Equal Energy Spacing

Why the spacing is constant

The energy eigenvalues of the quantum harmonic oscillator are

The gap between level $n+1$ and level $n$ is

Because the $n$ cancels out exactly, every adjacent pair of levels is separated by the same energy $\hbar\omega$, regardless of which rung of the ladder you are on. This uniform spacing is a characteristic signature of the harmonic oscillator and does not hold for other potentials such as the infinite square well or the hydrogen atom.

Reading the uniform ladder

The spectrum looks like equally-spaced rungs:

| Transition | Energy absorbed | |---|---| | $n=0 \to n=1$ | $\hbar\omega$ | | $n=1 \to n=2$ | $\hbar\omega$ | | $n=2 \to n=3$ | $\hbar\omega$ | | $n=k \to n=k+1$ | $\hbar\omega$ |

Every upward transition requires exactly one quantum $\hbar\omega$ of energy, whether it is supplied by a photon, a phonon, or any other excitation. This is why the quantum harmonic oscillator is the natural model for photon modes in a cavity: absorbing or emitting one photon changes the mode's quantum number by exactly one, and the photon carries energy $\hbar\omega$.

A concrete example: CO vibrational mode

Carbon monoxide (CO) has a strong infrared absorption band near wavenumber $\tilde{\nu} \approx 2143\,\text{cm}^{-1}$. Converting to energy using $E = hc\tilde{\nu}$ with $hc = 1.240 \times 10^{-4}\,\text{eV·cm}$:

The spectrum is dominated by a single strong fundamental absorption at this energy. For an ideal harmonic oscillator the dipole selection rule is strictly $\Delta n = \pm 1$, so only the fundamental would appear; the weaker overtones seen in a real molecule near integer multiples ($2\hbar\omega, 3\hbar\omega, \ldots$) arise precisely because the true potential is slightly anharmonic. Anharmonicity both relaxes the selection rule (allowing the overtones at all) and pulls them below exact integer multiples, but the leading harmonic estimate for the $n=0\to k$ overtone is $k\hbar\omega$.

Connection to ladder operators

The algebraic proof of equal spacing follows directly from the ladder-operator algebra. The raising operator $\hat{a}^\dagger$ adds exactly one quantum: $\hat{a}^\dagger|n\rangle = \sqrt{n+1}\,|n+1\rangle$. Because each application of $\hat{a}^\dagger$ raises the eigenvalue of the number operator $\hat{N} = \hat{a}^\dagger\hat{a}$ by one, and because $\hat{H} = \hbar\omega(\hat{N} + \frac{1}{2})$, each step up the ladder costs exactly $\hbar\omega$ of energy. No approximation is involved; the result is exact for any $n$.

Try it

This is a numerical exercise — your code should return a number. For a harmonic oscillator with $\hbar\omega = 0.18\,\text{eV}$, return the energy gap between any two adjacent levels.

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