# Quantum Mutual Information scaling

Wikipedia provides a simple definition of Quantum Mutual Information:

$$I(\rho^{ab})= S(\rho^{a}) + S(\rho^{b}) - S(\rho^{ab})$$

where in terms of relative information we have:

$$I(\rho^{ab})= S(\rho^{ab}|| \rho^{a} \otimes \rho^{b})$$

what is the best notational approach for these expression when there are more than two systems?

UPDATE

If we take the classical case:

$$I(X;Y) = H(X \cap Y)$$

The form of Pascal's triangle is apparent when the relation is viewed as (1 2 1):

$$0=- H(X\cup Y) + \{ H(X) + H(Y) \}- H(X\cap Y)$$ and (1,3,3,1): $$0=- H(X\cup Y\cup Z) + \{ H(X) + H(Y) + H(Z)\}-\{ H(X\cap Y) + H(X \cap Z) + H(Y \cap Z)\}+ H(X\cap Y\cap Z)$$

Following the inclusion-exclusion principle we find that the number of terms that comprise mutual information are:

$$\sum_{k=2}^{m}(-1)^{k-1}\dbinom{m}{k}$$

Where we could write the complete expression as:

$$H\left(\bigcup_{i=1}^{n}X_i\right)=\sum_{i=1}^{n}H(X_i) - \sum_{i,j:i<j}H(X_i\cap X_j) + \sum_{i,j,k:i<j<k}H(X_i\cap X_j \cap X_k) -\dots + (-1)^{n-1}H\left(\bigcap_{i=1}^{n}X_i\right)$$

or as: $$H\left(\bigcup_{i=1}^{n}X_i\right)=\sum_{i=1}^{n}H(X_i) + \sum_{k=2}^{n}(-1)^{k-1}\sum_{I\subset \{1,\dots,n\}|I|=k} H\left(\bigcap_{i\in I}^{n}X_i\right)$$

which highlights that the mutual information is associated with terms:

$$I\left(\bigcap_{i\in I}^{n}X_i\right)= \sum_{k=2}^{n}(-1)^{k-1}\sum_{I\subset \{2,\dots,n\}|I|=k} H\left(\bigcap_{i\in I}^{n}X_i\right)$$

If we follow the same form then the apparent form for quantum mutual information would would also follow pascal's triangle's binomial structure, so what is the most efficient notation to use?

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Your question is non-trivial on several front : 1. Classical information theory becomes more complex when you have more then two agents involved. It turns into network information theory, and you have various scenarios where the (classical) mutual information generalize to different quantities. 2. For quantum system, you also have inequivalent kinds of entanglement when you have 3 parties or more (GHZ, W states), which should lead to different quantities. –  Frédéric Grosshans Sep 13 '12 at 13:47
Thanks for this point! I have toyed with inclusion-exclusion principles before in the classical context, so your point brought some of that back. I will probably modify the question here to include some of that discussion. –  Hal Swyers Sep 14 '12 at 11:09