I make some calculations for double square potential well and I want the values to make sense for a real physical system. The energy is in eV. What would be a convenient unit of distance?

On wikipedia one can find the unit $\hbar c/{\rm eV}=1.97\cdot10^{-7}\rm m$. However with this unit everything seems to get nonsensical. For demonstration I calculate ground state of an electron in an infinite potential well: $$E_n=\frac{n^2 h^2}{8mL^2}$$ $$E_1=\frac{h^2}{8mL^2}$$

SI units:

Planck constant $$h_{SI}=6.626\cdot10^{-34}\,\rm Js$$

electron mass $$m_{SI}=9.109\cdot10^{-31}\rm kg$$

I choose: $$L_{SI}=0.390\,\rm nm$$

After pluging into the formula above we obtain $$E_1=3.96\cdot10^{-19}\rm J=2.47\rm eV$$ Which corresponds to example.

Electron volts $$h_{e}=4.136\cdot10^{-15}\,\rm eV\cdot s$$

$$m_{e}=511000\,\frac{\rm eV}{c^2}$$

$$L_e=\frac{0.39\cdot10^{-9}}{1.97\cdot10^{-7}}\frac{\rm \hbar c}{\rm eV}=0.198\cdot10^{-2}\frac{\rm \hbar c}{\rm eV}$$

If I plug it in this time I get: $$E_1=\frac{(4.136\cdot10^{-15})^2}{8\cdot 511000\cdot(0.198\cdot10^{-2})^2}{\rm eV}=1.07\cdot10^{-30}{\rm eV}$$

which is absolute nonsense. I figured that I need $L_e$ to be: $$L_e=\frac{0.39\cdot10^{-9}}{3\cdot10^8}$$ So it should be just divided by the speed of light $c$ but I need the dimension to be correct.

Edit1: Flipped $\frac{\hbar c}{eV}$ (there used to be $\frac{eV}{\hbar c}$ instead), corrected numerical value of $L_e$

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    $\begingroup$ It seems to me you are going about this the wrong way around. The length scale is fixed by the size of your well, and the energy scale follows from this. $\endgroup$ Commented Apr 21, 2017 at 23:02
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    $\begingroup$ The $eV/\hbar c$ should be flipped in $L$, and anyway its numerical value doesn't make sense, you should get something around $10^{-2}$. $\endgroup$
    – Javier
    Commented Apr 21, 2017 at 23:23
  • $\begingroup$ @ZeroTheHero - you mean $[E]=eV/L^2$? That could actually solve another problem too. I hope it will work for my assymetric well. $\endgroup$ Commented Apr 22, 2017 at 10:46
  • $\begingroup$ Actually it's a unit system that is used in particle physics. There should be $\hbar=1 \implies h=2\pi$ then everything is OK $\endgroup$ Commented Apr 22, 2017 at 12:09

2 Answers 2


This is a good example of why it's never a good idea to ignore the constants and hope that they work out correctly. Here's how you avoid doing that kind of mistake.

Your values for $L=0.198\times10^{-2}\:\hbar c/\mathrm{eV}=0.39\:\mathrm{nm}$, $h=4.136\times 10^{-15}\:\mathrm{eV\:s}$, and $m=5.11\times 10^5\:\mathrm{eV}/c^2$ are all correct. Now let's put them into your formula for $E_1$ without forgetting the constants: \begin{align} E_1 & = \frac{h^2}{4mL^2} \\& =\frac{(4.136\times 10^{-15}\:\mathrm{eV\:s})^2}{8(5.11\times 10^5\:\mathrm{eV}/c^2)(0.198\times10^{-2}\:\hbar c/\mathrm{eV})^2} \\& =\frac{(4.136\times 10^{-15})^2}{8\times 5.11\times 10^5\times(0.198\times10^{-2})^2} \times \frac{(\mathrm{eV\:s})^2}{ \mathrm{eV}/c^2(\hbar c/\mathrm{eV})^2} \\& =1.07\times 10^{-30} \frac{(\mathrm{eV\:s})^2}{ \mathrm{eV}/c^2(\hbar c/\mathrm{eV})^2} . \end{align} As you can see, the numerical factor comes out about correct, but the rest of the units come out as one jangled soup. A few cancellations are easy, starting with $c$, but to take this apart you need to go deeper: \begin{align} E_1 & = 1.07\times 10^{-30} \frac{(\mathrm{eV\:s})^2}{ \mathrm{eV}/c^2(\hbar c/\mathrm{eV})^2} \\ & = 1.07\times 10^{-30} \frac{\mathrm{eV\:s^2}}{ (\hbar /\mathrm{eV})^2} \\ & = 1.07\times 10^{-30} \frac{\mathrm{eV^3\:s^2}}{ \hbar^2} \\ & = 1.07\times 10^{-30} \times(2\pi)^2\left(\frac{\mathrm{eV\:s}}{ h}\right)^2\mathrm{eV} \\ & = 1.07\times 10^{-30} \times\left(\frac{2\pi}{4.136\times 10^{-15}}\right)^2\mathrm{eV} \\ & = 2.46\:\mathrm{eV} \end{align} There's no contradiction - just missing numerical factors. Leaving out the units in intermediate steps is OK if all your calculation is in SI units (and if you're OK with forgoing one of the easiest sanity checks that can raise flags immediately if you've e.g. forgotten to put in some factor) but with anything fancier than that, it leaves you exposed to this kind of thing. Work out the unit cancellations explicitly - it's good for a number of reasons.


Your question is not very clear since it seems you have an answer and are looking for the actual question. I do not know if my example is pertinent but molecules have energies in the eV range, so a possible example of application of the double square well is to understand basic properties of the ammonia molecule.

Schematically the molecule looks like this:

enter image description here

In fact, there is another configuration where the nitrogen atom is below the plane defined by the three hydrogen atoms. There are two symmetrically-placed equilibrium positions. The exact potential is in the shape of a W but can be approximated by a double-square well with a bump in the middle:

enter image description here

In the problem the mass should be the reduced mass $$ \frac{1}{\mu}=\frac{3}{m_H}+\frac{1}{m_N} $$ which will work out to about $\sim 2000 \hbox{MeV}/c^2$. The width $2b$ of the well should be related to the distance between equilibrium positions: $b\sim 0.55\times 10^{-10}$. Take the width of the bump $a\sim 0.4b$. The height of the bump can be used to define an energy scale. The last thing required to solve is the dimensionless parameter $\xi=\sqrt{\frac{2\mu V_0a^2}{\hbar^2}}$ which is typically between $1$ and $10$ for molecules, so choosing $\xi=4$ will produce values of $V_0$ in the $10^{-1}$eV range, and a lowest energy level in the right range.

You can tune the parameters in a more sophisticated way to reproduce the known energies of the first two low-lying states: $E_0=0.056$eV and $E_1=0.206$eV. You can play a little more with this and check out that the lowest energy solution will be symmetric, while the solution with energy $E_1$ is antisymmetric. You can also superpose these two solutions to create an initial state which is "mostly" in the configuration where the nitrogen is above the $H$-plane, and compute the inversion rate at which the nitrogen goes from the "above" to the "below" position, i.e. oscillates between the two wells. It's a pretty crude model if you just want to trie things with a double-square well this will provide you with hours of entertainment.

The physics is clear and emphasizes that the length scale of the problem is defined by the molecule itself, i.e. the distance between the equilibrium positions.

  • $\begingroup$ Thanks very much for your answer. I am actually dealing with structure where $b\sim10{\rm\,nm}$, the walls of the well are finite and the bump is not in the middle, so I can't use $\xi$ however I can pretend that L=1 and scale everything to it. I thought previously that it was not possible. $\endgroup$ Commented Apr 22, 2017 at 13:50
  • $\begingroup$ Of course it would be done in the same way except the trancendental equation would be a little more complicated but the physics is the same. Since you have two energies in your problem (the depth of the well and the height of the bump), you should probably use the depth of the well to set the scale. The character of your solution will depend on the energy: if you are solving for E above the bump then it's purely oscillatory. In the example I gave the energies are below the bump so the solution is exponential in the middle part. $\endgroup$ Commented Apr 22, 2017 at 13:55

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