# Completeness in Quantum Mechanics

While studying Matrix representation I found a topic about completeness for a basis by the equation:$$∑_{n}|ψ_n⟩⟨ψ_n|=1$$

I don't understand the physical meaning behind it. Is it something to do with infinite basis vectors?

• I don't think you'll get much physical meaning from it, but if you take an arbitrary state $|\psi\rangle$ and act on it with the completeness relation you will find that you get back that same state since you are just rewriting it as a sum over the basis vectors $|\psi_n\rangle$. Apr 5, 2021 at 20:35

If you recall the projection operator, $$\mathcal{P}$$, which look like $$\mathcal{P}=|\phi_n\rangle \langle \phi_n|$$
where $$\{|\phi_i\rangle\}$$ are assumed to be the basis set for the LVS. Then what this operator does is project a component of any arbitrary vector along with the basis $$|\phi_n\rangle$$.
What I'm trying to say, Given a vector $$|\psi\rangle =\sum_i c_i|\phi_i\rangle$$ $$\mathcal{P}|\psi\rangle =\sum_ic_i|\phi_n\rangle \langle \phi_n|\phi_i\rangle=c_n|\phi_n\rangle$$ That explain the projection operator.
Now If I project the along all its component, $$\sum_i \mathcal{P_i}|\psi\rangle =\sum_i|\phi_i\rangle \langle\phi_i|\left(\sum_jc_j|\phi_j\rangle\right)=\sum_{i,j}c_j|\phi_i\rangle \langle \phi_i|\phi_j\rangle =|\psi\rangle$$ In other world, $$\sum_i|\phi_i\rangle \langle \phi_i|=I$$ This is saying nothing but If I project the vector along all its basis vectors, I get the vector back.