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 \(\lfloor A \times 10^{37}\rfloor?\)


  • 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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