Editing Rotational transition (section)

=== Rotational energy levels ===
The first term in the above nuclear wave function equation corresponds to [[kinetic energy]] of nuclei due to their radial motion. Term {{math|{{sfrac|{{bra|Φ<sub>''s''</sub>}} ''N''<sup>2</sup> {{ket|Φ<sub>''s''</sub>}}|2''μR''<sup>2</sup>}}}} represents rotational kinetic energy of the two nuclei, about their center of mass, in a given electronic state {{math|Φ<sub>''s''</sub>}}. Possible values of the same are different rotational energy levels for the molecule.

[[Angular momentum|Orbital angular momentum]] for the rotational motion of nuclei can be written as
<math display="block"> \mathbf N = \mathbf J - \mathbf L </math>
where {{math|'''J'''}} is the total orbital angular momentum of the whole molecule and {{math|'''L'''}} is the orbital angular momentum of the electrons.
If internuclear vector {{math|'''R'''}} is taken along z-axis, component of {{math|'''N'''}} along z-axis – {{math|''N''<sub>''z''</sub>}} – becomes zero as
<math display="block"> \mathbf N = \mathbf R \times \mathbf P  </math>
Hence
<math display="block"> J_z = L_z </math>
Since molecular wave function Ψ<sub>s</sub> is a simultaneous [[eigenfunction]] of {{math|''J''<sup>2</sup>}} and {{math|''J''<sub>''z''</sub>}},
<math display="block"> J^2 \Psi_s = J(J+1) \hbar^2 \Psi_s  </math>
where J is called [[quantum number|rotational quantum number]] and {{math|''J''}} can be a positive integer or zero.
<math display="block"> J_z \Psi_s = M_j\hbar \Psi_s </math>
where {{math|−''J'' ≤ ''M''<sub>''j''</sub> ≤ ''J''}}.

Also since electronic wave function {{math|Φ<sub>''s''</sub>}} is an eigenfunction of {{math|''L''<sub>''z''</sub>}},
<math display="block"> L_z \Phi_s = \pm \Lambda\hbar \Phi_s </math>
Hence molecular wave function {{math|Ψ<sub>s</sub>}} is also an eigenfunction of {{math|''L''<sub>''z''</sub>}} with eigenvalue {{math|±Λ''ħ''}}.
Since {{math|''L''<sub>''z''</sub>}} and {{math|''J''<sub>''z''</sub>}} are equal, {{math|Ψ<sub>''s''</sub>}} is an eigenfunction of {{math|''J''<sub>''z''</sub>}} with same eigenvalue {{math|±Λ''ħ''}}. As {{math|{{abs|'''J'''}} ≥ ''J''<sub>''z''</sub>}}, we have {{math|''J'' ≥ Λ}}. So possible values of rotational quantum number are
<math display="block"> J = \Lambda, \Lambda +1, \Lambda+2, \dots </math>
Thus molecular wave function {{math|Ψ<sub>''s''</sub>}} is simultaneous eigenfunction of {{math|''J''<sup>2</sup>}}, {{math|''J''<sub>''z''</sub>}} and {{math|''L''<sub>''z''</sub>}}.
Since molecule is in eigenstate of {{ math|''L''<sub>''z''</sub>}}, expectation value of components perpendicular to the direction of z-axis (internuclear line) is zero. Hence
<math display="block"> \langle \Psi_s|L_x|\Psi_s\rangle = \langle L_x \rangle = 0 </math>
and
<math display="block"> \langle \Psi_s|L_y|\Psi_s\rangle = \langle L_y \rangle = 0 </math>
Thus
<math display="block"> \langle \mathbf J . \mathbf L \rangle = \langle J_z L_z \rangle = \langle {L_z}^2 \rangle </math>

Putting all these results together,
<math display="block"> \begin{align}
\langle \Phi_s |N^2|\Phi_s \rangle F_s(\mathbf R) &= \langle \Phi_s | \left(J^2 + L^2 - 2 \mathbf J \cdot \mathbf L\right) |\Phi_s \rangle F_s(\mathbf R) \\
&= \hbar^2 \left[J(J+1)-\Lambda^2\right] F_s(\mathbf R) + \langle \Phi_s | \left({L_x}^2 + {L_y}^2\right) |\Phi_s \rangle F_s(\mathbf R)
\end{align}</math>

The Schrödinger equation for the nuclear wave function can now be rewritten as
<math display="block">- \frac{\hbar^2}{2\mu R^2}\left[ \frac{\partial}{\partial R} \left(R^2 \frac{\partial}{\partial R}\right)- J(J+1)\right]F_s(\mathbf R)+[{E'}_s(R)-E]F_s(\mathbf R) = 0 </math>
where
<math display="block"> {E'}_s(R) = E_s(R) - \frac{\Lambda^2 \hbar^2}{2\mu R^2} + \frac{1}{2\mu R^2} \langle \Phi_s |\left({L_x}^2 + {L_y}^2\right)|\Phi_s \rangle </math>
E&prime;<sub>s</sub> now serves as effective potential in radial nuclear wave function equation.

==== Sigma states ====
Molecular states in which the total orbital momentum of electrons is zero are called [[molecular orbitals|sigma states]]. In sigma states {{math|1=Λ = 0}}. Thus {{math|1=''E''&prime;<sub>s</sub>(''R'') = ''E''<sub>s</sub>(''R'')}}. As nuclear motion for a stable molecule is generally confined to a small interval around {{math|''R''<sub>0</sub>}} where {{math|''R''<sub>0</sub>}} corresponds to internuclear distance for minimum value of potential {{math|''E''<sub>s</sub>(''R''<sub>0</sub>)}}, rotational energies are given by,
<math display="block"> E_r = \frac{\hbar^2}{2\mu {R_0}^2} J(J+1) = \frac{\hbar^2}{2I_0} J(J+1) = BJ(J+1) </math>
with
<math display="block"> J = \Lambda, \Lambda +1, \Lambda+2, \dots </math>
{{math|''I''<sub>0</sub>}} is [[moment of inertia]] of the molecule corresponding to [[mechanical equilibrium|equilibrium]] distance {{math|''R''<sub>0</sub>}} and {{math|''B''}} is called '''rotational constant''' for a given electronic state {{math|Φ<sub>''s''</sub>}}.
Since reduced mass {{math|''μ''}} is much greater than electronic mass, last two terms in the expression of {{math|''E''&prime;<sub>''s''</sub>(''R'')}} are small compared to {{math|''E''<sub>s</sub>}}. Hence even for states other than sigma states, rotational energy is approximately given by above expression.