Double well model of a molecule
The double well model is a simplified representation used in quantum mechanics to describe certain molecular systems, particularly those with bistable potential energy surfaces. In this model, the potential energy of the molecule is depicted as having two distinct minima separated by a barrier.
In order to explain the double well model of a molecule, we consider an ammonia molecule. In ammonia, there are three H atoms at the corner of a pyramid whose apex is nitrogen atom as shown in figure.
In the figure, the heavier atom N is regarded as fixed and lighter H atom form a rigid equilateral triangle with its axis passing through N. Then the potential energy depends upon only x, the distance between N atom and the plane defined by three H atom.
The shape of the potential V(x) is shown in the figure. Two minima of V(x) (i.e -b and +b) represents the symmetrical configuration of ammonia molecule in which it is classically stable.
At x = 0, the potential barrier is finite. This corresponds to repulsion of N by H atom in the respective plane.
We can deal four atom problem to a single atom problem introducing reduced mass,
Now we consider the bound state i.e E<0.
This double well potential can be approximated by the square well potential. For the convenience, the square well potential can be approximated by the delta potentials as shown in figure. Her the finite potential barrier at x = 0 corresponds to the repulsion of n by H atom when it is in their plane.
In this approximation, the double square well potential becomes
2m/ ℏ² V(x) = - λ/ d [ δ(x+d)+ δ(x-d)] ...i
where λ represents the strength of delta potential. The negative sign indicates the nature if the well is attractive.
The Schrodinger wave equation for E < 0 is
u''(x) - k² u(x) = - λ/ d δ(x)u(x)....ii with
k² = 2mE/ ℏ² ......iii
Single attractive delta function well
The bound state wave function are,
The boundary conditions for the delta potential are given by
i. u(x) is continuous
ii. u'(x) has a jump i.e
This discontinuity of u'(x) can be shown by integrating eq ii
Using eq iv we get
- KA' - KA = - λ/ d u(0)
- 2KA = - λ/ d
K = λ/ 2d ....iv
Thus eigen value equations
E = ℏ²K²/ 2m = ℏ² λ² / 8md² .....vii
In the eq i if x is replaced by -x then we can see the existence of parity. thus we discuss two separate cases here (even and odd solutions)
Even solutions:
The continuity of u(x) gives,
e^(-kd) = c cos hkd ......ix
And applying du(x)/dx at x=d
Substituting from ix
-kc cos hkd - kc sin hkd = - c ( λ/ d) cos hkd
k + k sin hkd/ cos hkd = λ/ d
k [ 1 + tan hkd ] = λ/ d
tan hkd = λ/ kd - 1
Put α = kd and β = λ/α - 1
tan h α = β .....x
For λ = α, (λ/α - 1) -> 0
But at λ = α, tan hα > 0
This indicates that the intersection occurs for α < λ and at intersection point (q1)
λ/α < 2
λ/ k d < 2 => λ < 2 kd
K _even > λ/2 d ....xi
Thus the energy eigen value,
l E l > ℏ² λ² / 8md² ....xii
Odd solutions
The continuity of u(x) gives
e^-kd = C sin hkd ....xiv
Using eqn v for x= d, we get
- ke^-kd - KC cos hkd = - λ/ d e^-kd
Substituting from xiv we obtain
cot hkd = ( λ/k d - 1)
cot h α = (λ/α - 1) = β
tan h α = β^−1 ....xv
To get the eigen value, we should plot eq xv. In the figure, it can be seen that
α = λ/2 , (λ/2 -1 ) -> 0
This suggests that the point of intersection occurs at α < λ/2 (q2)
Thus K _odd < λ/2d ....xvi
l E l < ℏ² λ² / 8md² ....xvii
Thus the odd bound state is less strongly bounded than the even state we get the doublet (Ee, Eo) with the even state deeper than the odd state as shown below. For two walls at the larger separation Eeven and Eodd become identical ( i. e degeneracy). This degeneracy can be removed by bringing the walls close to each other. The difference of energies Eo - Ee gives the Bohr frequency i.e Eo - Ee/ ℏ = w .....xviii
The significance of the double well model lies in its ability to provide a simplified yet insightful description of complex molecular behaviors, enabling researchers to understand, predict, and manipulate molecular processes across various disciplines.
This note is a part of the Physics Repository.