## Normalized and Orthogonal wave function

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.