I have a question about a question in the book Introduction à la théorie quantique by Michèle Desouter, Yves Justum and Xavier Chapuisat (ISBN 9782340-016675).

In this book, the exercise IV page 63 presents the modelisation of pi-electrons inside a 2D infinite well of potential. The well is rectangular with length $L/ \alpha$ and $\alpha L$. We find in the first questions that $E_{n_x,n_y} = \left( \left( \frac{n_x}{\alpha} \right)^2 + \left( \alpha n_y \right)^2 \right)\cdot \varepsilon$ with $\varepsilon = h^2 / 8mL^2$ with $m$ the mass of the electron.

The authors then introduce the molecule of cyclobutadiene, here is a picture below:

Cyclobutadiene's geometry, provided by http://www.chem.ucla.edu

Then they make us notice about the fact that each state can be occupied by at most 2 electrons. They also tell us about the fact that we can reuse the previous formula for a square ( $\alpha = 1$ ) then a rectangle shape because we assume the electrons are independent.

We need to find the total energies of the electrons $\pi$.

I was a bit lost so I looked to the solutions and it says:

Dans le cas du cyclobutadiène carré, on doit placer 4 électrons $\pi$. L'énergie totale est $E_{carré} = 2 \times 2 \varepsilon + 2 \times 5 \varepsilon = 14 \varepsilon$. Dans le cas de la forme rectangle, elle est $E_{rectangle} = 2 \left( \frac{1}{\alpha^2} + \alpha^2 \right) \varepsilon + 2 \left( \frac{4}{\alpha^2} + \alpha^2 \right) \varepsilon $

which translates to:

In the case of a square cyclobutadiene, we must fot 4 $\pi$ electrons. The total energy is $E_{Square} = 2 \times 2 \varepsilon + 2 \times 5 \varepsilon = 14 \varepsilon$. In the case of a rectangle, it is $E_{rectangle} = 2 \left( \frac{1}{\alpha^2} + \alpha^2 \right) \varepsilon + 2 \left( \frac{4}{\alpha^2} + \alpha^2 \right) \varepsilon $

The problem here is that I do not understand where the result come from.

My reasoning and questions:

It seems to fit the fact that 2 electrons have the state $(n_x,n_y) = (1,1)$ and two have the state $(2,1)$. Why are these states occupied? Why can't it be two $(1,1)$ and two $(1,2)$? Why can't it be two $(1,1)$ and one $(1,2)$ and one $(2,1)$?

Thanks in advance

  • 1
    $\begingroup$ This is a really cute pedagogical example. Do Desouter et al. say anything about whether this description actually works well? I would think that a good empirical test of the model would be to look for the excited states using visible-light or infrared spectroscopy. $\endgroup$ – Ben Crowell May 23 at 17:25
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    $\begingroup$ @BenCrowell They qualify this model as it's a "good first approximation" in the question. But this is an exercise for an introduction to quantum theory, so it doesn't have to be a realistic model mandatorily, just a good pedagogical example, as you said. $\endgroup$ – PackSciences May 23 at 17:33

Assuming that $\alpha>1$, the state shown in the answer is the lowest-energy state. Since $\frac{1}{\alpha}<\alpha$ in this case, this means that increasing $n_x$ raises the energy less than increasing $n_y$. Therefore, the state with two electrons with increased $n_x$ is lower in energy than any other possibility.

If instead $\alpha<1$, then the argument is reversed and putting electrons in $(1,2)$ is preferred. In the case $\alpha=1$, it doesn't matter.

  • $\begingroup$ Thank you very much. I forgot to mention that we assume that $\alpha > 1$. This answer convinces me, thank you very much. $\endgroup$ – PackSciences May 23 at 15:57

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