# why does a particle in a box work like an electron in an atom orbit?

The solution for the Schrödinger equation for a particle in a box, e.g. an electron, is a standing wave with the length $\lambda/2*n$ with $n$ as the number of the orbit or the so called state. Furthermore $n$ is called the principal quantum number.

This standing wave moves up and down with its amplitude but the knots of the standing wave do not move. Now imagine to take these waves and their discrete lengths and wrap them around a circle. What you get, in the opinion of my university professor, is the following:

If I got it right, an electron in a box, considering the Schrödinger equation, has the same matter wave as an electron on an atom orbit. So it moves in the same way if you hit the standing wave.

But I have to say, I do not know how that should work. Why should the particle in the box and the particle on the orbit act identically? And as you can see, there are twice as many half-periods of the wave for each $n$ in the second picture along the orbit as calculated before.

• The boundary conditions of a de Broglie wave-on-a-circle are those of a periodic box, so the states for a circumference $L$ represent only half those for a rigid box of width $L$. – dmckee Feb 5 '18 at 21:19
• Why is that so? Why is $L$ now twice the value it was before? – Kutsubato Feb 6 '18 at 15:41
• Because the boundary conditions are different for the two cases. – dmckee Feb 6 '18 at 18:51
• What are they for the orbit instead of the box? How are they? – Kutsubato Feb 7 '18 at 11:35
• For the rigid box the boundary conditions are $\psi(0) = \psi(L) = 0$, while periodic boundary conditions require $\psi(0) = \psi(L)$ and $\frac{\partial \psi (0)}{\partial t)}= \frac{\partial \psi (L)}{\partial t}$. In each case you have two conditions, but they are different. – dmckee Feb 7 '18 at 21:41

• I do not get that. Aren't the solutions of the Schrödinger equation matter waves? What do they represent if they are not equal to what de-Broglie described, e.g. via $$\lambda =\frac{h}{p}$$ and aren't the solutions of the Schrödinger equation made for electrons (and not entire atoms) as well? – Kutsubato Feb 6 '18 at 8:24
• @Kutsubato - The solutions of the Schrödinger equations are de Broglie waves, i.e., one-dimensional plane waves $$\psi=\psi_0 \exp{ i(\vec k \vec r -\omega t)}$$ with the de Broglie wavelength $\lambda=h/p=2\pi/k$ only in the case of a spatially and temporally constant (zero) potential. Colloquially, also the solutions of the Schrödinger equation, the wave functions, are sometimes called "matter waves". – freecharly Feb 6 '18 at 13:02