As we know the d'Alembert Equation is

$$ \frac{\partial^2\psi}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2} $$

for an undimentional string.

Now if we seek standing wave solution, we put $\psi(x,t)=f(x)g(t)$ can you tell me a physical argument which shows if $\psi$ is a standing wave then $\psi(x,t)=f(x)g(t)$.

I was speaking of this with my maths teacher when he said me he never understood why we use that. My physics teacher has no idea and I don't find anything about it in Feynman lessons.

And I have no idea for tags so I take "waves"

Thank you


1 Answer 1


A very simple way is to look at the nodes of the standing wave. As the name suggests, the nodes are not supposed to move as time goes on.

Suppose $x_0$ is one of the nodes at a particular time instant $t_0$, then we have $\psi(x_0,t_0)=0$. If position and time can be separated, then we have $f(x_0)g(t_0)=0$, which gives us two cases:

  1. $f(x_0)=0$, then we have $\psi(x_0,t)=f(x_0)g(t)=0$ for all $t$, i.e. $x_0$ is always a node of the wave. It doesn't move with time.
  2. $f(x_0)\neq 0$, then $g(t_0)=0$, then we have $\psi(x,t_0)=f(x)g(t_0)=0$ for all $x$, including $x_0$. This only tells us that at this particular time $t_0$ all the $x$ are nodes. Since we need to look at the time evolution characterized by a function $g(t)$ that is not constantly zero, this case is not so interesting.

Using the same argument, you can easily check that if we have functions where $x$ and $t$ are combined in such way that they are not able to be separated, a typical example is the general travelling wave solution $f(x\pm vt)$, then the nodes of the wavefunction will move with time.

  • $\begingroup$ you only show that $f(x)g(t)$ describes standing wave, but not that standing wave is described by $f(x)g(t)$ $\endgroup$ May 14, 2015 at 4:13
  • $\begingroup$ yes, that's why I mentioned in the end that if space and time part in a certain combination that cannot be separated, then the wave will appear to be travelling. Of course it can be elaborated to be slightly more rigorous, but the idea should be pretty clear. $\endgroup$
    – M. Zeng
    May 14, 2015 at 5:13

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