Significance of imaginary momentum in the finite potential well In the case of the finite potential well, when the particle enters the classically forbidden region its wavenumber, $k$ is imaginary. However $k=\frac{2\pi}{\lambda}=\frac{p}{\hbar}$ so we get an imaginary momnetum/wavelength.
I vaguely remember my lecturer mentioning that there was some significance to this; perhaps he said that the particle doesn't actually enter the region because it has imaginary momentum, but it is needed to make the mathematics work out? But that can't be right, because there are experimental effects that work only because the particles actually enter the classically forbidden region, and tunneling has been detected. So I must not be remembering him properly...
In either case, what is the significance of the imaginary momentum?
 A: The equation that you have written down is true of planes waves, that would oscillate. Wavelengths are sensibly defined only when there is spatial periodicity in the value of the function. Therefore, the equation $k=\frac{2\pi}{\lambda}=\frac{p}{\hbar}$ is not true for a wave function with no spatial periodicity, since lambda is not a well defined quantity for such a function. 
In the specific case of your step potential, with the incident wave have lesser energy that the barrier height, the solutions to the schrodinger equation, are:
$$\psi_{I}=Ae^{ik_1x}+Be^{-ik_1x}\\ \psi_{II}=Ce^{k_2x}+De^{-k_2 x}\\ \psi_{III}=Ee^{ik_1x}+Fe^{-ik_1x}$$ 
Where $I,II,III$ are the regions where the wave is incident, the region of the barrier and the region where of the outgoing wave respectively. Here $k_1=\sqrt{\frac{2mE}{\hbar^2}}$ and $k_2=\sqrt{\frac{2m(V-E)}{\hbar^2}}$. Notice that $A,B,C,D,E,F$ are determined with appropriate boundary conditions. Let us suppose the barrier is erected from $x=0$ to $x=a$ 
In region $I$ the solutions are naturally periodic and hence we interpret $k_1$ as being equal to $\frac{p}{\hbar}$. Stated another way, $e^{ik_1x}$is an eigenstate of the momentum operator, and hence the state has a definite momentum eigenvalue of $\hbar k_1$. Further, in accordance with the probability interpretation of quantum mechanics, if you observed the particle in region $I$, it would be moving towards the right with probability $|B|^2$ and toward the left with probability $|A|^2$, always carrying a momentum $\hbar k_1 $.
In the region $II$ the solutions exponentially decay with $x$ (When you put in the boundary conditions, you will find $C=0$). This is interpreted as the probability of finding the particle, deeper in the barrier decaying exponentially. This means that quantum mechanics predicts that the probability of finding the particle at a distance $x_1$ from the point where barrier where the barrier is erected(i.e from $x=0$) to be $|D|^2 e^{-2k_2x_1}$. 
If you insisted on measuring momentum in region $II$, you would certainly not measure imaginary momentum! instead notice that the exponentially decaying solutions do not have spatial periodicity. Hence, there are not waves of fixed momentum. They are actually in a superposition of many momenta. Quantum mechanics instead predicts what the probability of measuring a particular momentum would be. This is merely found by expanding $\psi_{II}$ in the $e^{ipx}$ (Often called a fourier transform), which are momentum eigenstates of fixed, real momentum and then squaring the coefficient to get the probability. 
$$\psi_{II}(x)=\sum_{P} a_p e^{ipx}$$
In such an expansion, $|a_p|^2$ is the probability of observing momentum $p$.
Tl;DR: There is no imaginary momentum. Since momentum is a hermitian operator, it must have real eigenvalues. Instead the exponential is interpreted as a reduction in probability of finding the particle deeper in the well. 
