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We've a potential given as: $V(x)=\left\{\begin{array}{ll}0, & x<0 \\ V_{0}, & x \geq 0\end{array}\right.$. We've got particles coming in from the left towards the step and getting reflected from it. Further, if we assume the energy of the particles coming towards the barrier have lesser energy than the step i.e $E<V$, then it's said that

$T = 0$, The transmission coefficient at a potential step with $E < V$ is zero.

Further it's said that,

The wavefunction in the second region, where $V(x)=V_0$ is $ψ_2(x) = e^{αx}$ where $\alpha=2m(V−E)$.

So we have a probability of finding a particle in the second region even though the transmission at the step is zero!? But this is contradictory. Can anyone please help.

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  • $\begingroup$ @ohneVal please see the bottom of this page electron6.phys.utk.edu/qm1/modules/m2/step.htm and this article.sciencepublishinggroup.com/html/… $\endgroup$
    – Kashmiri
    Commented Sep 2, 2020 at 11:33
  • $\begingroup$ @descheleschilder thank you. This is an example of tunneling but as my question said, it appears to contradict that $T=0$ $\endgroup$
    – Kashmiri
    Commented Sep 2, 2020 at 11:39
  • $\begingroup$ Maybe $T=0$ is meant in a classical sense. $\endgroup$ Commented Sep 2, 2020 at 11:44
  • $\begingroup$ In this case it means that probabilty current in transmitted region is zero and hence, probabilty does not change with time in this region. It does not mean that probabilty is zero but rather it is constant. But can constant probabilty represent a moving particle in transmitted region? $\endgroup$
    – Mridul
    Commented Dec 15, 2021 at 14:35

2 Answers 2

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This has to do with the definition of the transmission coefficient $T$ in quantum scattering experiment. What we want to know is how a particle will behave when it's "shot" at the barrier. In a general situation, part of the wave function probability will reflect off the barrier, and part will be transmitted. However, for your potential, as $t\xrightarrow{}\infty$, all of the probability will be reflected. Note that we are NOT talking about a static situation. The wave function changes with time, so there will be some points in time when there is a non-zero probability for the particle to be found in the region where $V(x)=V_0$. However, after a long time has past, all of the probability will be reflected.

If the barrier were not infinitely thick, there would be some probability of transmission. For example, a particle certainly could tunnel through the potential $$V=\begin{cases} 0 & x<0 \\ V_0 & 0\leq x < a \\ 0 & a\leq x. \end{cases}$$ Therefore, part of the probability could escape toward large $x$.

Edit: For those who want more information, I'll do an explicit calculation. $R$ and $T$ are defined in the following way. We take a particle, localized on one side of the barrier and sharply peaked in momentum space, and allow it to travel toward the barrier. We then define $T$ as the probability of finding the particle on the opposite side of the barrier after waiting an infinitely long time. $R$ is defined as the probability that the particle will be found on the original side.

A standard way to approach the problem of calculating $R$ and $T$ (which is covered in most introductory textbooks) is to consider a slightly different situation. We look at the stationary state solutions for this potential (the energy eigenstates), and use $R=\frac{j_R}{j_I}$ and $T=\frac{j_T}{j_I}$, where $j_I$ is the probability current of the part of the wave function that is incident on the barrier, $j_R$ is the current of the reflected wave function, and $j_T$ is the current of the transmitted wave function.

The solution in the region where $V=0$ contains both $e^{ikx}$ and $e^{-ikx}$ terms, which give probability currents moving both to the left and to the right. The solution in the region where $V=V_0$ is a falling exponential $\psi_T(x)=e^{-\alpha x}$. The associated probability current is proportional to $$j_T\propto \psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}=\psi\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi}{\partial x}=0$$ because the wave function is purely real ($\psi_T=\psi^*_T$). If you add time dependence, $\psi_T$ will rotate in phase space and acquire an imaginary part, but this imaginary part has no dependence on $x$, so it passes through the derivatives and does not affect the identity above. Therefore, $T=0$.

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  • $\begingroup$ So if I wait long enough the wavefunction in region $2$ drops to $0$? $\endgroup$
    – Kashmiri
    Commented Sep 2, 2020 at 14:42
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    $\begingroup$ @Joshua you said "as t→∞, all of the probability will be reflected" I believe that's incorrect. Time won't change $\psi^{2}$ $\endgroup$
    – Kashmiri
    Commented Sep 2, 2020 at 15:32
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    $\begingroup$ @Joshua ,does $ψ2$ vary with time on the right side? $\endgroup$
    – Kashmiri
    Commented Sep 2, 2020 at 16:11
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    $\begingroup$ So $ψ^2$ doesn't vary with time. If that's so then the probability on the right side will never die out. $\endgroup$
    – Kashmiri
    Commented Sep 3, 2020 at 5:46
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    $\begingroup$ I Agree with descheleschilder. $\endgroup$
    – Kashmiri
    Commented Sep 3, 2020 at 7:46
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Filling in the $V$, as given in the problem (the "one step up $V$"), in the Schrödinger equation, you'll get the wavefunction $ψ_2(x)=e^{αx}$ for the positive values of $x$. This means there is a quantum mechanical probability to find a particle with less kinetic energy than the height of the potential ($E\lt{V}$). The classical transmission coefficient will be zero obviously.

And also, $T$ will be zero in the quantum mechanical case. There is no transmission, as the particle doesn't travel up to infinity with a constant velocity. It will reside near the boundary between $V=0$ and $V=V_0$ and probably bounce back. Probably, which QM is all about.

I don't know what more I should write, so...I hope it helped you!

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  • $\begingroup$ You said "It will reside near the boundary between $V=0 $and$ V=V_0$ and probably bounce back " if it resides only near the boundary them how come we got a probability of finding it deep into the second region? I'm sorry I am finding it hard to follow your answer :( $\endgroup$
    – Kashmiri
    Commented Sep 2, 2020 at 14:39
  • $\begingroup$ @YasirSadiq The probability to find the particle deep into the second region is nearly zero. That's why it will probably stay near the boundary. Of course, there is a chance (the squared absolute value of the wavefunction) the particle will reach for infinity, but this chance approaches zero. The state of the particle evolves in time and that's the cause of a probable bounce back (it is also possible that the particle stays on the other side forever). There is no stationary solution for the particle's motion. $\endgroup$ Commented Sep 2, 2020 at 14:56
  • $\begingroup$ So basically ' because the particle doesn't travel to infinity on the right, that makes$T$=$0$ and also gives a finite probability of finding it on the right? Is this what you meant? $\endgroup$
    – Kashmiri
    Commented Sep 3, 2020 at 7:48
  • $\begingroup$ @YasirSadiq You've got it! $\endgroup$ Commented Sep 3, 2020 at 7:49
  • $\begingroup$ In my humble opinion it would be a clearer answer if you delete the 'aside' . Thank you again for your time. $\endgroup$
    – Kashmiri
    Commented Sep 3, 2020 at 8:11

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