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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?

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    $\begingroup$ 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 $|\psi_n\rangle$. $\endgroup$
    – Charlie
    Apr 5, 2021 at 20:35

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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.

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  • $\begingroup$ Ok so if I'm not wrong what they mean by complete is given a state I can completely expand it into its basis vectors kind of like fourier series and they needed to clarify it because the basis chosen is sometimes infinite. $\endgroup$
    – aakash
    Apr 5, 2021 at 20:50
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This is just the statement that any state can be written as a linear combination of the elements in the basis.

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