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

Imagine discrete orthonormal basis made of infinite set of kets $$|\phi_1\rangle , ..., |\phi_n\rangle,...$$ 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 $$|\phi_m\rangle\langle\phi_m|$$ are all zeros except one which is the square of norm of $$|\phi_m\rangle$$. 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.

• Any state $|\psi >$ can be written as $\sum a_n |\phi_n>$, as the set is a basis. Now apply your operator and use orthonormality. Nov 23, 2020 at 19:46
• The kronecker Delta forces the summation to take only one value and we can take another summation from both sides to get our $\psi$ again. Thank you. Nov 23, 2020 at 19:59

You can see that an operator of this type behaves as the identity in $$\Bbb R^3$$ with a basis $$\{|i\rangle\}$$:

$$\left(\sum_{i=1}^3|i\rangle\langle i|\right)|v\rangle=\sum_{i=1}^3|i\rangle\left(\langle i|v\rangle\right)=\sum_{i=1}^3|i\rangle v_i=\sum_{i=1}^3 v_i|i\rangle=|v\rangle \tag{1},$$

the generalisation to infinite dimensions means replacing the finite sum with an infinite one. In each term $$\langle i|v\rangle$$ gives the $$i$$th component of the vector $$|v\rangle$$. Going even further we can make a similar definition that occurs frequently in quantum mechanics:

$$\int dx\text{ }|x\rangle\langle x|\psi\rangle=\int dx\text{ }\psi(x)|x\rangle=|\psi\rangle. \tag{2}$$

In which $$|x\rangle$$ are the eigenstates of the position operator and $$\psi(x)$$ is the associated wavefunction of $$|\psi\rangle$$, therefore: $$\sum_{i=1}^3|i\rangle\langle i|=\Bbb I\quad \text{and}\quad \int dx\text{ }|x\rangle\langle x|=\Bbb I \tag{3},$$ in which $$\Bbb I$$ is the identity operator on the respective vector spaces.

• Thanks. So we always define $|\psi(x)> = \int dx \psi(x) |x>$? I assume this is what we mean by wave function in position space? I think you just answered my second question too. Thank you. Nov 23, 2020 at 20:13
• Yes the value of the wavefunction at the point $x$ is essentially a component of the Hilbert space state vector $|\psi\rangle$. Nov 23, 2020 at 20:14
• Okay out of curiosity (and this is the last one, I promise!), For the particle in infinite square well we got $\psi(x) = \sqrt(2/a) Sin(nπx/a)$. How would you derive $|\psi>$ from this? I guess you'll need the |x> eigenstates in integration but what are they? If I use $\hat{R_x} |x> = X_x |x>$ eigenvalue equation I get nowhere because I don't know what the position operator is. I hope my questions make sense... Nov 23, 2020 at 20:44
• Well technically you would get $|\psi\rangle$ from $|\psi\rangle=\int dx\text{ }\psi(x)|x\rangle$, but we generally work with the wavefunctions $\psi(x)$ rather than the vectors $|\psi\rangle$ since it's easier to extract information from them. Nov 23, 2020 at 21:25
• @AbdulQadeer the expression $|\psi(x)\rangle$ does not make sense and it is wrong we have rather $\psi(x)=\langle\psi|x\rangle$ Nov 23, 2020 at 21:49