# If the bound states to the infinite square well are stationary waves, why does the ground state consist of half a wavelength?

For a stationary wave, you need two counter-propagating travelling waves that 'reflect' off the boundaries of the well at either end then superimpose in the middle to produce a stationary wave.

If this is the case, then, for the infinite square well why is the ground state a travelling wave?

I call it a travelling wave as it doesn't have a long enough wavelength to 'reflect' of the walls and produce it's own standing wave. The image below may put my confusion into context: The text in green at the end says it all, and it confuses me even more as it just raises yet another question: Do all of the bound states (1st excited, 2nd excited etc.) need 2 counter-propagating (travelling) waves to produce their standing waves?

This question has been bothering me for some time now, and I have searched the internet but can't get a straight answer. I even looked at Born-Von Karman boundary condition in solid state; the density of states argument along with the Karman boundary condition tells us the ground state cannot be as it is shown in the image above.

But, that was solid state physics, not quantum mechanics.

Is there perhaps something else I am missing here?

• The ground state is a standing, not a travelling wave. Analogously, the lowest frequency wave of a guitar string also is a standing wave with a wavelength of half the length of the string. – my2cts Jul 8 '18 at 8:23
• The half wavelength state is long enough to "reflect". You seem to be generating some false expectations in your mind. Also, solid state physics could be thought of a sub-branch, or application of QM. Though they do sometime use semi-classical models to describe the state of phonons, etc. – ggcg Jul 8 '18 at 11:49
• @ggcg Hi, when I wrote that I was speaking loosely. What I meant was 'not long enough to reflect then superimpose with the part of the wave that hasn't been reflected yet'. I can't think of a way of explaining this properly, sorry. – BLAZE Jul 8 '18 at 17:07

## Classical Waves

It seems you are mixing up some things here, so let's start at the beginning: with classical waves. Imagine, for example, a sound wave in a one-dimensional pipe. The wave equation describing the situation is $$k^2 \partial_t^2 A(x,t) = \omega^2 \partial_x^2 A(x,t) ,$$ where $A$ is the wave amplitude, $\omega$ its frequency and $k = \omega / c_s$. The solution of the wave equation describing an incoming (left-to-right) wave is $$A_{\text{in}}(x,t) = A_0 \sin(\omega t + kx + \varphi) ,$$ where $\varphi$ is some phase shift.

Imagine now that the pipe has a closed end at $x=0$. The wave is reflected in such a way that $A(0,t) = 0$ and we have an outgoing wave $$A_{\text{out}}(x,t) = A_0 \sin(\omega t - kx + \varphi_2) .$$ We see that the boundary condition $A(0,t)=0$ requires $\varphi_2 = \varphi + \pi$ so that \begin{align} A(x,t) &= A_{\text{in}}(x,t) + A_{\text{out}}(x,t)\\ &= A_0 ( \sin(\omega t + kx + \varphi) - \sin(\omega t - kx + \varphi)) \\ &= 2A_0 \cos(\omega t + \varphi) \sin(kx) . \end{align}

We have found a standing wave, where $A(x,t)$ is a product of a function of $x$ only, and a function of $t$ only. Finally, if we add another closed end to the pipe at $x=-\ell$, we only get a standing wave if also $\sin(-k\ell) = 0$. This makes only certain frequencies / wavelengths admissible: the length $\ell$ of the pipe will have to be a multiple of half the wavelength.

## Quantum Mechanics

I want to show you that the situation in your case is very analogous. First of all, note that the Schrödinger equation solved in the picture is the time-independent Schrödinger equation (TISE), meaning that the time-dependence of the wave is already separated off. If $\psi(x)$ is a solution of the TISE with energy $E = \hbar\omega$, then the full solution of the Schrödinger equation is actually $$\Psi(x,t) = \psi(x)\, \mathrm e^{-\mathrm i\omega t} .$$ The $t$- and $x$-dependence is already separated, this is a standing wave!

The next point is that yes, this standing wave can again be written as the sum of an incoming and an outgoing wave, as shown on the slide. Plugging in $\psi(x) = A \sin(kx)$, we get $$\Psi(x,t) = \frac{A}{2\mathrm i} \left( \mathrm e^{\mathrm ikx - \mathrm i\omega t} - \mathrm e^{-\mathrm ikx - \mathrm i\omega t} \right) ,$$ which should remind you of the sound wave above. Only because we are talking about the Schrödinger equation instead of the classical wave equation, the real $\sin(\omega t + kx)$ is replaced by $\mathrm e^{\mathrm ikx - \mathrm i\omega t}$.

Finally, the boundary conditions here are the same as for the pipe with two closed ends, and the lowest energy solution is only half a wavelength. This is no reason not to call this a standing wave. In other physical situations, possibly sometimes in solid state physics, the boundary conditions will be different and the ground state will contain a full wavelength. The infinite square well is not one of these situations ;)

• Thanks for your answer, just to be clear; when you say "this is a standing wave!", you are referring to $\psi(x)$ and not the formula above, right? – BLAZE Jul 8 '18 at 9:07
• I think $\psi(x)$ and $\Psi(x,t)$ could both be called "standing wave". – Noiralef Jul 8 '18 at 9:23