Why should the quantum channels be described by linear maps?

  • $\begingroup$ This paper might be helpful, the available pdf is in this link. Mind that the B-map in the paper is exactly the choi operator. $\endgroup$
    – narip
    Sep 1, 2021 at 13:58

1 Answer 1


More precisely, they are described by affine maps. This is also not special to quantum mechanics, but rather a more general fact in physical theories (keyword: generalized probabilistic theory). Let me elaborate.

Suppose you can prepare two states $s_1$ and $s_2$ of a physical system, then it is always possible to prepare a probabilistic mixture of those. Simply throw a (biased) coin (say head has probability $p$) and prepare state $s_1$ when it shows head and $s_2$ when it's tail.

Mathematically, this means that it is always meaningful to consider convex combinations of states, i.e. $$ s = p s_1 + (1-p) s_2, $$ describes another valid state. This means that the set of physical states has a natural convex structure associated. For instance, states of a classical system are probability measures on phase space, while states of a quantum system are described by density matrices. Both are convex sets.

Now, any physical dynamics should preserve state space, in particular they should preserve convex combinations, i.e. we require that the physical mapping obeys $$ \Phi ( p s_1 + (1-p) s_2) = p \Phi(s_1) + (1-p) \Phi(s_2). $$ Such a map is called affine.

If we embed a convex set into a vector space, these maps are essentially given by linear maps on this vector space (actually affine transformations). For quantum states, we embed state space into the set of complex matrices, thus quantum channels are given by linear maps on complex matrices (with additional constraints, of course).


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