Quick Potential Barrier tunnelling question

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SUMMARY

The discussion centers on the calculation of transmission and reflection coefficients for a particle tunneling through a finite-width potential barrier, specifically when the energy (E) is less than the potential (V). The wavefunctions are defined for three regions: before the barrier (Aexp(ikx) + Bexp(-ikx)), within the barrier (Cexp(k1x) + Dexp(-k1x)), and after the barrier (Fexp(kx)). The presence of the imaginary unit (i) in the exponentials is attributed to the solutions of the second-order ordinary differential equation derived from the Schrödinger equation.

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I am dealing with the classic problem of a potential barrier of finite width, with a particle tunneling through, in the case of E < V.

I am to calculate the transmission/reflection coefficients, and we first start off with the wavefunctions for the three regions.

Before the barrier, we have Aexp(ikx) + Bexp(-ikx)

In the barrier, we have Cexp(k1x) + Dexp(-k1x)

After the barrier, we have Fexp(kx).

My only question about this is where the i term comes from. It is probably a very simple answer but I want to know exactly its reason before I go on past this point. In the case of E > V, the i term appears in all the exponentials. Thank you.
 
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It comes from the soloution to the second order ordninary differential equation.

Try to see how these wave functions comes, when solving the shrodinger equation for the different regions for your potentia:

"Before the barrier, we have Aexp(ikx) + Bexp(-ikx)

In the barrier, we have Cexp(k1x) + Dexp(-k1x)

After the barrier, we have Fexp(kx)."
 

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