How to prove that $sum_{n=1}^{infty} | phi_n rangle langle phi_n | = hat {I}$?

Question

Imagine discrete orthonormal basis made of infinite set of kets $|phi_1rangle , ..., |phi_nrangle,...$
Completeness or closure of the basis is given by:
$sum_{n=1}^{infty} | phi_n rangle langle phi_n | = hat {I}$
(From Zetilli's Quantum mechanics book)
So what I did manage is that the eigenvalues of the operator $|phi_mranglelanglephi_m|$ are all zeros except one which is the square of norm of $|phi_mrangle$. That means the matrix can be diagonalized but I don't really understand how you diagonalize matrices like this. Since the norm is 1, all entries should be zero except the $(n,n)$ one?
Also doesn't diagonalizing a matrix change it? I'm a bit confused, please guide me.

Charlie · Accepted Answer

You can see that an operator of this type behaves as the identity in $Bbb R^3$ with a basis ${|irangle}$:
$$left(sum_{i=1}^3|iranglelangle i|right)|vrangle=sum_{i=1}^3|irangleleft(langle i|vrangleright)=sum_{i=1}^3|irangle v_i=sum_{i=1}^3 v_i|irangle=|vrangle tag{1},$$
the generalisation to infinite dimensions means replacing the finite sum with an infinite one. In each term $langle i|vrangle$ gives the $i$th component of the vector $|vrangle$. Going even further we can make a similar definition that occurs frequently in quantum mechanics:
$$int dxtext{ }|xranglelangle x|psirangle=int dxtext{ }psi(x)|xrangle=|psirangle. tag{2}$$
In which $|xrangle$ are the eigenstates of the position operator and $psi(x)$ is the associated wavefunction of $|psirangle$, therefore: $$sum_{i=1}^3|iranglelangle i|=Bbb Iquad text{and}quad int dxtext{ }|xranglelangle x|=Bbb I tag{3},$$ in which $Bbb I$ is the identity operator on the respective vector spaces.

How to prove that $sum_{n=1}^{infty} | phi_n rangle langle phi_n | = hat {I}$?

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