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Quantum mechanics problem

  1. Apr 25, 2005 #1
    "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.
     
  2. jcsd
  3. Apr 25, 2005 #2

    OlderDan

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    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.
     
  4. Apr 25, 2005 #3

    dextercioby

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

    [tex] \langle x,y,z|n_{1},n_{2},n_{3}\rangle =\langle\vec{r}|n_{1},n_{2},n_{3}\rangle [/tex]

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

    Daniel.
     
  5. Apr 25, 2005 #4
    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.

    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.
     
  6. Apr 25, 2005 #5

    dextercioby

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    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 [itex] |n_{1},n_{2},n_{3}\rangle [/itex] uniparticle state is given in the occupation number-energy representation.

    Daniel.
     
  7. Apr 25, 2005 #6
    I thought I'd done that with my working above :confused:.

    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.
     
  8. Apr 25, 2005 #7

    dextercioby

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    I didn't say the reverse... :rolleyes:




    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)...

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

    Daniel.
     
  9. Apr 25, 2005 #8

    OlderDan

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    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: Apr 25, 2005
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