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Alice uses conjugate-basis steering of a Bell pair. Write the desired orthonormal basis for Bob as $|\psi ⟩=a|0⟩+b|1⟩$ and $|{\psi }^{\perp }⟩=c|0⟩+d|1⟩$. Entrywise complex conjugation in the computational basis gives Alice's orthonormal basisFor either $|u⟩$ in Bob's chosen orthonormal basis, the corresponding unnormalized conditional quantum state isIts squared norm is $1/2$. Thus Alice measures in the complex-conjugate basis, obtaining the required Bob pure states with probability $1/2$ each. Averaging over Alice's undisclosed outcome leaves Bob's density matrix equal to $I/2$, consistent with the no-communication theorem.

Initialize one qubit in $|0⟩$, then apply a Hadamard gate, a T gate, and a Hadamard gate, in that temporal order. Defining the Pauli X eigenstates $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, the successive quantum states are

We prove the one-query constancy obstruction for phase oracles while allowing arbitrary quantum ancillas. By purification of a density operator, the quantum state immediately before the sole query can be taken asThe ancilla vectors need not be normalized or orthogonal. Classical randomness and earlier quantum measurements can be included in additional ancillas; allowing access to these systems only gives the proposed algorithm more power.

For the zero constant function, the phase oracle is $I$, leaving $|\Psi ⟩$ unchanged. For each $j$, consider the Boolean function that is one at $j$ and zero elsewhere. Since $M>2$, it is nonconstant, and its phase oracle producesAny later fixed unitary transformation preserves these inner products. An arbitrary final quantum measurement cannot distinguish the constant output from each nonconstant output with certainty unless they are orthogonal, by perfect discrimination of pure states requires orthogonality. Explicitly, an accepting POVM effect $E$ would satisfy $E|\Psi ⟩=|\Psi ⟩$ and $E|{\Psi }_j⟩=0$, forcing $⟨\Psi |{\Psi }_j⟩=0$.

Therefore exact classification requires $p_j=1/2$ for every $j$. Summing giveswhich contradicts $M>2$. Thus one query cannot decide constancy with certainty. The other constant function gives $-I$, differing only by a global phase, so it does not change the argument. For comparison, $M=2$ permits $p_0=p_1=1/2$; the impossibility arises from demanding the test work for every nonconstant Boolean function.

A bipartite pure state is entangled exactly when it is not a product state: there are no normalized quantum states $|u{⟩}_A,|v{⟩}_B$ such thatEquivalently, its Schmidt rank exceeds one, or either reduced density matrix is a mixed state. This definition concerns quantum entanglement across the specified division into systems $A$ and $B$.

The sign factor separates as $(-1{)}^{x+y}=(-1{)}^x(-1{)}^y$. Consequently the quantum state factors into a tensor product of quantum systems:It is a product state, so it is not entangled.

Combine the hypothetical protocol with superdense coding. Before a message is chosen, Alice and Bob share a further Bell pair $|{\varphi }_{+}{⟩}_{QR}$, with $Q$ at Alice and $R$ at Bob, in addition to any resources needed for the hypothetical quantum teleportation protocol. To encode two classical bits $u,v$, Alice applies the Pauli gates $Z_Q^u{X}_Q^v$, preparingThe four choices are the four mutually orthogonal Bell states. Alice now uses the hypothetical protocol to teleport $Q$ to Bob using only one classical bit. Its assumed preservation of the joint quantum state gives Bob both $B$ and $R$ inBob performs a Bell-basis measurement. Explicitly, a controlled-NOT gate $B\to R$, followed by a Hadamard gate on $B$, givesA computational-basis measurement therefore recovers both message bits with certainty after only one transmitted classical bit. The initial Bell pairs are shared before the message is chosen and carry no message themselves. This proves the classical communication lower bound for exact qubit teleportation: a perfect protocol preserving an external reference needs at least two transmitted classical bits.

Write the arbitrary initial pure state asThe reference vectors need not be normalized or orthogonal. By linearity, the same quantum teleportation calculation, with the identity operator on $R$, gives the post-gate quantum stateThus each outcome still has probability $1/4$. Bob's Pauli gate correction leaves the normalized joint quantum stateAll operations act trivially on $R$, so the result is independent of its holder or location. This establishes teleportation as an identity channel on a reference, preserving quantum entanglement with the untouched reference. Since every corrected branch is the same, discarding the classical outcomes also leaves the required joint density matrix.

Alice holds the input qubit $Q$ in $|\psi {⟩}_Q=a|0⟩+b|1⟩$ and her half $A$ of a shared Bell pair $|{\varphi }_{+}{⟩}_{AB}$; Bob holds $B$. Alice applies a controlled-NOT gate $Q\to A$, then a Hadamard gate on $Q$. In the system order $Q,A,B$, the initial quantum state isAfter these two quantum gates, direct expansion yields the quantum teleportation identityAlice performs computational-basis measurements on $Q,A$, and sends the two classical bits $m,n$ to Bob. Each outcome has probability $1/4$, independently of the input pure state. Bob applies the indicated Pauli gates: first $X^n$, then $Z^m$, so his total correction is $Z^m{X}^n$. In every branch,This proves exact quantum teleportation using one Bell pair and two classical bits. Alice's original qubit has been measured; the protocol transfers the quantum state without leaving her another copy.

Let the sample area be $A$, set the band touching energy to zero, and write $m^{\ast }=m_e^{\ast }=m_h^{\ast }$. Assume one electronic valley in each energy band with two spin states. The near-edge parabolic dispersions areUnder periodic boundary conditions, each spin occupies reciprocal-space area $(2\pi {)}^2/A$. Counting the quantum states inside a circle givesIts derivative is the constant two-dimensional free-electron density of states. Defining $D=A{m}^{\ast }/(\pi {\hslash }^2)$, the signed-energy density of states isThe total electronic density of states is therefore $D$ on either side of zero, not $2D$ at every signed energy. If instead one counts positive-energy excitations, the electron and hole species each have density of states $D$, giving $2D$ when summed. A common additional valley degeneracy multiplies all these quantities by that factor. The effective mass approximation assumes thermally sampled energies remain within the near-edge regions of both energy bands.

For $ϵ\to 0^{+}$, the Taylor expansions needed in the exact energy density areSubstitution yieldsThereforeHere Big O notation describes the neglected small-coupling terms, and the logarithm is natural. The nonanalytic ${ϵ}^2\ln ϵ$ term comes from the formerly degenerate quantum states near the Brillouin zone boundary; a uniform nondegenerate perturbation theory expansion of the integrand would fail there. This is the same mechanism as the Peierls energy gain.

For a long chain of length $L$ with periodic boundary conditions, each spin has $Ldk/(2\pi )$ quantum states in a wavenumber interval $dk$. Including two spin states, a full energy band contains $2L/a$ electrons. The state counting in a Bloch band therefore gives the energy densityThe integral of $1+y^2$ is $8/3$. For $ϵ>0$, the even square-root integral isFor example, this follows by setting $u=2y$, applying the supplied antiderivative, and evaluating both endpoints; the lower-end logarithm contributes $\ln ϵ$. HenceThe continuous zero-coupling limit is $\mathcal{U}(0)=2E_{\mathrm{ZB}}/(3a)$. This agrees with directly integrating the one-dimensional free electron model up to $k_F=\pi /a$ at two electrons per unit cell.

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.

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.

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.

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.

Let the unknown input qubit be $|\psi {⟩}_S=\alpha |0⟩+\beta |1⟩$, with $|\alpha {|}^2+|\beta {|}^2=1$. Alice holds $S,A$, Bob holds $B$, and the shared Bell pair is $|{\varphi }^{+}{⟩}_{AB}$. All three-qubit strings below use the order $S,A,B$. Initially,The Bell-basis teleportation identity givesThus a perfect Bell state analyzer on Alice's two qubits identifies the Pauli operator affecting Bob's quantum state.

To display the intermediate three-qubit states explicitly, realize that Bell-basis measurement by a controlled-NOT gate $S\to A$, a Hadamard gate on $S$, and two computational-basis measurements. After the controlled-NOT gate,After the Hadamard gate,Measuring $S,A$ gives $(m,n)$ with probability $1/4$ for each outcome, independently of $\alpha ,\beta$. Conditional on that result, the normalized three-qubit post-measurement state isIf the Bell state analyzer is instead regarded as an ideal direct projective measurement preserving the measured qubits, replace $|mn{⟩}_{SA}$ here by the corresponding Bell state; Bob's factor is identical. A destructive optical analyzer records the same label but absorbs Alice's photons.

Alice sends the two classical bits $(m,n)$ to Bob. During transmission the above conditional three-qubit state is unchanged. Bob then applies the inverse Pauli operator $Z^m{X}^n$:The final conditional three-qubit state is therefore $|mn{⟩}_{SA}\otimes |\psi {⟩}_B$. The protocol transmits two classical bits per teleported qubit, consuming one shared Bell pair. Without Alice's record Bob averages over four Pauli operators and has density matrix $I/2$, so quantum teleportation respects the no-communication theorem. Alice's input has been measured and is no longer an independent copy, consistent with the no-cloning theorem.

For teleportation with an ambiguous Bell-state phase label, write $\rho =|\psi ⟩⟨\psi |$. The Bell-basis measurement outcomes ${\varphi }^{+}$ and ${\varphi }^{-}$ each occur with probability $1/4$. In either case a correct guess gives Bob $\rho$, while the wrong guess differs by the Pauli Z gate and gives $Z\rho Z$. The other two Bell states are identified correctly. Since Alice guesses each ambiguous label with probability $1/2$, the overall wrong-correction probability is $(1/2)(1/2)=1/4$. Averaging over the actual quantum measurement outcome and her guess gives the phase-flip channelConditional only on an ambiguous analyzer result, the output is $(\rho +Z\rho Z)/2$; that conditional channel has weight $1/2$ in the displayed unconditional result. There is no rejection of those rounds.

Use the teleportation fidelity for a pure state target, namely the squared quantum fidelityThis compares the input with the resulting mixed state; averaging state vectors before forming a density matrix would give a different, incorrect answer. Direct substitution givesThus the maximum is $1$, attained by $|0⟩$ and $|1⟩$ up to a global phase, because these are eigenstates of the Pauli Z gate. The minimum is $3/4$, attained by every equal-weight quantum superpositionon the equator of the Bloch sphere. The unsquared quantum fidelity is $\sqrt{F_{\psi }}$ if that alternative convention is used. No average over different input quantum states is needed for the requested state-dependent answer.