## 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.

## 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 $\psi^{*}$.

## 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 $(1.2.3)$

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

## 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 $\frac{h}{2L}$