Degeneracy of each of energies - Quantum

[Dassinia]

Hello, I don't understand something in this exercice and i have another question:

Homework Statement

Use separation of variables in Cartesian coordinates to solve the infinite cubical well (or "particle in a box"):

V (x, y, z) = { O,if x, y, z are all between 0 and a;
∞,otherwise.

(a) Find the stationary state wave functions and the corresponding energies.

(b) Cali the distinct energies E 1, E1, E3, ... , in order of increasing energy. Find E1, E2, E3 , E4 , Es, and E6. Determine the degeneracy of each ofthese energies (that is, the number of different states that share the same energy). Recall (Problem 2.42) that degenerate bound states do not occur in one dimension, but
they are common in three dimensions.

(c) What is the degeneracy of E 14, and why is this case interesting?

II/ I was wondering for the lowering and increasing operators a- and a+
When we have a-a+=1/(hω) H-1/2
H is considered as an operator ?

Homework Equations

The Attempt at a Solution

a. The energy : E=h²π²/(2ma²) (n_x²+n_y²+n_z²)
ψ(x,y,z)=(2/a)^3/2sin(k_xx)sin(k_yy)sin(k_zz)
with k_i=n_i²π²/a²

b. The thing that I don't get is that according to the correction that we can find here:
http://physicspages.com/2013/01/05/infinite-square-well-in-three-dimensions
Why the ground state is n_x=n_y=n_z=1 so n=3?
Whyt not for n=n_x²+n_y²+n_z²=1
the combinations 1,0,0 0,1,0 0,0,1 ?

Thanks !

 

[DrClaude]

Why the ground state is n_x=n_y=n_z=1 so n=3?
Whyt not for n=n_x²+n_y²+n_z²=1
the combinations 1,0,0 0,1,0 0,0,1 ?

What happens to the wave function when $n_\alpha=0$ ($\alpha \in \{x,y,z\}$)?

 

[Dassinia]

Hi,
When what ?

 

[DrClaude]

Hi,
When what ?

When any of $n_x$, $n_y$ or $n_z$ is equal to zero.

You ask why we can't take, for instance, $n_x = 1$, $n_y = 0$, $n_z = 0$ as the ground state. What does the wave function for that state look like?

 

[Dassinia]

Oh right, it's equal to 0 everywhere, and this is not possible !
Thank you !