# Is the monochromatic solution of the free Schroedinger equation a travelling wave? [closed]

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.

• You just wrote down the unnormalizable, unphysical single mode; all free wave packets are superpositions of these, as your text details. A single traveling phase will not give you probability waves, of course. What is the question? – Cosmas Zachos Dec 28 '19 at 22:46
• @ Cosmas Zachos Normally the solution of a differential equation is the solution for the physical situation. It can be undetermined, requiring boundary conditions. But this case is different: the solution is, as you said, unphysical. I suppose that boundary conditions (limiting the region of space for t = 0 for example) are what is missing, but I am not sure. – Claudio Saspinski Dec 28 '19 at 23:33
• No, solving the TDSE is not enough. You need normalizable probability amps. – Cosmas Zachos Dec 29 '19 at 5:16
• I'm voting to reopen because I had no problem understanding the question here. It's a simple conceptual question (how can we interpret a standing wave solution of the Schrödinger equation as a sum of travelling wave solutions?), backed up with technical information about the OP's attempt to do so. – Nathaniel Dec 29 '19 at 9:52

If the spatial part were to be multiplied by (for example) $$\cos \left( \dfrac{\hbar k^2}{2m}t \right)$$ this would be true. In this case you get a standing wave. But multiplying by $$e^{-i\frac{\hbar k^2}{2m}t}$$ yields a travelling wave.
• True. But $f(t)$ being complex yields a travelling wave, regardless of the fact that it's also an oscillation of each point in space. – Milan Dec 29 '19 at 9:39