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Chemistry 352

Energy Levels of the Harmonic Oscillator

Solving the Schrödinger equation for the harmonic oscillator Hamiltonian reveals a remarkable result: only a discrete set of energy values leads to physically acceptable (normalizable) wavefunctions. These allowed energies are the vibrational energy levels of the system.

Normalizable wavefunctions and allowed energies

The time-independent Schrödinger equation for the harmonic oscillator,

\[ \hat{H}\psi(x) = \left( -\frac{\hbar^2}{2\mu}\frac{d^2}{dx^2} + \frac{1}{2}kx^2 \right)\psi(x) = E\psi(x), \]

admits solutions for many mathematical values of \(E\). However, most of these solutions are not physically acceptable: they diverge as \(|x|\rightarrow\infty\) and cannot be normalized.

Requiring the wavefunction to remain finite and normalizable restricts the allowed energies to a discrete set. Only for these special values of \(E\) do the solutions decay to zero at large displacements.

Quantized vibrational energy levels

The allowed energy levels of the quantum harmonic oscillator are

\[ E_v = \hbar\omega\left(v + \frac{1}{2}\right), \qquad v = 0, 1, 2, \ldots \]

where \(v\) is the vibrational quantum number and \( \omega = \sqrt{k/\mu} \) is the angular vibrational frequency.

The appearance of discrete energy levels is a direct consequence of the boundary conditions imposed by normalizability, not by any explicit restriction placed on the energy in advance.

Zero-point energy

Even in the lowest vibrational state (\(v=0\)), the energy is not zero:

\[ E_0 = \frac{1}{2}\hbar\omega. \]

This zero-point energy reflects the fact that a quantum particle cannot be completely at rest. The simultaneous requirements of confinement and the uncertainty principle prevent both the position and momentum from being exactly zero.

Big idea: the vibrational energy levels of the harmonic oscillator arise because only certain energies produce wavefunctions that remain finite and normalizable. Quantization is therefore a consequence of the mathematical and physical requirements placed on the solutions of the Schrödinger equation.

Worked example: reduced mass of HCl (with \(^{35}\)Cl)

For a diatomic molecule, the reduced mass is

\[ \mu=\frac{m_1m_2}{m_1+m_2}. \]

For HCl, take the isotope \(^{35}\mathrm{Cl}\) and use atomic masses (in atomic mass units, u):

Substitute into the reduced-mass formula:

\[ \mu = \frac{(1.007825)(34.968853)}{1.007825+34.968853}\,\text{u} = \frac{35.2425}{35.976678}\,\text{u} \approx 0.9796\,\text{u}. \]

If you want the reduced mass in SI units, use \(1\,\text{u}=1.66054\times10^{-27}\,\text{kg}\):

\[ \mu \approx (0.9796\,\text{u})(1.66054\times10^{-27}\,\text{kg/u}) \approx 1.63\times10^{-27}\,\text{kg}. \]

Interpretation: because chlorine is much heavier than hydrogen, the reduced mass is close to the mass of hydrogen (but slightly smaller). This means most of the vibrational motion can be thought of as hydrogen moving relative to an almost stationary chlorine atom.

Worked example: force constant of HCl from \(\tilde{\nu}=2990\ \text{cm}^{-1}\)

For a harmonic oscillator, the vibrational angular frequency is \(\omega=\sqrt{k/\mu}\), so the force constant is

\[ k=\mu\omega^2. \]

Spectroscopic vibrational frequencies are usually reported as a wavenumber \(\tilde{\nu}\) (in \(\text{cm}^{-1}\)), which is related to the ordinary frequency \(\nu\) by

\[ \nu = c\,\tilde{\nu}, \qquad \omega = 2\pi\nu = 2\pi c\,\tilde{\nu}. \]

Use the reduced mass from the previous example: \(\mu \approx 1.63\times10^{-27}\ \text{kg}\) (for H\(^{35}\)Cl), and \(c=2.998\times10^{10}\ \text{cm s}^{-1}\).

Step 1: Convert \(\tilde{\nu}\) to \(\omega\)

\[ \omega = 2\pi c\tilde{\nu} = 2\pi(2.998\times10^{10}\ \text{cm s}^{-1})(2990\ \text{cm}^{-1}) \approx 5.63\times10^{14}\ \text{s}^{-1}. \]

Step 2: Solve for the force constant

\[ k = \mu\omega^2 \approx (1.63\times10^{-27}\ \text{kg})(5.63\times10^{14}\ \text{s}^{-1})^2 \approx 5.16\times10^{2}\ \text{N m}^{-1}. \]

Result: \(k \approx 5.2\times10^{2}\ \text{N m}^{-1}\).

Interpretation: this value reflects a fairly stiff (strong) chemical bond. Stronger bonds have larger \(k\) and therefore higher vibrational frequencies.

Your turn

Problem 1
For a quantum harmonic oscillator, what is the energy difference \(\Delta E\) between adjacent vibrational levels?
\(\hbar\omega(v+1)\)
\(\hbar\omega\)
\(\tfrac{1}{2}\hbar\omega\)
\(\hbar\omega^2\)
Problem 2
What is the physical origin of the zero-point energy \(\tfrac{1}{2}\hbar\omega\)?
The uncertainty principle prevents complete rest
The bond potential is anharmonic
The molecule absorbs thermal energy
The force constant is nonzero
Problem 3
Which expression correctly relates the force constant \(k\) to the vibrational wavenumber \(\tilde{\nu}\)?
\(k = \mu c\tilde{\nu}\)
\(k = \mu(2\pi\tilde{\nu})^2\)
\(k = \mu\tilde{\nu}^2\)
\(k = \mu(2\pi c\tilde{\nu})^2\)
Problem 4 (Qualitative)
Give a reason why you would expect the force constant \(k\) in \(\mathrm{O_2}\) to be larger than that in \(\mathrm{I_2}\).
Iodine atoms are lighter than oxygen atoms
I\(_2\) has a smaller reduced mass
O\(_2\) has a stronger, shorter bond
I\(_2\) vibrates more slowly
Problem 5 (Qualitative)
If two diatomic molecules have the same force constant, which one will have the higher vibrational frequency?
The one with the larger reduced mass
The one with the smaller reduced mass
The one with the higher zero-point energy
They will have the same frequency

Key points (one glance)