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Bell reduced state

Example

Let $V = W = C^{2}$ with the standard inner product and orthonormal basis $e_{+} = (\binom{1}{0})$ , $e_{-} = (\binom{0}{1})$ (Hilbert tensor construction: inner product, completion and product bases), and let

$Ψ : = \frac{1}{\sqrt{2}} (e_{+} \otimes e_{-} - e_{-} \otimes e_{+}) \in V \otimes W .$

Here a product vector means an elementary tensor $v \otimes u$ with $v \in V$ and $u \in W$ ; $T_{Ψ, Ψ}$ is the rank-one density operator determined by the unit vector $Ψ$ .

Then:

(a) $∥ Ψ ∥ = 1$ , and the coefficient matrix of $Ψ$ in the bases $(e_{+}, e_{-})$ is $C = \frac{1}{\sqrt{2}} (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ ;

(b) the partial traces satisfy ${Tr}_{W} T_{Ψ, Ψ} = C C^{*} = \frac{1}{2} I_{V}$ and ${Tr}_{V} T_{Ψ, Ψ} = C^{*} C = \frac{1}{2} I_{W}$ , so both reduced states are the maximally mixed matrix $\frac{1}{2} I$ ;

(c) the reduced density $ρ = \frac{1}{2} I$ has eigenvalues $\frac{1}{2}, \frac{1}{2}$ and trace-square $tr ρ^{2} = \frac{1}{2}$ , while $tr (T_{Ψ, Ψ}^{2}) = 1$ ; the Schmidt coefficients of $Ψ$ are $s_{1} = s_{2} = \frac{1}{\sqrt{2}}$ , and $Ψ$ is not a product vector.

Facts & Assumptions

Given: $V = W = C^{2}$ with the standard inner product; the orthonormal basis $(e_{+}, e_{-})$ ; the vector $Ψ$ above; the operators $T_{u, v} x = ⟨ x, v ⟩ u$ on the finite-dimensional spaces.

[A1]

On $V \otimes W$ the pairing is $⟨ u \otimes v, u^{'} \otimes v^{'} ⟩ = ⟨ u, u^{'} ⟩_{V} ⟨ v, v^{'} ⟩_{W}$ ; the four elementary tensors $e_{\pm} \otimes e_{\pm}$ form an orthonormal basis, each $Ψ \in V \otimes W$ has a unique expansion $\sum_{i j} c_{i j} e_{i} \otimes e_{j}$ with $∥ Ψ ∥^{2} = \sum_{i j} ∣ c_{i j} ∣^{2}$ , and the basis-free conventions (linearity in the first argument, conjugate symmetry) are those of Hilbert tensor construction: inner product, completion and product bases.

[A2]

For endomorphisms $R$ of $V \otimes W$ there is a unique ${Tr}_{W} R$ in $End (V)$ with $tr (({Tr}_{W} R) A) = tr (R (A \otimes I_{W}))$ for all $A$ ; it is linear and positive, ${Tr}_{W} T_{Ψ, Ψ} = C C^{*}$ where $C : W \to V$ is the coefficient map determined by $Ψ = \sum_{j} C f_{j} \otimes f_{j}$ , and ${Tr}_{V}$ is defined symmetrically; the Schmidt decomposition $Ψ = \sum_{r = 1}^{k} s_{r} v_{r} \otimes u_{r}$ has orthonormal families and positive $s_{1} \geq \dots \geq s_{k}$ equal to the positive singular values of $C$ , with $∥ Ψ ∥^{2} = \sum_{r} s_{r}^{2}$ , and for $∥ Ψ ∥ = 1$ the reduced density $ρ = {Tr}_{W} T_{Ψ, Ψ}$ satisfies $tr ρ^{2} = \sum_{r} s_{r}^{4}$ , which equals $1$ exactly when $k = 1$ (Finite partial trace, Schmidt decomposition and finite no-signalling(a),(b),(c)).

[A3]

$T_{u, v}$ is trace class with $tr (T_{u, v}) = ⟨ u, v ⟩$ , and $tr$ is linear (Rank-one operator calculus).

[A4]

In an orthonormal basis of a finite-dimensional inner product space the matrix of $T_{u, v}$ is the column-times-row matrix $u v^{*}$ , so $tr (T_{u, v}) = ⟨ u, v ⟩$ ; $T_{x, z} T_{w, y} = ⟨ w, z ⟩ T_{x, y}$ for rank-one operators.

Verification

1.1A1algebra

[Coefficient matrix and norm] By [A1], $Ψ = \frac{1}{\sqrt{2}} e_{+} \otimes e_{-} - \frac{1}{\sqrt{2}} e_{-} \otimes e_{+}$ is the expansion with $c_{+ -} = \frac{1}{\sqrt{2}}$ , $c_{- +} = - \frac{1}{\sqrt{2}}$ and the other coefficients $0$ , so the coefficient map $C : W \to V$ is the matrix $C = \frac{1}{\sqrt{2}} (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ in the bases $(e_{+}, e_{-})$ (columns indexed by $W$ ), and $∥ Ψ ∥^{2} = ∣ c_{+ -} ∣^{2} + ∣ c_{- +} ∣^{2} = 1$ .

2.1step 1.1A2algebra

[Partial traces] Computing $C C^{*} = \frac{1}{2} (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}) (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}) = \frac{1}{2} I_{V}$ and $C^{*} C = \frac{1}{2} I_{W}$ , [A2] gives ${Tr}_{W} T_{Ψ, Ψ} = C C^{*} = \frac{1}{2} I_{V}$ and, by symmetry, ${Tr}_{V} T_{Ψ, Ψ} = C^{*} C = \frac{1}{2} I_{W}$ ; both are positive with trace $1$ , as they must be for the reduction of a unit vector.

3.1step 1.1step 2.1A2A3A4algebra∎

[Purity and Schmidt coefficients] The matrix $\frac{1}{2} I_{V}$ has the single eigenvalue $\frac{1}{2}$ with multiplicity $2$ (in any basis its matrix is diagonal with entries $\frac{1}{2}$ ), so $tr ρ^{2} = 2 \cdot (\frac{1}{2})^{2} = \frac{1}{2}$ for $ρ = {Tr}_{W} T_{Ψ, Ψ}$ ; the positive singular values of $C$ are the square roots of the eigenvalues of $C^{*} C = \frac{1}{2} I_{W}$ , namely $s_{1} = s_{2} = \frac{1}{\sqrt{2}}$ , so the Schmidt decomposition of $Ψ$ has $k = 2$ terms and $\sum_{r} s_{r}^{2} = 1$ , in agreement with the norm formula of [A2]; since $k = 2 > 1$ , $Ψ$ is not a product vector. Finally $T_{Ψ, Ψ}^{2} = T_{Ψ, Ψ}$ because $T_{Ψ, Ψ} T_{Ψ, Ψ} = ⟨ Ψ, Ψ ⟩ T_{Ψ, Ψ}$ by [A4] and $∥ Ψ ∥ = 1$ , so $tr (T_{Ψ, Ψ}^{2}) = tr (T_{Ψ, Ψ}) = ⟨ Ψ, Ψ ⟩ = 1$ by [A3].

Bell reduced state

Example

Facts & Assumptions

Verification

Depends on

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