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Quantum Mechanics }Probability Density Current
If one has forgotten the expression for the probability density current, then one needs not despair! If one remembers the vaguest definition of a ``probability current," one can solve this problem without the usage of the forgotten formula:
Recall that the probability current density J is the difference between incoming P(in) and outgoing P(out) probability densities. This is just the difference between the probability densities of a rightwards moving plane wave and a leftwards moving plane wave, since the probability density is related to the wave function by 
-P(out)=|f(e^{-i k x})|^2 - |f(e^{i k x})|^2)
The given wave function can be written in terms of plane waves,
 + \frac{\beta}{2i}(e^{ikx}-e^{-ikx}) \right)\\
&=&\left( (\alpha+\beta/i)\frac{e^{ikx}}{2} + (\alpha-\beta/i)\frac{e^{-ikx}}{2} \right)
\end{eqnarray})
) gives the coefficient for the leftwards (negative-direction) wave, while ) gives the coefficient for the rightwards traveling wave. The probability density for each wave is given by,
|^2 &=& \frac{1}{4}\left(|\alpha|^2+|\beta|^2 + \frac{\alpha\beta^{*}}{-i} +\frac{\alpha^{*}\beta}{i}\right)\\
|f(e^{-ikx})|^2 &=& \frac{1}{4}\left(|\alpha|^2+|\beta|^2 + \frac{\alpha\beta^{*}}{i} -\frac{\alpha^{*}\beta}{i}\right)
\end{eqnarray})
Plugging these quantities into the formula for the probability current density above (J), one gets,
,)
which is choice (E).
Alternatively:
The formal expression for the probability current density can be effortlessly derived from recalling the definition of probability and Schrodinger's Equation---both of which every physics (or engineering) major should know by heart.
Probability is defined (in the Born Interpretation) as |^2 dx) . One should recall that  in general (to wit: the absolute value squared of a complex expression is itself times its complex conjugate).
The time-dependent Schrodinger's Equation is
 ,
where  has the form of the familiar time-independent Hamiltonian. From this, one finds that ) .
Generalizing the idea of a current from classical physics to the idea of a probability current, one takes the time derivative of the probability to get dx) , where the product-rule for baby-math derivatives has been used and the derivative has been taken inside the integral because the integral and derivative are with respect to different variables.
Plugging in the expression for  from the Schrodinger's Equation, one gets
+\Psi^* \frac{-i}{\hbar}\left( -\frac{\hbar^2}{2m}\Psi^{''}+V\Psi \right)\right)dx) ,
where the terms involving V's cancel out, and thus,
dx = \frac{i\hbar}{2m}\int\left( -\Psi\Psi^{*''}+ \Psi^*\Psi^{''}\right)dx)
Rewriting dx = \int \frac{\partial}{\partial x} \left( -\Psi\Psi^{*'}+ \Psi^*\Psi^{'}\right)dx) , one can eliminate the integral in the probability current by applying the fundamental theorem of calculus (to wit: -\psi(a)) ),
) . But, since the probability current is usually define as -J(b)) , one has
) .
(Aside:) One can print-out a cool poster or decent T-shirt iron-on to remember the Schrodinger's Equation (among other miscellanai) at a site the current author made several years ago,
\begin{quote}
http://anequationisforever.com/ds.php
\end{quote}
One can remember the general form of the probability current by recalling that it has to do with the difference of  times its conjugate.
Right, so onwards with the problem:
The problem gives the wave function, so one needs just chunk out the math to arrive at the final answer,
 +\beta \cos(kx) \right))
 +\beta \cos(kx) \right))
Thus,
 +\beta^{*} \sin(kx) \right) \left( -\alpha \sin(kx) +\beta \cos(kx) \right)) and,
 +\beta \sin(kx) \right) \left( -\alpha^{*} \sin(kx) +\beta^{*} \cos(kx) \right)) , where one notes that the imaginary terms go to unity from the complex conjugate.
Plugging this into the probability current, one arrives at the expression for choice (E).
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