Linear operator and linear vector space?

Its to do with the axioms of QM.

At the starting level we say the states form a vector space - and by definition vector spaces are linear - as hopefully you have learnt in a course of linear algebra.

The first axioms of QM is that observables are hermitian linear operators such that the eigenvalues (necessarily real since the operator is hermitian) are the possible outcomes of the observation associated with the observable.

The second axiom is the so called Born rule. If a system is in state u and it is observed with observable O the expected value of the outcome is <u|O|u>.

There is more that can be said at an advanced level - especially the very important Gleason's Theorem - see post 137:
https://www.physicsforums.com/threads/the-born-rule-in-many-worlds.763139/page-7

But basically linear is associated the vector space language QM is expressed in.

Thanks
Bill

 

An operator R defined on a set S of functions or vectors over a field F (with + and ×)[ with multiplication * between elements of F and elements of S] is linear if, for all f, g in S and all a in F, R(f⊕g) = R(f) ⊕ R(g), and R(s*f) = s*R(f).

A linear vector space is a set S of vectors closed under addition ⊕ and closed under multiplication ⊗ between scalars [from a field F (with +, ×)] and vectors is defined, such that vector addition is associative and commutative, there is a null vector and every vector has an additive inverse in S, and scalar multiplication is distributive: (a+b)*v = a*v⊕v*b, a(v⊕w) = a*v⊕a*w, and finally (a×b)*v = a*(b*v)