Normalization constant of a planar wave As we know for the plane waves ( $ae^{i k x}+b e^{-i k x}$), the normalization constant can be easily obtained from the integral $\int^{x_{2}}_{x_{1}}\psi^{*}\psi dx=1$ by the relation $|a|^{2}+|b|^{2}=1$. But what happens if the parameter $k$ is imaginary, i.e. $k=i \kappa$ where $\kappa$ is real. Do we have the same relation for the normalization?
 A: Using your parameterization, the wave is $ae^{-\kappa x}+be^{\kappa x}$. Note that this particular wavefunction blows up at $x=+\infty,-\infty$; so that it cannot be normalized unless we impose $a=0$ for $x<0$ and $b=0$ for $x>0$.  If you do this, you can simply carry out an integration to find out the relation between $a$ and $b$ that will normalize the wave.
Remember that $k=\sqrt{2m(E-V)}/\hbar$, so that it will be imaginary in regions where $E<V$. In particular consider a wave incident in $x<0$ on a step potential barrier of height $V_{0}$ for all $x>0$. If $E<V_{0}$, it will have the form $ae^{-\kappa x}$ at $x>0$, so that the wave actually exists inside the barrier even though the incident energy was less than the barrier height. This is how tunneling happens.
A: Plane waves can't be normalised, because they don't represent physically realisable states. It doesn't make sense to normalise a function like $ \psi = ae^{ikx} + be^{-ikx} $ over the boundary $(x_1, x_2)$ unless the particle is bounded, in which case the wavefunction will have a different solution. Another way to think about this: "There's no such thing as a free particle with a definite energy." See Griffiths intro to QM section 2.4
