# Symmetry $$\equiv$$ Degeneracy

A classic example in quantum mechanics is the infamous particle in a box. This basically refers to a particle trapped in a region of space bound by insurmountable potentials.

Because it is a bound state, a particle trapped in a 3 dimensional cubic box will have a given quantized energy based on three quantum numbers (three integers > 0) : $$n_x$$,$$n_y$$ and $$n_z$$. This energy is given by

$\large{ E_{n_x,n_y,n_z} = \frac{\pi^2 \hbar^2 }{2mL^2}(n_x ^ 2 + n_y ^ 2 + n_z ^ 2 ) }$

where $$L$$ is the side length of the cube and $$m$$ is the mass of the particle.

Because of this, a given energy can be described by different quantum numbers. For example $$E_{1,2,1}$$ and $$E_{2,1,1}$$ will have the same energy for different values of $$n_x$$,$$n_y$$ and $$n_z$$. Such a state is called degenerate. Our $$3D$$ box has many of these degenerate energies because of the symmetry of a cube.

Suppose we have an electron trapped in a $$0.5m$$ box . Suppose the energy under $$2.26536\times 10^{-34}$$ Joules that has the most degenerate states is A. What is $$\lfloor A \times 10^{37}\rfloor?$$

Examples

• For energies under $$3.61494 \times10^{-36}J$$, $$3.3739 \times 10^{-36} J$$ is the most degenerate because it has six degenerate states : $$E_{1,2,3},E_{1,3,2},E_{1,2,3},E_{2,1,3},E_{2,3,1},E_{3,2,1},E_{3,1,2}$$

Details and assumptions

• For physics reasons, only $$positive$$ quantum numbers are allowed: $$n_x, n_y, n_z \geq 1$$.

• 0 is not a quantum number for a particle in a box as quantum numbers are always positive.

• No knowledge of QM is needed to solve this problem

• The mass of an electron is $$9.109383 \times 10^{-31} kg$$

• $$\hbar$$ is the reduced Plank constant $$\hbar = \frac{h}{2\pi} = 1.054572 \times 10^{-34}$$ where $$h$$ is Plank's constant.

• Picture is not to scale.

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