The fundamental schrodinger's equation is,
∂²ψ + 8π²m (E−V) ψ = 0 .................. (1)
h²
The laplacian operator is,
∇² = ∂² + ∂² + ∂²
∂x² ∂y² ∂z²
Moreover for the particle inside the box,
The potential energy is zero [ V=0 ]
Therefore,
∂²ψ + ∂²ψ + ∂²ψ + 8π²m E−ψ = 0..(2)
∂x² ∂y² ∂z² h²
The wave function ψ is a function of x, y, z. This ψ cannot be solved directly but it can be solved using separation of variables method.
ψ (x, y, z) = Xx. Yy. Zz......................... (3)
Here, X= f(x)
Y= f(y)
Z=f(z)
Differentiating equation (3) write x,
∂ ψ = YZ ∂x
∂x ∂x
Again differentiating write x,
∂²ψ = YZ ∂²X , similarly
∂x² ∂x²
∂²ψ = XZ ∂²y
∂y² ∂y²
∂²ψ = XY ∂²z
∂z² ∂z²
Eqution (2) becomes,
YZ ∂²x + XZ ∂²y + XY ∂²z + 8π²mEψ =0
∂x² ∂y² ∂z² h²
Dividing the above eqution by XYZ
1 ∂²x + 1 ∂²y + 1 ∂²z = 8π²mEψ (4)
x ∂x² y ∂y² z ∂z² h²
let,
1 ∂²x = −∝₁² ; 1 ∂²y = −∝²₂; 1 ∂²z =−∝₃²
x ∂x² y ∂y² z ∂z²
−∝₁² −∝²₂−∝₃² = 8π²mEψ
h²
−(∝²₁ + ∝²₂ + ∝²₃) = − 8π²mEψ
h²
∝²₁ + ∝²₂ + ∝₃² = 8π²mEψ ........(5)
h²
Since,
1 ∂²x = − ∝²₁
x ∂x²
(or)
∂²x = − ∝₁²x
∂x²
(or)
∂²x + ∝²₁ x = 0 ...................(6)
∂x²
This equation is similar to the problem of the particles in one dimensional box.
For this problem the solutions are already known.
Solution of equation (5) is,
Xx = √2/a sin ( nx π) x
a
Similarly,
Yy = √2/b sin ( nyπ ) y
b
Zz = √2/c sin ( nzπ ) z
c
Therefore,
Equation (3) becomes
ψ = √ 8 sin ( nxπ)x.sin(nyπ) y.sin (nzπ) z
abc a b c
....................(7)
Using the solution of the particle in one dimensional box the ∝ values are defined as,
∝₁ = nxπ
a
∝₂ = nyπ
b
∝₃ = nzπ
c
Equation (5) becomes,
nx²π² + ny²π² + nz²π² = 8π²mE
a² b² c² h²
π² [ nx² + ny² + nz² ] = 8π²mE
a² b² c² h²
E=(nx² + ny² + nz² ) h² ..............(8)
a² b² c² 8m
This equation gives the energy of the particle inside the three dimensional box.
Summary :-
Eigen function ;
ψ(x,y, z) = √ 8 sin ( nxπ)x. sin( nyπ) y. sin
ʋ a b
( nzπ ). z
c
[ ʋ = a,b,c ]
Eigen value ;
E=(nx² + ny² + nz² ) h²
a² b² c² 8m
Good information
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