Question
Find the energy and wave function, ψ of a particle inside a 3D-box
Z
Inside; v=0 LX
LZ
Y
Outside; v= ∞
LY
X
- Consider a particle confined to a 3-Dimensional box by infinitely high potential barriers.
- We assume that inside the box, the potential is constant and this constant is equal to zero i.e.
V=0
- The barriers have infinite potential meaning that the particle cannot escape.
- Therefore the boundary conditions imposed are;
V (x, y, z) = 0 if
{ 0<x<Lx
0<y<Ly
0<z<Lz
…………………………………. 1
, - Otherwise V=Infinite, when the particle is outside the box
- Schrodinger wave equation in 3-D is given by;
2 2 2
∂ψ ∂ψ ∂ψ 8π²m
2 + 2 + 2 + (E-V) ψ =0
∂x ∂y ∂z h²
1. When particle is outside the box; V=∞
2 2 2
∂ψ ∂ψ ∂ψ 8π²m
2 + 2 + 2 + (E-∞) ψ =0 ……………….. 2
∂x ∂y ∂z h²
This is possible only when ψ=0 i.e. particle is not outside the box
2. When the particle is inside the box; V=0
2 2 2
∂ψ ∂ψ ∂ψ 8π²m
2 + 2 + 2 + (E-0) ψ =0
∂x ∂y ∂z h²
2 2 2
∂ψ ∂ψ ∂ψ 8π²m
2 + 2 + 2 + E ψ =0 ………………………. 3
∂x ∂y ∂z h²
∇.∇ ψ + K2 ψ = 0 {2nd order differential equation}
According to 1D box:
1. X- component
2
∂ψ 2 nxπ
2 + {K ψ}x = 0 Kx = ……………………….. 4
∂x Lx
Find the energy and wave function, ψ of a particle inside a 3D-box
Z
Inside; v=0 LX
LZ
Y
Outside; v= ∞
LY
X
- Consider a particle confined to a 3-Dimensional box by infinitely high potential barriers.
- We assume that inside the box, the potential is constant and this constant is equal to zero i.e.
V=0
- The barriers have infinite potential meaning that the particle cannot escape.
- Therefore the boundary conditions imposed are;
V (x, y, z) = 0 if
{ 0<x<Lx
0<y<Ly
0<z<Lz
…………………………………. 1
, - Otherwise V=Infinite, when the particle is outside the box
- Schrodinger wave equation in 3-D is given by;
2 2 2
∂ψ ∂ψ ∂ψ 8π²m
2 + 2 + 2 + (E-V) ψ =0
∂x ∂y ∂z h²
1. When particle is outside the box; V=∞
2 2 2
∂ψ ∂ψ ∂ψ 8π²m
2 + 2 + 2 + (E-∞) ψ =0 ……………….. 2
∂x ∂y ∂z h²
This is possible only when ψ=0 i.e. particle is not outside the box
2. When the particle is inside the box; V=0
2 2 2
∂ψ ∂ψ ∂ψ 8π²m
2 + 2 + 2 + (E-0) ψ =0
∂x ∂y ∂z h²
2 2 2
∂ψ ∂ψ ∂ψ 8π²m
2 + 2 + 2 + E ψ =0 ………………………. 3
∂x ∂y ∂z h²
∇.∇ ψ + K2 ψ = 0 {2nd order differential equation}
According to 1D box:
1. X- component
2
∂ψ 2 nxπ
2 + {K ψ}x = 0 Kx = ……………………….. 4
∂x Lx