makes this a Schmidt decomposition with nonnegative coefficients. It also checks both partial traces and exhibits the entanglement responsible for the mixed local descriptions.

For a Hermitian matrix $A$, the spectral theorem gives $A=U\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_d)U^{†}$, with real eigenvalues and a unitary matrix $U$. HenceThe square root here is the unique positive square root of an operator. Taking the trace, which is unchanged by unitary similarity, provesFor a density matrix, all eigenvalues are nonnegative and sum to one. Its trace norm is thereforeThe absolute values cannot be omitted for a general Hermitian matrix, since positive and negative eigenvalues can cancel in its trace but never in its trace norm.

With no access to the discarded qubit, the two preparations are exactly the reduced density matrices in c, with equal prior probabilities. Their difference isSince $\mathrm{Tr}\Delta =0$, its eigenvalues are $\pm \sqrt{2}/6$. The trace norm and trace distance are consequentlyThe Holevo–Helstrom theorem gives the optimumTo establish attainability, rather than only an upper bound, measure the Hermitian operatorwhere $X,Z$ are Pauli matrices. Since $\Delta =(\sqrt{2}/6)B$, report first qubit for the positive outcome; second qubit for the negative outcome. The corresponding orthogonal projections are $E_{\pm }=(I\pm B)/2$. Their success probability iswhich reaches the bound. Identical eigenvalues of the two density matrices do not make them indistinguishable: their eigenvectors differ, and this quantum measurement detects that difference.

Use the conventional phase gate $P_{\phi }=\mathrm{diag}(1,e^{i\phi })$ in computational basis order $0,1$. The single-qubit Hadamard gate and the successive quantum states areThe final Hadamard gate recombines the two components, givingThe leading factor in the last expression is a global phase. The Born rule therefore yieldsAt $\phi =0$ the output is certainly zero, whereas at $\phi =\pi$ it is certainly one. The phase gate changes no intermediate outcome probabilities; the final Hadamard gate converts a relative quantum phase into an observable quantum interference fringe.

Write $c=⟨e_0|e_1⟩=v{e}^{i\alpha }$. The Cauchy-Schwarz inequality gives $0\le v\le 1$ because the two environment states are normalized. Immediately after the interaction, the joint quantum state isApplying the last Hadamard gate givesSince the environment is unobserved, the Born rule uses the squared norm of the environment-valued coefficient of $|0⟩$:Thus the Hadamard interference with environment overlap formula isThe plus sign follows from the explicitly chosen phase gate and the order $⟨e_0|e_1⟩$ of the inner product. Equivalently, before the last Hadamard gate the partial trace over the environment givesThe complex environment overlap in qubit dephasing both reduces the quantum coherence by $v$ and adds a relative quantum phase $\alpha$. Orthogonal environment states have $v=0$ and give $P_0=1/2$ for every $\phi$. If $v=1$, the quantum interference fringe has full contrast and is only shifted; the qubit has not become mixed. For $v=0$, $\alpha$ is immaterial.

Moving the interaction to between the first Hadamard gate and the phase gate givesthe identical joint quantum state. Therefore the expression is unchanged. This is the commutation of computational-basis dephasing and phase gates; it uses the specified population-preserving interaction with the same environment states, rather than an arbitrary model of quantum decoherence.