Question:
What is the probability current density of a particle with wavefunction $\psi(x, t) = e^{\frac{i}{\hbar}(p x - E t)} $.
Recall that the probability current density can be computed as: $ j = \frac{\hbar}{2mi}[\psi^{*}\frac{\partial}{\partial x}\psi - \psi \frac{\partial}{\partial x}\psi^{*}] $
Solution:
The problem is in some sense easy, because it really just ask for computing an expression. All I have to do is simply do it.
$ \begin{eqnarray*} \frac{\partial}{\partial x} \psi &=& \frac{\partial}{\partial x} e^{\frac{i}{\hbar}(p x - E t)} \\ &=& \frac{\partial}{\partial x} e^{\frac{i}{\hbar}p x} e^{\frac{-i}{\hbar} E t} \\ &=& e^{\frac{-i}{\hbar} E t} \frac{\partial}{\partial x} e^{\frac{i}{\hbar}p x} \\ &=& e^{\frac{-i}{\hbar} E t} \frac{ip}{\hbar} e^{\frac{i}{\hbar}p x} \\ &=& \frac{ip}{\hbar} e^{\frac{i}{\hbar}p x} e^{\frac{-i}{\hbar} E t} \\ &=& \frac{ip}{\hbar} e^{\frac{i}{\hbar}(p x - E t)} \\ &=& \frac{ip}{\hbar} \psi \end{eqnarray*} $
Similarly, we have
$ \begin{eqnarray*} \frac{\partial}{\partial x} \psi^{*} &=& \frac{\partial}{\partial x} e^{\frac{-i}{\hbar}(p x - E t)} \\ &=& \frac{\partial}{\partial x} e^{\frac{-i}{\hbar}p x} e^{\frac{i}{\hbar} E t} \\ &=& e^{\frac{i}{\hbar} E t} \frac{\partial}{\partial x} e^{\frac{-i}{\hbar}p x} \\ &=& e^{\frac{i}{\hbar} E t} \frac{-ip}{\hbar} e^{\frac{-i}{\hbar}p x} \\ &=& \frac{-ip}{\hbar} e^{\frac{-i}{\hbar}p x} e^{\frac{i}{\hbar} E t} \\ &=& \frac{-ip}{\hbar} e^{\frac{-i}{\hbar}(p x - E t)} \\ &=& \frac{-ip}{\hbar} \psi^{*} \end{eqnarray*} $
With all these preparation, we can get the final answer as follow:
$ \begin{eqnarray*} j &=& \frac{\hbar}{2mi}[\psi^{*}\frac{\partial}{\partial x}\psi - \psi \frac{\partial}{\partial x}\psi^{*}] \\ &=& \frac{\hbar}{2mi}[\psi^{*}\frac{ip}{\hbar}\psi - \psi \frac{-ip}{\hbar}\psi^{*}] \\ &=& \frac{p}{2m}[\psi^{*}\psi + \psi \psi^{*}] \\ &=& \frac{p}{m}[\psi \psi^{*}] \\ &=& \frac{p}{m} \end{eqnarray*} $
The last equality follow from the fact the $ \psi $ is just a complex exponential, which means its norm is 1.
What is the probability current density of a particle with wavefunction $\psi(x, t) = e^{\frac{i}{\hbar}(p x - E t)} $.
Recall that the probability current density can be computed as: $ j = \frac{\hbar}{2mi}[\psi^{*}\frac{\partial}{\partial x}\psi - \psi \frac{\partial}{\partial x}\psi^{*}] $
Solution:
The problem is in some sense easy, because it really just ask for computing an expression. All I have to do is simply do it.
$ \begin{eqnarray*} \frac{\partial}{\partial x} \psi &=& \frac{\partial}{\partial x} e^{\frac{i}{\hbar}(p x - E t)} \\ &=& \frac{\partial}{\partial x} e^{\frac{i}{\hbar}p x} e^{\frac{-i}{\hbar} E t} \\ &=& e^{\frac{-i}{\hbar} E t} \frac{\partial}{\partial x} e^{\frac{i}{\hbar}p x} \\ &=& e^{\frac{-i}{\hbar} E t} \frac{ip}{\hbar} e^{\frac{i}{\hbar}p x} \\ &=& \frac{ip}{\hbar} e^{\frac{i}{\hbar}p x} e^{\frac{-i}{\hbar} E t} \\ &=& \frac{ip}{\hbar} e^{\frac{i}{\hbar}(p x - E t)} \\ &=& \frac{ip}{\hbar} \psi \end{eqnarray*} $
Similarly, we have
$ \begin{eqnarray*} \frac{\partial}{\partial x} \psi^{*} &=& \frac{\partial}{\partial x} e^{\frac{-i}{\hbar}(p x - E t)} \\ &=& \frac{\partial}{\partial x} e^{\frac{-i}{\hbar}p x} e^{\frac{i}{\hbar} E t} \\ &=& e^{\frac{i}{\hbar} E t} \frac{\partial}{\partial x} e^{\frac{-i}{\hbar}p x} \\ &=& e^{\frac{i}{\hbar} E t} \frac{-ip}{\hbar} e^{\frac{-i}{\hbar}p x} \\ &=& \frac{-ip}{\hbar} e^{\frac{-i}{\hbar}p x} e^{\frac{i}{\hbar} E t} \\ &=& \frac{-ip}{\hbar} e^{\frac{-i}{\hbar}(p x - E t)} \\ &=& \frac{-ip}{\hbar} \psi^{*} \end{eqnarray*} $
With all these preparation, we can get the final answer as follow:
$ \begin{eqnarray*} j &=& \frac{\hbar}{2mi}[\psi^{*}\frac{\partial}{\partial x}\psi - \psi \frac{\partial}{\partial x}\psi^{*}] \\ &=& \frac{\hbar}{2mi}[\psi^{*}\frac{ip}{\hbar}\psi - \psi \frac{-ip}{\hbar}\psi^{*}] \\ &=& \frac{p}{2m}[\psi^{*}\psi + \psi \psi^{*}] \\ &=& \frac{p}{m}[\psi \psi^{*}] \\ &=& \frac{p}{m} \end{eqnarray*} $
The last equality follow from the fact the $ \psi $ is just a complex exponential, which means its norm is 1.
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