Classical world and Quantum world

Classical world vs Quantum world:

The classical world and the quantum world are two fundamentally different ways of describing the behavior of matter and energy.

In the classical world, the laws of physics are described by classical mechanics, which is based on the concepts of position, velocity, and acceleration of objects. Classical mechanics is deterministic, meaning that if you know the initial conditions of a system, you can predict its future behavior with complete accuracy. This is the world we experience in our everyday lives, and it is characterized by a continuous, smooth flow of events.

In contrast, the quantum world is described by quantum mechanics, which is based on the behavior of particles on a subatomic scale. In the quantum world, particles do not have well-defined positions and velocities but rather exist in a superposition of many possible states. Moreover, measurements of quantum particles do not give deterministic results, but rather give probabilities of various outcomes. This probabilistic nature of quantum mechanics is known as the uncertainty principle.

Another important feature of the quantum world is entanglement, which occurs when two particles become linked in such a way that the state of one particle depends on the state of the other particle, even if they are separated by large distances. This has important implications for the way we understand the nature of reality itself.

While the classical and quantum worlds may seem very different, they are not entirely separate from each other. Classical mechanics can be seen as an approximation of quantum mechanics for macroscopic objects, and quantum mechanics can be used to explain phenomena that cannot be explained by classical mechanics alone.

Overall, the classical world and the quantum world are both valid ways of describing the behavior of matter and energy, and they each have their own unique properties and characteristics.

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Intensity of a wave

Definition of Intensity of a wave:

In a medium, the energy per unit area per unit time delivered perpendicuar to the direction of the wave propagation s caled the intensity of the wave. It is denoted by $I$.

If the energy $E$ is delivered in the time $t$ rom area $A$ perpendicular to the wave propagation, then

$I=\frac{E}{At} \qquad{1}$

Unit: $Joule/m^{2}-sec$ or $watt/m^{2}$

Dimensional formula: $[MT^{-3}]$

We know that the total mechanical energy of a vibrating particle is

$E=\frac{1}{2}m \omega^{2} a^{2}$

Where $\omega$ is the angular frequency and $a$ is the amplitude of the wave.

$E=\frac{1}{2}m (2\pi n)^{2} a^{2} \qquad \left( \omega=2\pi n \right)$

$E=2 \pi^{2} m n^{2} a^{2} \qquad(2)$

Where $m$ is the mass of the vibrating particle.

Now substitute the value of $E$ from equation $(2)$ to equation $(1)$. So the intensity of the wave

$I=\frac{2 \pi^{2} m n^{2} a^{2}}{At} \qquad(3)$

If the wave travels the distance $x$ in time $t$ with velocity $v$, Then

$t=\frac{x}{v}$

Now substitute the value of the above equation in equation $(3)$

$I=\frac{2 \pi^{2} m v n^{2} a^{2}}{Ax}$

$I=\frac{2 \pi^{2} m v n^{2} a^{2}}{V}$

Where $V$ is the volume of the corresponding medium during the wave propagation in time $t$.

$I=2 \pi^{2} \rho v n^{2} a^{2} \qquad \left(\because \rho=\frac{m}{V} \right)$

It is clear that for wave propagation in a medium with a constant velocity, i.e. wave's intensity is directly proportional to the square of amplitude and frequency both.

$I\propto a^{2}$ and $I \propto n^{2}$

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Difference between sound waves and light waves

Sound Waves:

Sound waves are mechanical waves in nature.

They need a medium to propagate and so cannot be produced in a vacuum.

They can move in all types of mediums as solid liquid or gas whether they are transparent or not.

They propagate in the form of longitudinal waves.

The particle of the medium vibrates along the direction of the propagation and so contraction and rarefaction are formed there.

These are three-dimensional waves.

These waves do not show a polarization effect.

For a normal human being the audible frequency range is $20$ to $20000 Hertz$.

The speed of the sound waves is more in a dense medium than in a rare medium.

Light waves:

light waves are electromagnetic waves in nature.

They don't need any medium and so can produce and propagate in a vacuum.

Their velocity in a vacuum is the maximum of value $3 \times 10^{8} m/s$

They can move in a transparent medium only.

They propagate in the form of transverse waves.

The electric field and magnetic field vibration are perpendicular to the direction of wave propagation.

These wave waves are also three-dimensional waves.

These waves source the polarization effect.

For a normal human being the visible frequency range is $4 \times 10^{18} Hz$ to $8 \times 10^{14} Hz$.

The speed of light is more in rare mediums than in dense ones.

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Momentum wave function for a free particle

A non-relativistic free particle of mass $m$ moving in the positive $x$-direction with speed $v_{x}$ has kinetic energy

$E=\frac{1}{2} m v^{2}_{x}$

and momentum

$p_{x}=mv_{x}$

The energy and momentum are associated with a wave of wavelength $\lambda$ and frequency $\nu$ given by

$\lambda = \frac{h}{p_{x}}$

and

$\nu=\frac{E}{h}$

The propagation constant $k_{x}$ of the wave is

$k_{x}= \frac{2\pi}{\lambda}=\frac{2\pi}{\left(\frac{h}{p_{x}} \right)}=\frac{p_{x}}{\left(\frac{h}{2\pi} \right)}=\frac{p_{x}}{\hbar}$

and the angular frequency $\omega$ is

$\omega = 2\pi \nu = \frac{2\pi E}{\hbar}=\frac{E}{\hbar}$

A plane wave traveling along the $x$ axis in the positive direction may be represented by

$\psi(x,t)=A e^{-i\left(k_{x} \: x - \omega t\right)}$

Now subtitute the value of $\omega$ and $k_{x}$ in above equation then we get

$\psi(x,t)=A e^{i\left( \frac{p_{x}}{\hbar} \: x - \frac{E}{\hbar} \: t\right)}$

$\psi(x,t)=A e^{\frac{i}{\hbar}\left( p_{x} \: x - E \: t\right)}$

The superposition of a number of such waves of propagation number slightly different from an average value traveling simultaneously along the same line in the positive $x$- direction forms a wave packet of small extension. By Fourier's theorem the eave packet may be expressed by

$\psi(x,t) = \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{+\infty} A (p_{x}) e^{\frac{i}{\hbar}\left( p_{x} \: x - E \: t \right)} \: \: dp_{x} \qquad(1)$

The function $\psi(x,t)$ is called the momentum wave function for the motion of the free particle in one dimension.

The amplitude $A(p_{x})$ of the $x$-component of the momentum is given by the Fourier tranform

$A(p)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{+\infty} \psi (x,t) e^{-\frac{i}{\hbar}\left( p_{x} \: x - E \: t \right)} \: \: dx \qquad(2)$

In three dimension the wave function is represented by

$\psi(\overrightarrow{r},t) = \frac{1}{(2 \pi \hbar)^{3/2}} \int_{-\infty}^{+\infty} A (\overrightarrow{p}) e^{\frac{i}{\hbar}\left( \overrightarrow{p} . \overrightarrow{r} - E \: t \right)} \: \: d^{3}\overrightarrow{p} \qquad(3)$

Where $d^{3}\overrightarrow{p}=dp_{x} \: dp_{y} \: dp_{z}$ is the volume element in the momentum space. In equation $(1)$, equation $(2)$ and equation $(3)$ $\frac{1}{\sqrt{2 \pi \hbar}}$ and $\frac{1}{(2 \pi \hbar)^{3/2}}$ are normalization constants.

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Schrodinger's equation for the complex conjugate waves function

Derivation:

The time dependent Schrodinger quation for the wave function function $psi(x,y,z,t)$ is

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

Since wave function, $\psi$ is complex quantity i.e.

$\psi=\psi_{1}+i \: \psi_{2} \qquad(2)$

Where $\psi_{1}$ and $\psi_{2}$ are real functions of $x,y,z,t$. Substituting this form for $\psi$ in equation $(1)$, we get

$-\frac{\hbar^{2}}{2m}\nabla^{2} \left( \psi_{1}+i \: \psi_{2} \right) + V\left( \psi_{1}+i \: \psi_{2} \right) \\ \qquad = i\hbar\frac{\partial }{\partial t} \left( \psi_{1}+i \: \psi_{2} \right)$

Equation real and imaginary parts on either side of this equation, we obtain the following two equations:

$-\frac{\hbar^{2}}{2m}\nabla^{2} \psi_{1} + V\psi_{1}=-\hbar\frac{\partial \psi_{2}}{\partial t} \qquad(3)$

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

Mutiplying equation $(4)$ by $-i$ and adding it to equation $(3)$, we get

$-\frac{\hbar^{2}}{2m}\nabla^{2} \left( \psi_{1} - i \: \psi_{2} \right) + V\left( \psi_{1} - i \: \psi_{2} \right) \\ \qquad = -i\hbar\frac{\partial }{\partial t} \left( \psi_{1} - i \: \psi_{2} \right) \qquad(5)$

The complex conjugate of wave function $\psi^{*}$ is

$\psi^{*}=\psi^{*}_{1} - i \: \psi^{*}_{2} \qquad(6)$

Therefore, The equation $(5)$ can be written as

$-\frac{\hbar^{2}}{2m}\nabla^{2} \psi^{*} + V \psi^{*} =-i \hbar \frac{\partial \psi^{*}}{\partial t}$

This is the equation for complex conjugate wave function $\psi^{*}$.

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Probability Current Density for a free particle in Quantum Mechanics

1.) Derivation of Probability Current Density for a free particle:

Let a particle of mass $m$ is moving in the positive $x$- direction in the region from $x_{1}$ to $x_{2}$.

For the one-dimensional motion of the particle, the wave function is $psi(x,t)$ Let $dA$ be the area of the cross-section of the region.

The probability of finding a particle in the region is

$\int_{x_{1}}^{x_{2}} \psi(x,t) \: \psi^{*}(x,t) \: dx \: dA \qquad(1)$

and the probability density of finding the particle in the region is

$P=\psi(x,t) \: \psi^{*}(x,t) \qquad(2)$

If the probability of finding the particle in the region decreases with time, the rate of decrease of the probability that the particle is in the region from $x_{1}$ to $x_{2}$ per unit area is called the probability current density out of the region. Therefore, the probability current density $S_{2} - S_{1}$ out of the region in the positive $x$-direction is given by

$S_{2} - S_{1} = - \frac{1}{dA} \left[- \frac{d}{dt} \int_{x_{1}}^{x_{2}} P \: dx \: dA \right]$

$S_{2} - S_{1} = - \frac{\partial}{\partial t} \int_{x_{1}}^{x_{2}} P \: dx $

$S_{2} - S_{1} = - \frac{\partial}{\partial t} \int_{x_{1}}^{x_{2}} \psi(x,t) \: \psi^{*}(x,t) \: dx \qquad(3)$

And the probability of current density at position $x$ is

$S = - \frac{\partial}{\partial t} \int \psi(x,t) \: \psi^{*}(x,t) \: dx \qquad(4)$

1.1) Show that: $S = \frac{i \hbar}{2m} \frac{\partial}{\partial x} \left[ \psi^{*} \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^{*} }{\partial x} \right]$

Proof:

According to the Schrodinger equation for wave function $\psi(x,t)$ and $\psi^{*}(x,t)$ are

$i \hbar \frac{\partial \psi}{\partial t} =- \frac{\hbar^{2}}{2m} \frac{\partial^{2} \psi}{\partial x^{2}} + V \psi \qquad(1.1.1)$

The complex conjugate of the wave function

$-i \hbar \frac{\partial \psi^{*}}{\partial t} = -\frac{\hbar^{2}}{2m} \frac{\partial^{2} \psi^{*}}{\partial x^{2}} + V \psi^{*} \qquad(1.1.2)$

Multiplying equation $(1.1.1)$ by $\psi^{*}$ and equation $(1.1.2)$ by $\psi$, we get

$i \hbar \psi^{*} \frac{\partial \psi}{\partial t} =- \frac{\hbar^{2}}{2m} \psi^{*} \frac{\partial^{2} \psi}{\partial x^{2}} + \psi^{*} V \psi \quad(1.1.3)$

$-i \hbar \psi \frac{\partial \psi^{*}}{\partial t} = -\frac{\hbar^{2}}{2m} \psi \frac{\partial^{2} \psi^{*}}{\partial x^{2}} + \psi V \psi^{*} \quad(1.1.4)$

Now subtracting equation $(1.1.4)$ and equation $(1.1.3)$, we get

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

$i \hbar \frac{\partial}{\partial t} \left( \psi \psi^{*} \right) =-\frac{\hbar^{2}}{2m} \frac{\partial}{\partial x} \left[ \psi^{*} \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^{*}}{\partial x} \right]$

$ \frac{\partial}{\partial t} \left( \psi \psi^{*} \right) =\frac{i\hbar}{2m} \frac{\partial}{\partial x} \left[ \psi^{*} \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^{*}}{\partial x} \right] \quad(1.1.5)$

We know that

$S = - \frac{\partial}{\partial t} \int \psi(x,t) \: \psi^{*}(x,t) \: dx $

Now substitute the value of equation $(9)$ in the above equation that can be written as

$ S = -\frac{i\hbar}{2m} \int \frac{\partial}{\partial x} \frac{\partial}{\partial x} \left[ \psi^{*} \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^{*}}{\partial x} \right] dx $

$ S = -\frac{i\hbar}{2m} \left[ \psi^{*} \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^{*}}{\partial x} \right] $

1.2) Show That The probability current density for a free particle is equal to the product of its probability density and its speed.

Proof:

For a free particle that is moving in the positive $x$-axis direction and the momentum $p_{x}$ at position $x$ is given by

$\frac{\hbar}{i} \frac{\partial \psi}{\partial x} = p_{x} \psi$

$ \frac{\partial \psi}{\partial x} = \frac{i}{\hbar} p_{x} \psi \qquad(1.2.1)$

and

$-\frac{\hbar}{i} \frac{\partial \psi^{*}}{\partial x} = p_{x} \psi^{*}$

$ \frac{\partial \psi^{*}}{\partial x} = - \frac{i}{\hbar} p_{x} \psi^{*} \qquad(1.2.2)$

We know that

$ S = -\frac{i\hbar}{2m} \left[ \psi^{*} \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^{*}}{\partial x} \right] $

Now substitute the value of equation $(1.2.1)$ and equation $(1.2.2)$ in the above equation, we get

$ S = -\frac{i\hbar}{2m} \left[ \psi^{*} \frac{i}{\hbar} p_{x} \psi + \psi \frac{i}{\hbar} p_{x} \psi^{*}\right] $

$ S = \frac{1}{2m} \left[ \psi^{*} p_{x} \psi + \psi p_{x} \psi^{*}\right] $

$ S = \frac{1}{m} \left( \psi \psi^{*} p_{x} \right)$

$ S = \frac{ p_{x} }{m} \left( \psi \psi^{*}\right) \qquad(1.2.3) $

Now put $p_{x}= m v_{x}$ in equation $(1.2.3)$

$ S = \frac{m v_{x} }{m} \left( \psi \psi^{*}\right)$

$ S = \left( \psi \psi^{*}\right) v_{x}$

Now put $p_{x}= \hbar k_{x}$ in equation $(1.2.3)$

$ S = \frac{ \hbar \: k_{x} }{m} \left( \psi \psi^{*}\right) $

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