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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?

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Similar if not same to quations/60228 –  Angel Joaniquet Tukiainen Apr 11 '13 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.

EDIT:

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 $\rho=|\psi(x)|^2$

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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. –  71GA Apr 12 '13 at 7:41
    
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. –  71GA Apr 12 '13 at 14:25
    
Now that i know $j$, how can i now get equations for $T$ and $R$? –  71GA Apr 12 '13 at 14:55
    
@71GA: Both our definitions are the same, bear in mind that $-i = \frac{1}{i}$ –  Angel Joaniquet Tukiainen Apr 12 '13 at 17:18
    
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. –  Angel Joaniquet Tukiainen Apr 12 '13 at 17:21

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