The Simple Molecular Vibrations:
The Harmonic Oscillator

The potential energy of diatomic molecule has a minimum on definite distance between the two atoms. The potential energy V(x) expansion in series relative to state of equilibrium x0(for which dV(x)/dx|x0 = 0 ) gives V(x) = V(0) + ½ d²V(x)/dx²|x0 x2. If we choose the zero point of energy V(0) and assign d²V(x)/dx²|x0 as constant k then we'll obtain the potential approximation

V(x) = ½ kx² (x is a value that corresponds to deviation from equilibrium position)

The Schroedinger equation

(−h²/2md²/dx² + ½ kx²) y = E y

Since A and B vibrate in the opposite directions then we will change m on the reduced mass µ = mAmB/mA+mB. If we introduce new distance variable quantity z² = x² (kµ/h²)½ then we'll finally obtain the Schroedinger equation:

(−d²/dz² + z²) y(z) = 2 E/h w·y(z)

where the vibration frequency is as follows w = (k/µ)½ . A reader who is interested in more detailed mathematical solution can find it here.

Result:

Generally it's considered for possible energy levels:

En = h w (n + ½)

where n can take the following values n = 0, 1, 2, .... .

The energy ground state Eo = ½ h w is assigned as the zero point energy. The wavefunction of this lowest energy state is the Gauss function:

yo(z) = p-¼ e−z²/2

The highest possibility of system location in the ground state is for equalibrium that z=0. And in general one can obtain for the wavefunction of n-state

yn(z) = Cn Hn(z) e−z²/2

where constants Cn can be obtained from the normalization condition -¥ò+∞yn* yn dz = 1 in the following way:

Cn = 1/(π½n!2n )½

and the functions Hn(z) are the Hermitian polynomials:

H0(z) = 1
H1(z) = 2z
H2(z) = 4z2 − 2
H3(z) = 8z3 − 12z
H4(z) = 16z4 − 48z2 + 12
H5(z) = 32z5 − 160z3 + 120z
H6(z) = 64z6 - 480z4 + 720z2 − 120

This natural quantum number of vibrations is assigned by v in the spectroscopy (against assignment by n) and therefore the energy levels are as follows Ev = h w (v + ½) where v = 0,1,2,... . And nothing changes by its meaning, that's for sure!

The wavefunction (to the left) and the detection probability (to the right) for four lowest vibrational states. | The detection probability for the vibrational state n=10. The dashed line gives us the classical detection probability.

The highest detection probability of the ground state is in the equilibrium position. The quantity |yn|2 would be higher in the classical turning points region E = V(x).
The dependence if classical oscillator detection probability is given by 1/w(xm²−x²)½. This curve (see dashed line on the right illustration) is assigned to probability density in the range [x, x+dx]; i.e. the highest chance is in the classical turning points xm. One can find that quantum description reaches classical description for higher states.