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My understanding was that the standing wave solution is that of a free particle in the region before it enters the classically forbidden region. Does multiplying the wave function by its complex conjugate not give a constant probability density? The spatial part has the exponential terms which cancel out, and so does the time dependent part. The high energy potential barrier is from x=0 to x=L and the total energy of the particle is less than the potential energy of the barrier. Also outside of the forbidden region the potential energy is zero.

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  • $\begingroup$ Please credit the source of the figure. $\endgroup$ – Ben Crowell Nov 24 '19 at 13:42
  • $\begingroup$ as written now this question is incomplete. $\endgroup$ – ZeroTheHero Nov 24 '19 at 13:57
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The Schrödinger equation gives a standing wave solution at a hard boundary, just like a water wave equation does. A second quantum phenomenon is that the wave exponentially penetrates - tunnels into - the barrier.

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tunneling

Let's try it with this diagram.

My understanding was that the standing wave solution is that of a free particle in the region before it enters the classically forbidden region. Does multiplying the wave function by its complex conjugate not give a constant probability density?

Yes, for a plane wave.

But it is not the probability density shown in the tunneling diagrams, but the wavefunction itself. So on the left a plane wave impinges on a barrier and classically it would reflect or be absorbed. In the quantum mechanical framework a new function describes the particle within the barrier, that can be joined mathematically continuously to the incoming plane wave, and then to an outgoing plane wave of much smaller ampltude. Doing the $Ψ^*Ψ$ gives a smaller probability for the particle to be outside the barrier, with the same energy. It is a schematic of the boundary conditions that allow penetration of a barrier quantum mechanically.

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