# Quantum:- SHO potential and its energy

1. Jan 10, 2010

### indie452

1. The problem statement, all variables and given/known data

the energy levels of a particle mass in a symmetric 3d SHO potential are:

E = (nx + ny + nz + 3/2)*h-bar*$$\sqrt{\frac{C}{m}}$$

C=constant
n=principal quantum number = nx + ny + nz

A) If 10 electrons are in the potential whats the lowest possible value for the total energy of all the electrons?

B) If instead 10 pi- mesons are placed in th same potential what is the lowest possible value of the total energy of the mesons?

3. The attempt at a solution

A)
for 10 electrons with + or - 1/2 spin the l quantum number is 2 as if the electrons were in an atom there would be s, p, d orbitals
This means that n is and integer greater than 2 [lowest n is 3]

so E = (3+3/2)*h-bar*$$\sqrt{\frac{C}{m}}$$)
me = mass of electron

= (9/2)*h-bar*$$\sqrt{\frac{C}{10me}}$$)

i did think that maybe the l quantum number should be 3 but it doesn't need to be in this example it just has 2 as the lowest possible. if all electrons had same spin direction then it would need l=3 due to PEP

B)
would the answer be the same as above? as l=2 but the mass would be 2730me as pi meson is approx 273 times mass of electron
= (9/2)*h-bar*$$\sqrt{\frac{C}{2730me}}$$)

2. Jan 10, 2010

### Matterwave

Consider that the electron is a Fermion and the pi meson is a Boson. The energy levels they can occupy are different.

3. Jan 10, 2010

### vela

Staff Emeritus
The atomic quantum numbers $nlm$ don't apply at all here. That's for an electron in a Coulomb potential. In the 3D SHO, the states are labeled by $n_x$, $n_y$, and $n_z$. The ground state would be $(n_x,n_y,n_z)=(0,0,0)$. At the next energy level, there are three degenerate states, $(1,0,0), (0,1,0), (0,0,1)$, and so on.