# Exercise 2

## from Example Sheet 4

### Exercise 2.

Let $$|\psi\rangle_{ABE}$$ be a pure state of a tripartite system $$ABE$$. define the coherent information from $$A$$ to $$B$$ of $$\psi$$ to be $I_c^{A>B}(\psi) = - S(A|B)_\psi.$ Here $$S(A|B)_\psi$$ denotes the conditional entropy of the subsystem $$A$$ with respect to subsystem $$B$$, given that the composite system $$ABE$$ is in the pure state $$|\psi\rangle_{ABE}$$. Henceforth we shall omit $$\psi$$.

Prove the following identities:

1. $$\frac{1}{2} I(A:B) + \frac{1}{2} I(A:E)= S(A)$$

2. $$\frac{1}{2} I(A:B) - \frac{1}{2} I(A:E)= I_c^{A>B}$$

1. By definition of the quantum mutual information, \begin{aligned} \frac{1}{2} I(A:B) + \frac{1}{2} I(A:E) &= \frac{1}{2} [ S(A) + S(B) - S(AB) + S(A) + S(E) - S(AE)] \\ &= S(A) + \frac{1}{2} [ S(B) - S(AB) + S(E) - S(AE)]. \end{aligned} Since $$| \psi \rangle_{ABE}$$ is pure, $$S(B) = S(AE)$$ and $$S(E) = S(AB)$$ (since for any bipartition of the systems ABE, the entropies of both reduced density matrices are equal). Thus, the term in brackets vanishes and we obtain $$S(A)$$ as desired.
2. Likewise, \begin{aligned} \frac{1}{2} I(A:B) - \frac{1}{2} I(A:E) &= \frac{1}{2} [ S(A) + S(B) - S(AB) - S(A) - S(E) + S(AE)] \\ &= \frac{1}{2} [ S(B) - S(AB) - S(E) + S(AE)] \\ &= \frac{1}{2} [ S(B) - S(AB) - S(AB) + S(B)] \\ &= S(B) - S(AB) = - S(A|B) = I_c^{A>B}. \end{aligned}