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Does it make sense to think of $e^{ikx}\equiv $cos$(kx)+i$sin$(kx)$ as a right propagating wave? I am rather confused by the imaginary term here.

Context:

\begin{equation} \phi(x)=\begin{cases} Ae^{ik_1x}+Be^{-ik_1x}, & \text{if $x<0$}.\\ Ce^{ik_2x}+De^{-ik_2x}, & \text{if $0 \leq x \leq a$}.\\ Ee^{ik_3x}+Fe^{-ik_3x}, & \text{$x>a$}. \end{cases} \end{equation}

Here we assume the particle is travelling to the right from $x=-\infty$ and $\phi(x)$ are the eigenstates

usainlightning
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1 Answers1

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Propagation is only well-defined if you have a time-dependent component. For the Schrodinger equation, you would always have to multiply by something like $e^{-i \omega t}$ to find the time-dependent part. You would get a wave such as $e^{i(kx-\omega t)}$, which shows that for increasing $t$, we need to have increasing $x$ to keep the same phase and thus the wave is right-propagating. The imaginary term does not matter for this definition.

Troy
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  • Why does $e^{i(kx-\omega t)}$ represent a wave though? How would I sketch this given the expression contains a complex number – usainlightning Feb 15 '16 at 13:18
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    @usainlightning Take a look at circular polarisation: https://en.wikipedia.org/wiki/Circular_polarization – CAGT Feb 15 '16 at 13:34
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    @usainlightning As CAGT says, it can be thought of as circular polarization. Also take a look at David Z's answer: http://physics.stackexchange.com/questions/75363/how-is-the-schroedinger-equation-a-wave-equation. Any function of the form $f(x - \frac{\omega}{k} t)$ satisfies the wave equation with speed $v = \omega/k$, so it can be regarded as a wave. – Troy Feb 15 '16 at 20:22