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Particle in a box in cartesian coordinates

  1. May 13, 2015 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    upload_2015-5-13_14-52-38.png

    2. Relevant equations


    3. The attempt at a solution
    a) The schrödinger equation
    $$i \hbar \frac {\partial \Psi}{\partial t} = - \frac {\hbar^{2}}{2m} \nabla^{2} \psi + V \psi $$
    For the case ##0 \le x,y,z \le a##, ##V = 0##
    $$i \hbar \frac {\partial \Psi}{\partial t} = - \frac {\hbar^{2}}{2m} \Big [ \frac {\partial^{2} \psi}{\partial x^{2}} + \frac {\partial^{2} \psi}{\partial y^{2}}+ \frac {\partial^{2} \psi}{\partial z^{2}} \Big ]$$
    But with the sum of the second derivates, is it possible to set each of these equal to the time derivative separately? If so, then why?
     
  2. jcsd
  3. May 13, 2015 #2

    Maylis

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    ImageUploadedByPhysics Forums1431503226.695426.jpg ImageUploadedByPhysics Forums1431503246.977950.jpg

    Okay I have made some progress on this problem, however I am curious, is it the case that ##n_{x} = n_{y} = n_{z}##? If they are allowed to be different numbers, I suspect that is what allows for degenerate energies.
     
    Last edited: May 13, 2015
  4. May 13, 2015 #3

    PeroK

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    Re your first question. If you have separate functions of each variable, then consider holding every spatial variable constant. This shows that the time function must be constant. In general:

    ##T(t) = F(x, y, z) \ \Rightarrow \ T(t) = C = F(x, y, z)##

    And:

    ##X(x) + Y(y) + Z(z) = C \ \Rightarrow \ X(x) = C_x, \ Y(y) = C_y, \ Z(z) = C_z## where ##C_x + C_y + C_z = C##

    But, there's no reason that ##C_x = C_y = C_z = C/3##
     
  5. May 13, 2015 #4

    Maylis

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    Is there any easy way to find a given ##E_{n}##? Or do you need to do every combination to get ##E_{14}##?
     
  6. May 13, 2015 #5

    BvU

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    Not too many possibilities to get 14 as the sum of three squares ...
     
  7. May 13, 2015 #6

    Maylis

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    That is not what is meant by ##E_{14}##, they mean the 14th highest energy level. Not that the sum of the squares is 14
     
  8. May 13, 2015 #7

    BvU

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    From the problem statement wording I'm inclined to say you are right.
    Usually, we call them EN with EN = N E0 -- so I got sidetracked.

    But then why Griffiths thinks E14 is so interesting is a mystery to me. Let us know if you find something !
     
  9. May 13, 2015 #8

    PeroK

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    There's a whole sequence here of the number of ways that each integer can be expressed as a sum of 3 non-zero squares:

    https://oeis.org/A025427

    https://oeis.org/A025427/b025427.txt

    For example:

    66 is the first integer that can be expressed in 3 different ways; 129 in 4 different ways; 194 in 5 different ways; 209 in 6 ways.
     
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