Since the energy now depends on K, these levels are only 2⁢J+1 degenerate due to the 2⁢J+1 different M values that arise for each J value. The eigenfunctions |J,M,K> are the same rotation matrix functions as arise for the spherical-top case.

The rotational eigenfunctions and energy levels of a molecule for which all three principal moments of inertia are distinct (a asymmetric top) can not easily be expressed in terms of the angular momentum eigenstates and the J, M, and K quantum numbers. However, given the three principal moments of inertia Ia, Ib, and Ic, a matrix representation of each of the three contributions to the general rotational Hamiltonian in Equation 4.3.5 can be formed within a basis set of the {|J,M,K⟩} rotation matrix functions. This matrix will not be diagonal because the |J,M,K⟩ functions are not eigenfunctions of the asymmetric top Hr⁢o⁢t. However, the matrix can be formed in this basis and subsequently brought to diagonal form by finding its eigenvectors {C n, J,M,K } and its eigenvalues {En}. The vector coefficients express the asymmetric top eigenstates as ψn⁡(θ,φ,χ)=∑J,M,KCn,J,M,K|J,M,K⟩ Because the total angular momentum J2 still commutes with Hr⁢o⁢t, each such eigenstate will contain only one J-value, and hence Ψn can also be labeled by a J quantum number: ψn,J⁡(θ,φ,χ)=∑M,KCn,J,M,K|J,M,K⟩ To form the only non-zero matrix elements of Hr⁢o⁢t within the |J,M,K⟩ basis, one can use the following properties of the rotation-matrix functions: ⟨j,⟩=⟨j,⟩=1/2<j,⟩=h⁡2⁢[J⁢(J+1)−K⁢2], ⟨j,⟩=h2⁡K2 ⟨j⟩=−⟨j⟩=h2⁡[J⁢(J+1)−K⁢(K±1)]⁢1/2⁢[J⁢(J+1)−(K±1)⁢(K±2)]⁢1/2⁢⟨j⟩=0 Each of the elements of Jc2, Ja2, and Jb2 must, of course, be multiplied, respectively, by 1/2⁢Ic, 1/2⁢Ia, and 1/2⁢Ib and summed together to form the matrix representation of Hr⁢o⁢t. The diagonalization of this matrix then provides the asymmetric top energies and wavefunctions.

The fraction FJ is obtained from the rotational partition function. (4.2.22)FJ=(2⁢J+1)⁢(h⁡Bk⁢T)⁢(e−2⁢h⁡B⁢(J+1)k⁢T) The exponential is the Boltzmann factor that accounts for the thermal population of the energy states. The factor 2⁢J+1 in this equation results from the degeneracy of the energy level. The more states there are at a particular energy, the more molecules will be found with that energy. The (h⁡B/k⁢T) factor results from normalization to make the sum of FJ over all values of J equal to 1. At room temperature and below only the ground vibrational state is occupied; so all the molecules (nt⁢o⁢t⁢a⁢l) are in the ground vibrational state. Thus the fraction of molecules in each rotational state in the ground vibrational state must add up to 1. Exercise 4.2.7 Show that the numerator, J⁢(J+1)⁢h⁡B in the exponential of Equation 4.2.22 is the energy of level J. Exercise 4.2.8: Hydrogen Chloride Calculate the relative populations of the lowest ( J=0) and second ( J=1) rotational energy level in the HCl molecule at room temperature. Do the same for the lowest and second vibrational levels of HCl. Compare the results of these calculations. Are Boltzmann populations important to vibrational spectroscopy? Are Boltzmann populations important for rotational spectroscopy? Now we put all these pieces together and develop a master equation for the maximum absorption coefficient for each line in the rotational spectrum, which is identified by the quantum number, J, of the initial state. Start with Equation 4.2.15 and replace μT using Equation 4.2.2. (4.2.23)γmax=C⁡(μ2⁢J+12⁢J+1)⋅Δ⁢n Then replace Δ⁢n using Equation 4.2.20. (4.2.24)γmax=C⁡(μ2⁢J+12⁢J+1)⁢(e−2⁢h⁡B⁢(J+1)k⁢T)⁢nJ Finally replace nJ using Equations 4.2.21 and 4.2.22 to produce (4.2.25)γmax=C⁡[μ2⁢J+12⁢J+1]⁢[e−2⁢h⁡B⁢(J+1)k⁢T]⁢[(2⁢J+1)⁢(h⁡Bk⁢T)⁢(e−2⁢h⁡B⁢(J+1)k⁢T)]⁢nt⁢o⁢t⁢a⁢l Equation 4.2.25 enables us to calculate the relative maximum intensities of the peaks in the rotational spectrum shown in Figure 4.2.2, assuming all molecules are in the lowest energy vibrational state, and predict how this spectrum would change with temperature. The constant C includes the fundamental constants ϵo, c and h, that follow from a more complete derivation of the interaction of radiation with matter. The complete theory also can account for the line shape and width and includes an additional radiation frequency factor.

Equation 4.2.18 can be rewritten as

A linear combination that describes an appropriately antisymmetrized multi-electron wavefunction for any desired orbital configuration is easy to construct for a two-electron system. However, interesting chemical systems usually contain more than two electrons. For these multi-electron systems, a relatively simple scheme for constructing an antisymmetric wavefunction from a product of one-electron functions is to write the wavefunction in the form of a determinant. John Slater introduced this idea so the determinant is called a Slater determinant. The Slater determinant for the two-electron wavefunction of helium is (3.9.27)|ψ⁡(r1,r2)⟩=12⁢|ϕ1⁢s⁡(1)⁢α⁡(1)ϕ1⁢s⁡(1)⁢β⁡(1)ϕ1⁢s⁡(2)⁢α⁡(2)ϕ1⁢s⁡(2)⁢β⁡(2)| We can introduce a shorthand notation for the arbitrary spin-orbital (3.9.28)ϕi⁢α⁡(r)=ϕi⁡α or (3.9.29)ϕi⁢β⁡(r)=ϕi⁡β as determined by the ms quantum number. A shorthand notation for the determinant in Equation 3.9.27 is then (3.9.30)|ψ⁡(r1,r2)⟩=2−12⁢D⁢e⁢t|⁢ϕ1⁢s⁢α⁡(r1)⁢ϕ1⁢s⁢β⁡(r2)| The determinant is written so the electron coordinate changes in going from one row to the next, and the spin orbital changes in going from one column to the next. The advantage of having this recipe is clear if you try to construct an antisymmetric wavefunction that describes the orbital configuration for uranium! Note that the normalization constant is (N!)−12 for N electrons. The generalized Slater determinant for a multi-electron atom with N electrons is then