Description:
We know that $\psi^{*}\psi$ or $\left|\psi \right|^{2} d\tau $ represent the probability of finding the particle in volume element $d\tau$.
The total probability of finding the particle in the entire space is 1 so
$ \int \left|\psi(r,t) \right|^{2} d\tau=1 $
Where integral extends overall space.
$\int \psi^{*}(r,t) \psi(r,t) d\tau=1$
A wave function satisfies the above equation so it is called normalized to unity.
For any wave function that is a solution of the time-dependent Schrodinger equation
$\int \psi^{*} \psi d\tau=N$
$\frac{1}{N} \int \psi^{*} \psi d\tau=1$
$\int \frac{\psi{*}}{\sqrt{N}} \frac{\psi}{\sqrt{N}} d\tau = 1$
Where
$\sqrt{N}$ → Normalized Factor
$\frac{\psi}{N} $ → Normalised wave function
If independent coordinate $x$,$y$,$z$, and $\psi$ satisfy the Schrodinger wave equation. Then it is evident that $\frac{\psi}{\sqrt {N}}$ also satisfies the Schrodinger wave equation.
If $\psi_{i}$ and $\psi_{j}$ are two different wave functions both the satisfactory solution of the wave equation for a given system. Then these functions will be normalized if
$\psi_{i}^{*} \psi_{i} d\tau=1 \quad and \quad \psi_{j}^{*} \psi_{j} d\tau=1$
If the two wave function $\psi_{i}$ and $\psi_{j}$ are the satisfactory solution of the wave equation for a given system. Then these functions will be mutually orthogonal if
$\psi_{i}^{*} \psi_{j} d\tau=0 \qquad Where \: $i \neq j$
$\psi_{j}^{*} \psi_{i} d\tau=0 \qquad Where \: $i \neq j$
These integral vanishes over the entire space.