Derivation of time independent Schrodinger wave equation

Time independent Schrodinger wave equation:

We know the time dependent Schrodinger wave equation:

$i \hbar \frac{\partial \psi(x,t)}{\partial t}= -\frac{\hbar^{2}}{2m} \frac{\partial^{2} \psi(x,t)}{\partial x^{2}}+ V(x) \psi(x,t) \qquad(1)$

The wave function $\psi(x,t)$ is the product of space function $\psi(x)$ and time function $\psi(t)$. So

$\psi(x,t)=\psi(x) \psi(t) \qquad (2)$

Now apply the wave function form of equation$(2)$ to time dependent Schrodinger wave equation $(1)$

$i \hbar \psi(x) \frac{d \psi(t)}{d t}= -\frac{\hbar^{2}}{2m} \psi(t) \frac{d^{2} \psi(x)}{d x^{2}}+ V(x) \psi(x) \psi(t) \qquad(3)$

In the above equation $(3)$ ordinary derivatives is used in place of partial derivatives because each of function $\psi(x)$ and $\psi(t)$ depends on only one variable.

Now divide the above equation $(3)$ by $\psi(x)\psi(t)$ so

$i \hbar \frac{1}{\psi(t)} \frac{d \psi(t)}{d t}= -\frac{\hbar^{2}}{2m} \frac{1}{\psi(x)} \frac{d^{2} \psi(x)}{d x^{2}}+ V(x) \qquad(4)$

The above equation is known as the separation of time-independent part and time-independent part of the wave equation. The time-independent part is known as the energy function operator. i.e

$E=i \hbar \frac{1}{\psi(t)} \frac{d \psi(t)}{d t} \qquad(5)$

So from equation $(4)$ and equation$(5)$

$E= -\frac{\hbar^{2}}{2m} \frac{1}{\psi(x)} \frac{d^{2} \psi(x)}{d x^{2}}+ V(x)$

$E \psi(x)= -\frac{\hbar^{2}}{2m} \frac{d^{2} \psi(x)}{d x^{2}}+ V(x) \psi(x)$

$\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0$

This is time-independent Schrodinger wave equation.

Now for a free particle i.e, there is no force acting on the particle then the potential energy of a particle will be zero i.e. $V(x)=0$. Therefore time independent Schrodinger equation can be written as:

$\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2mE}{\hbar^{2}}\psi(x)=0$

This is time-independent Schrodinger wave equation for a free particle.

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Derivation of time dependent Schrodinger's wave equation

Time-dependent Schrodinger wave equation:

Let a particle of mass $m$ is moving along the positive $x$-direction. So the total energy $E$ of the particle is:

$E=\frac{1}{2}mv^{2}+V(x)$

$E=\frac{(mv)^{2}}{2m}+V(x)$

$E=\frac{(P_{x}^{2})^{2}}{2m}+V(x) \qquad(1)$

Since moving particles are associated with the wave function $\psi(x,t)$. So multiply $\psi(x,t)$ on both sides of equation$(1)$

$E\psi(x,t) =\frac{(P_{x}^{2})^{2}}{2m} \psi(x,t)+V(x) \psi(x,t) \qquad(2)$

The wave function $\psi(x,t)$ representing the plane wave associated with the particle is given by:

$\psi(x,t)=A e^{\frac{i}{\hbar}(P_{x}.x-Et)} \qquad(3)$

Differentiate with respect to $x$ the above equation $(3)$

$\frac{\partial \psi(x,t)}{\partial x}= A e^{\frac{i}{\hbar}(P_{x}.x-Et)} (\frac{i}{\hbar})P_{x} \qquad(4)$

Again differentiate the above equation$(4)$

$\frac{\partial^{2} \psi(x,t)}{\partial x^{2}}= A e^{\frac{i}{\hbar}(P_{x}.x-Et)} (\frac{i}{\hbar})^{2}P_{x}^{2}$

$\frac{\partial^{2} \psi(x,t)}{\partial x^{2}}= - \frac{P_{x}^{2}}{\hbar ^{2}} \psi(x,t)$

$P_{x}^{2}\psi(x,t) =- \hbar^{2} \frac{\partial^{2} \psi(x,t)}{\partial x^{2}} \qquad(5)$

Now differentiate the equation $(3)$ with respect to $t$

$\frac{\partial \psi(x,t)}{\partial t}= A e^{\frac{i}{\hbar}(P_{x}.x-Et)} (\frac{i}{\hbar})(-E)$

$\frac{\partial \psi(x,t)}{\partial t}= - (\frac{i}{\hbar})E\psi(x,t)$

$\frac{\partial \psi(x,t)}{\partial t}= - (\frac{i}{\hbar})E\psi(x,t)$

$ E\psi(x,t)= -(\frac{\hbar}{i}) \frac{\partial \psi(x,t)}{\partial t}$

$ E\psi(x,t)= i^{2}(\frac{\hbar}{i}) \frac{\partial \psi(x,t)}{\partial t} \qquad (\because i^{2}=-1)$

$ E\psi(x,t)= i \hbar \frac{\partial \psi(x,t)}{\partial t} \qquad(6)$

Now substitute the value $E\psi(x,t)$ and $P_{x}^{2} \psi(x,t)$ in equation $(2)$. Then

$i \hbar \frac{\partial \psi(x,t)}{\partial t}= -\frac{\hbar^{2}}{2m} \frac{\partial^{2} \psi(x,t)}{\partial x^{2}}+ V(x) \psi(x,t)$

Thus above equation is time-dependent Schrodinger equation. Another form of the above equation can be written as:

$(i \hbar \frac{\partial}{\partial t}) \psi(x,t)= [-\frac{\hbar^{2}}{2m} \frac{\partial^{2}}{\partial x^{2}}+ V(x) ] \psi(x,t)$

Where
$i \hbar \frac{\partial}{\partial t}$→ Energy operator (E)

$-\frac{\hbar^{2}}{2m} \frac{\partial^{2}}{\partial x^{2}}+ V(x)$→ Hamiltonian operator(H)

Now above equation:

$E \psi(x,t)=H \psi(x,t)$

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Significance of Compton effect

Description of Significance of Compton effect:

There are the following significance of the Compton effect→

1. The greatest significance of the Compton effect is that is to provide final and deciding proof for Planck-Einstein's visualization of the quantum nature of radiation. The particle nature of light was established after the discovery of the Compton effect.
2. The discovery of the Compton effect led to the formulation of quantum mechanics by W. Heisenberg and E. Schrodinger and provided the basis for the beginning of the theory of quantum electrodynamics.
3. It is most important to radiobiology, as it happens to be the most probable interaction of high energy x-ray with atomic nuclei in living beings and is applied in radiation therapy.
4. It is used to prove the wave function of electrons in the matter in the momentum representation.
5. It is the most effective in Gamma spectroscopy that gives rise to Compton edge, as it is possible for gamma rays to scatter out of the detectors used.
6. The Compton effect has played a significant role in diverse scientific areas such as nuclear engineering, experimental and theoretical nuclear physics, atomic physics, plasma physics, x-ray crystallography, etc.
7. The Compton effect provides an important research tool in some branches of medicine, including molecular chemistry, solid-state physics, etc.
8. The Compton effect has an appropriate application in the measurement of lungs density in living organisms.
9. The Compton effect is useful in putting large detectors in orbit above the earth's atmosphere.
10. The development of a high-resolution semiconductor radiation detector opened a new area for the application of Compton scattering.

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