Lets say we have a potential step with regions 1 with zero potential $W_p\!=\!0$ (this is a free particle) and region 2 with potential $W_p$. Wave functions in this case are:

\begin{align} \psi_1&=Ae^{i\mathcal L x} + B e^{-i\mathcal L x} & \mathcal L &\equiv \sqrt{\frac{2mW}{\hbar^2}}\\ \psi_2&=De^{-\mathcal K x} & \mathcal K &\equiv \sqrt{\frac{2m(W_p-W)}{\hbar^2}} \end{align}

Where $A$ is an amplitude of an incomming wave, $B$ is an amplitude of an reflected wave and $D$ is an amplitude of an transmitted wave. I have sucessfuly derived a relations between amplitudes in potential step:

\begin{align} \frac{A}{D} &= \frac{i\mathcal L-\mathcal K}{2i\mathcal L} & \frac{A}{B}&=-\frac{i \mathcal L - \mathcal K}{i \mathcal L + \mathcal K} \end{align}

I know that if i want to calculate transmittivity coefficient $T$ or reflexifity coefficient $R$ i will have to use these two relations that i know from wave physics:

\begin{align} T &= \frac{j_{trans.}}{j_{incom.}} & R &= \frac{j_{trans.}}{j_{incom.}} \end{align}

Question 1: I know that $j = \frac{dm}{dt} = \frac{d}{dt}\rho V \propto \rho v \propto \rho k$ But what is a density $\rho$ equal to?

Question 2: I noticed that $\mathcal L$ and $\mathcal K$ are somehow (i dont know how) connected to the wavevector $k$ from the equation in 1st question but how? How can i make it obvious?

  • 1
    $\begingroup$ Similar if not same to quations/60228 $\endgroup$
    – iiqof
    Apr 11, 2013 at 6:58

1 Answer 1


Are you shore thats the definition of current you want? In quantum mechanics you have tho related concepts:

  • Probability density defined by :$\rho = |\psi(x)|^2$
  • Probability current: $\vec{j} = \frac{1}{m}\Re(\psi^*\hat{p}\psi) = \frac{\hbar}{2mi}(\psi^*\nabla \psi - \psi\nabla\psi^*) $
  • Both concepts are related by a continuity equations: $\frac{\partial \rho}{\partial t} + \nabla \vec{j} =0$ expresing the local conservation of probability. It's eassily derivable froma the top definitions and Schrödinger's equation.

Try using this concepts in the first part. The second part shoud follow.


The operator nabla is definied as $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})$, so operationg over $\psi$ it would look as: $\nabla\psi = (\frac{\partial\psi}{\partial x},\frac{\partial\psi}{\partial y},\frac{\partial\psi}{\partial z})$. In the one dimensional problem you have, the operator is also one dimensional, so it is symply the position derivative $\frac{\partial}{\partial x}$.

Then the 1_D current is defined just as before, changin $\nabla$ for $\frac{\partial}{\partial x}$. Thats the definition. The 'physical' picture is the flow of probability.

With continous estates (in the base of an operator with continous eigenvalues), the inner product is defined as :

$$\left<\psi\right|\phi\left. \right> = \int_{-\infty}^\infty \psi^*(x)\phi(x) dx$$

so the probability of the state in $[a,b]$ is $$P([a,b])=\left<\phi [a,b]\right|\phi [a,b]\left. \right> = \int_{a}^b |\phi(x)|^2 dx$$

so you could define a local density so that integrating the density over $[a,b]$ gives the probability. With that we have that the density can be defined as $\rho=|\psi(x)|^2$

  • $\begingroup$ Isn't $|\psi|^2$ only a probability? I know jet nothing about probability current $j$. How is it defined? If possible i would perfer if you avoid $\nabla$ operator. I am not familiar with it yet. $\endgroup$
    – 71GA
    Apr 12, 2013 at 7:41
  • $\begingroup$ I found a link on how to derive the probability current here: physics.ucdavis.edu/Classes/Physics115A/probcur.pdf It seems to me that he somehow messed a sign. It should be: $$\vec{j} = -\frac{\hbar i }{2m}(\psi^*\nabla \psi - \psi\nabla\psi^*)$$ Please confirm. $\endgroup$
    – 71GA
    Apr 12, 2013 at 14:25
  • $\begingroup$ Now that i know $j$, how can i now get equations for $T$ and $R$? $\endgroup$
    – 71GA
    Apr 12, 2013 at 14:55
  • $\begingroup$ @71GA: Both our definitions are the same, bear in mind that $-i = \frac{1}{i}$ $\endgroup$
    – iiqof
    Apr 12, 2013 at 17:18
  • $\begingroup$ I suggest you to take some book in intruductory quantum mechanics, like Mandl's Quantum Mechanics or Shankar's Princinciples of Quantum Mechanics. Both have the solution of the problem. $\endgroup$
    – iiqof
    Apr 12, 2013 at 17:21

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