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Suppose you have the following quantum state: $$\frac{1}{\sqrt{2^n}}\sum_{i}|\bar{i}\rangle_A|i\rangle_S$$ where $\{|i\rangle_A\}$ and $\{|i\rangle_S\}$ are orthonormal bases for an $n$-qubit ancilla and $n$-qubit quantum system respectively. The system $S$ stores a quantum code of distance $d$. This setup comes from Preskill's lecture notes.

Since the code has distance $d$, there is no observable on $< d$ qubits that gives you any information about the encoded state. Since the ancilla and system are maximally entangled, learning about $S$ tells you about $A$; hence, there can be no correlations between $A$ and any $\leq d-1$-qubit subsystem of $S$. This implies, for any subsystem $Q$ of $d-1$ qubits, $$S(AQ) = S(A) + S(Q)$$ But the subadditivity of Von Neumann entropy is saturated iff $\rho_{AQ} = \rho_A \otimes \rho_Q$.

How can one show in this setup that the joint state of $A$ and any $Q$ is a product state (specifically, how do we mathematically connect the intuitive argument that $Q$ and $A$ are uncorrelated to saturating the subadditivity condition)? More generally, for $\textit{any}$ joint state of some (not necessarily maximally) entangled systems $X$ and $Y$, does the condition that the density matrix of a $k$-qubit subsystem of $Y$ (call it $Q_Y$) is $\propto$ identity imply that $S(XQ_Y) = S(X) + S(Q_Y)$, i.e. $\rho_{XQ_Y} = \rho_X \otimes \rho_{Q_Y}$? How do we show it? As far as I understand, this is different from proving the decoupling inequality.

ssd42
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1 Answers1

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Let, $$|\phi\rangle=\frac{1}{\sqrt{d}}\sum_{i=0}^{d-1}|i\rangle_{X}|ii\rangle_{Q_YR_Y} \ ,$$ where $Q_YR_Y=Y$.

Notice, that the state on $XY$ is entangled, as can be seen by noting $$S(XY) \leq S(X),S(Y) \ .$$ Also, notice reduced $\phi$ state on $Q_Y$ is maximally mixed, same for reduced state on $X$.

However, the state on $XQ_Y$ is not, therefore the entropy equality is not satisfied.

So the answer is no.

Brendan Darrer
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aliayt atyte
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