$\require{braket}$

In a function space $X$, the completeness of a set of functions ${\psi_k}$ means $\mathrm{span}{\psi_k:k}=X$. If we add the orthonormality, we can write a beautiful equation as a equivalent condition (The inner product here treats the second element linearly $\langle a,\lambda b\rangle=\lambda\langle a,b\rangle$)

\[\psi=\sum_k \langle \psi_k,\psi\rangle \psi_k\]

For a proof you can see the 18th chapter of Mathematical Analysis by Zorich.

However, as a student major in physics, by using Dirac’s notion, the equation can be written more beautifully:

\[\ket{\psi}=\sum_k \ket{k}\bra{k}\ket{\psi},\forall\psi\in X\]

This means an operator $\sum_k \ket{k}\bra{k}=\mathrm{id}_X=I$, this is the completeness introduced by your Quantum Mechanics teacher. To emphasize how beautiful it is, let me write it again

\[\sum_k \ket{k}\bra{k}=\mathrm{id}_X=I\]

However, this notion is more than just beautiful looking. By insert this identity map to your formula, you can see it’s power.

Parseval Equation:

Just insert the identity map in a self inner product,

\[\lVert\psi\rVert^2=\braket{\psi}=\mel**{\psi}{I}{\psi}=\sum_k \bra{\psi}\ket{\psi_k}\bra{\psi_k}\ket{\psi} =\sum_k \lvert\bra{\psi_k}\ket{\psi}\rvert^2\]
To construct Dirac delta function:

Just insert the identity map in $\bra{x}\ket{y}=\delta(x-y)$

\[\sum_k \psi_k^*(x)\psi_k(y)=\sum_k\bra{x}\ket{\psi}\bra{\psi}\ket{y}=\bra{x}\ket{y}=\delta(x-y) \tag{1}\]

The property $\sum_k \psi_k^*(x)\psi_k(y)=\delta(x-y)$ for a orthonormal complete basis of function space is not easy to proof. However the Dirac’s notion help us to neglect those analytical parts and pay more attention on algebraic parts.

To prove eq(1) without Dirac’s notion

Note: in fact the proof is extract from Dirac notion’s one-step proof.

\[\begin{align*} \sum _{k} \psi _{k}^{*}( a) \psi _{k}( b) & =\sum _{k}\int \mathrm{d} x\ \delta ( x-a) \psi _{k}^{*}( x) \ \int \mathrm{d} y\ \delta ( y-b) \psi _{k}( y)\\ & =\int \mathrm{d} y\ \delta ( y-b)\sum _{k}\int \mathrm{d} x\ \delta ( x-a) \psi _{k}^{*}( x) \psi _{k}( y)\\ & \xlongequal{F( x) :=\delta ( x-a)}\int \mathrm{d} y\ \delta ( y-b)\sum _{k}\int \mathrm{d} x\ F( x) \psi _{k}^{*}( x) \psi _{k}( y)\\ & =\int \mathrm{d} y\ \delta ( y-b)\sum _{k}\bra{\psi _{k}}\ket{F} \psi _{k}( y)\\ & =\int \mathrm{d} y\ \delta ( y-b)\left[\sum _{k}\bra{\psi _{k}}\ket{F} \psi _{k}\right]( y)\\ & =\int \mathrm{d} y\ \delta ( y-b) F( y)\\ & =F( b)\\ & =\delta ( a-b) \end{align*}\]