- #1
Dassinia
- 144
- 0
Hello, I don't understand something in this exercice and i have another question:
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 ?
a. The energy : E=h2π2/(2ma2) (nx2+ny2+nz2)
ψ(x,y,z)=(2/a)3/2sin(kxx)sin(kyy)sin(kzz)
with ki=ni2π2/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 nx=ny=nz=1 so n=3?
Whyt not for n=nx2+ny2+nz2=1
the combinations 1,0,0 0,1,0 0,0,1 ?
Thanks !
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) (nx2+ny2+nz2)
ψ(x,y,z)=(2/a)3/2sin(kxx)sin(kyy)sin(kzz)
with ki=ni2π2/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 nx=ny=nz=1 so n=3?
Whyt not for n=nx2+ny2+nz2=1
the combinations 1,0,0 0,1,0 0,0,1 ?
Thanks !