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

$P=\psi(x,t) \: \psi^{*}(x,t) \qquad(2)$
Motion of Particle in One Dimensional Region of a Cross Section Area
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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