Particle in a box in cartesian coordinates

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The discussion focuses on solving the Schrödinger equation for a particle in a three-dimensional box with zero potential. Participants explore whether the second derivatives in the equation can be treated separately and discuss the implications of varying quantum numbers n_x, n_y, and n_z on energy levels and degeneracy. There is curiosity about efficiently determining specific energy levels, particularly E_14, and how they relate to combinations of quantum numbers. The conversation also references the mathematical challenge of expressing integers as sums of three squares, highlighting the complexity of finding energy states. The thread emphasizes the intricacies of quantum mechanics and the mathematical relationships involved in energy calculations.
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Homework Statement


upload_2015-5-13_14-52-38.png


Homework Equations

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?
 
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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:
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##
 
Is there any easy way to find a given ##E_{n}##? Or do you need to do every combination to get ##E_{14}##?
 
Not too many possibilities to get 14 as the sum of three squares ...
 
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
 
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 !
 
Maylis said:
Is there any easy way to find a given ##E_{n}##? Or do you need to do every combination to get ##E_{14}##?

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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