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Consider the following quantum well: enter image description here

Region 1 is a classically forbidden region, and hence the WKB wave-function will take the form of equation

$$\psi(x) = \frac{C}{\sqrt{q(x)}}e^{+\int_b^a q(x')dx'/\hbar} + \frac{D}{\sqrt{q(x)}}e^{-\int_b^a q(x')dx'/\hbar}. $$

Let us assume that there are no more classical turning points between $x = a$ and $x = -\infty$. In this case, we need to neglect the term in which will “blow up” at minus infinity:

According to my notes the solution in region 1 is as follows:

$$\psi_1(x) = \frac{A_1}{\sqrt{q(x)}}e^{\int_x^a -q(x')dx'/\hbar},\tag{1}$$

whilst the solution in region 3 is as follows: $$\psi_3(x) = \frac{A_2}{\sqrt{q(x)}}e^{\int_b^x -q(x')dx'/\hbar}.\tag{3}$$

Note that $$q(x) = \sqrt{(2m(V(x)- E))}.$$

My question is why that the term for $\psi_1$ contains the negative exponential; surely that will be the one to blow up as negative * negative is positive. Surely if we go to negative infinity then we would want to discard the term with the negative exponential and keep the one with the positive exponential.

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1 Answer 1

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  1. Notice first of all that $q>0$ is positive in the forbidden regions, so the integrand is positive.

  2. Since $x<a$ in region 1, the integral in eq. (1) is positive, so the exponential is damped $<1$, as it should be.

  3. Since $b<x$ in region 3, the integral in eq. (3) is positive, so the exponential is damped $<1$, as it should be.

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