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What happens if we have $E=V$, where $E$ is the energy of a incoming particle and $V$ is the height of a square potential barrier? This wiki page actually gives a finite transmission probability for this case. But what does the wave function look like in the barrier region?

Edit:

I just realized that the potential barrier case can be easily solved and the transmission can be calculated to be the one given in wiki. However, things are slightly different if we have a step potential at the origin instead of a square barrier. Even though a step potential is just a square barrier with infinite width, we look at the situation separately.

If we look at the schroedinger equation for the barrier region, which goes from 0 to $\infty$, then we have $$\psi''(x)=0$$ which means $\psi(x)=ax+b$, where $a$ and $b$ are undetermined constants. Suppose for the no-barrier region, the wave function is given by $e^{ikx}+re^{-ikx}$, where $r$ is the reflection coefficient. Then after matching the boundary conditions, we have $1+r=b$ and $1-r=-ia/k$.

If we require that the wave function does not blow up on the potential side, then we must have $a=0$ and consequently we have $r=1$ and $b=2$, which means that even though it becomes all reflected the wave function at the potential region is a non-zero constant function.

So how to explain this peculiarity? Is it because that transmission is not necessarily related to probability density?

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  • $\begingroup$ Hi have you tried to put in the numbers and let E = V in the scattering equation. I would be genuinely interested in the result.... probably like you I have studied it where E < V and the opposite, but never as stated in your post regards $\endgroup$ – user74893 Apr 7 '15 at 18:15
  • $\begingroup$ @irishphysics, thanks for reminding me this, please refer to the edited version of the question. $\endgroup$ – M. Zeng Apr 8 '15 at 9:52
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As you guessed, the probability density is not the same as transmission. In order to have non-zero transmission there has to be nonzero gradient: $$j=\frac{1}{2m}(\psi^\dagger \hat{p}\psi - \psi \hat{p}\psi^\dagger)$$

In the case of a constant probability, there is no gradient. You only have particle "buildup". The reason for this is that you are solving the time-independent Schrodinger equation, which gives the steady-state solutions. This means the system has had a long time to stabilize and reach a steady state. Since the time is so long (essentially infinite), the particle built up is really large (integral of probability from zero to +infinity). If E was lower than V, this build up would be limited to regions near x=0 and have a negative exponential form (also known as the evanescent part), but E=V is an edge case.

Had you solved the time-dependent Schrodinger equation (TDSE), you would observe the build up forms gradually in time. Specifically, assume a particle source at $-\infty$ (which makes this an open system), providing a constant particle flow at fixed energy $E=k^2/2m$ and a potential barrier $V(x)=V_0 u(x)$, where $u(x)$ is the step function. Start solving the TDSE with an initial condition at $t=0$ of $\psi(t=0,x)=\exp(ikx)u(-x)$, which is a crude approximation of the shape of the wavefunction right before it hits the potential barrier.

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  • $\begingroup$ I don't quite get the connection between time-dependence and the particle build up. If the time-dependence is added in, then the only difference is that all the coefficients I have solved will gain a unitary phase factor. How does the phase factor indicate build-up? $\endgroup$ – M. Zeng Jun 20 '15 at 3:27
  • $\begingroup$ I edited my post a bit. You can see that my proposed initial condition is not an eigenstate, therefore it cannot be associated with a single phase factor. Moreover, the system in the example is an open system, because of the source at $-\infty$. $\endgroup$ – Pooya Jun 22 '15 at 3:17

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