Rigid Rotator

In many cases the molecular rotation spectra of molecules can be described successfully with the assumption that they rotate as rigid rotors. In these cases the energies can be modeled in a manner parallel to the classical description of the rotational kinetic energy of a rigid object. From these descriptions, structural information can be obtained (bond lengths and angles). The most straightforward examples are those of diatomic molecules. Energy calculations in quantum mechanics involve the solution of the Schrodinger equation with a properly formulated Hamiltonian to represent the energy operator. The form of the Hamiltonian can often be implied from the nature of the classical energy of such a physical system. The process involves finding the quantum mechanical operators associated with the constituents of the system energy. The energy of a freely rotating rigid rotor is simply the rotational kinetic energy, which can be expressed in terms of the angular momentum.

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Things to Remember

When the distance between tow point particle is always constant and can rotate about on axis passing through it's centre of mass is called as rigid rotators.
Then the moment of inertia of atomic particle or diatomic molecules about an axis passing through it's centre of mass perpendicular to the axis of rotation is given by.

$I= m_1r_1^2 + m_2 r_2^2$
$E=\frac{1}{2I} \hbar^2k^2$

Which is the energy of rigid rotator.

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Rigid Rotator

Rigid Rotator:

When the distance between tow point particle is always constant and can rotate about on axis passing through it's centre of mass is called as rigid rotators.

Consider the two point masses $m_1$ and $m_2$ separately by a distance 'r'. Almost all of the atomic particle or diatomic molecues at low temperature can act as a rigid rotator.

Then the moment of inertia of atomic particle or diatomic molecules about an axis passing through it's centre of mass perpendicular to the axis of rotation is given by.

$I= m_1r_1^2 + m_2 r_2^2\dotsm(1)$

Since for the particle rotating about centre of mass we have,

$m_1r_1= m_2r_2\dotsm(2)$

Also,

$r= r_1+ r_2\dotsm(3)$

From equation (2)

$r_1=\frac{m_2}{m_1 } r_2$

So, from equation (3)

$r=\frac{m_2}{m_1}r_2+ r+2$

$or,\; r_2= \frac{m_1}{m_1+m_2}r\dotsm(4)$

Now, from equation (1),

$I= m_1\biggl(\frac{m_2 r}{m_1+ m_2}\biggr)^2+ m_2\biggl(\frac{m_2r}{m_1+m_2}\biggr)^2$

$or,\; I= \biggl(\frac{m_1m_2}{m_1+m_2}\biggr)r^2$

$or,\; I=\mu r^2\dotsm(5)$

Where, $\mu= \frac{m_1m_2}{m_1+m_2}$ is the reduced mass of the system for two point masses of the rigid rotator.

Now the potential energy and the energy of rigid rotator is

$P=I\omega$

and

$E=\frac12 I\omega^2\dotsm(6)$

Also we have,

$P=\hbar k$

So, $P=\hbar k= I\omega$

$or,\omega= \frac{\hbar k}{I}$

So from equation (6)

$E= E_k= \frac12 I \biggl(\frac{\hbar k}{I}\biggr)^2$

$E=\frac{1}{2I} \hbar^2k^2$

Which is the energy of rigid rotator.

Reference:

Mathews, P.M and K Venkatesan.A Text Book of Quantum Mechanics.New Delhi: Tata McGraw Hill Publishing Co. Ltd, 1997.
Merzbacher, E.Quantum Mechanics .New York: John Wiley, 1969.
Prakash, S and S Salauja.Quantum Mechanics.Kedar Nath Ram Nath Publishing Co, 2002.
Singh, S.P, M.K and K Singh.Quantum Mechanics.Chand & Company Ltd., 2002.

Lesson

Central Potential Problems

Subject

Physics

Grade

Bachelor of Science

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