This is the following excerpt from 13–2States of definite energy where Feynman tells why $k$ must be real:
[...] Notice that we have been assuming that the number $k$ that we put in our trial solution, Eq. (13.10), was a real number. We can see now why that must be so if we have an infinite line of atoms. Suppose that $k$ were an imaginary number, say $ik′.$ Then the amplitudes an would go as $e^{k′x_n}$, which means that the amplitude would get larger and larger as we go toward large $x$’s—or toward large negative $x$’s if $k′$ is a negative number. This kind of solution would be O.K. if we were dealing with line of atoms that ended, but cannot be a physical solution for an infinite chain of atoms. It would give infinite amplitudes—and, therefore, infinite probabilities—which can’t represent a real situation.
I've few confusions in this: why did he say 'the amplitude would get larger and larger as we go toward large $x$’s'? He writes in previous para
[...] the amplitude goes as a complex oscillation—the magnitude is the same at every atom, but the phase at a given time advances by the amount ($ikb$) from one atom to the next.
Even if $x$ increases, it would increase by an increment of $b$'s the width between two adjacent atoms. But this is just a phase & the magnitude that is $e^{ikx_n}$ remains smae though, as argued by Feynman. But here, it seems to me that instead of taking $e^{ikb}$ as phase, he is including it in the magnitude & thus writes 'the amplitude would get larger and larger as we go toward large $x$’s'. Why is it so? Why did Feynman say the amplitude would increase as $x$ increases when $k$ is complex, after all, the amplitude would also increase with increase in $x$ when $k$ is real, isn't it? In the case of $k$ being imaginary, he includes $e^{ikb}$ as the magnitude while he considers it only as phase with magnitude of amplitude being same at all atoms when $k$ is real. Why is it so? Can anyone please explain this?
