The particle is moving in x axis only, for this particle the wave function ψ and the eigen value E has to be determined.
The fundamental schrodinger equation is,
∇²ψ + 8π²m (E−V) ψ= 0................... (1)
h²
The laplacian operator ∇² is,
∇² = ∂² + ∂² + ∂²
∂x² ∂y² ∂z²
For the particle moving in x axis alone,
∇² = ∂²
∂x²
Moreover ,the potential energy inside the box is zero,
V= 0
Equation (1) becomes,
∂²ψ + 8π²m E/ ψ =0
∂x² h²
Let ∝² = 8π²m E ...........................(1a)
h²
∴ ∂²ψ + ∝²ψ = 0............................. (2)
∂x²
The solution for equation (2) is
ψ = Ae i∝ˣ
Expanding,
ψ = A {cos∝x + i sin x } eiθ = cos + i sin θ
ψ = A cos∝x + i A sin ∝x [ iA = A’ ]
ψ = A cos ∝x +A′ sin ∝x ................(3)
Now we can consider, the first boundary condition,
When x=0 ; ψ=0
Equation (3) becomes,
0 = A cos ∝ 0 + A′ sin ∝ 0 [cos 0 = 1] [sin 0 = 0 ]
A = 0 ..................(4)
Therefore equation (3) becomes,
ψ = A' sin∝ x ................... (5)
Consider the secondary boundary conditions,
When x = 0 ; ψ = 0
Equation (5) becomes,
O = A' sin ∝ a
This means,
A' ≠ 0 and sin ∝ a = 0
This solution is,
∝ = nπ ........ (6)
a
Equation (5) becomes,
ψ = A' sin ( nπ ) x ........ (7)
Solving the above equation the solution is,
A' = √ 2/a .................. (8)
Substituting equation (8) in (7)
Ψ = √2/a sin ( nπ ) x .......(9)
a
This equation explains the wave nature of the particle moving in the one dimensional box.
The ∝² and ∝² are ∝² = n²π²
a²
∝ = nπ → ∝² = n²π²
a a²
Comparing equation (1a)
∝² = 8π²m E
h²
Comparing n²π² = 8π²m E
a² h²
E = n²h² ...........(10)
8ma²
The above the equation is the energy of the particle moving in one dimensional box.
Summary :-
Eigen function, ψ = √2/a sin ( nπ) x
a
Eigenvalue, E = n²h²
8ma²
Super
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