Question

Question: Find the probability current, $\mathbf{J}$ (Problem 1.14) for the free particle wave function Equation $\mathbf{2}\mathbf{.}\mathbf{94}$. Which direction does the probability flow?

Short answer

The probability current$\left(\mathrm{J}\right)$ is

$j=\frac{\hat{\hslash }k}{m}|A{|}^2$

Step-by-step solution

The probability current density (J)

The required formula of probability current density$\left(\mathrm{J}\right)$ is,

role="math" localid="1657786971189" $\mathit{J}\mathbf{=}\frac{\mathbf{i}\mathbf{\cap }}{\mathbf{2}\mathbf{m}}\left(\Psi \frac{\partial {\Psi }^{\text{'}}}{\partial x}-{\Psi }^{*}\frac{\partial \Psi }{\partial x}\right)$

Where $y$ is the wave function of the particle and complex conjugate of $y^{\text{is}}y°$.

Compute probability current density

Equation $2.94$from the chapter is

${\Psi }_k(X,t)=A{e}^{ik{x}^{jk{k}^2}}2m^2$

${\psi }^{\text{'}}=A{e}^{ik\frac{k{m}^2}{2m^2}?},$

Thus, it can written that,

$J=\frac{i\circlearrowright }{2m}\left(\Psi \frac{\partial {\Psi }^{*}}{\partial x}-{\Psi }^{*}\frac{\partial \Psi }{\partial x}\right)$

$\left.=\frac{i\times }{2m}\left|A{|}^2\left[e^{i\left(kx-\frac{x{k}^2}{2m}t\right.}(-ik)e^{-i\left(kx-\frac{a{k}^2}{2m}t\right.}-e^{\left(\left\{kx\frac{x{k}^{2}t}{2m}t\right.\right.}\right)\right(ik)e^{\left(kx-\frac{m{k}^2}{2m}t\right)}\right]$

$J=\frac{i\circlearrowright }{2m}\left|A{|}^2\right(-2ik)$

$J=\frac{\geqq k}{m}|A{|}^2$

It flows in the positive $\left(\mathrm{x}\right)$direction.