Postulate of wave mechanics or Quantum Mechanics

Postulate of Wave or Quantum Mechanics (or Operator formalism in Quantum mechanics) →

The formulation of mathematical equations of quantum mechanics is based on the linear operator. This operator formulation of quantum mechanics is known as postulates of quantum mechanics. These postulates are given below:-

1. For a system consisting of particles moving in a field of a conservative force, there is an associated complex wave function $\psi(x, y, z, t)$ where $x$,$y$,$z$ space coordinates, and $t$ is the time. This wave function enables us to obtain a description of the behavior of the system, consistent with the principle of uncertainty.


2. There is an operator with every observable dynamical quantity. The operator corresponding to the pertinent dynamical quantities is:-
   
   
   

   Dynamical Variable	Symbol	Quantum Mechanical Operator
   Position	$x$ $y$ $z$	$x$ $y$ $z$
   Momentum	$P_{x}$ $P_{y}$ $P_{z}$ Generalised Form $\overrightarrow{P}$	$\frac{\hbar}{i}\frac{\partial}{\partial x}$ $\frac{\hbar}{i}\frac{\partial}{\partial y}$ $\frac{\hbar}{i}\frac{\partial}{\partial z}$ Generalised Form $\frac{\hbar}{i}\overrightarrow{\nabla}$
   Total Energy	$E$	$i\hbar \frac{\partial}{\partial t}$
   Total Energy	$E$	$-\frac{\hbar^{2}}{2m} \nabla^{2}+V(x)$ This is also known as Hamiltonian Operator $H$.
   Kinetic Energy	$K$	$-\frac{\hbar^{2}}{2m} \nabla^{2}$
   Potential Energy	$V(x,y,z)$	$V(x,y,z)$

   
   
   All the operators have eigen functions and eigen values.


3. The wave function $\psi(x,y,z,t)$ and its partial derivatives $\frac{\partial \psi}{\partial x}$, $\frac{\partial \psi}{\partial y}$, $\frac{\partial \psi}{\partial z}$ must be finite, continuous and single-valued for all values of $x$, $y$, $z$ and $t$


4. The product of $\psi(x,y,z,t)$ and $\psi^{*}(x,y,z,t)$ is always a real quantity. The product is called the probability density and $\psi\psi^{*} d\tau $ is interpreted as a probability that the particle will be found in volume element $d\tau$ at $x$,$y$,$z$ and time $t$. Since the total probability of finding particles somewhere in the entire space must be equal to 1.
   
   
   $\int_{-\infty}^{+\infty} \psi \: \psi^{*}\: d\tau = 1$
   
   
   The integral is taken overall space.


5. The average or expectation value of an observable quantity $\alpha$ with which an operator $\hat{\alpha}$ is associated is defined by

$\left< \alpha \right> = \int_{-\infty}^{+\infty} \psi^{*} \hat{\alpha} \psi d\tau$


The integral being taken overall space.

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Eigenfunction, Eigenvalues and Eigenvectors

Eigenfunction and Eigenvalues → If $\psi$ is a well-behaved function, then an operator $\hat{P}$ may operate on $\psi$ in two different ways depending upon the nature of function $\psi$ `

1. When an operator $\hat{P}$ operates on any function $\psi$ then this function $\psi$ changes into another function $\phi$. i.e.
   
   
   $\hat{P} \psi =\phi$
   
   
   Where $\phi$ is a new function linearly depending upon the initial function $\psi$.
   
   
   Example:
   
   
   Let us consider a function $f(x)=x^{2}$ and an operator ie. differential operator $\frac{d}{dx}$ is operate on the function. Then we get
   
   
   $\frac{d}{dx}f(x)= \frac{d}{dx} (x^{2})$
   
   
   $\frac{d}{dx}f(x)= 2x$
   
   
   Now the given function $f(x)=x^{2}$ change into another function $f(x)=x$.


2. When an operator $\hat{P}$ operates on any function $\psi$ then this function $\psi$ does not change into another function but now this function $\psi$ may be with multiples of complex or real numbers(or values).i.e
   
   
   $\hat{P} \psi =\lambda \phi$
   
   
   Where $\lambda$ is Real OR Complex Number. This number or value is known as Eigenvalues.
   
   
   In this case, the function $\psi$ is a member of the class of physically meaningful functions called the eigen function of the operator $\hat{P}$. The number $\lambda$ is called the eigen value of operator $\hat{P}$ associated with eigen function $\psi$ and this equation is known as the eigenvalue equation.
   
   
   Example: Let us consider a function $f(x)=e^{2x}$ and an operator ie. differential operator $\frac{d}{dx}$ is operate on the function. Then we get
   
   
   $\frac{d}{dx}f(x)= \frac{d}{dx} (e^{2x})$
   
   
   $\frac{d}{dx}f(x)= 2e^{2x}$
   
   
   Now the given function $f(x)=e^{2x}$ change into another function $f(x)=x$.

Eigenvalues and Eigenvectors →

Let a linear transformation equation

$AX=\lambda X \qquad(1)$

Where
A → Square matrix of $n$ order (where $n=1,2,3,.....$)
$\lambda$ → Scalar Factor

The equation $(1)$ may be written as

$AX=I\lambda X$

$AX-I\lambda X=0$

$(A-I\lambda)X=0 \qquad(2)$

Where $I$ is unit matrix

Any value of $\lambda$ for which equation $(1)$ or equation $(2)$ has non zero (i.e $X \neq 0$) solution is called eigenvalues or characteristic roots or latent root of the matrix $A$ and corresponding non zero solution of $X$ is called eigenvectors or characteristic vectors or latent vectors of the matrix $A$

The matrix $\left| A- \lambda I \right|$ is called characteristic matrix of $A$. The determinant $\phi(\lambda) = \left| A- \lambda I \right|$ is called the characteristic polynomial of $A$. So

$\phi(\lambda) = \left| A- \lambda I \right| =0$

And $\phi(x)= a_{0}+a_{1} \lambda a_{2} \lambda^{2}+ .......+a_{n} \lambda^{n}=0 $

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Physical interpretation of the wave function

We have assumed that the wave associated with a particle in motion is represented by a complex variable quantity called the wave function $\psi(x,t)$. Therefore, it can not have a direct physical meaning. Since it is a complex quantity, it may be expressed as

$\psi(x,y,z,t)=a+ib \qquad(1)$

Where $a$ and $b$ are real functions of the variable $(x,y,z,t)$. The complex conjugate of wave function $\psi(x,y,z,t)$

$\psi^{*}(x,y,z,t)=a-ib \qquad(2)$

Multiply equation $(1)$ and equation $(2)$

$\psi(x,y,z,t).\psi^{*}(x,y,z,t)=a^{2}+b^{2} \qquad(3)$

$ \left| \psi(x,y,z,t) \right|^{2}=a^{2}+b^{2} \qquad(4)$

If $\psi \neq 0$ Then the product of $\psi$ and $\psi^{*}$ is real and positive. Its positive square root is denoted by $\left|\psi(x,y,z,t) \right|$, and it is called the modulus of $\psi$.

The quantity $ \left| \psi(x,y,z,t) \right|^{2}$ is called the probability density $(P)$. So for the motion of a particle, the probability of finding the particle in the region $d\tau$ will be:

$\int {P d\tau}= \int {\psi(x,y,z,t).\psi^{*}(x,y,z,t).d\tau}=\int {\left| \psi(x,y,z,t) \right|^{2}d\tau}$

Here $P$ are the probability that tells us that the particle will be found in a volume element $d\tau(=dx.dy.dz)$ surrounding the point at position $(x,y,z)$ at time $t$.

For the motion of a particle in one dimension, the probability of finding the particle in the region $dx$ will be:

$\int{P dx}= \int {\psi(x,t).\psi^{*}(x,t).dx}=\int {\left| \psi(x,t) \right|^{2}dx}$

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Particle in one dimensional box (Infinite Potential Well)

Let us consider a particle of mass $m$ that is confined to one-dimensional region $0 \leq x \leq L$ or the particle is restricted to move along the $x$-axis between $x=0$ and $x=L$. Let the particle can move freely in either direction, between $x=0$ and $x=L$. The endpoints of the region behave as ideally reflecting barriers so that the particle can not leave the region. A potential energy function $V(x)$ for this situation is shown in the figure below.

Particle in One-Dimensional Box(Infinite Potential Well)

The potential energy inside the one -dimensional box can be represented as

$\begin{Bmatrix} V(x)=0 &for \: 0\leq x \leq L \\ V(x)=\infty & for \: 0> x > L \\ \end{Bmatrix}$

$\frac{\partial^{2} \psi(x)}{\partial x^{2}}+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0 \qquad(1)$

If the particle is free in a one-dimensional box, Schrodinger's wave equation can be written as:

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

$\frac{\partial^{2} \psi(x)}{\partial x^{2}}+\frac{8 \pi^{2} mE}{h^{2}}\psi(x)=0 \quad (\because \hbar=\frac{h}{2 \pi}) \quad(2)$

$\frac{\partial^{2} \psi(x)}{\partial x^{2}}+ k^{2}\psi(x)=0 \quad (\because k^{2}=\frac{8 \pi^{2} mE}{h^{2}}) \quad(3)$

The general solution of the above differential equation $(2)$

$\psi(x)= A sin(kx)+ B cos(kx) \qquad(4)$

The wave function $\psi(x)$ should be zero everywhere outside the box since the probability of finding the particle outside the box is zero. Similarly, the wave function $\psi(x)$ must also be zero at walls of the box because the probability density $[\psi(x)]^{2}$ must be continuous. Thus, the boundary conditions for this problem is that

(i) $\psi(x)=0$ For $x=0$

(ii) $\psi(x)=0$ For $x=L$

Now applying the boundary condition in equation$(4)$ i.e.

(i) At $x=0$ the wave function $\psi(0)=0$

Now we get

$\psi(0)= A sin(k.0)+ B cos(k.0)$

$A sin(k.0)+ B cos(k.0)=0 \qquad (\because \psi(0)= 0)$

$B=0$

Hence substitute the value of $B$ in equation$(4)$ ,

$\psi(L)= A sin(kx) \qquad(5)$

Now applying the second boundary condition:

(ii) At $x=L$ the wave function $\psi(L)=0$, we get

$\psi(x)= A sin(kL) \qquad(6)$

This equation will satisfy only for certain values of $k$, say $k_{n}$. Since $A$ can not be taken zero hence

$sin(k_{n}L)=0 $

$sin(k_{n}L)=sin(n\pi) $

$k_{n}L=n\pi $

$k_{n}=\frac{n\pi}{L} \qquad(7)$

Thus for each allowed values of $k_{n}$ there is a wave function $\psi(x)$ given as, using equation$(5)$ and equation$(7)$

$\psi_{n}(x)=A sin(\frac{n\pi x}{L})$

This is the expression of the wave function or eigen function for a particle in a box.

Now, from equation $(3)$ and equation$(7)$, we get

$k^{2}=\frac{8 \pi^{2} mE}{h^{2}}= (\frac{n \pi}{L})^{2}$

$E=\frac{n^{2} h^{2}}{8mL^{2}}$

This is the expression of energy or eigen value for a particle in a box.

In general, the expression for this energy is written as:

$E_{n}=\frac{n^{2} h^{2}}{8mL^{2}}$

For different values of $n$ energy values can be written as

For $n=1$

$E_{1}=\frac{h^{2}}{8mL^{2}}$

It is known as zero-point energy or ground energy state

For $n=2$

$E_{2}=\frac{2^{2} h^{2}}{8mL^{2}}=2^{2}E_{1}$

For $n=3$

$E_{3}=\frac{3^{2} h^{2}}{8mL^{2}}=3^{2}E_{1}$

For $n=4$

$E_{4}=\frac{4^{2} h^{2}}{8mL^{2}}=4^{2}E_{1}$

So generalized form of the above equation can be written as

$E_{n}=n^{2}E_{1}$

Some of the possible energies for a particle in a box are shown on an energy-level diagram in the figure below.

Possible Energies for a particle in a box

The energy levels have a spacing that increases with increasing $n$ and thus the particle in a box can take only certain discrete energy values, called Eigen-values. This means that the energy levels of a particle in a box are quantized but according to classical mechanics, the particle may take any continuous range of energy values between zero and infinity.

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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-independent Schrodinger wave equation $(1)$

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

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

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

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

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

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

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

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

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

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:

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

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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