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### Wave function of a particle in free state

The wave function of a free particle:

Suppose, A particle of mass $m$ is in motion along the x-axis. Suppose no force is acting on the particle so that the potential energy of the particle is constant. For convenience, the constant potential energy is taken to b zero. Therefore, the time-independent Schrodinger equation for a free particle is:

$-\frac{\hbar^{2}}{2m} \frac{d^{2} \psi}{dx^{2}}=E \psi \qquad(1)$

Since the particle is moving freely with zero potential energy, its total energy $E$ is the kinetic energy given by

$E=\frac{p^{2}_{x}}{2m}$

Where $p_{x}$ is the momentum of the particle which is moving along the x-axis.

$\frac{d^{2} \psi}{dx^{2}} + \frac{2mE}{\hbar^{2}} \psi =0$

Let $k^{2}=\frac{2mE}{\hbar^{2}} \qquad(2)$, Now substitute this value in the above equation that can be written as

$\frac{d^{2} \psi}{dx^{2}} + k^{2} \psi =0 \qquad(3)$

The solution of the above equation $(3)$

$\psi (x) = A e^{ikx} + B e^{-ikx} \qquad(4)$

Here $A$ and $B$ are constants.

The above equation$(4)$ gives the time independent of the wave function. The complete wave function (i.e for both time-dependent and independent ) for a particle is given by

$\psi(x,t)=\psi(x) e^{-i\omega t}$

$\psi(x,t)=\left( A e^{ikx} + B e^{-ikx} \right) e^{-i\omega t}$

$\psi(x,t)=A e^{\left(ikx-i\omega t \right)} + B e^{\left(-ikx -i\omega t \right)}$

$\psi(x,t)= A e^{-i\left(\omega t - kx \right)} + B e^{-i\left(\omega t + kx\right)} \quad(5)$

The above equation $(5)$ represents a continuous plane simple harmonic wave. The first term on the right side of the above equation $(5)$ represents the wave traveling in the positive x-direction, and the second term represents the wave traveling in the negative x-direction. Therefore, the wave function for the motion of a particle in the positive x-direction, we have

$\psi(x,t)= A e^{-i\left(\omega t - kx \right)} \qquad(6)$

The complex conjugate of the above wave function of free particle:

$\psi^{*}(x,t)= A e^{i\left(\omega t - k x \right)} \qquad(7)$

Eigenfunction and Eigen Value of linear momentum operator:

Now the momentum operator $\frac{\hbar}{i} \frac{\partial}{\partial x}$, operating on the equation $(6)$ i.e wave function of a free particle moving along the positive x-axis:

$\frac{\hbar}{i} \frac{\partial \psi(x,t)}{\partial x}= \frac{\hbar}{i} \frac{\partial }{\partial x} \left[ A e^{-i\left(\omega t - k x \right)} \right]$

$\frac{\hbar}{i} \frac{\partial \psi(x,t)}{\partial x}= \frac{\hbar}{i} \left[ A \left( ik \right)e^{-i\left(\omega t - kx \right)} \right]$

$\frac{\hbar}{i} \frac{\partial \psi(x,t)}{\partial x}= \hbar k \left[ A e^{-i\left(\omega t - kx \right)} \right]$

$\frac{\hbar}{i} \frac{\partial \psi(x,t)}{\partial x}= \hbar k \psi(x,t) \qquad(8)$

This equation shows that the wave function $\psi(x,t)$ for the particle is an eigenfunction of the linear momentum operator, and the momentum $p_{x}$ is the eigenvalue of the operator. Hence the momentum remains sharp with the value $p_{x}$

Probability of finding the free particle:

The probability of finding the position a particle in the region between $x$ and $x+dx$ is given by

$P \: dx =\psi(x,t) \psi^{*}(x,t) dx \qquad(8)$

Now substitute the value of $\psi(x,t)$ and $\psi^{*}(x,t)$ from equation $(6)$ and equation $(7)$ in above equation $(8)$, then we get

$P \: dx = A^{2} dx \qquad(9)$

Therefore the probability density $P$ for the position of the particle with the definite value of momentum is constant over the x-axis, i.e., All positions of the particle are equally probable. This conclusion is also obtained from the principle of uncertainty.

According to the interpretation of the wave function, the probability of finding the particle somewhere in space must be equal to $1$. i.e

$\int_{-\infty}^{+\infty} \psi(x,t) \psi^{*}(x,t) dx=1 \qquad(9)$

In this case $\psi(x,t) \psi^{*}(x,t) = A^{2}$ ,

There, the integral on the left side of the equation $(9)$ is infinite. Hence the wave function for the free particle cannot be normalized and $A$ must remain arbitrary. The difficulty arises because we are dealing with an ideal case. In practice, we can not have an absolutely free particle. The particle will always be confined within an enclosure in the laboratory, and hence its position can be determined. This means that its momentum cannot be determined with absolute accuracy.

### Numerical Aperture and Acceptance Angle of the Optical Fibre

Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater than the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of the incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less than the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of lig

### Fraunhofer diffraction due to a single slit

Let $S$ be a point monochromatic source of light of wavelength $\lambda$ placed at the focus of collimating lens $L_{1}$. The light beam is incident normally from $S$ on a narrow slit $AB$ of width $e$ and is diffracted from it. The diffracted beam is focused at the screen $XY$ by another converging lens $L_{2}$. The diffraction pattern having a central bright band followed by an alternative dark and bright band of decreasing intensity on both sides is obtained. Analytical Explanation: The light from the source $S$ is incident as a plane wavefront on the slit $AB$. According to Huygens's wave theory, every point in $AB$ sends out secondary waves in all directions. The undeviated ray from $AB$ is focused at $C$ on the screen by the lens $L_{2}$ while the rays diffracted through an angle $\theta$ are focussed at point $p$ on the screen. The rays from the ends $A$ and $B$ reach $C$ in the same phase and hence the intensity is maximum. Fraunhofer diffraction due to

### Electromagnetic wave equation in free space

Maxwell's Equations: Maxwell's equation of the electromagnetic wave is a collection of four equations i.e. Gauss's law of electrostatic, Gauss's law of magnetism, Faraday's law of electromotive force, and Ampere's Circuital law. Maxwell converted the integral form of these equations into the differential form of the equations. The differential form of these equations is known as Maxwell's equations. $\overrightarrow{\nabla}. \overrightarrow{E}= \frac{\rho}{\epsilon_{0}}$ $\overrightarrow{\nabla}. \overrightarrow{B}=0$ $\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$ $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}$ Modified Form: $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \left(\overrightarrow{J}+ \epsilon \frac{ \partial \overrightarrow{E}}{\partial t} \right)$ For free space