Normalization of the wave function of a particle in one dimension box or infinite potential well
Description of Normalization of the wave function of a particle in one dimension box or infinite potential well:
We know that the wave function for the motion of the particle along the x-axis is
$\psi_{n}(x)= A \: sin \left( \frac{n \pi x}{L} \right) \quad \left\{ Region \quad 0 \lt x \lt a \right\}$
$\psi_{n}(x)= 0 \quad \left\{ Region \quad 0 \gt x \gt a \right\}$
The total probability that the particle is somewhere in the box must be unity. Therefore,
$\int_{0}^{L} \left| \psi_{n}(x)\right|^{2}dx =1$
Now substitute the value of the wave function in the above equation. Then
$\int_{0}^{L} \left| A \: sin \left( \frac{n \pi x}{L} \right) \right|^{2}dx =1$
$\int_{0}^{L} A^{2} \: sin^{2} \left( \frac{n \pi x}{L} \right) dx =1$
$ \frac{A^{2}}{2}\int_{0}^{L} \left[ 1- cos \left( \frac{2n \pi x}{L} \right) \right] dx =1$
$ \frac{A^{2}}{2} \left[ x - \left( \frac{L}{2n\pi} \right) sin \left( \frac{2n \pi x}{L} \right) \right]_{0}^{L} =1$
$ \frac{A^{2}}{2} \left[ L - \left( \frac{L}{2n\pi} \right) sin \left( \frac{2n \pi L}{L} \right) \right] =1$
$ \frac{A^{2}}{2} \left[ L - \left( \frac{L}{2n\pi} \right) sin \left( 2n \pi \right) \right] =1$
$ \frac{A^{2}}{2} \left[ L - \left( \frac{L}{2n\pi} \right) sin \left( 2n \pi \right) \right] =1$
$ \frac{A^{2}}{2} \left[ L - 0 \right] =1 \qquad(\because sin2n\pi =0)$
$ \frac{A^{2} L}{2} =1$
$ A= \sqrt{\frac{2}{L}}$
Hence, the normalized wave function
$\psi_{n}(x)=\sqrt{\frac{2}{L}} sin \left( \frac{n \pi x}{L} \right)$
The absolute square $\left| \psi_{n}(x) \right|^{2}$ of the wave function $\psi_{n}(x)$ gives the probability density. Hence
$\left| \psi_{n}(x) \right|^{2} = \frac{2}{L} sin^{2} \left( \frac{n \pi x}{L} \right)$
The wave function for the particle in a box can be viewed in analogy with standing waves on a string. The wave function for a standing wave that has nodes at endpoints is of the form $\psi_{n}(x)= A \: sin \left( \frac{n \pi x}{L} \right)$. The condition for a standing wave can also be expressed in terms of wavelength.
$\lambda_{n}=\frac{2 \pi}{k_{n}}$
$\lambda_{n}=\frac{2 \pi}{\frac{n \pi}{L}} \qquad \left( \because k_{n}=\frac{n \pi}{L} \right)$
$\lambda_{n}=\frac{2 L}{n}$
$L= \frac{n \: \lambda_{n}}{2}$
So,
$L= \frac{\: \lambda_{1}}{2} \qquad \left( for \: n=1 \right)$
$L= \lambda_{2} \qquad \left( for \: n=2 \right)$
$L= \frac{3 \: \lambda_{3}}{2} \qquad \left( for \: n=3 \right)$
$L= 2 \lambda_{4} \qquad \left( for \: n=4 \right)$
Geo structure of wave function $\psi_{n}(x)$ and wave function's density $\left| \psi_{n}(x) \right|^{2}$.
Variation of the wave function and probability of finding the particle in a one-dimensional box:
We know that normalised wave function $\psi_{n}(x)$
$\psi_{n}(x)=\sqrt{\frac{2}{L}} sin \left( \frac{n \pi x}{L} \right)$
The probability density of wave function $\left| \psi_{n}(x) \right|$
$\left| \psi_{n}(x) \right|^{2} = \frac{2}{L} sin^{2} \left( \frac{n \pi x}{L} \right)$
Maximum Condition:
The values of $\psi_{n}(x)$ and $\left| \psi_{n}(x) \right|^{2}$ will be maximum. When
$sin \left( \frac{n \pi x}{L} \right)=1$
$sin \left( \frac{n \pi x}{L} \right )=sin \frac{\left( 2m+1 \right) \pi}{2}$
$ \frac{n \pi x}{L} =\left( 2m+1 \right) \frac{ \pi}{2}$
$ x =\left( 2m+1 \right) \frac{ L}{2n}$
Minima Condition:
The values of $\psi_{n}(x)$ and $\left| \psi_{n}(x) \right|^{2}$ will be minima. When
$sin \left( \frac{n \pi x}{L} \right)=0$
$sin \left( \frac{n \pi x}{L} \right)= \sin \: m\pi$
$ \frac{n \pi x}{L} = \: m\pi$
$x=m\left( \frac{L}{n} \right)$
