I am reading Randall D. Knight's "Physics for Scientists and Engineers" textbook, and he claims that if matter has wave properties, then a particle in a box will form a standing wave. He then uses the fact that standing waves only exist for certain frequencies (for a given length) to show why energy must be quantized. I tried to find another explanation, but they all seem to rely on the Schrodinger equation, which has not been introduced in the text.

So my question is whether it is true that a particle in a box necessarily forms a standing wave, or is this another assumption that has to be made?

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    $\begingroup$ Anything that follows a wave equation (quantum mechanics, air soundwaves, electromagnetic waves etc, will only have time independent solutions (that is, equilibrium solutions) at specific frequencies that can sustain standing waves. en.wikipedia.org/wiki/Standing_wave $\endgroup$
    – user338734
    Jun 27, 2022 at 0:36
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    $\begingroup$ The question just resolves into the solving a wave equation in the box, where the box walls are non-transparent or have a certain transparency. The solution of those equations are standing waves. $\endgroup$ Jun 27, 2022 at 13:06

4 Answers 4


The particle in a box does not necessarily form a standing wave. In fact in quantum mechanics these are states of definite energy that have trivial time evolution, but superpositions of them have non-trivial time evolution, and so you can have a particle inside the box that is relatively well-localized and bouncing between the two walls, using an appropriate superposition.

The case in point about the Schrödinger equation is just that for a free particle (as we assume it is between the walls) we have $E=p^2/(2m)$ and states of definite energy for such particles are thus states of definite momentum, which are plane waves in quantum mechanics. But we have a slight freedom: $p^2$ means that $\pm p$ have the same energy. When you combine plane waves moving forward and backward in equal measure, you get a standing wave, and this equal mixture is the only way to guarantee, with definite energy, the boundary conditions that the probability goes to 0 at the boundary. But you have to stipulate, to get this solution, that you are seeking one of these solutions of fixed energy and trivial time dependence.


It’s important to make a distinction between two different “particle in a box” setups.

1.) Infinite Potential Well

Imagine that the “box” is a region with $0$ potential energy, and everywhere outside the region has infinite potential energy. Because particle waves describe probability, and it cannot have infinite energy to escape the box, we know that at any given time the particle will be in our zero-potential region. Thus, we will not find the particle on the boundaries of our box, so the probability there is zero. These ends are fixed across all time, making a standing wave. This gives rise to certain discrete energies (waves and energy go hand in hand).

2.) Finite Potential Well

If we have a box with a zero-potential region and then some finite potential energy walls around it, then it does not need to be a standing wave. Quantum mechanics allows for the particle to be found inside of or on the other side of the wall (this is known as quantum tunneling). This is because the particle can “jump” to any finite energy value.

Correction: In 1.), the fixed ends do not imply a standing wave. It tells us the particle COULD be a standing wave. At this point, the equations of quantum mechanics can be solved, and these yield standing wave equations. Before the particle’s energy is observed, however, the wave does not have to be a standing wave.

  • $\begingroup$ This would imply that if indeed the condition is that the particle be in the "box" forever (and therefore never tunnel out of it), it must be a standing wave. That would probably apply to, say, the electron of a hydrogen atom. $\endgroup$ Jun 27, 2022 at 9:37
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    $\begingroup$ Your answer 1.) is not correct. Everything is correct up to "these ends are fixed across all time, making a standing wave". But not all fixed-end solutions are standing waves, and in fact there are infinitely more non-standing-wave solutions than there are standing wave solutions. $\endgroup$ Jun 27, 2022 at 10:39
  • $\begingroup$ @doublefelix thank you for pointing that out! I see where and why I was wrong, and I hope the edits I made clearer that up. $\endgroup$ Jun 27, 2022 at 12:58

It is completely possible for a particle in a box not to be in a standing wave state. In fact there are infinitely many more non-standing wave solutions to the particle in the box than there are standing wave solutions. Any time you add two standing-wave solutions together, with any combination of weights, you get a non-standing-wave solution.

The other possible interpretation of the author is that eventually the particle will transition to a standing wave state. But this is also not true. If you have a superposition of two different energies, they will just change their "relative phase" to one another (basically the angle that their complex numbers form when plotted in the complex plane) and you will never equilibrate to just one energy, which would be a standing wave.

So the author's reasoning is not sound, assuming it was summarized accurately. You can start from the Schrödinger equation to show that energy is quantized (in certain systems, since it isn't ALWAYS) and that is much more sound reasoning. But of course it is more mathematical work.


No. Any "suitably well-behaved" function $\psi_x$ can be the wave function of the particle in a box (infinite well version) provided that it is zero both at and outside the walls of the box. The "suitably well-behaved" part means it must be Lebesgue square-integrable, which basically means that besides having no singularities, it must also not be "too badly" discontinuous. Fortunately, it is impossible to explicitly construct a singularity-free function that is not Lebesgue integrable in standard mathematics (ZFC set theory), and they are likely irrelevant physically because they would require an infinitely elaborate preparation procedure to even try to get.

The only thing that will happen if one has such a wave function is that the energy will no longer be eigenized. That is, it will now acquire a non-trivial probability distribution, or "wave function", just like the position has a wave function. This "energetic wave function" can be found by taking the Hilbert inner product with the standing wave states:

$$\psi_E(E_n) = \langle n|\psi\rangle$$

or more explicitly, if the box is in, say, $[0, L]$ and the positional wave function is $\psi_x$, then

$$\psi_E(E_n) = \int_{0}^{L} \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\ \psi_x(x)\ dx.$$

where one must note that the domain of definition of this energetic wave function (at least in this rendition) is only the permitted energy levels $E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}$.

For example, suppose the particle had a uniform distribution in the box, i.e. no position information at all, beyond that "it is in the box":

$$\psi_x(x) = \begin{cases} \sqrt{\frac{1}{L}} && \text{if $x \in (0, L)$}\\ 0 && \text{otherwise} \end{cases}$$


$$ \begin{align} \psi_E(E_n) &= \int_{0}^{L} \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right) \sqrt{\frac{1}{L}}\ dx\\ &= \frac{\sqrt{2}}{L} \int_{0}^{L} \sin\left(\frac{n\pi x}{L}\right)\ dx\\ &= \begin{cases} 0 && \text{if $n$ is even} \\ \frac{2\sqrt{2}}{n\pi} && \text{if $n$ is odd} \end{cases}\end{align}$$

is the energetic wave function.


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