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Calculate the probability current density vector $\vec{j}$ for the wave function : $$\psi = Ae^{-i(wt-kx)}.$$

From my very poor and beginner's understanding of probability density current it is :

$$\frac{d(\psi \psi^{*})}{dt}=\frac{i\hbar}{2m}[\frac{d\psi}{dx}\psi^{*}-\frac{d\psi^{*}}{dx}\psi]$$

By applying the RHS of the above equation :

$$\frac{i\hbar}{2m}[-A^{2}ikxe^{-i(ωt-kx)}e^{i(ωt-kx)}-A^{2}ikxe^{i(ωt-kx)}e^{-i(ωt-kx)}]$$

This gives :

$$\frac{-2iA^{2}ik\hbar}{2m}=\frac{k \hbar A^{2}}{m}$$

This is not the correct answer. :( What have I done wrong ?

In the model workings instead of A in the complex conjugate of the wave function they have written $A^{*}$. Why is this necessary since $A$ is likely to be a real number anyways ?

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    $\begingroup$ $A$ is not necessarily a real number, so it is more appropriate to use $A^*$. $\endgroup$ – Kyle Kanos Apr 3 '14 at 12:43
  • $\begingroup$ Oh, okay, I didn't know that. So technically I should be writing $ \lvert{A} \rvert^{2}$ versus just $A^{2}$ ? $\endgroup$ – Ari Ben Canaan Apr 3 '14 at 12:44
  • $\begingroup$ Yes, but note that it can be written in shorter notation as $|A|^2$, rather than throwing in the $\rm mod$ in there $\endgroup$ – Kyle Kanos Apr 3 '14 at 12:45
  • $\begingroup$ Yup, was just looking for the right LaTeX command. What about my answer below ? Does it make sense ? $\endgroup$ – Ari Ben Canaan Apr 3 '14 at 12:47
  • $\begingroup$ I commented on the answer that it should be $|A|^2$ (though I realize now that I forgot the modulus bars), but it does look reasonable. $\endgroup$ – Kyle Kanos Apr 3 '14 at 12:49
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The current density formula you consider is not right, because $\frac{d(\psi\psi^*)}{dt}$ is always 0.

The current density is defined as $\bf{j}=(i\hbar/2m)(\psi\nabla\psi^*-\psi^*\nabla\psi)$. In your one dimensional case change $\nabla$ to $\partial/\partial x$, you will get your answer.

A more physical definition of current density operator is this: $$\frac{d}{dt}\int|\psi|^2dV=-\int \nabla\cdot \bf{j}\ \mathrm{d}V=-\int\bf{j}\cdot d\bf{S}$$ The two integral are integrated within the same finite volume. The explain is that the increased probability for find a particle within a finite volume with time is equal to the probability current density flow into that volume.

From above definition it is easy to see that actually this equation(rather than yours) holds: $$\partial |\psi|^2/\partial t +\nabla\cdot\bf{j}=0$$ Which is the analog of continuity equation $\partial \rho/\partial t +\nabla\cdot\bf{j}=0$ in classical mechanics.

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Oh, I think I've figured it out... I sort of realised this just as I posted my question :

From my earlier intros to quantum mechanics I know that p = mv = $\hbar k$. By simple substitution I can then obtain a final answer of :

j = $v\lvert{A}\rvert^{2}$

I'll wait for someone to confirm my line of thinking.

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    $\begingroup$ I think that it should be $A^2$, not $A$. $\endgroup$ – Kyle Kanos Apr 3 '14 at 12:44
  • $\begingroup$ yes, of course. No idea how I managed to lose that square when i had it previously. Does it make sense now ? $\endgroup$ – Ari Ben Canaan Apr 3 '14 at 12:49

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