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Consider a particle moving through a potential of the form $$V(x)=\begin{cases} 0 & x<0\\ V & 0\leq x < L\\ 0 & x\geq L \end{cases}.$$ We may write out a solution to the time-independent Schrodinger equation for each of these regions $$\psi(x)=Ae^{ikx} + Be^{-ikx} \quad k\hbar = \sqrt{2mE} \quad \quad \text{Region 1}$$ $$\psi(x)=A'e^{ikx} + B'e^{-ikx} \quad k\hbar = \sqrt{2m(E-V)} \quad \quad \text{Region 2}$$ $$\psi(x)=A''e^{ikx} + B''e^{-ikx} \quad k\hbar = \sqrt{2mE} \quad \quad \text{Region 3}$$ and use the fact that the wave function and its first derivative must be continuous at $x=0,L$ along with the normalization requirement to find the coefficients for the exponential terms in the wavefunction. So far, this all makes sense.

However, when we consider the transmission probability $T$ of moving through the barrier, it is defined as $$T = \frac{|A''|^2}{|A|^2}.$$ The justification for this is that this is the ratio of the flux density ($|A''|^2$) of particles moving away form the wall in Region 3 to that of the particles moving toward the wall in Region 1 (which is proportional to |A|^2). This is the part that is confusing to me. The wavefunctions for Regions 1, 2, and 3 all are parts of a single wavefunction representing the stationary state of a molecule in this system. This seems distinct from saying: prepare a particle moving to the right in region 1 and then measure the flux being transmitted through the wall. Perhaps I am understanding the idea of flux density incorrectly. I apologize if my question is poorly phrased, but I was hoping someone could clarify how to correctly think about flux density and $T$ in this case.

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You are mistaken in the setup. In order for $T=\frac{|A^{\prime\prime}|^2}{|A|^2}$, it is necessary for $B^{\,\prime\prime}=0$.

I also recently answered another question, very related to this one, about why this definition makes sense. The introductory textbooks are being sloppy (really, they are trying to avoid maths when they can avoid maths), and that makes for these confusions.

In short, yes, you are correct that all three parts are one single wavefunction. We have another criteria that gives us the correct behaviour, but it is hidden from you at this point. If you use the correct criteria, it will make sense why we can just deduce the results using these parts.

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    $\begingroup$ Thank you! Just to be clear, the correct criteria that you are referring to is $B''=0$, right? I am still however, a little bit confused. Even if we do assume $B'' = 0$, we still have solved for a stationary state of the system that gives the probability density of finding the particle in a particular place at a single instant of time. We don't actually follow the process of scattering, starting from the particle moving toward the wall and then checking whether transmission occurred? Why are these two processes the same? $\endgroup$
    – Neel
    Commented Nov 3 at 18:09
  • $\begingroup$ You are getting confused over something insane. Consider a free particle. Then the momentum eigenstate is also a stationary state. It simply is the case that input and output are still well-defined when considering stationary states. $\endgroup$ Commented Nov 4 at 1:07
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In response to your comment, note that if we set $B''=0$ and include the time evolution factor $e^{-iEt/ \hbar}$, then your equations model the situation you want since $A$ is the amplitude of a wave travelling to the right in region 1 (towards the barrier) and there is no wave travelling to the left in region 3. The amplitude $A''$ can be seen as telling us how much of the incoming wave has been transmitted, so it is interesting to quantify $A''$ in terms of $A$.

To see that $T = {|A''|^2}/{|A|^2}$ can be seen as the probability that the incoming wave has been transmitted, note that the "probability" to find the particle in any interval of length $\ell$ in region 3 is $|A''|^2 \cdot\ell$ (carry out the very simple integral), whereas if there was no potential barrier (i.e. $V=0$) then the probability to find the incoming matter wave in the same interval would be $|A|^2 \cdot\ell$. So the ratio $|A'' |^2/|A|^2 =|A'' |^2 \cdot\ell/(|A|^2 \cdot\ell)$ can be seen as the transmission probability.

Note that the argument above is not literally correct, which is why I put quotation marks around "probability": the solutions to this problem cannot be normalized, and so it doesn't make sense to talk about probabilities of finding states in some interval. But this is a different matter - hopefully this argument still helps a little bit.

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Quantum tunnelling is a quasi static process: the wave packet is a lot wider than the width of the barrier. As a result for most of the wave packet the time independent solution is a good approximation apart from some short transients near the start and end of the wavepacket. For more explanation see Herbert Winful's review of about tunnelling time and references therein, especially Section 5

https://winful.engin.umich.edu/wp-content/uploads/sites/376/2018/01/physics_reports_review_article__2006_.pdf

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