Particle in a box in cartesian coordinates

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gfd43tg
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Homework Statement


Textbook problem 4.2 on the infinite cubical well: potential zero inside the cube 0 to a in x, y, z and infinite otherwise, with parts (a) stationary states and energies, (b) energies E1 to E6 and degeneracies, (c) degeneracy of E14.


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?
 
Handwritten derivation of the 3D particle-in-a-box Schrödinger equation separated into X(x), Y(y) and Z(z), with boundary conditions and quantised wavenumbers k_x = nπ/a.
Handwritten derivation of a particle in a three-dimensional box: separated sine wavefunctions in x, y and z, normalisation integral, values of Ax, Ay, Az, and energy expressions with quantum numbers nx, ny, nz = 1, 2, 3, ...


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.
 
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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##
 
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.