Entanglement versus mutual information in quantum spin chains

To obtain the mutual information of classical spin configurations in the ground state, we have to compute the probability for each classical configuration after a measurement of local spin orientations. If we choose the z axis for the spin orientations, measuring ${\sigma }_{1}^{z},{\sigma }_{2}^{z},\ldots ,{\sigma }_{L}^{z}$, these probabilities can be computed from the coefficients of the ground state wavefunction |ψ₀〉 represented in the product basis of the eigenvectors of ${\sigma }_{i}^{z}$. For example, if we consider an Ising chain with two spins in the ground state

$$\begin{eqnarray} &&\vert {\psi }_{0}\rangle ={a}_{1}\vert {\uparrow }{\uparrow }\rangle +{a}_{2}\vert {\uparrow }{\downarrow }\rangle +{a}_{3}\vert {\downarrow }{\uparrow }\rangle +{a}_{4}\vert {\downarrow }{\downarrow }\rangle , \end{eqnarray} \tag{ 5 }$$

where |↑↓〉 ≡ |↑〉₁⊗|↓〉₂ is the product basis of the eigenvectors |↑〉_i and |↓〉_i of ${\sigma }_{i}^{z}$, the probability P({σ^z}) to obtain the classical configuration {σ^z} is given by $P(\uparrow \uparrow )={a}_{1}^{2},P(\uparrow \downarrow )={a}_{2}^{2},P(\downarrow \uparrow )={a}_{3}^{2}$, and $P(\downarrow \downarrow )={a}_{4}^{2}$. To determine the ground state of the Hamiltonian (4) at finite L, we use the Lanczos method  [7].

Each classical configuration obtained from the measurement is now divided into two segments, $\{ {\sigma }^{z}\} =\{ {\sigma }_{\mathrm{A}}^{z},{\sigma }_{\mathrm{B}}^{z}\}$. Determining the Shannon entropy of the probability distribution in the segments and in the whole chain,

$$\begin{eqnarray} &&{H}_{\mathrm{A}}=-\mathop{\sum }\limits _{\{ {\sigma }_{\mathrm{A}}^{z}\} }P(\{ \sigma _{\mathrm{A}}^{z}\} )\ln P(\{ {\sigma }_{\mathrm{ A}}^{z}\} ), \end{eqnarray} \tag{ 6 }$$

$$\begin{eqnarray} &&{H}_{\mathrm{B}}=-\mathop{\sum }\limits _{\{ {\sigma }_{\mathrm{B}}^{z}\} }P(\{ \sigma _{\mathrm{B}}^{z}\} )\ln P(\{ {\sigma }_{\mathrm{ B}}^{z}\} ), \end{eqnarray} \tag{ 7 }$$

$$\begin{eqnarray} &&{H}_{\mathrm{AB}}=-\mathop{\sum }\limits _{\{ {\sigma }_{\mathrm{A}}^{z},{\sigma }_{\mathrm{B}}^{z}\} }P(\{ \sigma _{\mathrm{A}}^{z},{\sigma }_{\mathrm{ B}}^{z}\} )\ln P(\{ {\sigma }_{\mathrm{ A}}^{z},{\sigma }_{\mathrm{ B}}^{z}\} ), \end{eqnarray} \tag{ 8 }$$

where $P(\{ {\sigma }_{\mathrm{A}}^{z}\} )={\sum }_{\{ {\sigma }_{\mathrm{B}}^{z}\} }P(\{ {\sigma }_{\mathrm{A}}^{z},{\sigma }_{\mathrm{B}}^{z}\} )$ and $P(\{ {\sigma }_{\mathrm{B}}^{z}\} )={\sum }_{\{ {\sigma }_{\mathrm{A}}^{z}\} }P(\{ {\sigma }_{\mathrm{A}}^{z},{\sigma }_{\mathrm{B}}^{z}\} )$, one can obtain the classical mutual information I_A:B(L,l) = H_A + H_B − H_AB for a given length l of section A.

On the other hand, to compute the entanglement E_A:B(L,l) between sections A and B, we perform the partial trace on the density matrix ρ = |ψ₀〉〈ψ₀| obtained from the Lanczos method, leading to the reduced density matrices ρ_A and ρ_B. Since the entanglement of a pure state is given by the von Neumann entropy of the reduced density matrices (see equation (1)), we can determine the entanglement E_A:B(L,l) by numerically diagonalizing ρ_A or ρ_B.

Using this method, we calculate E_A:B(L,l) and I_A:B(L,l) for various values of g, up to L = 24. As expected, at the critical point g = 1 the entanglement E_A:B(L,l) is found to obey the scaling form (2), as shown in figure 2 with f(ξ) = 1/πsin(πξ). Surprisingly, the mutual information I_A:B(L,l) also obeys the same type of scaling form. To illustrate this finding, we plot two scaling forms

$$\begin{eqnarray} &&{E}_{\mathrm{A:B}}(L,l)=a+\frac{1}{6}\ln \left (\frac{L}{\pi }\sin \frac{l \pi }{L}\right ), \end{eqnarray} \tag{ 9 }$$

$$\begin{eqnarray} &&{I}_{\mathrm{A:B}}(L,l)={a}^{\prime}+\frac{{c}^{\prime}}{6}\ln \left (\frac{L}{\pi }\sin \frac{l \pi }{L}\right ), \end{eqnarray} \tag{ 10 }$$

as a solid and a dashed line in figure 2, together with the numerical data obtained by the Lanczos method. Here, we have used a ≈ 0.478, a' ≈ 0.329, and c' ≈ 0.715. This means that the initial entanglement and the mutual information after the measurement differ only by a factor at the critical point.

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Note that the classical mutual information after the measurement is always smaller than or equal to the initial entanglement in the ground state. The same observation holds in the off-critical case g ≠ 1 (not shown here).

The inequality I_A:B ≤ E_A:B is valid not only in the ground state but also in time-dependent pure states of the quantum Ising chain. The time evolution of the density matrix is given by

$$\begin{eqnarray} &&\rho (t)={\mathrm{e}}^{-\mathrm{i}H t}\vert \psi (0)\rangle \langle \psi (0)\vert {\mathrm{e}}^{\mathrm{i}H t}, \end{eqnarray} \tag{ 11 }$$

where H is the quantum Ising Hamiltonian (4) and |ψ(0)〉 is the initial state. Since ρ(t) is also a pure state, one can obtain E_A:B(L,l) and I_A:B(L,l) using the same methods as in the ground state. In figure 3 we show results for g = 0.1 and 1.0 in a chain with L = 10 and l = 5, using an initial state that is fully magnetized in the z-direction. As can be seen, the quantities oscillate irregularly but satisfy the inequality I_A:B ≤ E_A:B at any time.

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It turns out that this inequality holds generally for arbitrary local measurements on entangled pure states in bipartite system, as will be proved in section 3.

In the following let us consider an arbitrary bipartite system on _AB = _A⊗_B which is in a pure state ρ_AB = |ψ〉〈ψ| so that the von Neumann entropy S[ρ_AB] =− tr[ρ_ABlnρ_AB] vanishes. In such a pure state the entanglement between the subsystems A and B is given by

$$\begin{eqnarray} &&{E}_{\mathrm{A:B}}=S[{\rho }_{\mathrm{A}}]=S[{\rho }_{\mathrm{B}}], \end{eqnarray} \tag{ 12 }$$

where ρ_A = tr_Bρ_AB and ρ_B = tr_Aρ_AB denote the reduced density matrices. On both sides let us now perform local projective measurements

$$\begin{eqnarray} &&{M}_{\mathrm{A}}=\mathop{\sum }\limits _{a}a\vert {\phi }_{a}\rangle \langle {\phi }_{a}\vert =\mathop{\sum }\limits _{a}a{\Pi }_{a},\tqs {M}_{\mathrm{B}}=\mathop{\sum }\limits _{b}b\vert {\phi }_{b}\rangle \langle {\phi }_{b}\vert =\mathop{\sum }\limits _{b}b{\Pi }_{b} \end{eqnarray} \tag{ 13 }$$

with the projectors Π_a = |ϕ_a〉〈ϕ_a| and Π_b = |ϕ_b〉〈ϕ_b|, which may be thought of as spin configuration measurements, as in the example given above. This measurement converts the pure state ρ_AB into a mixed state

$$\begin{eqnarray} &&{\rho }_{\mathrm{AB}}^{\prime}=\mathop{\sum }\limits _{a,b}{\Pi }_{a b}{\rho }_{\mathrm{AB}}{\Pi }_{a b}, \end{eqnarray} \tag{ 14 }$$

where Π_ab = Π_a⊗Π_b. Moreover, the measurement completely destroys the entanglement between the two subsystems. To see this, let us consider the entanglement of formation

$$\begin{eqnarray} &&{E}_{\mathrm{F}}({\rho }_{\mathrm{AB}}^{\prime})=\inf \left \{ \mathop{\sum }\limits _{k}{q}_{k}S[{\text{tr}}_{\mathrm{B}}\vert k\rangle \langle k\vert ]\left \vert \rho _{\mathrm{AB}}^{\prime}=\mathop{\sum }\limits _{k}{q}_{k}\vert k\rangle \langle k\vert \right .\right \} \end{eqnarray} \tag{ 15 }$$

defined as the infimum of the entropy over all possible statistical ensembles represented by ${\rho }_{\mathrm{AB}}^{\prime}$. Since ${E}_{\mathrm{F}}({\rho }_{\mathrm{AB}}^{\prime})$ is non-negative and the entropy of the particular representative

$$\begin{eqnarray} &&{\rho }_{\mathrm{AB}}^{\prime}=\mathop{\sum }\limits _{a b}{\Pi }_{a b}\text{tr}\bigl [{\Pi }_{a b}\hspace{0.167em} {\rho }^{\prime}\hspace{0.167em} {\Pi }_{a b}\bigr ] \end{eqnarray} \tag{ 16 }$$

vanishes, we can conclude that ${E}_{\mathrm{F}}({\rho }_{\mathrm{AB}}^{\prime})=0$ after the measurement. This means that the measurement destroys the original entanglement and converts it to some extent into classical correlations which can be quantified by the mutual information I_A:B in equation (3). Note that the classical mutual information I_A:B is equivalent to the quantum mutual information $I({\rho }_{\mathrm{AB}}^{\prime})$ of the post-measurement state ${\rho }_{\mathrm{AB}}^{\prime}$, i.e., 

$$\begin{eqnarray} &&{I}_{\mathrm{A:B}}=I({\rho }_{\mathrm{AB}}^{\prime})=S[{\rho }_{\mathrm{ A}}^{\prime}]+S[{\rho }_{\mathrm{ B}}^{\prime}]-S[{\rho }_{\mathrm{ AB}}^{\prime}], \end{eqnarray} \tag{ 17 }$$

where ${\rho }_{\mathrm{A}}^{\prime}={\text{tr}}_{\mathrm{B}}[{\rho }_{\mathrm{AB}}^{\prime}]$ and ${\rho }_{\mathrm{B}}^{\prime}={\text{tr}}_{\mathrm{A}}[{\rho }_{\mathrm{AB}}^{\prime}]$.

According to the Stinespring theorem [8], the measurement process (16) can be carried out by embedding ρ_AB in a higher-dimensional Hilbert space, performing a unitary transformation on it, and finally tracing out the additional degrees of freedom. To this end let us extend the Hilbert space _AB = _A⊗_B by an auxiliary space ${\mathcal{H}}_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}}={\mathcal{H}}_{\tilde {\mathrm{A}}}\otimes {\mathcal{H}}_{\tilde {\mathrm{B}}}$ whose task will be to store the measurement outcome encoded in the form of orthonormal basis vectors $\vert {\tilde {\phi }}_{a b}\rangle =\vert {\tilde {\phi }}_{a}\rangle \otimes \vert {\tilde {\phi }}_{b}\rangle$. Moreover, let us define the linear map

$$\begin{eqnarray} &&V:{\mathcal{H}}_{\mathrm{AB}}\rightarrow {\mathcal{H}}_{\mathrm{AB}}\otimes {\mathcal{H}}_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}}:\vert \psi \rangle \rightarrow V\vert \psi \rangle =\mathop{\sum }\limits _{a b}\langle {\phi }_{a b}\vert \psi \rangle (\vert {\phi }_{a b}\rangle \otimes \vert {\tilde {\phi }}_{a b}\rangle ) \end{eqnarray} \tag{ 18 }$$

which obeys the condition VV^† = 1 so that it maps the initial state isometrically onto a subset of the extended space like a unitary (entropy-preserving) transformation. With the corresponding extended density matrix

$$\begin{eqnarray} &&{\rho }_{\mathrm{AB}\tilde {\mathrm{A}}\tilde {\mathrm{B}}}:= V{\rho }_{\mathrm{AB}}{V}^{\dagger }=\mathop{\sum }\limits _{a b}\mathop{\sum }\limits _{{a}^{\prime}{b}^{\prime}}\langle {\phi }_{a b}\vert {\rho }_{\mathrm{AB}}\vert {\phi }_{{a}^{\prime}{b}^{\prime}}\rangle (\vert {\phi }_{a b}\rangle \langle {\phi }_{{a}^{\prime}{b}^{\prime}}\vert \otimes \vert {\tilde {\phi }}_{a b}\rangle \langle {\tilde {\phi }}_{{a}^{\prime}{b}^{\prime}}\vert ) \end{eqnarray} \tag{ 19 }$$

the measurement process (16) can now be written as

$$\begin{eqnarray} &&{\rho }_{\mathrm{AB}}^{\prime}={\text{tr}}_{\tilde {\mathrm{ A}}\tilde {\mathrm{B}}}[{\rho }_{\mathrm{AB}\tilde {\mathrm{A}}\tilde {\mathrm{B}}}]. \end{eqnarray} \tag{ 20 }$$

According to Schmidt's theorem, any quantum state |ψ〉∈_AB can be decomposed into

$$\begin{eqnarray} &&\vert \psi \rangle =\mathop{\sum }\limits _{i}\sqrt{{\lambda }_{i}}\vert {i}_{\mathrm{A}}\rangle \otimes \vert {i}_{\mathrm{B}}\rangle \end{eqnarray} \tag{ 21 }$$

with certain vectors |i_A〉∈_A, |i_B〉∈_B and probabilities λ_i∈[0,1], called Schmidt coefficients, which sum up to 1. This means that the initial state can be written as

$$\begin{eqnarray} &&{\rho }_{\mathrm{AB}}=\vert \psi \rangle \langle \psi \vert =\mathop{\sum }\limits _{i,j}\sqrt{{\lambda }_{i}{\lambda }_{j}}\vert {i}_{\mathrm{A}}\rangle \langle {j}_{\mathrm{A}}\vert \otimes \vert {i}_{\mathrm{B}}\rangle \langle {j}_{\mathrm{B}}\vert . \end{eqnarray} \tag{ 22 }$$

Calculating the partial traces we obtain

$$\begin{eqnarray} &&{\rho }_{\mathrm{A}}={\text{tr}}_{\mathrm{B}}[{\rho }_{\mathrm{AB}}]=\mathop{\sum }\limits _{i,j}\sqrt{{\lambda }_{i}{\lambda }_{j}}\vert {i}_{\mathrm{A}}\rangle \langle {j}_{\mathrm{A}}\vert \mathop{\underbrace{{\text{tr}}_{\mathrm{B}}[\vert {i}_{\mathrm{B}}\rangle \langle {j}_{\mathrm{B}}\vert ]{}}}\limits _{={\delta }_{i j}}=\mathop{\sum }\limits _{i}{\lambda }_{i}\vert {i}_{\mathrm{A}}\rangle \langle {i}_{\mathrm{A}}\vert \end{eqnarray} \tag{ 23 }$$

and a similar expression for ρ_B, meaning that the Shannon entropy of the Schmidt coefficients is equal to the initial entanglement:

$$\begin{eqnarray} &&{E}_{\mathrm{A:B}}=-\mathop{\sum }\limits _{i}{\lambda }_{i}\ln {\lambda }_{i}. \end{eqnarray} \tag{ 24 }$$

Let us now introduce another auxiliary Hilbert space _C with canonical basis vectors |ij〉 whose task will be to store pairs of Schmidt coefficients. On the combined Hilbert space _ABC = _AB⊗_C we define the density operator

$$\begin{eqnarray} &&{\rho }_{\mathrm{ABC}}:= \mathop{\sum }\limits _{i j}\sqrt{{\lambda }_{i}{\lambda }_{j}}\vert {i}_{\mathrm{A}}\rangle \langle {j}_{\mathrm{A}}\vert \otimes \vert {i}_{\mathrm{B}}\rangle \langle {j}_{\mathrm{B}}\vert \otimes \vert i j\rangle \langle i j\vert . \end{eqnarray} \tag{ 25 }$$

If we compare this operator with its own diagonal part

$$\begin{eqnarray} &&{\hat {\rho }}_{\mathrm{ABC}}:= \mathop{\sum }\limits _{i}{\lambda }_{i}\vert {i}_{\mathrm{A}}\rangle \langle {i}_{\mathrm{A}}\vert \otimes \vert {i}_{\mathrm{B}}\rangle \langle {i}_{\mathrm{B}}\vert \otimes \vert i i\rangle \langle i i\vert . \end{eqnarray} \tag{ 26 }$$

It is easy to check that integer powers of these operators have always the same trace, i.e. $\text{tr}[{\rho }_{\mathrm{ABC}}^{k}]=\text{tr}[{\hat {\rho }}_{\mathrm{ABC}}^{k}]$ for all k∈N, meaning that they have the same entropy

$$\begin{eqnarray} &&S[{\rho }_{\mathrm{ABC}}]=S[{\hat {\rho }}_{\mathrm{ABC}}]=-\mathop{\sum }\limits _{i}{\lambda }_{i}\ln {\lambda }_{i}={E}_{\mathrm{A:B}}. \end{eqnarray} \tag{ 27 }$$

We now use the map V defined in equation (18) to extend this operator even further to the space ${\mathcal{H}}_{\mathrm{AB}\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}={\mathcal{H}}_{\mathrm{AB}}\otimes {\mathcal{H}}_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}}\otimes {\mathcal{H}}_{\mathrm{C}}$ by defining

$$\begin{eqnarray} &&\fl {\rho }_{\mathrm{AB}\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}=(V \hspace{0.167em} \otimes \hspace{0.167em} {\mathbf{1}}_{\mathrm{C}}){\rho }_{\mathrm{ABC}}({V}^{\dagger }\otimes {\mathbf{1}}_{\mathrm{ C}})\nonumber\\ &&=~\mathop{\sum }\limits _{a b}\mathop{\sum }\limits _{{a}^{\prime}{b}^{\prime}}\mathop{\sum }\limits _{i j}\sqrt{{\lambda }_{i}{\lambda }_{j}}\langle {\phi }_{a}\vert {i}_{\mathrm{A}}\rangle \langle {j}_{\mathrm{A}}\vert {\phi }_{{a}^{\prime}}\rangle \langle {\phi }_{b}\vert {i}_{\mathrm{B}}\rangle \langle {j}_{\mathrm{B}}\vert {\phi }_{{b}^{\prime}}\rangle \nonumber\\ &&\times ~\mathop{\underbrace{\vert {\phi }_{a b}\rangle \langle {\phi }_{{a}^{\prime}{b}^{\prime}}\vert {}}}\limits _{\mathrm{AB}}\otimes \underbrace{\vert {\tilde {\phi }}_{a b}\rangle \langle {\tilde {\phi }}_{{a}^{\prime}{b}^{\prime}}\vert {}}_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}}\otimes \mathop{\underbrace{\vert i j\rangle \langle i j\vert {}}}\limits _{\mathrm{C}}. \end{eqnarray} \tag{ 28 }$$

As V is an isometry, this extension does not change the entropy, hence

$$\begin{eqnarray} &&S[{\rho }_{\mathrm{AB}\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}]={E}_{\mathrm{A:B}}. \end{eqnarray} \tag{ 29 }$$

Tracing out the original Hilbert we have tr_AB[|ϕ_ab〉〈ϕ_a'b'|] = δ_aa'δ_bb', leading to

$$\begin{eqnarray} &&\fl {\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}={\text{tr}}_{\mathrm{AB}}[{\rho }_{\mathrm{AB}\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}]=\mathop{\sum }\limits _{a b}\mathop{\sum }\limits _{i j}\sqrt{{\lambda }_{i}{\lambda }_{j}}\langle {\phi }_{a}\vert {i}_{\mathrm{A}}\rangle \langle {j}_{\mathrm{A}}\vert {\phi }_{a}\rangle \langle {\phi }_{b}\vert {i}_{\mathrm{B}}\rangle \langle {j}_{\mathrm{B}}\vert {\phi }_{b}\rangle \nonumber\\ &&\times ~\vert {\tilde {\phi }}_{a b}\rangle \langle {\tilde {\phi }}_{a b}\vert \otimes \vert i j\rangle \langle i j\vert . \end{eqnarray} \tag{ 30 }$$

Since the density matrix in equation (30) has only diagonal elements, the von Neumann entropy of ${\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}$ and its reduced density matrices can be obtained easily as follows:

$$\begin{eqnarray} &&\fl S[{\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}]=-\mathop{\sum }\limits _{i}{\lambda }_{i}\ln {\lambda }_{i}-\mathop{\sum }\limits _{a b}\mathop{\sum }\limits _{i j}(\sqrt{{\lambda }_{i}{\lambda }_{j}}\langle {\phi }_{a}\vert {i}_{\mathrm{A}}\rangle \langle {j}_{\mathrm{A}}\vert {\phi }_{a}\rangle \langle {\phi }_{b}\vert {i}_{\mathrm{B}}\rangle \langle {j}_{\mathrm{B}}\vert {\phi }_{b}\rangle \nonumber\\ &&\times ~\ln \langle {\phi }_{a}\vert {i}_{\mathrm{A}}\rangle \langle {j}_{\mathrm{A}}\vert {\phi }_{a}\rangle \langle {\phi }_{b}\vert {i}_{\mathrm{B}}\rangle \langle {j}_{\mathrm{B}}\vert {\phi }_{b}\rangle ), \end{eqnarray} \tag{ 31 }$$

$$\begin{eqnarray} &&\fl S[{\rho }_{\tilde {\mathrm{A}}\mathrm{C}}]=-\mathop{\sum }\limits _{i}{\lambda }_{i}\ln {\lambda }_{i}-\mathop{\sum }\limits _{a,i}{\lambda }_{i}\langle {\phi }_{a}\vert {i}_{\mathrm{A}}\rangle \langle {i}_{\mathrm{A}}\vert {\phi }_{a}\rangle \ln \langle {\phi }_{a}\vert {i}_{\mathrm{A}}\rangle \langle {i}_{\mathrm{A}}\vert {\phi }_{a}\rangle , \end{eqnarray} \tag{ 32 }$$

$$\begin{eqnarray} &&\fl S[{\rho }_{\tilde {\mathrm{B}}\mathrm{C}}]=-\mathop{\sum }\limits _{i}{\lambda }_{i}\ln {\lambda }_{i}-\mathop{\sum }\limits _{b,i}{\lambda }_{i}\langle {\phi }_{b}\vert {i}_{\mathrm{B}}\rangle \langle {i}_{\mathrm{B}}\vert {\phi }_{b}\rangle \ln \langle {\phi }_{b}\vert {i}_{\mathrm{B}}\rangle \langle {i}_{\mathrm{B}}\vert {\phi }_{b}\rangle , \end{eqnarray} \tag{ 33 }$$

where ${\rho }_{\tilde {\mathrm{A}}\mathrm{C}}={\text{tr}}_{\tilde {\mathrm{B}}}[{\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}]$ and ${\rho }_{\tilde {\mathrm{B}}\mathrm{C}}={\text{tr}}_{\tilde {\mathrm{A}}}[{\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}]$. Combining equations (31)–(33), one obtains a useful relation

$$\begin{eqnarray} &&S[{\rho }_{\tilde {\mathrm{A}}\mathrm{C}}]+S[{\rho }_{\tilde {\mathrm{B}}\mathrm{C}}]-S[{\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}]={E}_{\mathrm{A:B}}. \end{eqnarray} \tag{ 34 }$$

Tracing out the auxiliary space _C we obtain the reduced density matrices ${\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}},{\rho }_{\tilde {\mathrm{A}}}$, and ${\rho }_{\tilde {\mathrm{B}}}$ in which the outcome of the measurement is stored. Therefore, the classical mutual information after the measurement can be expressed in terms of these density matrices by

$$\begin{eqnarray} &&S[{\rho }_{\tilde {\mathrm{A}}}]+S[{\rho }_{\tilde {\mathrm{B}}}]-S[{\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}}]={I}_{\mathrm{A:B}}. \end{eqnarray} \tag{ 35 }$$

To prove the inequality I_A:B ≤ E_A:B, we use the strong subadditivity of the von Neumann entropy [10], which is known to hold in the two following equivalent forms [11]:

$$\begin{eqnarray} &&S[{\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}]+S[{\rho }_{\tilde {\mathrm{B}}}]\leq S[{\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}}]+S[{\rho }_{\tilde {\mathrm{B}}\mathrm{C}}] \end{eqnarray} \tag{ 36 }$$

$$\begin{eqnarray} &&S[{\rho }_{\tilde {\mathrm{A}}}]+S[{\rho }_{\tilde {\mathrm{B}}}]\leq S[{\rho }_{\tilde {\mathrm{A}}\mathrm{C}}]+S[{\rho }_{\tilde {\mathrm{B}}\mathrm{C}}]. \end{eqnarray} \tag{ 37 }$$

Using the first relation the mutual information can be written as

$$\begin{eqnarray} &&{I}_{\mathrm{A:B}}=S[{\rho }_{\tilde {\mathrm{A}}}]+S[{\rho }_{\tilde {\mathrm{B}}}]-S[{\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}}]\leq S[{\rho }_{\tilde {\mathrm{A}}}]+S[{\rho }_{\tilde {\mathrm{B}}\mathrm{C}}]-S[{\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}]. \end{eqnarray} \tag{ 38 }$$

Replacing $\tilde {A}\leftrightarrow \tilde {B}$ one obtains in the same way

$$\begin{eqnarray} &&{I}_{\mathrm{A:B}}\leq S[{\rho }_{\tilde {\mathrm{B}}}]+S[{\rho }_{\tilde {\mathrm{A}}\mathrm{C}}]-S[{\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}]\hspace{0.167em} . \end{eqnarray} \tag{ 39 }$$

Adding (38) and (39) gives an inequality, where we again apply the second version of strong subadditivity (37)

$$\begin{eqnarray} &2{I}_{\mathrm{A:B}}&\leq S[{\rho }_{\tilde {\mathrm{A}}}]+S[{\rho }_{\tilde {\mathrm{B}}}]+S[{\rho }_{\tilde {\mathrm{A}}\mathrm{C}}]+S[{\rho }_{\tilde {\mathrm{B}}\mathrm{C}}]-2 S[{\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}]\nonumber\\ &&\leq 2 S[{\rho }_{\tilde {\mathrm{A}}\mathrm{C}}]+2 S[{\rho }_{\tilde {\mathrm{B}}\mathrm{C}}]-2 S[{\rho }_{\tilde {\mathrm{A}}\tilde {\mathrm{B}}\mathrm{C}}]=2{E}_{\mathrm{A:B}}, \end{eqnarray} \tag{ 40 }$$

proving the initial assertion.