Ehrenfest's Theorem and Derivation

Ehrenfest's Theorem Statement:

The theorem states that

Quantum mechanics gives the same results as classical mechanics for a particle for which average or expectation values of dynamical quantities are involved.

Proof of theorem:

The proof of the theorem for one-dimensional motion of a particle by showing that

1) $\frac{d \left < x \right >}{dt} = \frac{\left < p_{x} \right > }{m}$

2) $\frac{d \left < p_{x} \right >}{dt} = \left < F_{x} \right >$

1.) To Show that: $\frac{d \left < x \right > }{dt} = \frac{\left < p_{x} \right > }{m}$

Let $x$ is the position coordinate of a particle of mass $m$, at time $t$

The expectation value of position $x$ of a particle is given by

$\left < x \right > = \int_{- \infty}^{+ \infty} \psi^{*} (x,t) . x \: \psi (x,t) dx \qquad (1)$

Differentiating the above equation $(1)$ with respect to $t$

$\frac{d \left < x \right > }{dt} = \int_{- \infty}^{+ \infty} x \frac{\partial (\psi \psi^{*})}{\partial t} dx \qquad(2)$

We know the probablity current density

$\frac{\partial (\psi \psi^{*})}{\partial t} = \frac{i \hbar}{2m} \frac{\partial}{\partial x} \left[ \psi^{*} \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^{*} }{\partial x} \right] \qquad(3)$

Now substitute the above eqaution$(3)$ in eqaution $(2)$

$\frac{d \left < x \right > }{dt} = \frac{i \hbar}{2m} \int_{- \infty}^{+ \infty} x \frac{\partial}{\partial x} \left[ \psi^{*} \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^{*} }{\partial x} \right] dx $

Integrating the right-hand side by parts of the above equation, we get

$\frac{d \left < x \right > }{dt} = \frac{i \hbar}{2m} \left[ x \left( \psi^{*} \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^{*} }{\partial x} \right) \right]^{+\infty}_{-\infty} \\ \qquad - \frac{i \hbar}{2m} \int_{- \infty}^{+ \infty} \left( \psi^{*} \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^{*} }{\partial x} \right) dx $

As $x$ approaches either $+ \infty$ or $-\infty$, $\psi$ and $\frac{\partial \psi}{\partial x}$ approach zero, and therefore the first term becomes zero.

Hence we get

$\frac{d \left < x \right > }{dt} = - \frac{i \hbar}{2m} \int_{- \infty}^{+ \infty} \left( \psi^{*} \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^{*} }{\partial x} \right) dx \\ \qquad\qquad\qquad\qquad\qquad\qquad ---(4)$

The expectation value of $p_{x}$ is given by

$ \left < p_{x} \right > = \int_{- \infty}^{+ \infty} \psi^{*} \frac{\hbar}{i} \frac{\partial \psi}{\partial x} $

$ \int_{- \infty}^{+ \infty} \psi^{*} \frac{\partial \psi}{\partial x} dx = \frac{i}{\hbar}\left < p_{x} \right > \qquad(5)$

Similarly

$ \int_{- \infty}^{+ \infty} \psi \frac{\partial \psi^{*}}{\partial x} dx = - \frac{i}{\hbar}\left < p_{x} \right > \qquad(6)$

Substituting the values of these integrals in equation $(4)$

$\frac{d \left < x \right > }{dt} = - \frac{i \hbar}{2m} \left[ \frac{i}{\hbar}\left < p_{x} \right > + \frac{i}{\hbar}\left < p_{x} \right >\right] $

$\frac{d \left < x \right > }{dt} = - \frac{\left < p_{x} \right >}{m} \qquad(7)$

This is the first result of Ehrenfest's Theorem.

2) To show that: $\frac{d \left < p_{x} \right >}{dt} = \left < F_{x} \right >$

We know that the expectation value of the momentum $p_{x}$ is given by

$ \left < p_{x} \right > = \int_{- \infty}^{+ \infty} \psi^{*} \frac{\hbar}{i} \frac{\partial \psi}{\partial x} $

$ \left < p_{x} \right > =\frac{\hbar}{i} \int_{- \infty}^{+ \infty} \psi^{*} \frac{\partial \psi}{\partial x} \qquad(8)$

Differentiating the equation $(8)$ with respect to $t$, we get

$\frac{d \left < p_{x} \right >}{dt} = \frac{\hbar}{i} \int_{- \infty}^{+ \infty} \left[ \frac{\partial \psi^{*}}{\partial t} \frac{\partial \psi}{\partial x} + \psi^{*} \frac{\partial^{2} \psi}{\partial x \partial t} \right] $

$\frac{d \left < p_{x} \right >}{dt} = \int_{- \infty}^{+ \infty} \left[-i \hbar \frac{\partial \psi^{*}}{\partial t} \frac{\partial \psi}{\partial x} - i\hbar \psi^{*} \frac{\partial^{2} \psi}{\partial x \partial t} \right] \\ \qquad\qquad\qquad\qquad\qquad\qquad ---(9)$

Now the time-dependent Schrodinger equations for $\psi$ and $\psi^{*}$ are

$i \hbar \frac{\partial \psi}{\partial t} =- \frac{\hbar^{2}}{2m} \frac{\partial^{2} \psi}{\partial x^{2}} + V \psi \qquad(10)$

The complex conjugate of Schrodinger function

$-i \hbar \frac{\partial \psi^{*}}{\partial t} = -\frac{\hbar^{2}}{2m} \frac{\partial^{2} \psi^{*}}{\partial x^{2}} + V \psi^{*} \qquad(11)$

Differentiating the equation $(10)$ with respect to $x$

$i \hbar \frac{\partial^{2} \psi}{\partial x \partial t} = - \frac{\hbar^{2}}{2m} \frac{\partial^{3} \psi}{\partial x^{3}} + \frac{\partial (V \psi)}{\partial x} \qquad(12)$

Now substitute the value of $-i \hbar \frac{\partial \psi^{*}}{\partial t}$ and $i \hbar \frac{\partial^{2} \psi}{\partial x \partial t}$ in the equation $(9)$, we get


$\frac{d \left < p_{x} \right >}{dt} = \int_{- \infty}^{+ \infty} \left[ \left( -\frac{\hbar^{2}}{2m} \frac{\partial^{2} \psi^{*}}{\partial x^{2}} + V \psi^{*} \right) \frac{\partial \psi}{\partial x} - \psi^{*} \left( - \frac{\hbar^{2}}{2m} \frac{\partial^{3} \psi}{\partial x^{3}} + \frac{\partial (V \psi)}{\partial x} \right) \right]$


$\frac{d \left < p_{x} \right >}{dt} = \int_{- \infty}^{+ \infty} \left[-\frac{\hbar^{2}}{2m} \left( \frac{\partial^{2} \psi^{*}}{\partial x^{2}}\frac{\partial \psi}{\partial x} - \psi^{*} \frac{\partial^{3} \psi}{\partial x^{3}} \right) - \left( V \psi^{*} \frac{\partial \psi}{\partial x} - \psi^{*} \frac{\partial (V \psi)}{\partial x} \right) \right] dx$


$\frac{d \left < p_{x} \right >}{dt} = -\frac{\hbar^{2}}{2m} \int_{- \infty}^{+ \infty} \left( \frac{\partial^{2} \psi^{*}}{\partial x^{2}}\frac{\partial \psi}{\partial x} - \psi^{*} \frac{\partial^{3} \psi}{\partial x^{3}} \right) dx + \int_{- \infty}^{+ \infty} \left( V \psi^{*} \frac{\partial \psi}{\partial x} - \psi^{*} \frac{\partial (V \psi)}{\partial x} \right) dx$


$\frac{d \left < p_{x} \right >}{dt} = -\frac{\hbar^{2}}{2m} \int_{- \infty}^{+ \infty} \frac{\partial}{\partial x} \left( \frac{\partial \psi^{*}}{\partial x}\frac{\partial \psi}{\partial x} - \psi^{*} \frac{\partial^{2} \psi}{\partial x^{2}} \right) dx + \int_{- \infty}^{+ \infty} \left( V \psi^{*} \frac{\partial \psi}{\partial x} - \psi^{*} \frac{\partial (V \psi)}{\partial x} \right) dx$

Now put $\frac{\partial (V \psi)}{\partial x}= \left\{ \psi \frac{\partial V }{\partial x}+ V\frac{\partial \psi}{\partial x} \right\}$ in above equation:


$\frac{d \left < p_{x} \right >}{dt} = -\frac{\hbar^{2}}{2m} \int_{- \infty}^{+ \infty} \frac{\partial}{\partial x} \left( \frac{\partial \psi^{*}}{\partial x}\frac{\partial \psi}{\partial x} - \psi^{*} \frac{\partial^{2} \psi}{\partial x^{2}} \right) dx + \int_{- \infty}^{+ \infty} \left( V \psi^{*} \frac{\partial \psi}{\partial x} - \psi^{*} \left\{ \psi \frac{\partial V }{\partial x}+ V\frac{\partial \psi}{\partial x} \right\} \right) dx$


$\frac{d \left < p_{x} \right >}{dt} = -\frac{\hbar^{2}}{2m} \left[ \frac{\partial \psi^{*}}{\partial x}\frac{\partial \psi}{\partial x} - \psi^{*} \frac{\partial^{2} \psi}{\partial x^{2}} \right]_{- \infty}^{+ \infty} + \int_{- \infty}^{+ \infty} \left( V \psi^{*} \frac{\partial \psi}{\partial x} - \psi^{*} \left\{ \psi \frac{\partial V }{\partial x}+ V\frac{\partial \psi}{\partial x} \right\} \right) dx$

As $x$ approaches either $+ \infty $ or $-\infty$ and $\frac{\partial \psi}{\partial x}$ is zero. Therefore the first term of the above equation on the right-hand side will be zero.


$\frac{d \left < p_{x} \right >}{dt} = \int_{- \infty}^{+ \infty} \left( V \psi^{*} \frac{\partial \psi}{\partial x} - \psi^{*} \left\{ \psi \frac{\partial V }{\partial x}+ V\frac{\partial \psi}{\partial x} \right\} \right) dx$


$\frac{d \left < p_{x} \right >}{dt} = \int_{- \infty}^{+ \infty} \left( V \psi^{*} \frac{\partial \psi}{\partial x} - \psi \frac{\partial V }{\partial x} \psi^{*} - V \psi^{*} \frac{\partial \psi}{\partial x} \right) dx$

$\frac{d \left < p_{x} \right >}{dt} = \int_{- \infty}^{+ \infty} - \psi \frac{\partial V }{\partial x} \psi^{*} dx$

$\frac{d \left < p_{x} \right >}{dt} = - \int_{- \infty}^{+ \infty} \psi \frac{\partial V }{\partial x} \psi^{*} dx$

$\frac{d \left < p_{x} \right >}{dt} = -\left < \frac{\partial V }{\partial x} \right > $

Here the $\left < \frac{\partial V }{\partial x} \right >$ is the average value or expectation value of potential gradient and the negative value of the potential gradient is equal to the average value or expectation value of force $\left < F_{x} \right >$ along the $x$ direction.

$\frac{d \left < p_{x} \right >}{dt} = \left < F_{x} \right > $

This is the second result of Ehrenfest theorem and it represents Newton's second law of motion. Thus if the expectation values of dynamical quantities for a particle are, considered, quantum mechanics given the equations of classical mechanics.

Eigen value of the momentum of a particle in one dimension box or infinite potential well

Equation of eigen value of the momentum of a particle in one dimension box:

The eigen value of the momentum $P_{n}$ of a particle in one dimension box moving along the x-axis is given by

$P^{2}_{n} = 2 m E_{n}$

$P^{2}_{n} = 2 m \frac{n^{2} \pi^{2} \hbar^{2}}{2 m L^{2}} \qquad \left( \because E_{n}= \frac{n^{2} \pi^{2} \hbar^{2}}{2 m L^{2}} \right)$

$P^{2}_{n} = \frac{n^{2} \pi^{2} \hbar^{2}}{L^{2}}$

$P_{n} = \pm \frac{n \pi \hbar}{L}$

$P_{n} = \pm \frac{n h}{2L} \qquad \left( \hbar = \frac{h}{2 \pi} \right)$

The $\pm$ sign indicates that the particle is moving back and forth in the infinite potential box.

The above equation shows that eigen value of the momentum of the particle is discrete and the difference between the momentum corresponding to two consecutive energy levels is always constant and equal to $\frac{h}{2L}$

Electric field intensity due to uniformly charged solid sphere (Conducting and Non-conducting)

A.) Electric field intensity at different points in the field due to the uniformly charged solid conducting sphere:

Let us consider, A solid conducting sphere that has a radius $R$ and charge $+q$ is distributed on the surface of the sphere in a uniform manner. Now find the electric field intensity at different points due to the solid-charged conducting sphere. These different points are:
  1. Electric field intensity outside the solid conducting sphere
  2. Electric field intensity on the surface of the solid conducting sphere
  3. Electric field intensity inside the solid conducting sphere

1.) Electric field intensity outside the solid conducting sphere:

If $O$ is the center of solid conducting spherical then the electric field intensity outside of the sphere can be determined by the following steps →

  1. First, take the point $P$ outside the sphere
  2. Draw a spherical surface of radius r which passes through point $P$. This hypothetical surface is known as the Gaussian surface.
  3. Now take a small area $\overrightarrow {dA} $ around point $P$ on the Gaussian surface to find the electric flux passing through it.
  4. Now find the direction between the electric field vector and a small area vector.

Due to uniform charge distribution, the electric field intensity will be the same at every point on the Gaussian surface. So from the figure,
Electric field intensity outside the uniformly charged solid conducting sphere
The direction of electric field intensity on the Gaussian surface is radially outward which is in the direction of the area vector of the Gaussian surface. i.e. ($\theta=0^{\circ}$). Here $\overrightarrow {dA}$ is a small area around point $P$ so the small electric flux $d\phi_{E}$ will pass through this small area $\overrightarrow {dA}$. so this flux can be found by applying Gauss's law in question given below:

$ d\phi_{E}= \overrightarrow {E}\cdot \overrightarrow{dA}$

$ d\phi_{E}= E\:dA\: cos\: 0^{\circ} \quad \left \{\because \theta=0^{\circ} \right \}$

$ d\phi_{E}= E\:dA \quad (1) \quad \left \{\because cos\:0^{\circ}=1 \right \}$

The electric flux passes through the entire Gaussian surface, So integrate the equation $(1)$ →

$ \oint d\phi_{E}= \oint E\:dA $

$\phi_{E}=\oint E\:dA\qquad (2)$

According to Gauss's law:

$ \phi_{E}=\frac{q}{\epsilon_{0}}\qquad (3)$

From equation $(2)$ and equation $(3)$, we can write as

$ \frac{q}{\epsilon_{0}}=\oint E\:dA$

$ \frac{q}{\epsilon_{0}}= E\oint dA$

Now substitute the area of the entire Gaussian spherical is $\oint {dA}=4\pi r^{2}$ in the above equation. So the above equation can be written as:

$ \frac{q}{\epsilon_{0}}= E(4\pi r^{2})$

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{q}{r^{2}}$

From the above equation, we can conclude that the behavior of the electric field at the external point due to the uniformly charged solid conducting sphere is the same as the entire charge is placed at the center, point charge

If the surface charge density is $\sigma$, Then the total charge $q$ on the surface of a solid conducting sphere is→

$ q=4\pi R^{2}\: \sigma$

Substitute this value of charge $q$ in the above equation, so we can write the equation as:

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{4\pi R^{2}\: \sigma}{r^{2}}$

$ E=\frac{\sigma}{\epsilon_{0}}\frac{R^{2}}{r^{2}}$

This equation describes the electric field intensity at the external point of the solid conducting sphere.

2.) Electric field intensity on the surface of the solid conducting sphere:
Electric field intensity on the surface of the uniformly charged solid conducting sphere
If point $P$ is placed on the surface of the solid conducting sphere i.e. ($r=R$). so electric field intensity on the surface of the solid conducting sphere can be found by putting $r=R$ in the formula of electric field intensity at the external point of the solid conducting sphere:

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{q}{R^{2}}$

$ E=\frac{\sigma}{\epsilon_{0}}$

3.) Electric field intensity inside the solid conducting sphere:
Electric field intensity inside the uniformly charged solid conducting sphere
If point $P$ is placed inside the solid conducting sphere then electric field intensity will be zero because the charge is distributed uniformly on the surface of the solid sphere and there will not be any charge on the Gaussian surface. So the electric flux will be zero inside the solid sphere. i.e.

$ \phi_{E}=\oint E\:dA$

$ 0=E\oint dA \qquad\quad \left \{ \because \phi_{E}=0 \right \}$

$ E=0$

Electric field intensity distribution with distance for Conducting Solid Sphere:
Electric field intensity distribution with distance for conducting Solid
Electric field intensity distribution with distance shows that the electric field is maximum on the surface of the sphere and zero inside the sphere. Electric field intensity distribution outside the sphere reduces with the distance according to $E=\frac{1}{r^{2}}$.

B.) Electric field intensity at different points in the field due to the uniformly charged solid non-conducting sphere:

Let us consider, A solid non-conducting sphere of radius R in which $+q$ charge is distributed uniformly in the entire volume of the sphere. So electric field intensity at a different point due to the solid charged non-conducting sphere:

  1. Electric field intensity outside the solid non-conducting sphere
  2. Electric field intensity on the surface of the solid non-conducting sphere
  3. Electric field intensity inside the non-solid conducting sphere

1.) Electric field intensity outside the solid non-conducting sphere:

Let us consider, An external point $P$ which is at a distance $r$ from the center point $O$ of the sphere. The electric flux is radially outward in the sphere. So the direction of the electric field vector and the small area vector will be in the same direction i.e. ($\theta =0^{\circ}$). Here $\overrightarrow {dA}$ is a small area, the small amount of electric flux will pass through this area i.e. →

$ d\phi_{E}= \overrightarrow {E}\cdot \overrightarrow{dA}$

$ d\phi_{E}= E\:dA\: cos\: 0^{\circ} \quad \left \{\because \theta=0^{\circ} \right \}$

$ d\phi_{E}= E\:dA \qquad (1) \quad \left \{\because cos\:0^{\circ}=1 \right \}$
Electric field intensity outside the uniformly charged solid non-conducting sphere
The electric flux passes through the entire Gaussian surface, So integrate the equation $(1)$ →

$ \phi_{E}=\oint E\:dA\qquad (2)$

According to Gauss's law:

$ \phi_{E}=\frac{q}{\epsilon_{0}}\qquad (3)$

From equation (1) and equation (2), we can write as

$ \frac{q}{\epsilon_{0}}=\oint E\:dA$

$ \frac{q}{\epsilon_{0}}= E\oint dA$

Now substitute the area of the entire Gaussian spherical surface is $\oint {dA}=4\pi r^{2}$ in the above equation. So the above equation can be written as:

$ \frac{q}{\epsilon_{0}}= E(4\pi r^{2})$

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{q}{r^{2}}$

From the above equation, we can conclude that the behavior of the electric field at the external point due to the uniformly charged solid non-conducting sphere is the same as the point charge i.e. like the entire charge is placed at the center.

Since the sphere is a non-conductor so the charge is distributed in the entire volume of the sphere. So charge distribution can calculate by volume charge density →

$q=\frac{4}{3} \pi R^{3} \rho $

Substitute this value of charge $q$ in the above equation, so we can write the equation as:

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{4\pi R^{3}\: \rho}{3r^{2}}$

$ E=\frac{\rho}{\epsilon_{0}}\frac{R^{3}}{3r^{2}}$

This equation describes the electric field intensity at the external point of the solid non-conducting sphere.

2.) Electric field intensity on the surface of the solid non-conducting sphere:
Electric field intensity on the surface of the uniformly charged solid non-conducting sphere
If point $P$ is placed on the surface of a solid non-conducting sphere i.e. ($r=R$). so electric field intensity on the surface of a solid non-conducting sphere can be found by putting $r=R$ in the formula of electric field intensity at the external point of the solid non-conducting sphere:

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{q}{R^{2}}$

$ E=\frac{\rho R}{3\epsilon_{0}}$

3.) Electric field intensity inside the solid non-conducting sphere:
Electric field intensity inside the uniformly charged solid non-conducting sphere
If point $P$ is placed inside the sphere and the distance from the origin $O$ is $r$, the electric flux which is passing through the Gaussian surface

$ \phi_{E}= E.4\pi r^{2}$

Where $\phi_{E}=\frac{q'}{\epsilon_{0}}$

$ \frac{q'}{\epsilon_{0}}=E.4\pi r^{2}$

Where $q'$ is part of charge $q$ which is enclosed with Gaussian Surface

$ E=\frac{1}{4 \pi \epsilon_{0}} \frac{q'}{r^{2}} \qquad \qquad (4)$

The charge is distributed uniformly in the entire volume of the sphere so volume charge density $\rho$ will be the same as the entire solid sphere i.e.

$ \rho=\frac{q}{\frac{4}{3}\pi R^{3}}=\frac{q'}{\frac{4}{3}\pi r^{3}}$

$ \frac{q}{\frac{4}{3}\pi R^{3}}=\frac{q'}{\frac{4}{3}\pi r^{3}}$

$ q'=q\frac{r^{3}}{R^{3}}$

$ q'=q\left (\frac{r}{R} \right)^{3}$

Put the value of $q'$ in equation $(4)$, so

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{q}{r^{2}}\left(\frac{r}{R} \right)^{3}$

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{qr}{R^{3}}$

Where $q=\frac{4}{3} \pi R^{3} \rho $. So above equation can be written as:

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{\frac{4}{3} \pi R^{3} \rho r}{R^{3}}$

$E=\frac{ \rho r}{3 \epsilon_{0}}$

Electric field intensity distribution with distance for non-conducting Solid Sphere:
Electric field intensity distribution with distance for non-conducting solid sphere
Electric field intensity distribution with distance shows that the electric field is maximum on the surface of the sphere and zero at the center of the sphere. Electric field intensity distribution outside the sphere reduces with the distance according to $E=\frac{1}{r^{2}}$.

Derivation of Planck's Radiation Law

Derivation: Let $ N$ be the total number of Planck’s oscillators and $E$ be their total energy, then the average energy per Planck’s oscillator is

$ \overline{E}=\frac{E_{N}}{N} \qquad (1)$

Let there be $ N_{0}, N_{1} ,N_{2} ,N_{3},---N_{n}$ oscillator having energy $ E_{0}, E_{1}, E_{2}, -- E_{n}$ respectively.

According to Maxwell’s distribution, the number of oscillators in the $ n^{th}$ energy state is related to the number of oscillators in the ground state by

$ N_{n}=N _{0} e^{\tfrac{-nh\nu }{kt} }\qquad (2)$

Where $ n$ is a positive integer. So put $ n= 1,2,3,…….$. The above equation can be written for different energy states. i.e.

$ N_{1}= N _{0} e^{\tfrac{-h\nu }{kt} }$

$ N_{2}= N _{0} e^{\tfrac{-2h\nu }{kt} }$

$ N_{3}= N _{0} e^{\tfrac{-3h\nu }{kt} }$

$.............$

$.............$

So, the total number of Planck’s Oscillators –

$ N= N _{0} +N_{1}+N_{2}+N_{3}+.... N_{n}$

$ N= N _{0} + N_{0} e^{\tfrac{-h\nu }{kt}}+ N_{0} e^{\tfrac{-2h\nu }{kt}}+...+ N_{0}e^{\tfrac{-nh\nu }{kt}}$

$ N= N _{0}[1+e^{\tfrac{-h\nu }{kt}}+e^{\tfrac{-2h\nu }{kt}}+.... e^{\tfrac{-nh\nu }{kt}}] \qquad(3)$

Let $ x = e^{\frac{-h\nu }{kt}} \qquad (4) $

Then $ N = N_{0}[1+x+x^{2}+x^{3}+...+ x^{n}]$

$ N = \frac{N_{0}}{1-x} \qquad(5)$ [ from Binomial theorem]

Now the total energy of oscillators –

$ E_{N} = E_{0}N_{0}+ E_{1}N_{1}+ E_{2}N_{2}+...+ E_{n}N_{n}$

$ E_{N } = 0.N_{0}+ h\nu N_{0} e^{\tfrac{-h\nu }{kt}}+...+nh\nu N_{0} e^{\tfrac{-nh\nu }{kt}}$

$ E_{N } = N_{0}h\nu (e^{\tfrac{-h\nu }{kt}}+2e^{\tfrac{-2h\nu }{kt}}+...+ ne^{\tfrac{-nh\nu }{kt}})$

From equation $(4)$ put $ x = e^{\tfrac{-h\nu }{kt}}$ in above equation. i.e

$ E_{N } = N_{0}h\nu (x+2x^{2}+3x^{3}+...+nx^{n}) $

$ E_{N } = N_{0}h\nu x(1+2x+3x^{2}+....)$

$ E_{N } = \frac{N_{0}h\nu x}{(1-x)^{2}} \qquad(6)$ (from Bionomical theorem)

Now substituting the value of $N$ from equation $ (5)$ and $ E_{n}$ from equation $ (6)$  in equation $ (1)$ –

$ \overline{E}= \frac{\frac{N_{0}h\nu x}{(1-x)^{2}}}{\frac{N_{0}}{(1-x)}}$

$ \overline{E}= \frac{h\nu }{(\frac{1}{x}-1)}$

$ \overline{E}= \frac{h\nu }{e^{\tfrac{h\nu }{kt}}-1} \qquad(7)$

The number of oscillators per unit volume in wavelength range to $ ( \lambda + d\lambda )$ is $\frac{8\pi }{\lambda ^{4}} d\lambda$.

The energy per unit volume $ (E_{\lambda }d\lambda )$ in the wavelength range to $( \lambda +d\lambda ) $ is –

$ E_{\lambda}d\lambda = \frac{8\pi }{\lambda ^{4}}d\lambda \overline{E} \qquad(8)$

From equation $ (7)$ and $ (8)$ –

$ E_{\lambda}d\lambda = \frac{8\pi }{\lambda ^{4}}d\lambda \frac{h\nu }{(e^{\tfrac{h\nu}{kt}}-1)}$

$ E_{\lambda}d\lambda = \frac{8\pi hc}{\lambda ^{5}} \frac{d\lambda}{(e^{\tfrac{hc}{\lambda kt}}-1)}$

The above equation describes Planck’s radiation law and this law was able to thoroughly explain the black body radiation spectrum.

Wien’s Displacement law from Planck’s Radiation Law:

Planck’s radiation law gives the energy in wavelength region $ \lambda to \lambda +d\lambda $ as –

$ E_{\lambda}d \lambda = \frac{8\pi hc}{\lambda ^{5}}(\frac{1}{e^{\tfrac{hc}{\lambda kt}}-1})d\lambda \qquad(1)$

For shorter wavelength $ \lambda T$ will be small and hence

$ e^{\tfrac{hc}{\lambda kt}}> > 1$

Hence, for a small value of $\lambda T$ Planck’s formula reduces to -

$ E_{\lambda}d \lambda = \frac{8\pi hc}{\lambda ^{5}}(\frac{1}{e^{\tfrac{hc}{\lambda kt}}}) d\lambda$

$ E_{\lambda}d\lambda = \frac{8\pi hc}{\lambda ^{5}}e^{\tfrac{-hc}{\lambda kt}}d\lambda$

$E_{\lambda}d\lambda = A \lambda ^{-5} e^{\tfrac{-hc}{\lambda kt}}d\lambda$

Where $ A = 8\pi hc$

The above equation is Wien’s law of energy distribution verified by Planck radiation law.

Rayleigh-Jeans law from Planck’s Radiation Law:

According to Planck’s radiation law –

$ E_{\lambda}.d\lambda = \frac{8\pi hc}{\lambda ^{5}}\frac{1 }{e^{\tfrac{hc}{\lambda kt}}-1}.d\lambda$

For longer wavelength $ e^{\frac{hc}{\lambda kt}}$ is small and can be expanded as-

$ e^{\tfrac{hc}{\lambda kt}} = 1+\frac{hc}{\lambda kt}+\frac{1}{2!}(\frac{hc}{\lambda kt})^{2}+....$

Neglecting the higher-order term –

$ e^{\tfrac{hc}{\lambda kt}} = 1+\frac{hc}{\lambda kt}$

Hence for longer wavelength, Planck’s formula reduces to –

$ E_{\lambda}.d\lambda = \frac{8\pi kt}{\lambda ^{5}}[\frac{1}{1+\frac{hc}{\lambda kt}-1}]$

$ E_{\lambda}.d\lambda = \frac{8\pi kt}{\lambda ^{4}}.d\lambda$

This is Rayleigh Jean’s law verified by Planck Radiation Law.

Orthogonality of the wave functions of a particle in one dimension box or infinite potential well

Description of Orthogonality of the wave functions of a particle in one dimension box or infinite potential well:

Let $\psi_{n}(x)$ and $\psi_{m}(x)$ be the normalized wave functions of a particle in the interval $(0, L)$ corresponding to the different energy level $E_{n}$ and $E_{m}$ respectively. These wave functions are:

$\psi_{n}(x)= \sqrt{\frac{2}{L}} sin \frac{n \pi x}{L}$

$\psi_{m}(x)= \sqrt{\frac{2}{L}} sin \frac{m \pi x}{L}$

Where $m$ and $n$ are integers.

In this function are real. Therefore

$\psi_{n}^{*}(x) = \psi_{n}(x)$

$\psi_{m}^{*}(x) = \psi_{m}(x)$

Where $m=n$,

$\int_{0}^{L} \psi_{n}^{*}(x) \psi_{m}^{*}(x) dx = \frac{2}{L} \int_{0}^{L} sin \frac{m \pi x}{L} . sin \frac{n \pi x}{L} dx$

$\int_{0}^{L} \psi_{n}^{*}(x) \psi_{m}^{*}(x) dx =\frac{1}{L} \int_{0}^{L} \left[ cos \left\{ \frac{(m-n) \pi x}{L} \right\} - cos \left\{ \frac{(m+n) \pi x}{L} \right\} \right] dx $

$\int_{0}^{L} \psi_{n}^{*}(x) \psi_{m}^{*}(x) dx =\frac{1}{L} \left[ \frac{L}{\pi(m-n)} sin \left\{ \frac{(m-n) \pi x}{L} \right\} - \frac{L}{\pi(m+n)} sin \left\{ \frac{(m+n) \pi x}{L} \right\} \right]_{0}^{L} $

$\int_{0}^{L} \psi_{n}^{*}(x) \psi_{m}^{*}(x) dx =\frac{1}{L} \left[ \frac{L}{\pi(m-n)} sin \left\{ \frac{(m-n) \pi x}{L} \right\} - \frac{L}{\pi(m+n)} sin \left\{ \frac{(m+n) \pi x}{L} \right\} \right]_{0}^{L} $

$\int_{0}^{L} \psi_{n}^{*}(x) \psi_{m}^{*}(x) dx =0$

Hence, The function is mutually orthogonal in the interval $(0, L)$. These functions $\psi_{n}(x)$ and $\psi_{m}(x)$ are also normalized in this interval. The wave function, which is normalized and mutually orthogonal in an interval is said to form an orthogonal set in this interval. Since the wave function are zero outside the interval $(0, L)$, they are also orthogonal wave function in the whole range of $x$ axis in the interval $(-\infty, +\infty)$.

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