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How many states

  1. Oct 18, 2006 #1
    The energy levels of a quantum particle confined to a cubical box are:

    [tex] E = \frac{\hbar^2 \pi^2}{2mL^2} (n_{x}^2 + n_{y}^2 + n_{z}^2) [/tex]
    where nx, ny, nz are postiive integers

    Define the dimensionless energy
    [tex] \epsilon = E \frac{2mL^2}{\hbar^2 \pi^2} [/tex]

    and define S(e) to be the number of states less than or equal to e


    a)Compute S(e) (im going to call it S(e) for all integer values of e from 3 to 300
    suppose e was 3, then S(3) is 1. because if epsilon is 3, then all the n's are 1, and that is ground state
    suppose e was 4, then S(4) is 4
    e=9, S(5) = 7
    e=27, S(6) = 8

    do i have to calculate them manually or is there an easier way??

    b)Plot S(e) for all integer values of e from 3, to 300
    trying to figire out the function right now...

    c) One can fit this with a function which is the volume of the positive octant. Derive this expression. It is given by:
    [tex] S (e) = \frac{\pi}{6} e^{3/2} [/tex]

    what is a positive octant?? Not sure where to start here...

    *Note* I have used epsilon and e intechageably... i am sorry if it causes any confusion.*
     
    Last edited: Oct 18, 2006
  2. jcsd
  3. Oct 18, 2006 #2

    quasar987

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    question doesn't make sense.... Do you mean "S(e) is the number of states for wich epsilon is less than or equal to e"? Or something else? ...
     
  4. Oct 18, 2006 #3

    OlderDan

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    I think you are missing the point here.

    [tex] \epsilon = E \frac{2mL^2}{\hbar^2 \pi^2} = (n_{x}^2 + n_{y}^2 + n_{z}^2)[/tex]

    It is impossible for e to ever be 4. It can be 3, the minimum value. It can also be 6, and there are three states that have this energy. The next possible value is 9 and there are three states that have that energy. Etc., etc. If S(e) is the number of states with energy less than or equal to e, then S(5) = S(4) = S(3) = 1
     
  5. Oct 18, 2006 #4
    i edited the question.. it should make sense now...

    and no e is not exponential, e is the dimensionless energy
     
  6. Oct 18, 2006 #5
    ya i was thinking i made a mistake with that

    so minimum e =3 for all n = 1
    then e = 6, one n is 2, others are 1 , degeneracy 3, S(e) = 4
    then e = 9, degen = 3, S(e) = 7
    e = 12, degen = 1, S(e) = 8

    am i on the right track now

    is there an easier way to compute S(e) though?
     
  7. Oct 18, 2006 #6

    StatusX

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    I don't think there's a nice formula for the exact value. But you can approximate it by noting that the number of states is the same as the number of points with integer coefficients in the ball of radius sqrt(e) in R^3.
     
    Last edited: Oct 18, 2006
  8. Oct 18, 2006 #7

    quasar987

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    question still doesn't make sense to me. One that would is

    Given [itex]\epsilon_0[/itex], find the number of states (i.e. the number of ordered triplets [itex](n_x,n_y,n_z)[/itex]) for which [itex]\epsilon \leq \epsilon_0[/itex].
     
  9. Oct 18, 2006 #8
    ok so i got S(e) = 64 for epsilo from 3 to 300

    i manually computed it all

    didnt take too long

    now how do i fit this function... with te given
    [tex] S(e) = \frac{\pi}{6} e^{3/2} [/tex]

    do i assume that [tex] \Sigma(\epsilon) = x \epsilon^{a} [/tex]

    and solve for x and a??
     
    Last edited: Oct 18, 2006
  10. Oct 18, 2006 #9

    quasar987

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    If S(e) is defined as I guessed in post #7, then the question asks you to compute [itex]S(\epsilon_0)[/itex] for epsilon_0 going from 3 to 300. So you have to count 297 of them.

    S(300) = 64? This is an excessively low number provided that for any triplet [itex](n_x,n_y,n_z)[/itex] satisfying [itex]\epsilon(n_x,n_y,n_z) \leq 300[/itex], there are 3!=6 permutations, leading to 5 other (possible distinc) solutions.

    For instance, S(3) = 1, but S(6) = 4 (there's e(1,1,1)=1, and then there's e(1,1,2), e(1,2,1), e(2,1,1)=6.

    Did you count the permutations? They all representent distinct states.
     
  11. Oct 18, 2006 #10

    OlderDan

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    I still don't think the wording makes sense
    is probably supposed to say

    define S(e) to be the number of states with dimensionless energy less than or equal to e

    If in fact S(e) is the number of states with dimensionless energy less than or equal to e, then S(300) is greater than 1000. (10,10,10) would give you e = 300, and all combinations of ns up through 10 have energy 300 or less. There are still more with some n>10. One of the n could be as high as 17 (17^2 = 289)

    Please check your c) also. What you have written says that S(e) is a constant less than 1.
    The positive octant is the region of three dimensional space where all three position coordinates are positive.
     
    Last edited: Oct 18, 2006
  12. Oct 18, 2006 #11
    How OlderDan defined S(e) is waht is meant by the question

    so im not done wth e(10,10,10)??

    there's more huh...

    what do i do after i get to 10,10,10 though??
    im confused....
     
  13. Oct 18, 2006 #12

    quasar987

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    Since people have agreed that there is no way of finding an exact fomula for [itex]\Sigma(\epsilon)[/itex], you have to count all the possible triplets one by one. Work your way methodically from (1,1,1) to (10,10,10). (This HW is stupid)
     
  14. Oct 18, 2006 #13
    ok this is how i went
    state, degeneracy, e, S(e)
    1,1,1 , 1 , 3, 1
    1,2,1 3 6 4
    2,2,1 3 9 7
    2,2,2 1 12 8
    2,3,2 3 17 11
    an so on
    lik that
    and thats how i got 64...
     
  15. Oct 18, 2006 #14

    nrqed

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    About the octant part.
    Think of the n_x, n_y, n_z, giving the locations of points in a 3d grid. Then the energy is basically a constant times the square of the "distance" from the origin, n_x^2+n_y^2 +n_z^2. The points with n_x, n_y and n_z all positive are in the first octant (think of a 3-D coordinate system, the first octant is the 1/8 of the coordinate system for which the x, y and z axis are all positive. Another octant would be with y and z positive but y negative, and so on).

    So S(e) is basically the number of points fitting within a sphere of radius e.

    if you consider very large energies, the points that correspond to the integer values of n_x and n_y and n_z will become close together (relative to the radius of the sphere). Then, the number of states available is simply the volume of the first octant (which is 1/8 * 4/3 Pi R^3 = Pi/6 R^3) divided by the volume of each little cube coresponding to a change of one unit for n_x, n_y and n_z.

    That should give you enough infor to get their formula.

    Hope this helps

    Patrick
     
  16. Oct 18, 2006 #15

    quasar987

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    Did you cover the ones like

    17,1,1 , 3 , 291

    ??

    And after 2,3,2, did you "go back" to 2,3,1?
     
  17. Oct 18, 2006 #16
    but that would mena there are over a 1000 possibilites!!

    this is going to be tedious...
     
  18. Oct 18, 2006 #17

    StatusX

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    You want to find all integer solutions to [itex]{n_1}^2+{n_2}^2 +{n_3}^2 \leq 300[/itex]. All the terms are positive, so none can be greater than 300, so we must have [itex]|n_i| \leq 17[/itex].

    Start with [itex]n_3=17[/itex]. Then you want to count the integer solutions to [itex]{n_1}^2+{n_2}^2 \leq 300-17^2=11[/itex]. Again, you need [itex]n_j^2 \leq 11 => |n_j| \leq 3[/itex], j=1,2.

    So start with [itex]n_2=3[/itex]. Then count the integer solutions to [itex]{n_1}^2 \leq 300-17^2-3^2=2[/itex]. There's just [itex]n_1=\pm 1[/itex]. Next take [itex]n_2=
    2[/itex]. The solutions to [itex]{n_1}^2 \leq 300-17^2-2^2=7[/itex] are [itex]n_1=\pm 1, \pm 2[/itex]. Finally, for [itex]n_2=
    1[/itex], the solutions to [itex]{n_1}^2 \leq 300-17^2-1^2=10[/itex] are [itex]n_1=\pm 1, \pm 2, \pm 3[/itex].

    This counts all the solutions with [itex]n_3=17, n_2>0[/itex]. It's not hard to see there are the same number of solutions with [itex]n_2 <0 [/itex], so putting this all together, there are 24 solutions with [itex]n_3=17[/itex].

    This is the simplest case. You won't be able to do this all the way down to [itex]n_3=1[/itex]. Plus, they said to calculate not just S(300), but S(e) for all e up to 300. That isn't a reasonable request. It wouldn't be too hard to write up a computer program to do it though.
     
    Last edited: Oct 18, 2006
  19. Oct 19, 2006 #18

    OlderDan

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    With some help from Excel I got S(300)=2392 with 479 different combinations of n values. Using the volume of one octant of a sphere of radius sqrt(300) you get 2720.

    Anybody want to check up on me :yuck:
     
  20. Oct 25, 2006 #19
    im familiar with excel... but im not quite sure how you would have done that

    i dont wanna do this manually

    did you have excel make an array of integers and add one every time?
     
  21. Oct 26, 2006 #20

    OlderDan

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    It was at best a semi-automated process. I systematically listed all the possible combinations of three n^2 values whose sums are 300 or less, and then generated the number of permutations possible for each combination (1 or 3 or 6) and summed over the permutations. It was mostly cutting and pasting and sorting before adding up the permutations. Not very sophisticated.
     
    Last edited: Oct 26, 2006
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