# I Eigenvectors - eigenvalues mappings in QM

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1. Jan 1, 2018

### referframe

In non-relativistic QM, say we are given some observable M and some wave function Ψ. For each unique eigenvalue of M there is at least one corresponding eigenvector. Actually, there can be a multiple (subspace) eigenvectors corresponding to the one eigenvalue.

But if we are given a set of distinct eigenvectors to start with, then there is always just one unique eigenvalue for each of those distinct eigenvectors. There are never multiple eigenvalues associated with just one eigenvector. Is that a true statement?

2. Jan 1, 2018

### kith

Yes, see the wikipedia article on eigenvalues and eigenvectors. If you apply a linear transformation to a vector, you need to get a unique result. For example, if you apply a reflection across a given line to a vector, you don't get two reflections.

Last edited: Jan 1, 2018
3. Jan 1, 2018

### Staff: Mentor

Formally, we get for an eigenvector $x$ of $\psi$ with two eigenvalues $\lambda , \mu$ the equation $\psi.x=\lambda \cdot x = \mu \cdot x$ and so $(\lambda - \mu)\cdot x = 0$. So we have either the zero vector $x=0$, which is contained in all eigenspaces, or identical eigenvalues $\lambda = \mu\,$.