Rigid rotor - WikiHQ

In rotordynamics, the rigid rotor is a mechanical model of rotating systems. An arbitrary rigid rotor is a 3-dimensional rigid object, such as a top. To orient such an object in space requires three angles, known as Euler angles (ψ, θ, φ). A special rigid rotor is the linear rotor requiring only two angles to describe, for example a diatomic molecule such as HI, HCl, CO. More general molecules are 3-dimensional, such as water (asymmetric rotor), ammonia (symmetric rotor), or methane (spherical rotor).

Linear rotor

The linear rigid rotor model consists of two point masses located at fixed distances from their center of mass. The fixed distance between the two masses and the values of the masses are the only characteristics of the rigid model. However, for many actual diatomic molecules this model is too restrictive since distances are usually not fixed. Corrections on the rigid model can be made to compensate for small variations in the distance. Even in such a case the rigid rotor model is a useful point of departure (zeroth-order model).

Classical linear rigid rotor

The classical linear rotor consists of two point masses <math>m_1</math> and <math>m_2</math> (with reduced mass <math display="inline">\mu = \frac{m_1 m_2}{m_1 + m_2}</math>) at a distance <math>R</math> of each other. The rotor is rigid if <math>R</math> is independent of time. The kinematics of a linear rigid rotor is usually described by means of spherical polar coordinates (<math>\theta, \varphi, R</math> ), which form a coordinate system of R<sup>3</sup>. In the physics convention the coordinates are the co-latitude (zenith) angle <math>\theta \,</math>, the longitudinal (azimuth) angle <math>\varphi\,</math> and the distance <math>R</math>. The angles specify the orientation of the rotor in space. The kinetic energy <math>T</math> of the linear rigid rotor is given by

\begin{align}

2T &= \mu R^2 \left[\dot{\theta}^2 + (\dot\varphi\,\sin\theta)^2\right] \\[1ex]

\mu R^2 \begin{pmatrix} \dot{\theta} & \dot{\varphi} \end{pmatrix}

\begin{pmatrix}

1 & 0 \\

0 & \sin^2\theta \\

\end{pmatrix}

\begin{pmatrix} \dot{\theta} \\ \dot{\varphi} \end{pmatrix} \\[1ex]

\mu \begin{pmatrix}\dot{\theta} & \dot{\varphi}\end{pmatrix}

\begin{pmatrix}

h_\theta^2 & 0 \\

0 & h_\varphi^2 \\

\end{pmatrix}

\begin{pmatrix}\dot{\theta} \\ \dot{\varphi}\end{pmatrix},

\end{align}

</math>

where <math>h_\theta = R\, </math> and <math>h_\varphi= R\sin\theta\,</math> are scale (or Lamé) factors.

Scale factors are of importance for quantum mechanical applications since they enter the Laplacian expressed in curvilinear coordinates. In the case at hand (constant <math>R</math>)

\begin{align}

\nabla^2 &= \frac{1}{h_\theta h_\varphi}\left[

\frac{\partial}{\partial \theta} \frac{h_\varphi}{h_\theta} \frac{\partial}{\partial \theta}

+ \frac{\partial}{\partial \varphi} \frac{h_\theta}{h_\varphi} \frac{\partial}{\partial \varphi}

\right] \\

\frac{1}{R^2}\left[

\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}

\sin\theta\frac{\partial}{\partial\theta}

+ \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}

\right].

\end{align}

</math>

The classical Hamiltonian function of the linear rigid rotor is

H = \frac{1}{2\mu R^2}\left[p^2_{\theta} + \frac{p^2_{\varphi{\sin^2\theta}\right].

</math>

Quantum mechanical linear rigid rotor

The linear rigid rotor model can be used in quantum mechanics to predict the rotational energy of a diatomic molecule. The rotational energy depends on the moment of inertia for the system, <math>I </math>. In the center of mass reference frame, the moment of inertia is equal to:

where <math>\mu</math> is the reduced mass of the molecule and <math>R</math> is the distance between the two atoms.

According to quantum mechanics, the energy levels of a system can be determined by solving the Schrödinger equation:

where <math>\Psi</math> is the wave function and <math>\hat H</math> is the energy (Hamiltonian) operator. For the rigid rotor in a field-free space, the energy operator corresponds to the kinetic energy of the system:

<math display="block">\hat H = - \frac{\hbar^2}{2\mu} \nabla^2</math>

where <math>\hbar</math> is reduced Planck constant (<math>h\over2\pi</math>) and <math>\nabla^2</math> (del squared) is the Laplacian. The Laplacian is given above in terms of spherical polar coordinates. The energy operator written in terms of these coordinates is:

<math display="block">\hat H =- \frac{\hbar^2}{2I} \left [ {1 \over \sin \theta} {\partial \over \partial \theta} \left ( \sin \theta {\partial \over \partial \theta} \right) + {1 \over {\sin^2 \theta {\partial^2 \over \partial \varphi^2} \right]</math>

This operator appears also in the Schrödinger equation of the hydrogen atom after the radial part is separated off. The eigenvalue equation becomes

\hat H Y_\ell^m(\theta, \varphi) = \frac{\hbar^2}{2I} \ell(\ell+1) \, Y_\ell^m(\theta, \varphi).

</math>

The symbol <math>Y_\ell^m (\theta, \varphi)</math> represents a set of functions known as the spherical harmonics. Note that the energy does depend on <math>m \,</math>through <chem>\ell</chem>. The energy

<math display="block"> E_\ell = {\hbar^2 \over 2I} \ell \left (\ell+1\right)</math>

is <math>2\ell+1</math>-fold degenerate: the functions with fixed <math>\ell</math> and <math>m=-\ell,-\ell+1,\dots,\ell</math> have the same energy.

Introducing the rotational constant <math>B</math>, we write,

<math display="block"> E_\ell = B\; \ell \left (\ell+1\right)\quad

\textrm{with}\quad B \equiv \frac{\hbar^2}{2I}.

</math>

In the units of reciprocal length the rotational constant is

The dots over the time-dependent Euler angles on the right hand side indicate time derivatives. Note that a different rotation matrix would result from a different choice of Euler angle convention used.

Lagrange form

Back substitution of the expression of <math>\boldsymbol{\omega}</math> into T gives

the kinetic energy in Lagrange form (as a function of the time derivatives of the Euler angles). In matrix-vector notation,

2 T =

\begin{pmatrix}

\dot{\alpha} & \dot{\beta} & \dot{\gamma}

\end{pmatrix}

\; \mathbf{g} \;

\begin{pmatrix}

\dot{\alpha} \\ \dot{\beta} \\ \dot{\gamma}\\

\end{pmatrix},

</math>

where <math>\mathbf{g}</math> is the metric tensor expressed in Euler angles—a non-orthogonal system of curvilinear coordinates—

\mathbf{g}=

\begin{pmatrix}

I_1 \sin^2\beta \cos^2\gamma+I_2\sin^2\beta\sin^2\gamma+I_3\cos^2\beta &

(I_2-I_1) \sin\beta\sin\gamma\cos\gamma &

I_3\cos\beta \\

(I_2-I_1) \sin\beta\sin\gamma\cos\gamma &

I_1\sin^2\gamma+I_2\cos^2\gamma & 0 \\

I_3\cos\beta & 0 & I_3 \\

\end{pmatrix}.

</math>

Angular momentum form

Often the kinetic energy is written as a function of the angular momentum <math>\mathbf{L}</math> of the rigid rotor. With respect to the body-fixed frame it has the components <math>L_i</math>, and can be shown to be related to the angular velocity,

\mathbf{L} =

\mathbf{I}(0)\;

\boldsymbol{\omega}\quad\hbox{or}\quad L_i = \frac{\partial T}{\partial\omega_i},\;\; i=x,\,y,\,z.

</math>

This angular momentum is a conserved (time-independent) quantity if viewed from a stationary space-fixed frame. Since the body-fixed frame moves (depends on time) the components <math>L_i</math> are not time independent. If we were to represent <math>\mathbf{L}</math> with respect to the stationary space-fixed frame, we would

find time independent expressions for its components.

The kinetic energy is expressed in terms of the angular momentum by

T = \frac{1}{2} \left[ \frac{L_x^2}{I_1} + \frac{L_y^2}{I_2}+ \frac{L_z^2}{I_3}\right].

</math>

Hamilton form

The Hamilton form of the kinetic energy is written in terms of generalized momenta

\begin{pmatrix}

p_\alpha \\

p_\beta \\

p_\gamma \\

\end{pmatrix}

\mathrel\stackrel{\mathrm{def{=}

\begin{pmatrix}

\partial T/{\partial \dot{\alpha\\

\partial T/{\partial \dot{\beta \\

\partial T/{\partial \dot{\gamma \\

\end{pmatrix}

= \mathbf{g}

\begin{pmatrix} \; \,

\dot{\alpha} \\ \dot{\beta} \\ \dot{\gamma}\\

\end{pmatrix},

</math>

where it is used that the <math>\mathbf{g}</math> is symmetric. In Hamilton form the kinetic energy is,

2 T =

\begin{pmatrix}

p_{\alpha} & p_{\beta} & p_{\gamma}

\end{pmatrix}

\; \mathbf{g}^{-1} \;

\begin{pmatrix}

p_{\alpha} \\ p_{\beta} \\ p_{\gamma}\\

\end{pmatrix},

</math>

with the inverse metric tensor given by

\sin^2\beta\; \mathbf{g}^{-1} =

\begin{pmatrix}

\frac{1}{I_1}\cos^2\gamma + \frac{1}{I_2}\sin^2\gamma &

\left(\frac{1}{I_2} - \frac{1}{I_1}\right)\sin\beta\sin\gamma\cos\gamma &

-\frac{1}{I_1}\cos\beta\cos^2\gamma - \frac{1}{I_2}\cos\beta\sin^2\gamma \\

\left(\frac{1}{I_2} - \frac{1}{I_1}\right)\sin\beta\sin\gamma\cos\gamma &

\frac{1}{I_1}\sin^2\beta\sin^2\gamma + \frac{1}{I_2}\sin^2\beta\cos^2\gamma &

\left(\frac{1}{I_1} - \frac{1}{I_2}\right)\sin\beta\cos\beta\sin\gamma\cos\gamma \\

-\frac{1}{I_1}\cos\beta\cos^2\gamma - \frac{1}{I_2}\cos\beta\sin^2\gamma &

\left(\frac{1}{I_1} - \frac{1}{I_2}\right)\sin\beta\cos\beta\sin\gamma\cos\gamma &

\frac{1}{I_1}\cos^2\beta\cos^2\gamma + \frac{1}{I_2}\cos^2\beta\sin^2\gamma + \frac{1}{I_3}\sin^2\beta \\

\end{pmatrix}.

</math>

This inverse tensor is needed to obtain the Laplace-Beltrami operator, which (multiplied by <math>-\hbar^2</math>) gives the quantum mechanical energy operator of the rigid rotor.

The classical Hamiltonian given above can be rewritten to the following expression, which is needed in the phase integral arising in the classical statistical mechanics of rigid rotors,

<math display="block">\begin{align}

T ={} &\frac{1}{2I_1 \sin^2\beta}

\left( \left(p_\alpha - p_\gamma\cos\beta\right) \cos\gamma -

p_\beta\sin\beta\sin\gamma \right)^2 +{} \\

&\frac{1}{2I_2 \sin^2\beta}

\left( \left(p_\alpha - p_\gamma\cos\beta\right) \sin\gamma +

p_\beta\sin\beta\cos\gamma \right)^2 + \frac{p_\gamma^2}{2I_3}. \\

\end{align}</math>

Quantum mechanical rigid rotor

As usual quantization is performed by the replacement of the generalized momenta by operators that give first derivatives with respect to its canonically conjugate variables (positions). Thus,

p_\alpha \longrightarrow -i \hbar \frac{\partial}{\partial \alpha}

</math>

and similarly for <math>p_\beta</math> and <math>p_\gamma</math>. It is remarkable that this rule replaces the fairly complicated function <math>p_\alpha</math> of all three Euler angles, time derivatives of Euler angles, and inertia moments (characterizing the rigid rotor) by a simple differential operator that does not depend on time or inertia moments and differentiates to one Euler angle only.

The quantization rule is sufficient to obtain the operators that correspond with the classical angular momenta. There are two kinds: space-fixed and body-fixed

angular momentum operators. Both are vector operators, i.e., both have three components that transform as vector components among themselves upon rotation of the space-fixed and the body-fixed frame, respectively. The explicit form of the rigid rotor angular momentum operators is given here (but beware, they must be multiplied with <math>\hbar</math>). The body-fixed angular momentum operators are written as <math>\hat{\mathcal{P_i</math>. They satisfy anomalous commutation relations.

The quantization rule is not sufficient to obtain the kinetic energy operator from the classical Hamiltonian. Since classically <math>p_\beta</math> commutes with <math>\cos\beta</math> and <math>\sin\beta</math> and the inverses of these functions, the position of these trigonometric functions in the classical Hamiltonian is arbitrary. After

quantization the commutation does no longer hold and the order of operators and functions in the Hamiltonian (energy operator) becomes a point of concern. Podolsky (for the special case of the symmetric rotor). This is one of the few cases where the Schrödinger equation can be solved analytically. All these cases were solved within a year of the formulation of the Schrödinger equation.)

Nowadays it is common to proceed as follows. It can be shown that <math>\hat{H}</math> can be expressed in body-fixed angular momentum operators (in this proof one must carefully commute differential operators with trigonometric functions). The result has the same appearance as the classical formula expressed in body-fixed coordinates,

\hat{H} = \frac{1}{2}\left[ \frac{\mathcal{P}_x^2}{I_1} + \frac{\mathcal{P}_y^2}{I_2} +

\frac{\mathcal{P}_z^2}{I_3} \right].

</math>

The action of the <math>\hat{\mathcal{P_i</math> on the Wigner D-matrix is simple. In particular

\mathcal{P}^2\, D^j_{m'm}(\alpha,\beta,\gamma)^* = \hbar^2 j(j+1) D^j_{m'm}(\alpha,\beta,\gamma)^* \quad\hbox{with}\quad

\mathcal{P}^2 = \mathcal{P}^2_x + \mathcal{P}_y^2+ \mathcal{P}_z^2,

</math>

so that the Schrödinger equation for the spherical rotor (<math>I=I_1=I_2=I_3</math>) is solved with the <math> (2j+1)^2 </math> degenerate energy equal to <math>\tfrac{\hbar^2 j(j+1)}{2I}</math>.

The symmetric top (= symmetric rotor) is characterized by <math>I_1=I_2</math>. It is a prolate (cigar shaped) top if <math>I_3 < I_1 = I_2</math>. In the latter case we write the Hamiltonian as

\hat{H} = \frac{1}{2}\left[ \frac{\mathcal{P}^2}{I_1} + \mathcal{P}_z^2\left(\frac{1}{I_3}

-\frac{1}{I_1} \right) \right],

</math>

and use that

\mathcal{P}_z^2\, D^j_{m k}(\alpha,\beta,\gamma)^* = \hbar^2 k^2\, D^j_{m k}(\alpha,\beta,\gamma)^*.

</math>

Hence

\hat{H}\,D^j_{m k}(\alpha,\beta,\gamma)^* = E_{jk} D^j_{m k}(\alpha,\beta,\gamma)^*

</math>with

\frac{1}{\hbar^2}E_{jk} = \frac{j(j + 1)}{2I_1} + k^2\left(\frac{1}{2I_3} - \frac{1}{2I_1}\right).

</math>

The eigenvalue <math>E_{j0}</math> is <math>2j+1</math>-fold degenerate, for all eigenfunctions with <math>m = -j, -j+1, \dots, j</math> have the same eigenvalue. The energies with |k| > 0 are <math>2(2j+1)</math>-fold degenerate. This exact solution of the Schrödinger equation of the symmetric top was first found in 1927. At low temperatures, the rotations of molecules (or part thereof) can be frozen. This could be directly visualized by scanning tunneling microscopy, i.e., the stabilization could be explained at higher temperatures by the rotational entropy.

Applications

With the approximation, we can do the following:

Approximate the bond length using the formula above