# How Do You Calculate the Lowest Energy of 10 Electrons in a 3D Quantum Box?

• Nylex
In summary: If there are 10 pi^- mesons in the box, what is the lowest possible value for the summed energy of the pi^- mesons? Show your working.I think that the pi^- meson is a more massive particle and therefore it would have a higher energy.
Nylex
"The energy levels of a particle in a 3D box with sides of length a are:

E = (n_x^2 + n_y^2 + n_z^2)[(h-bar)^2.(pi)^2]/2ma^2

where n_x, n_y and n_z are integers and greater than zero.

If 10 electrons are placed in such a box, what is the lowest possible value for the summed energy of the electrons? Show your working".

I know that the electrons must have different values of n_x, n_y, n_z, but I don't know where to start, besides that. Which values of n_x, n_y and n_z am I meant to use and how do I vary them for different levels? If it was a one dimensional thing, I'd just use n = 1, 2, 3, .., right? For a 3D case, I'm not sure how to deal with having 3 n terms there. Do I need to do stuff with the quantum numbers l and m_l too?

Any help appreciated, thanks.

Nylex said:
"The energy levels of a particle in a 3D box with sides of length a are:

E = (n_x^2 + n_y^2 + n_z^2)[(h-bar)^2.(pi)^2]/2ma^2

where n_x, n_y and n_z are integers and greater than zero.

If 10 electrons are placed in such a box, what is the lowest possible value for the summed energy of the electrons? Show your working".

I know that the electrons must have different values of n_x, n_y, n_z, but I don't know where to start, besides that. Which values of n_x, n_y and n_z am I meant to use and how do I vary them for different levels? If it was a one dimensional thing, I'd just use n = 1, 2, 3, .., right? For a 3D case, I'm not sure how to deal with having 3 n terms there. Do I need to do stuff with the quantum numbers l and m_l too?

Any help appreciated, thanks.

I'm pretty sure I am interpreing this right.

A combination of the ns represents a quantum energy state, so {1,1,1} is the lowest energy state, {2,1,1}, {1,2,1}, and {1,1,2} are three distinct but equal energy states, etc. Electrons are Fermions, so there is a limit to how many electrons can occupy an energy state. You should know how many. Start filling up energy states with electrons from lowest level to higher levels.

Yes,u should identify the quantum # which label one quantum state...They are $n_{1},n_{2} \ \mbox{and} \ n_{3}$.Therefore,a quantum state is describes by the wavefunction

$$\langle x,y,z|n_{1},n_{2},n_{3}\rangle =\langle\vec{r}|n_{1},n_{2},n_{3}\rangle$$

The 10 electron system (presumably noninteracting) is described through a vector in the antisymetrized tensor product of uniparticle spaces ...

Daniel.

OlderDan said:
I'm pretty sure I am interpreing this right.

A combination of the ns represents a quantum energy state, so {1,1,1} is the lowest energy state, {2,1,1}, {1,2,1}, and {1,1,2} are three distinct but equal energy states, etc. Electrons are Fermions, so there is a limit to how many electrons can occupy an energy state. You should know how many. Start filling up energy states with electrons from lowest level to higher levels.

1 electron in each level. Ok, I think I understand. So for the first electron,

n_x^2 + n_y^2 + n_z^2 = 3 => E = [3(h-bar^2).pi^2]/2ma^2.

Second would be n_x^2 + n_y^2 + n_z^2 = 4 + 1 + 1 = 6, E = [6(h-bar^2).pi^2]/2ma^2

Third {3, 2, 1}: n_x^2 + n_y^2 + n_z^2 = 9 + 4 + 1 = 14, E = [14(h-bar)^2.pi^2]/2ma^2

.. and so on? Then just add up once I've done all 10.

dextercioby said:
Yes,u should identify the quantum # which label one quantum state...They are $n_{1},n_{2} \ \mbox{and} \ n_{3}$.Therefore,a quantum state is describes by the wavefunction

$$\langle x,y,z|n_{1},n_{2},n_{3}\rangle =\langle\vec{r}|n_{1},n_{2},n_{3}\rangle$$

The 10 electron system (presumably noninteracting) is described through a vector in the antisymetrized tensor product of uniparticle spaces ...

Daniel.

Honestly, I have no idea what "antisymetrized tensor product of uniparticle spaces" means. We did some of that bracket notation, but I don't really understand.

A single quantum state can be occupied by maximum one electron.Electrons are identical fermionic particles,so the 6-th postulate applies.U need to find 10 different uniparticle states with lowest energy...

The 6-th postulate describes,in this case,what will be the quantum state of the 10 noninteracting electrons.That $|n_{1},n_{2},n_{3}\rangle$ uniparticle state is given in the occupation number-energy representation.

Daniel.

dextercioby said:
A single quantum state can be occupied by maximum one electron.Electrons are identical fermionic particles,so the 6-th postulate applies.U need to find 10 different uniparticle states with lowest energy...

I thought I'd done that with my working above .

The 6-th postulate describes,in this case,what will be the quantum state of the 10 noninteracting electrons.That $|n_{1},n_{2},n_{3}\rangle$ uniparticle state is given in the occupation number-energy representation.

Daniel.

6th postulate?? What are n_1, n_2 and n_3?

Why is spin relevant? The next bit of the question asks about comparing this if there were 10 pi^- mesons in the box and says that the particle is 273 times the mass of an electron and has spin = 0.

Nylex said:
I thought I'd done that with my working above .

I didn't say the reverse...

Nylex said:
6th postulate?? What are n_1, n_2 and n_3?

Labels for the quantum state.Or if u prefer,quantum numbers,or if u prefer,weights of the irreducible representation(s) of the symmetry group(s)...

Nylex said:
Why is spin relevant? The next bit of the question asks about comparing this if there were 10 pi^- mesons in the box and says that the particle is 273 times the mass of an electron and has spin = 0.

Well,the fact that u're asked about $\pi^{-}$ mesons,instead of electrons,should give u a hint...

Daniel.

Nylex said:
1 electron in each level. Ok, I think I understand.

Not just one electron in each level. The energy levels given in the problem correspond to the classical notion of the kinetic energy of the particle, energy of motion. In addition, electrons are said to posses an "intrinsic angular momentum" which has only two states that are often referred to as "spin up" and "spin down". the restriction that no two Fermions be in the same quantum state, combined with the two spin possibilities permits two electrons to occupy any of the energy states described by {n1,n2,n3}

You do not have the proper combinations of ns in your calculations. There are 3 different ways to get "6", and then you should be able to get some "9"s

Last edited:

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