It consist of two point masses located at fixed distances from their centre of mass.
The fixed distances between the two masses and the value of the masses are the only two characteristics of the rigid model.
It's used to predict rotational energy of a diatomic molecules. The rotational energy depends on the moment of inertia,
I = μR²
Here, l = moment of inertia
μ - is called reduced mass of two point masses m₁ and m₂ .
Reduced mass μ = m₁ m²
m₁ +m₂
The fundamental schrodinger equation is,
∆²ψ + 8π²m (E−V) ψ = 0
h²
For rigid rotor v = 0 and the laplacian operator ( ∆²) should be in terms of ( x, y, z) or ( r, θ, φ)
∆² = ∂² + ∂² + ∂² [(x,y,z)]
∂x² ∂y² ∂z²
(or)
∆² = ∂² + 2 ∂ + 1² [ sinθ ∂ ]
∂r² γ ∂r r²sinθ ∂θ
(r,θ,φ) + 1 ∂²
r²sin²θ ∂φ²
In the case of rotating system, the laplacian operator should be expressed in terms of r, θ, φ.
Replacing the mass by moment of inertia, the schrodinger equation is,
∆²ψ + 8π² I E ψ = 0
h²
∂²ψ + 2 ∂ψ + 1 [sinθ ∂ψ ] +
∂r² r ∂r r²sinθ ∂θ
1 ∂²ψ + 8π² I E ψ = 0
r²sin²θ ∂φ² h²
In the case of rigid rotor, since r is constant the r part of ∆² should be omitted.
1 [sinθ ∂ψ ] + 1 ∂²ψ +
r²sinθ ∂θ r²sin²θ ∂φ²
8π² I E ψ = 0
h²
This is the Schrodinger equation for rigid rotor.
By solving this equation. We can get the wave function and eigen value for the rigid rotor problem.
The wave function in the above equation is the function of θ and φ.
Therefore, this equation should be solved by separation of variable method.
Ψ = Yθ ζφ
The solution is
ψ (θ, φ) = √( 2l + 1) l − m
4π l + m
Pₗᵐ cosθ, e± iᵐφ
l = 1, 2, 3,......................
m = (2l+1)
m = + l to - l
Pₗᵐ cosθ = degenerate polynomial.
The eigen value,
E = h² l ( l + 1)
8π² I
Super
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