# WKB approximation difficulty - deciding what term to neglect

Consider the following quantum well:

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.

1. Notice first of all that $$q>0$$ is positive in the forbidden regions, so the integrand is positive.
2. Since $$x in region 1, the integral in eq. (1) is positive, so the exponential is damped $$<1$$, as it should be.
3. Since $$b in region 3, the integral in eq. (3) is positive, so the exponential is damped $$<1$$, as it should be.