# Energy inside cyclobutadiene, a 2D infinite well

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: 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 $$(2,1)$$?

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