What is the minimum possible energy for five (non-interacting) spin $\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}$particles of massmin a one dimensional box of length L ? What if the particles were spin-1? What if the particles were spin $\mathbf{-}\frac{\mathbf{3}}{\mathbf{2}}$?

The given data is spin $-\frac{1}{2}$, spin $-1$, spin $-\frac{3}{2}$.

The electron electric dipole moment (EDM)$d_e$is an intrinsic property of an electron such that the potential energy is linearly related to the strength of the electric field.

The electron's EDM must be collinear with the direction of the electron's magnetic moment (spin).

Adding energies for the spin $-1/2$ for five non interacting particles in one dimensional box from the above table gives.

role="math" localid="1660059135505" $$\begin{gathered} E_{\text{int}}=2\frac{{\pi }^2{h}^2}{2m{L}^2}+2\frac{4{\pi }^2{h}^2}{2m{L}^2}+\frac{9{\pi }^2{h}^2}{2m{L}^2} \\ {E}_{\text{int}}=19\frac{{\pi }^2{h}^2}{2m{L}^2} \end{gathered}$$

For the spin-1 case, the particles are bosons and the minimum possible energy for five bosons from the above table is $E_{\text{int}}=5\frac{{\pi }^2{h}^2}{2m{L}^2}$.

For the spin $-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$case, there are up to $2S+1=4$different particles, each with different spin orientation.

So, the minimum possible energy for five fermions from the above table is shown below.

$$\begin{gathered} E_{\text{int}}=4\frac{{\pi }^2{h}^2}{2m{L}^2}+1\frac{4{\pi }^2{h}^2}{2m{L}^2} \\ {E}_{\text{int}}=8\frac{{\pi }^2{h}^2}{2m{L}^2} \end{gathered}$$

Hence, The resultant answer is$19\frac{{\pi }^2{h}^2}{2m{L}^2},5\frac{{\pi }^2{h}^2}{2m{L}^2}$and$8\frac{{\pi }^2{h}^2}{2m{L}^2}$.