Question

state and prove the princip
le ofblinear superposition of eigen states

Answer 1

This is very simple to prove, and I am rather taken aback that nobody has yet given you a proof. Quantum states are generally represented as vectors. The reason we use "eigenfunction" instead of say "eigenvector" if because it is often convenient to use functions which form a vector space, rather than, say, column vectors. However, you can reason about them in the same way as column vectors, and indeed for finite dimensional systems such as spins we do often use column vectors. So, I'll prove this using column vectors, but exactly the same argument can be applied with functions.

Take two solutions to an eigenvalue equation, u and v which correspond to the same eigenvalue e. So we have Mu = eu and Mv = ev. In this case, if we take any linear combination w=au+bv, we can see that it must also be an eigenvector of M with eigenvalue e: M(au+bv)= Mau + Mbv = aMu + bMv = aeu + bev = e(au+bv). Thus Mw = ew. So any linear combination of eigenvectors with the same eigenvalue simply results in another eigenvector with the same eigenvalue. This is not true if the eigenvalues are different, and this fact accounts for all dynamics.

This is, in fact, nothing to do with quantum physics, but rather a simple consequence of linear algebra, so it shows up in lots of areas of physics (such as electromagnetism). As Alexander mentions, however, in quantum mechanics, in order to preserve probability, the only physical solutions are linear combinations which have been normalised to be unit vectors.