THE VIBRATIONAL PARTITION FUNCTION

The partition function and the vibrational energy of a (H₂ N₂ O₂) diatomic molecule is given by,

qv = ∑ɡve ⁻∈v/kT    .............................(1)

For a diatomic molecule vibrating as a simple harmonic oscillator, the vibrational energy obtained by the solution of the schrodinger wave equation is given by

∈ʋ = ∑ᵥ₌₀∞ e⁻ (ʋ+¹/₂) hʋ kT .........(3)

=eʋh/2kT ∑ᵥ₌₀∞ e⁻ʋh/kT  ............(4)

In the case of simple harmonic oscillator, the spacing between the neighbouring energy level is very large compared with the translational or the rotational energy levels. In other words,

hʋ >> kT. Hence summation can't be replaced by integration. The summation can be carried out and  followed. Let x = hʋ/kT

∑ᵛᵥ ₌₀⁼∞ e⁻ᵛʰᵛ kT = ∑ᵛᵥ⁼₌₀∞ e⁻ᵛˣ =1+ e⁻ˣ + e⁻²ˣ + e⁻³ˣ + .........................

=       1        =         1          ............... (5)

1-e⁻ˣ                1-e⁻ hʋ /kT

If we define θvib, the characteristic vibrational temperature of the oscillator as θvib = hʋ/kT, then the vibrational partition function is given by,

               qv=    e⁻θvib/2T  .  ...................(6)

                        1-e⁻θvib/T

According to Heisenberg's uncertainty principle, the oscillator cannot have a total energy of zero because this would imply that both the position and the momentum of the oscillator can be precisely determined. Even in the ground vibrational state, the oscillator possesses an energy hv/2 called zero point energy. It's convenient to measure the vibrational energies from the ground state rather than from zero energy. Hence the vibrational energy expression for simple.

∈v = ʋhʋ (v = 0,1,2,3,.............0)

qv= ∑ᵥ₌₀∞ e⁻ᵛʰᵛ/ᵏᵀ where x = e⁻ʰᵛ/kᵀ

          = 1+ x + x² + x³ +...............

For x <1, 1+ x+ x²+ x³+............1   

                                                 1−x

Since e⁻ʰᵛ/kT <1, we have, 

qv =           1.          = (1-e⁻ʰᵛᵏᵀ⁻¹

          1- e⁻

qv =       1          =       1        = (1- e⁻ˣ)⁻¹

          1- e ⁻θvib/T  1- e⁻ˣ

      Where χ = hʋ/kT

Case(1) : T is very low so that θvib/T>>1. Then e⁻θvib/T is negligible compared to unity in the denominator

         qv = e⁻vib /2T at low temperature. 

Case(2): T is very high so that θvib/T<<1,

Then the expotential in the denominator can be expanded as a series retaining only the first two terms. 

           e⁻θvib/T = 1 - θvib /T +................. 

Hence qv =   T  e⁻θvib /2T at high 

                  θvib 

temperature. The values of υ and hence 

θvib is obtained from the vibrational spectrum of the molecule.