Answer to Question #178359 in Quantum Mechanics for Shraddha

Question

the normalized wave function for the first excited state of particle in one dimensional box

Answer 1

The wave function can be written as

$\Psi_n(x)=A_nsin(\dfrac{n\pi x}{L})$

The constant $A_n$ is determined by normalization condition

$\int\Psi^2_n dx=\int A_n^2sin^2(\dfrac{n\pi x}{L})dx=1$

The result of evaluating the integral and solving for $A_n=\sqrt{\dfrac{2}{L}}$ is independent from n.

The normalized wave function for a particle in a box are thus

$\Psi_n(x)=\sqrt{\dfrac{2}{L}}sin(\dfrac{n\pi x}{L})$

For first excited state n = 2

$\therefore\space \Psi_2(x)=\sqrt{\dfrac{2}{L}}sin(\dfrac{2\pi x}{L})$