# Quantum Chemistry/Example 8

Write a question and its solution that quantitatively demonstrates the Heisenberg uncertainty principle holds for the J=0 state of a quantum rigid rotator

## Background Information

### Rigid rotor model

Rigid rotor model for the example molecule HCl

In quantum chemistry, the rigid rotor model is used to describe the rotations in molecules, such as HCl. The assumptions used in the rigid rotor model is that the rotating molecules is rigid, and the changes in bond length that naturally occur in a molecule (such as vibrational) is insignificant compared to the bond length re, and thus, negligible. [1]

The total energy of the system is the summation of the potential energy and the kinetic energy. The potential energy of a rigid rotor is 0 given the assumption that the rigid rotor bond length is constant. As such, the total energy of the system is equal to KE, which is equal to the angular momentum. [1] The energy level (EJ) of a linear rigid rotor model (such as HCl) is given by the equation:

${\displaystyle E_{J}={\frac {\hbar }{2I}}J(J+1)}$  [1]

In which I is the inertia, based on the reduced mass of the diatomic and bond length, and J is the quantum energy level.

Rigid rotor model for reduced mass

The rigid rotor model is 3 dimensional, and for ease of calculations, rather than 2 sets of masses that are used (m1 and m2), 1 reduced mass is used instead (μ). [1]

And in a spherical model of the rigid rotor, there are 2 variables that are used to determine position, the angels θ and ϕ, given that r is the constant bond length re.

Thus, the position of the reduced mass is given by the wave function:

${\displaystyle Y(\theta ,\varphi )}$

### Heisenberg Uncertainty Principle

The Heisenberg uncertainty principle states that the exact position and momentum of a particle cannot be determines at one given point, and the more precisely either is determined, the less certain the other would be.[2]

However, even thou the exact position and momentum cannot be calculated at a given moment, they can be related, and that relation is: [2]

${\displaystyle \bigtriangleup \phi \bigtriangleup L_{z}\geq {\frac {\hbar }{2}}}$

In which ${\displaystyle \hbar }$  is the reduced Planck constant, ${\displaystyle \bigtriangleup \phi }$   is the uncertainty of position, and ${\displaystyle \bigtriangleup L_{z}}$  is the uncertainty of momentum.

However, in the rigid rotor model, this equation does not work. The Heisenberg inequality is recalculated: [2]

${\displaystyle \bigtriangleup \phi \bigtriangleup L_{z}\geq {\frac {1}{2}}|<[L_{z},\phi ]>|={\frac {\hbar }{2}}[1-2\pi |\psi (0)^{2}]}$

${\displaystyle \bigtriangleup \phi \bigtriangleup L_{z}\geq {\frac {\hbar }{2}}[1-2\pi |\psi (0)^{2}]}$

## Example

#### Question: using the rigid rotor model at J=0, what is the moment of angular momentum and position? Does the rigid rotor follow Heisenberg’s principle at J=0?

In the ground state zero-point, J = 0, and thus, the energy level of the rigid rotor is 0.

Given that the energy is 0, the angular momentum is known (Lz = 0).

The probability density of the position can be obtained using the wavefunction:

Average of position = 0, as such, the variance is calculated:

${\displaystyle \delta \phi ={\sqrt {<\varphi ^{2}>-<\varphi >^{2}}}}$

${\displaystyle <\varphi >=\int _{0}^{2\pi }(Y^{*})(\varphi )(Y)d\varphi }$

Y = ${\displaystyle i^{m+|m|}{\sqrt {{\frac {2J+1}{2}}{\frac {(J-|m|)!}{(J+|m|)!}}}}.P_{0}^{0}cos(\theta )}$

at m=0 J=0

Y = ${\displaystyle {\frac {1}{\sqrt {4\pi }}}}$

${\displaystyle <\varphi >=\int _{0}^{2\pi }({\frac {1}{\sqrt {4\pi }}})(\varphi )({\frac {1}{\sqrt {4\pi }}})d\varphi }$

${\displaystyle <\varphi >=({\frac {1}{\sqrt {4\pi }}})^{2}\int _{0}^{2\pi }(\varphi )d\varphi }$

${\displaystyle <\varphi >=({\frac {1}{4\pi }})2\pi ^{2}={\frac {\pi }{2}}}$

${\displaystyle <\varphi >^{2}={\frac {\pi ^{2}}{4}}}$

${\displaystyle <\varphi ^{2}>=\int _{0}^{2\pi }(Y^{*})(\varphi )^{2}(Y)d\varphi }$

${\displaystyle <\varphi ^{2}>=\int _{0}^{2\pi }({\frac {1}{\sqrt {4\pi }}})(\varphi ^{2})({\frac {1}{\sqrt {4\pi }}})d\varphi }$

${\displaystyle <\varphi ^{2}>=({\frac {1}{\sqrt {4\pi }}})^{2}\int _{0}^{2\pi }(\varphi ^{2})d\varphi }$

${\displaystyle <\varphi ^{2}>=({\frac {1}{4\pi }}){\frac {8\pi ^{3}}{3}}={\frac {2\pi ^{2}}{3}}}$

${\displaystyle \delta \phi ={\sqrt {<\varphi ^{2}>-<\varphi >^{2}}}={\sqrt {{\frac {2\pi ^{2}}{3}}-{\frac {\pi ^{2}}{4}}}}=\pm {\frac {\sqrt {15}}{6}}\pi }$

if ${\displaystyle \delta \phi }$  is plotted on a sphere

Where m = integers between J and – J.

|m| = number of longitudinal node

J = number of latitudinal nodes

At ground state, J = 0, |m| = 0. Thus, probability of position is equally spread across the sphere.

at J=0,

${\displaystyle {\frac {\hbar }{2}}[1-2\pi |\psi (0)^{2}]=0}$

thus

${\displaystyle \bigtriangleup \phi \bigtriangleup L_{z}\geq {\frac {\hbar }{2}}[1-2\pi |\psi (0)^{2}]}$

${\displaystyle (\pm {\frac {\sqrt {15}}{6}}\pi )(0)\geq 0}$

is satisfied

Thus, this shows that at ground state, J=0, the rigid rotor model follows the Heisenberg uncertainty.

## References

[1] Anderson, J.M. Introduction to Quantum Chemistry, 1969, W.A. Benjamin, Inc, pg.91-100.

[2] Fayngold & Fayngold, Quantum Mechanics and Quantum Information, pp 384-388.

1. Anderson, J.M. Introduction to Quantum Chemistry, 1969, W.A. Benjamin, Inc, pg.91-100.
2. a b c d Fayngold & Fayngold, Quantum Mechanics and Quantum Information, pp 384-388.