When we examine the monochromatic solution of the free Schroedinger equation:
$$\psi(t,\mathbf r) = e^{-i\left(\frac{ħk^2}{2m}\right)t}\left(Ae^{i\mathbf{k\cdot r}} + Be^{-i\mathbf{k\cdot r}}\right).$$
It is said that it is a linear combination of a left and a right moving wave.
But looking at the part independent of time, all we can see is a static pattern, where each point of space is associated to a complex number.
And the effect of the time dependent part is not more than making each point oscillates at a given frequency.
It becomes more clear changing the constants A and B:
$$A = \frac{C + D}{2}\; \text{and} \;B = \frac{C - D}{2}.$$
The solution turns to:
$$\psi(t,\mathbf r) = e^{-i\left(\frac{ħk^2}{2m}\right)t}\left(\frac{C}{2}(e^{i\mathbf{k\cdot r}} + e^{-i\mathbf{k\cdot r}}) + \frac{D}{2}(e^{i\mathbf{k\cdot r}} - e^{-i\mathbf{k\cdot r}})\right).$$
If we define:
$$X = \frac{C}{2}(e^{i\mathbf{k\cdot r}} + e^{-i\mathbf{k\cdot r}}),$$ $$Y = -i\frac{D}{2}(e^{i\mathbf{k\cdot r}} - e^{-i\mathbf{k\cdot r}}).$$
The spatial part can be translated to:
$$\psi(0,\mathbf r) = X + iY,$$
where:
$$X = C\cos(\mathbf{k\cdot r}),$$ $$Y = D\sin(\mathbf{k\cdot r}).$$
That is a parametric equation of an ellipse:
$$\frac{X^2}{C^2} + \frac{Y^2}{D^2} = 1.$$
So, the spacial part of the equation is a kind of mapping of each point of space over an ellipse.
If C = D the ellipse becomes a circle. In this case, we have $B = 0$, and the solution becomes:
$$\psi(t,\mathbf r) = e^{-i\left(\frac{ħk^2}{2m}\right)t}Ae^{i\mathbf{k\cdot r}}.$$
My question is: can such a function be called a travelling wave? Or a travelling wave can only be described using Fourier analysis?
I believe that it must have a starting and a final point in the space. And that points move along the time, defining what can be called the wave velocity. In this case it is necessary Fourier analysis to get zero behind and in front of the wave range.
