I I want to understand how the transmission coefficient is obtained

[hmparticle9]

I am going through "Introduction to Quantum Mechanics" by Griffiths and I am having trouble with a particular question. I have taken screenshots of the question and of the solutions.

Here they are:

In the book, the transmission coefficient is just stated and not derived.

I can answer parts a,b, and d of the above problem but I had major problems with c. I really could not even get off the ground. I made an attempt by linking the velocity of the particle before and after hitting the barrier. That is,

$$v_{before} = \sqrt\frac{E}{2m}, v_{after}=\sqrt\frac{E-V_0}{2m}$$

I would like a hand in two ways.

1) I would like to understand, using the mechanics of the text up to this point, how to derive the transmission coefficient:
$$T = \frac{|F|^2}{|A|^2}$$
2) I would like help understanding the solution to part c). That is, both versions. The first that uses the diagram and the second that uses the probability current. I would like to derive the transmission coefficient under this potential.

In order to understand the solutions I will provide the equations that they are referring to:

and the probability current is given as

$$J(x,t) = \frac{ih}{2m}(\Psi \frac{\partial \Psi^*}{\partial x} - \Psi^* \frac{\partial \Psi}{\partial x})$$

For instance, the opening line "From the diagram $T = P_t/P_i = |F|^2v_t/|A|^2v_i$..." I have no idea how that was taken from the diagram. None at all. Both of them equalitites.

Where it says that $$\frac{v_t}{v_i} = \frac{\sqrt{E-V_0}}{\sqrt{E}}$$ is obtained from (2.98), I understand that.

Finally,

Finally, when he says $T = \frac{J_t}{J_i}$ I really don't understand why that is the case either. From my above workings, if $l$ and $k$ are the same then we can obtain the first transmission coefficient by considering probability current. I just don't understand why we are taking $T = \frac{J_t}{J_i}$. What is the motivation?

EDIT: THE GREEN HIGHLIGHTED STATEMENT IS REFERRING TO THE FORM NOT THE ABOVE CALCULATIONS.

 

[div_grad]

@hmparticle9 In your last picture, does the sentence "But I don't understand why?" highlighted in green refer to the form of the probability current density $J_2(x,t)$? If so, the answer is clearly obtained by calculating the current according to the definition you provided in your post:

$$J(x,t) = \frac{ih}{2m}(\Psi \frac{\partial \Psi^*}{\partial x} - \Psi^* \frac{\partial \Psi}{\partial x})$$

where you plug in $\Psi = F \exp(\mathrm{i} l x)$.

Now, as to why the reflection/transmission coefficients (probabilities) are defined as ratios of probability currents (such as in $T=\frac{J_t}{J_i}$), the answer is provided by the appropriate continuity equation, with general three-dimensional form
$$
\frac{\partial \varrho}{\partial t} + \nabla\cdot\mathbf{j} = 0 \rm{.}
$$
In quantum mechanics, $\varrho = \Psi^* \Psi$ is the probability density, while $\mathbf{j}$ is the probability current density (in one-dimensional case, $\nabla \equiv \frac{\partial}{\partial x}$). This equation is obtained by considering the time-derivative of a "normalization" condition,
$$
\frac{d}{d t} \int |\Psi(x,t)|^2 {\rm{d}}x
$$
and if this time-derivative is zero, then that means that $\int |\Psi(x,t)|^2 {\rm{d}}x = {\rm{const}}$ does not change with time, and therefore you can normalize the wave function and interpret it as the probability amplitude. However, this requires (by continuity equation) that the wave function goes to zero at large distances sufficiently fast and the probability of finding a particle arbitrarily far from some localized region of space vanishes. This is true for bound-state wave functions, but for scattering wave functions (which you consider in your scattering example) such wave functions are non-zero even at very large distances, so that the associated probability currents $\mathbf{j}$ do not vanish. As a consequence, the time-derivative of the normalization condition above is not zero and you cannot normalize the wave function of a scattering state - but you can calculate the probability current density in some region of space, then calculate the relevant current in another region of space, and then obtain a relative probability by taking the ratio of the two calculated currents. By doing so, you will have a way to characterize the probability of finding a particle "rather here" than "there".

In general, you know the wave functions of incoming particles (those particles which will impinge on the potential and scatter off of it) and therefore you known the associated probability current density for incoming particles - $J_i$, in one dimension. The particles encounter the potential and in this one-dimensional case, there is a probability that they reflect back from the potential step, and the probability that they pass through the potential, i.e., that they will be transmitted. Now, you calculate the current densities associated with reflected ($J_r$) and transmitted ($J_t$) wave functions, and you take the aforementioned ratio $J_r/J_i$ and $J_t/J_i$: in this way you characterize the scattering processes by effectively comparing the probability of reflection/transition relative to the known initial current density of incoming particles.

You can try calculating $J_i$, $J_t$ and $J_r$ using the formula for the probability current that you wrote in your question, and then take the appropriate ratios to obtain the answers.

I have no idea how that was taken from the diagram. None at all.

Me neither. It seems that this "diagrammatic" example is based on classical-mechanical considerations, but in the example that you are considering, there is a key difference between classical and quantum mechanics: namely, a classical particle with energy $E > V_0$ would always pass above the potential - it would always be transmitted. Whereas the quantum particle with energy $E > V_0$ can also be reflected - there is a non-zero probability that the particle will reflect off the potential, despite its energy being larger than the barrier, and this is a quantum phenomenon.

 

[hmparticle9]

Thank you for your reply. The green sentence was in regard to the form. That is, why $T$ takes the form that it does. That is, the quotient of probability currents. The material that you are referring to in your post is nowhere to be seen in the book prior to this exercise. For example, this continuity equation.

I have done all the exercises and read all the material up to this point and nothing would have helped me answer this part of the question. I am going to re-read your post a few times for it to sink in.

 

[hmparticle9]

https://en.wikipedia.org/wiki/Probability_current

What you refer to in the beginning of your post is in the above link under "motivation". According to the wiki article $T$ and $R$ are just defined that way. I was hoping that there was more to it!

I hope that someone can make sense of the first solution (involving the diagram).

I am going to work through the problem again to test my understanding.

Thanks again!