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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.

The spin Hilbert space is a tensor product of two two-dimensional spaces and therefore has dimension four. Angular momentum addition decomposes it asThese are three orthogonal triplet states and one singlet state. For identical electrons the complete wavefunction must also satisfy fermionic exchange symmetry: a symmetric triplet spin factor accompanies an antisymmetric spatial factor, whereas the antisymmetric singlet spin factor accompanies a symmetric spatial factor. If both electrons occupy the same spatial orbital, only the singlet spin factor is allowed. The four-dimensional spin space does not assert four allowed spin states for that particular spatial occupation.

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.

The Part C inverse-transform display also prints $j$ as its summation index while the summand uses $k$; that sum must be over $k$, as in the corrected inverse above.

A pure bipartite quantum state is entangled if it cannot be written as a tensor product $|a⟩\otimes |b⟩$. More generally a mixed quantum state is entangled if its density matrix cannot be expressed as a convex mixture of product density matrices. Two applications are quantum teleportation, which transfers an unknown qubit using a shared entangled pair plus classical communication, and superdense coding, which uses a shared entangled pair and transmission of one qubit to communicate two classical bits. Neither permits faster-than-light communication.

A Pauli correlator is the expectation value of a tensor product of Pauli matrices, such as $⟨X\otimes Z⟩$. It measures a specified correlation between qubit observables.

Every linear map from a qubit into the tensor product of that qubit and an environment can be expanded in the Pauli matrix orthonormal basis of operators. If $V|0⟩=|0⟩|e_{00}⟩+|1⟩|e_{01}⟩$ and $V|1⟩=|0⟩|e_{10}⟩+|1⟩|e_{11}⟩$, then the four coefficient vectors areThese environment vectors need not be orthogonal, so this is an operator expansion rather than a claim that the quantum channel is a classical Pauli channel. Physical isometries additionally preserve the inner products of the two input basis states.

The $n$-qubit Pauli group consists of tensor products of $I,X,Y,Z$ multiplied by phases $\pm 1$ or $\pm i$.

An $n$-qubit Pauli operator is a tensor product of single-qubit Pauli matrices, possibly multiplied by a phase. Hermitian Pauli operators have eigenvalues $+1$ and $-1$.

For tensor products of Pauli factors from $\{I,X,Y,Z\}$, let $k$ count the sites where both factors are nonidentity and different. Distinct nonidentity Pauli matrices anticommute, and equal or identity factors commute. Reversing the product site by site therefore gives $PQ=(-1{)}^{k}QP$. Thus two Pauli strings commute exactly when $k$ is even. The criterion applies to the perfect-correlation observables in the GHZ theorem.

An orthogonal projection $P$ splits a one-particle Hilbert space into empty and filled modes. Take completed exterior Fock spaces of empty particles and of holes in the filled modes, with the graded tensor product enforcing the canonical anticommutation relations between both factors. The vacuum is killed by particle annihilation in $PH$ and by particle creation in $(I-P)H$. Only finite particle-hole excitations are needed for a dense basis. Normal ordering subtracts the filled-vacuum expectation from bilinear observables.

Such a continuous projective unitary representation admits strongly continuous rotations whose self-adjoint generator is bounded below and whose adjoint action rotates the loops. In the basic fermionic construction, integer one-particle modes are filled below zero. The rotation energy counts nonnegative particle energies and positive hole energies. It has finite-dimensional integer eigenspaces and implements circle rotations. Tensor products give higher nonnegative integer levels.

For an isolated quantum system, pure states are normalized rays in a complex Hilbert space, observables are self-adjoint operators, and their probabilities obey the Born rule. A spectral projector $P_a$ has outcome probability $⟨\psi |P_a|\psi ⟩$; an ideal repeatable projective measurement has conditional state $P_a\psi /\sqrt{⟨\psi |P_a|\psi ⟩}$. The Schrödinger equation specifies continuous unitary evolution between such ideal measurements, given a Hamiltonian operator. General measurements need not have the projective state-update rule.

Composite systems use the tensor product of their state spaces. Identical particles additionally obey the bosonic or fermionic exchange symmetry appropriate to their species. The photoelectric effect, Davisson–Germer experiment and double-slit experiment motivate energy quanta, matter-wave phases and probability interference, respectively; these experiments constrain the framework but do not separately prove every abstract postulate or specify every system's Hamiltonian.

For closed manifolds with the product Riemannian metric, the positive Laplace-Beltrami operator is the sum of the two factor operators. Products of their orthonormal eigenbases form a complete basis, and eigenvalues add. Hence the product eigenspace at $\lambda$ is the displayed finite orthogonal sum of tensor products. The discrete assertion requires a discrete spectral setting.

The K-theory transfer of a finite covering distributes over a pulled-back tensor product of vector bundles:For an $n$-sheeted covering map, $p_{!}(1)$ is the permutation vector bundle of rank $n$. Its difference from $n$ is a nilpotent element by nilpotence of rank-zero K-theory classes, so it is a unit after inverting $n$. Thus pullback on Topological K-theory becomes injective after this localization of a ring.

For a diffeomorphism, pull a type $(r,s)$ tensor field back by the inverse differential on vector slots and the dual differential on covector slots. Invertibility distinguishes this construction from pullback of purely covariant tensors along an arbitrary smooth map. This operation preserves tensor products and tensor contractions, supplying the flow definition of the Lie derivative of a tensor field.

A pure tensor is an element of a tensor product that can be written as $m\otimes n$. General tensors are finite sums of pure tensors and need not themselves be pure.

A quantum circuit is a finite composition of quantum gates acting on quantum registers. Reading the diagram from input to output gives the order in which its unitary operations are applied.

A quantum register is a named tensor product factor of a quantum circuit's Hilbert space, usually a collection of qubits. Registers separate data, control, output and workspace roles; uncomputation returns workspace to a fixed quantum state without destroying coherence in the data.