Controlled Gates, Phase, and Entanglement

For every $z\in H$, [F2] and [F3] give $P_u^{2}z=⟨z,u⟩⟨u,u⟩u=⟨z,u⟩u=P_{u}z$, since $⟨u,u⟩=\mathrm{\parallel }u{\mathrm{\parallel }}^2=1$. Thus $P_u$ is idempotent.

For all $z,w\in H$, [F1] and [F3] give $⟨P_{u}z,w⟩=⟨z,u⟩⟨u,w⟩=⟨z,⟨w,u⟩u⟩=⟨z,P_{u}w⟩$. The outer product is linear by [F3], so [F4] and [F5] imply $P_u^{\ast }=P_u$.

The image of $P_u$ lies in $\mathbb{C}u$, and [F2] and [F3] give $P_{u}u=⟨u,u⟩u=u$, so its range is exactly the one-dimensional span of $u$. Also [F1] and [F3] give $⟨P_{u}z,z⟩=⟨z,u⟩⟨u,z⟩=\mathrm{\mid }⟨z,u⟩{\mathrm{\mid }}^2\ge 0$. With self-adjointness from step 1.2, [F6] makes $P_u$ non-negative.

Choose an orthonormal basis $(e_j{)}_{j<n}$ of $H$ using [F8]. By [F3] and [F7], $\mathrm{tr}(P_u)={\sum }_{j<n}⟨P_u{e}_j,e_j⟩={\sum }_{j<n}\mathrm{\mid }⟨e_j,u⟩{\mathrm{\mid }}^2=\mathrm{\parallel }u{\mathrm{\parallel }}^2=1$, where [F9] gives the Parseval equality. Thus [L1] makes $P_u$ a density operator, using non-negativity from step 2.1. The basis is finite, so no choice axiom is used.

Choose ordered orthonormal bases of $H_A,H_B$ using [F5], and let $C$ be the coefficient matrix of $\psi$ in their product basis [L2]. For any other ordered orthonormal bases, let $A,B$ have the new basis vectors as columns in the old coordinates. Their columns are orthonormal, so the finite row-by-column identities $A^{\ast }A=B^{\ast }B=I$ hold; because they are basis changes, both matrices are invertible and hence unitary. Expanding the new product basis vectors in the old one and using conjugate-linearity in the second argument gives $c_{pq}^{\mathrm{\prime }}={\sum }_{i,k}\overline{A_{ip}}{c}_{ik}\overline{B_{kq}}$, so the new coefficient matrix is $C^{\mathrm{\prime }}=A^{\ast }C\bar{B}$. The factors $A^{\ast }$ and $\bar{B}$ are invertible, so the map represented by $C^{\mathrm{\prime }}=A^{\ast }C\bar{B}$ is $A^{\ast }\circ T_C\circ \bar{B}$, where $T_C$ is the map represented by $C$. Since $\bar{B}$ is onto, its image is $A^{\ast }(\mathrm{im}{T}_C)$; since $A^{\ast }$ is injective, this image has the same dimension as $\mathrm{im}{T}_C$. By [F2], the matrix rank is therefore unchanged, so it is basis-independent. Multiplying $\psi$ by any nonzero scalar multiplies $C$ by that scalar and also preserves its rank.

In any Schmidt decomposition $\psi ={\sum }_{j=1}^r{s}_j{u}_j\otimes v_j$ with all $s_j>0$ and orthonormal factor families, tensor coordinates multiply, so its coefficient matrix is $C=UD{V}^{\mathsf{T}}$, where the columns of $U,V$ are the coordinates of the $u_j,v_j$ and $D=\mathrm{diag}(s_1,\dots ,s_r)$. The orthonormal columns give $U^{\ast }U=V^{\ast }V=I_r$, so $U$ is injective; conjugating $V^{\ast }V=I_r$ gives $V^{\mathsf{T}}\bar{V}=I_r$, so $V^{\mathsf{T}}$ is surjective; and $D$ is invertible. Thus $\mathrm{rank}(C)=r$. Conversely, [L3] constructs such a Schmidt decomposition from the SVD, and [F3] and [F4] show that its number of positive coefficients is the rank of $C$. Therefore the number of terms is the same for every Schmidt decomposition and equals the defined Schmidt rank.

If $\psi =c(a\otimes b)$ for nonzero $a,b,c$, its coefficient matrix is $c\alpha {\beta }^{\mathsf{T}}$ for the coordinate columns $\alpha ,\beta$, so its rank is one. Conversely, if $C$ has rank one, choose a nonzero column $\alpha$; every column is a scalar multiple ${\beta }_k\alpha$, so $C=\alpha {\beta }^{\mathsf{T}}$ with $\beta \mathrm{\ne }0$. The coefficient expansion then gives $\psi =a\otimes b$ for the vectors with coordinates $\alpha ,\beta$. Thus the ray is a product ray exactly when the Schmidt rank is one.

Since $\psi \mathrm{\ne }0$, its coefficient matrix is nonzero and has positive rank. By the product-ray equivalence in step 2.1, failure to be a product ray is equivalent to Schmidt rank greater than one, which is precisely entanglement by the Definition.

Positive definiteness gives $\mathrm{\parallel }x\mathrm{\parallel }>0$, so $u$ is defined. By [F1] and [F2], $\mathrm{\parallel }u{\mathrm{\parallel }}^2=⟨x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel },x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel }⟩=⟨x,x⟩\mathrm{/}\mathrm{\parallel }x{\mathrm{\parallel }}^2=1$. Again using linearity in the first argument and conjugate-linearity in the second, $\mathrm{\parallel }\lambda u{\mathrm{\parallel }}^2=⟨\lambda u,\lambda u⟩=\mathrm{\mid }\lambda {\mathrm{\mid }}^2⟨u,u⟩=1$. Taking nonnegative square roots gives both norms equal to one.

For every $z\in H$, the root-first conversion [L3] gives $\mathrm{\mid }\lambda u⟩⟨\lambda u\mathrm{\mid }z=⟨z,\lambda u{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}\lambda u=\bar{\lambda }\lambda ⟨z,u{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}u=\mathrm{\mid }u⟩⟨u\mathrm{\mid }z$. Hence the two rank-one operators are equal.

By [L4], the common operator is a density operator, so the outcome law [L2] defines its weight for each effect $E_a$. Since the density operator is identical for $u$ and $\lambda u$, those weights agree; [L5] proves the finite family of weights is a probability distribution.

Since $x=\mathrm{\parallel }x\mathrm{\parallel }u$ with $\mathrm{\parallel }x\mathrm{\parallel }$ a nonnegative real, [F1] and [L3] give $\mathrm{\mid }x⟩⟨x\mathrm{\mid }=\mathrm{\parallel }x{\mathrm{\parallel }}^2\mathrm{\mid }u⟩⟨u\mathrm{\mid }$. In any basis, its matrix is therefore $\mathrm{\parallel }x{\mathrm{\parallel }}^2$ times the matrix of $\mathrm{\mid }u⟩⟨u\mathrm{\mid }$; summing the diagonal entries in that basis as in [F3] gives $\mathrm{tr}(\mathrm{\mid }x⟩⟨x\mathrm{\mid })=\mathrm{\parallel }x{\mathrm{\parallel }}^2\mathrm{tr}(\mathrm{\mid }u⟩⟨u\mathrm{\mid })=\mathrm{\parallel }x{\mathrm{\parallel }}^2$ by [L4]. If $\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\ne }1$, it is not trace one and hence is not a density operator by [L1].

Choose an ordered orthonormal basis of $H$ using [F2]. Together with the control basis in [L2], it supplies the factor bases required by [L1]; the pairing is independent of this choice by [L1]. Write an arbitrary vector of $C\otimes H$ as $\mathrm{\mid }0⟩\otimes v_0+\mathrm{\mid }1⟩\otimes v_1$. The orthonormal control basis and the product inner product from [L1] give squared norm $\mathrm{\parallel }{v}_0{\mathrm{\parallel }}^2+\mathrm{\parallel }{v}_1{\mathrm{\parallel }}^2$.

The block operator sends that vector to $\mathrm{\mid }0⟩\otimes v_0+\mathrm{\mid }1⟩\otimes U{v}_1$. By [F1], $\mathrm{\parallel }U{v}_1\mathrm{\parallel }=\mathrm{\parallel }{v}_1\mathrm{\parallel }$, so the squared norm is unchanged.

The block matrix $\mathrm{diag}(I_H,U^{-1})$ is a two-sided inverse for $C(U)$, so $C(U)$ is invertible as well as norm-preserving. By the unitary definition in [F1], $C(U)$ is unitary.

For any nonzero scalar $c$, the ratio of the two coordinates of $c\psi$ is $(c\beta )\mathrm{/}(c\alpha )=\beta \mathrm{/}\alpha$. Its unit-modulus part is therefore also unchanged, so the Definition assigns a property of the ray.

Since $\mathrm{\mid }\eta \mathrm{\mid }=1$ and $e_0,e_1$ are orthonormal, ${\psi }_{\eta }$ has norm one. In the qubit instance [L2] identifies this ordered pair with $\mathrm{\mid }0⟩,\mathrm{\mid }1⟩$. By [L3] and [L4], its rank-one projector has matrix $\frac{1}{2}\left(\begin{array}{cc}1& \bar{\eta }\\ \eta & 1\end{array}\right)$ in the ordered pair.

If $\eta \mathrm{\ne }{\eta }^{\mathrm{\prime }}$, the lower-left entries of these two projector matrices differ, so the projectors differ. Also, if ${\psi }_{{\eta }^{\mathrm{\prime }}}=c{\psi }_{\eta }$ for a nonzero scalar $c$, comparison of the $e_0$ coordinate forces $c=1$ and comparison of the $e_1$ coordinate forces ${\eta }^{\mathrm{\prime }}=\eta$, a contradiction. Hence changing the relative phase can change the ray.

Each $U_j$ is an invertible isometry by the unitary convention in [L1] and [F1]. Thus its adjoint exists in finite dimension and is its inverse, so the inverse call specified in the Definition has the stated mathematical meaning.

For $k>0$, repeated application of [F2] gives $(U_k\cdots U_1{)}^{\ast }=U_1^{\ast }\cdots U_k^{\ast }$. Acting on a vector from right to left, this applies $U_k^{\ast }$ first and $U_1^{\ast }$ last, exactly reversing the original chronology.

For $k=0$, the specified total map is $I_H$ and its inverse is also $I_H$; there are no calls or operations in either sequence. This covers the empty-sequence endpoint without interpreting an undefined product.

The matrix identity determines the inverse map but says nothing about whether the access model grants or charges its calls. The Definition therefore records inverse availability and cost explicitly instead of inferring either from forward access.

The target computational basis is orthonormal by [L4], and [F1] gives modulus one for both diagonal entries of $P_{\theta }$. Thus $P_{\theta }$ is an invertible isometry and hence unitary, so [L1] applies to its controlled block. The first control block applies the identity, so $\mathrm{\mid }00⟩$ and $\mathrm{\mid }01⟩$ are fixed. In the second block, $P_{\theta }$ fixes $\mathrm{\mid }10⟩$ and sends $\mathrm{\mid }11⟩$ to $e^{i\theta }\mathrm{\mid }11⟩$. These four actions give the diagonal matrix in the Definition.

Every diagonal entry of $CP(\theta )$ has modulus one by [F1]; its conjugate transpose is therefore its inverse, so it is unitary.

By [L3], $R_z(\theta )=\mathrm{diag}(e^{-i\theta \mathrm{/}2},e^{i\theta \mathrm{/}2})$. By [F2], multiplying it by $e^{i\theta \mathrm{/}2}$ gives $P_{\theta }=\mathrm{diag}(1,e^{i\theta })$. Therefore $CP(\theta )=\mathrm{diag}(I,e^{i\theta \mathrm{/}2}{R}_z(\theta ))$, which is the block phase $\mathrm{diag}(I,e^{i\theta \mathrm{/}2}I)$ applied after $C(R_z(\theta ))$. Dropping that control phase would change the defined operation.

By [F1] and [F2], $U$ satisfies $U{U}^{\ast }=U^{\ast }U=I$, hence is normal. The complex spectral theorem [F3] gives an orthonormal eigenbasis; since $H$ is nonzero, choose one unit eigenvector $\varphi$ and write $U\varphi =\mu \varphi$. Norm preservation and [F6], [F7] give $1=\mathrm{\parallel }U\varphi \mathrm{\parallel }=\mathrm{\mid }\mu \mathrm{\mid }\mathrm{\parallel }\varphi \mathrm{\parallel }=\mathrm{\mid }\mu \mathrm{\mid }$, so $\mathrm{\mid }\mu \mathrm{\mid }=1$. This chooses one vector from a finite nonempty basis and uses no choice axiom.

Choose an ordered orthonormal basis of $H$ using [L8]. Together with the control basis from [L5], it supplies the factor bases required by [L3], whose proof shows the pairing is independent of this choice. Put $P_{\pm }=\mathrm{\mid }\pm ⟩⟨\pm \mathrm{\mid }$ on the control and $E_{\pm }=P_{\pm }\otimes I_H$ on the joint register. By the Statement, $\mathrm{\mid }\pm ⟩=(\mathrm{\mid }0⟩\pm \mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$. Using [L3] and [L5], direct inner-product calculations give $⟨+,+⟩=⟨-,-⟩=1$ and $⟨+,-⟩=0$; also $\mathrm{\mid }0⟩=(\mathrm{\mid }+⟩+\mathrm{\mid }-⟩)\mathrm{/}\sqrt{2}$ and $\mathrm{\mid }1⟩=(\mathrm{\mid }+⟩-\mathrm{\mid }-⟩)\mathrm{/}\sqrt{2}$, so these vectors form an orthonormal basis. Every $v\in C\otimes H$ is a finite sum of elementary tensors; expanding each control factor in this basis and grouping terms writes $v=\mathrm{\mid }+⟩\otimes h_{+}+\mathrm{\mid }-⟩\otimes h_{-}$. The product pairing [L3] makes these two summands orthogonal, and the tensor-map definition [L4] gives $E_{+}v=\mathrm{\mid }+⟩\otimes h_{+}$ and $E_{-}v=\mathrm{\mid }-⟩\otimes h_{-}$. Hence $E_{+}+E_{-}=I$ and $⟨E_{\pm }v,v⟩=\mathrm{\parallel }{h}_{\pm }{\mathrm{\parallel }}^2\ge 0$. Applying the same decomposition to $v,w$ gives $⟨E_{\pm }v,w⟩=⟨v,E_{\pm }w⟩$; the adjoint characterization [F10] and finite-dimensional existence/uniqueness [F11] imply each effect is self-adjoint. By [F9], each effect is non-negative; together with the two-element outcome set and $E_{+}+E_{-}=I$, this is a finite POVM by [L2].

Set $c_{\mu }=(\mathrm{\mid }0⟩+\bar{\mu }\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$. By [L5], [F6], and [F7], its squared norm is $\frac{1}{2}(⟨0,0⟩+\mathrm{\mid }\mu {\mathrm{\mid }}^2⟨1,1⟩)=1$; the cross terms vanish by orthogonality. Since $\varphi$ is unit, the product pairing [L3] makes $c_{\mu }\otimes \varphi$ unit. The controlled block rule [L1] gives $C(U)(c_{\mu }\otimes \varphi )=\frac{\mathrm{\mid }0⟩\otimes \varphi +\bar{\mu }\mathrm{\mid }1⟩\otimes U\varphi }{\sqrt{2}}=\mathrm{\mid }+⟩\otimes \varphi$.

Put $\alpha =e^{i\theta }$. By [F4], $\mathrm{\mid }\alpha \mathrm{\mid }=1$. For every $v\in H$, [F6] and [F7] give $\mathrm{\parallel }\alpha Uv{\mathrm{\parallel }}^2=\mathrm{\mid }\alpha {\mathrm{\mid }}^2\mathrm{\parallel }Uv{\mathrm{\parallel }}^2=\mathrm{\parallel }v{\mathrm{\parallel }}^2$; since $\alpha \mathrm{\ne }0$ and $U$ is invertible, $\alpha U$ is invertible, so it is unitary by [F1]. For $c_{\mu }=(\mathrm{\mid }0⟩+\bar{\mu }\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$, [L5], [F6], [F7] and step 1.1 give $\mathrm{\parallel }{c}_{\mu }{\mathrm{\parallel }}^2=1$. The separately granted controlled block gives $C(e^{i\theta }U)(c_{\mu }\otimes \varphi )=\frac{\mathrm{\mid }0⟩+e^{i\theta }\mathrm{\mid }1⟩}{\sqrt{2}}\otimes \varphi$. Its control factor has squared norm $\frac{1}{2}(1+\mathrm{\mid }{e}^{i\theta }{\mathrm{\mid }}^2)=1$ by [F4], [F6], and [F7]. No access to this block is inferred from a call to $C(U)$; it is part of the explicit model assumption in the Given.

Let $\omega =\chi \otimes \varphi$ be either output in steps 2.1–2.2. Both factors are unit vectors, so the product norm [L3] makes $\mathrm{\parallel }\omega \mathrm{\parallel }=1$; [L6] makes ${\rho }_{\omega }=\mathrm{\mid }\omega ⟩⟨\omega \mathrm{\mid }$ a density operator by [L11]. Choose a finite orthonormal basis $(e_j)$ of $C\otimes H$ by [L8]. From [L7] and [F8], $\mathrm{tr}({\rho }_{\omega }{E}_{+})={\sum }_j⟨E_{+}{e}_j,\omega ⟩⟨\omega ,e_j⟩={\sum }_j⟨e_j,E_{+}\omega ⟩⟨\omega ,e_j⟩=⟨\omega ,E_{+}\omega ⟩$, where self-adjointness from step 1.2 gives the middle equality and Parseval [L9] gives the last equality. By the product pairing [L3] and tensor-map action [L4], $⟨\omega ,E_{+}\omega ⟩=⟨\chi ,P_{+}\chi ⟩⟨\varphi ,\varphi ⟩=\mathrm{\mid }⟨\chi ,+{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}{\mathrm{\mid }}^2$. The finite weights normalize by [L10], so this trace value is the probability of the plus outcome.

For the output in step 2.1, $\chi =\mathrm{\mid }+⟩$, so $p_{+}=1$. For the output in step 2.2, $\chi =(\mathrm{\mid }0⟩+e^{i\theta }\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$. Put $z=e^{i\theta \mathrm{/}2}$. By [F12], $z^2=e^{i\theta }$ and $z^{-1}=e^{-i\theta \mathrm{/}2}$; [F4] gives $\mathrm{\mid }z\mathrm{\mid }=1$ and $z=\cos (\theta \mathrm{/}2)+i\sin (\theta \mathrm{/}2)$, so $z^{-1}=\bar{z}$ and $z+z^{-1}=2\cos (\theta \mathrm{/}2)$. Hence $1+e^{i\theta }=z(z+z^{-1})=2z\cos (\theta \mathrm{/}2)$ and $p_{+}=\mathrm{\mid }1+e^{i\theta }{\mathrm{\mid }}^2\mathrm{/}4={\cos }^2(\theta \mathrm{/}2)$.

By [F5], $\theta \mathrm{\notin }2\pi \mathbb{Z}$ means $z=e^{i\theta }\mathrm{\ne }1$. Since $\mathrm{\mid }z\mathrm{\mid }=1$, $\mathrm{\mid }1+z{\mathrm{\mid }}^2=2+2\mathrm{Re}z<4$: equality would force $\mathrm{Re}z=1$ and then $\mathrm{Im}z=0$, hence $z=1$. Thus the probability from step 4.1 is strictly below one, while the other is one; the two outcome laws differ.

If $\theta \in 2\pi \mathbb{Z}$, [F5] gives $e^{i\theta }=1$, so the two controlled block maps are identical by [L1]. The operators therefore produce the same state for every common input and hence the same outcome law for the specified POVM.

By [L5], $X$ is unitary, so [L1] defines its controlled block. For control bit $a=0$, the first block of $C(X)$ is the identity, so $\mathrm{\mid }0,0⟩\mapsto \mathrm{\mid }0,0⟩$ and $\mathrm{\mid }0,1⟩\mapsto \mathrm{\mid }0,1⟩$.

For control bit $a=1$, [L2] flips the target, so $\mathrm{\mid }1,0⟩\mapsto \mathrm{\mid }1,1⟩$ and $\mathrm{\mid }1,1⟩\mapsto \mathrm{\mid }1,0⟩$. These are exactly $\mathrm{\mid }a,t\oplus a⟩$ for the two remaining basis inputs.

The columns in the stated ordered basis are therefore $e_{00},e_{01},e_{11},e_{10}$, giving $\mathrm{CNOT}=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\end{array}\right)$. By [L3] and [L4], this is a permutation of an orthonormal basis, so it preserves the sum of squared coordinate moduli and thus every vector norm. The permutation is its own inverse, so the operator is invertible and hence unitary. The displayed action fixes the stated control-target order.

By [L4], the vectors $\mathrm{\mid }x⟩\otimes v$ over the computational control basis span the joint register. The cases in the Definition therefore determine a unique linear operator and give the activation behavior on every superposition.

Let $n=(n_1,\dots ,n_m)\in \{0,1{\}}^m$ have $n_j=1$ exactly when $s_j=0$. By [L2] and [L4], $X_N\otimes I_T$ sends $\mathrm{\mid }x⟩\otimes v$ to $\mathrm{\mid }x\oplus n⟩\otimes v$. The middle positive-control block activates exactly when $x\oplus n=1^m$, which coordinatewise is equivalent to $x=s$. The leftmost $X_N$ restores the original control string because $n\oplus n=0$. Thus the conjugated operation has exactly the defining action of $C_s(U)$, including $N=\mathrm{\varnothing }$ and $N=\{1,\dots ,m\}$.

Write any joint vector uniquely as $v={\sum }_{x\in \{0,1{\}}^m}\mathrm{\mid }x⟩\otimes v_x$ using the ordered product basis in [L4]. The Definition maps each block to $W_x{v}_x$, where $W_s=U$ and $W_x=I_T$ otherwise. Each $W_x$ is an invertible isometry by [F1], so orthogonality of the control basis and the product pairing give $\mathrm{\parallel }{C}_s(U)v{\mathrm{\parallel }}^2={\sum }_x\mathrm{\parallel }{W}_x{v}_x{\mathrm{\parallel }}^2={\sum }_x\mathrm{\parallel }{v}_x{\mathrm{\parallel }}^2=\mathrm{\parallel }v{\mathrm{\parallel }}^2$. The block map with $W_s=U^{-1}$ and other blocks identity is a two-sided inverse. Hence $C_s(U)$ is an invertible isometry, therefore unitary by [F1].

Substituting $\theta =\pi$ into [L1] gives $CP(\pi )=\mathrm{diag}(1,1,1,e^{i\pi })=\mathrm{diag}(1,1,1,-1)$ by [F1].

SWAP fixes $\mathrm{\mid }00⟩$ and $\mathrm{\mid }11⟩$ and interchanges $\mathrm{\mid }01⟩$ with $\mathrm{\mid }10⟩$. The ordered product basis in [L2] spans the tensor product, so these actions determine a unique linear map. By [L2] and [L3] it permutes an orthonormal basis, hence preserves norms and is its own inverse; therefore it is unitary.

The preceding action changes the state vector in the fixed ordered tensor product, while the first and second factors remain the declared wires. Thus the definition is an actual linear gate action, not a change of names for the tensor factors.

The first column of the Hadamard matrix in [L3] gives $H\mathrm{\mid }0⟩=(\mathrm{\mid }0⟩+\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}=\mathrm{\mid }+⟩$. Thus before CNOT the state is $(\mathrm{\mid }00⟩+\mathrm{\mid }10⟩)\mathrm{/}\sqrt{2}$ in the product order of [L4] and [L5].

By [L1], CNOT fixes $\mathrm{\mid }00⟩$ and sends $\mathrm{\mid }10⟩$ to $\mathrm{\mid }11⟩$. Linearity therefore gives the output $(\mathrm{\mid }00⟩+\mathrm{\mid }11⟩)\mathrm{/}\sqrt{2}$ in the Statement.

In the ordered product basis, the coefficient matrix of the output is $\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$, whose determinant is $1\mathrm{/}2\mathrm{\ne }0$ and whose rank is two. If a nonzero product vector is $a\otimes b$, its coefficient in row $i$, column $j$ is $a_i{b}_j$, so its coefficient matrix is $a{b}^{\mathsf{T}}$. Since $a,b\mathrm{\ne }0$, at least one column is nonzero and every column is a scalar multiple of $a$; its rank is exactly one. A nonzero scalar multiple of a product vector has the same rank-one property, so the output ray is not a product ray. By [L2] it is entangled.