According to the National Radio Astronomy Observatory, the $21\,{\rm cm}$ line of hydrogen can be emitted by the neutral hydrogen atom due to the magnetic interaction between the spins of the proton and electron. However, it says the hydrogen molecule ${\rm H}_2$ does not posses a permanent dipole, and thus does not emit a spectral line at radio frequencies. Why is that the interaction between the spins of the proton and electrons no longer manifests itself when we are talking about the ${\rm H}_2$ molecule?
[Physics] Why doesn’t the ${\rm H}_2$ molecule have a permanent dipole while the neutral ${\rm H}\,{\rm\small I}$ has one
astronomyhydrogenmoleculesradio frequencyspectroscopy
Related Solutions
This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation? After all, there's no particular reason for an electron to be in an eigenstate.
Good question! The function $\psi$ does not need to be Hamiltonian eigenfunction. Whatever the initial $\psi$ and whatever the method used to find future $\psi(t)$, the time-dependent Schroedinger equation $$ \partial_t \psi = \frac{1}{i\hbar}\hat{H}\psi $$ implies that the atom will radiate EM waves with spectrum sharply peaked at the frequencies given by the famous formula $$ \omega_{mn} = \frac{E_m-E_n}{\hbar}, $$
where $E_m$ are eigenvalues of the Hamiltonian $\hat{H}$ of the atom.
Here is why. The radiation frequency is given by the frequency of oscillation of the expected average electric moment of the atom
$$ \boldsymbol{\mu}(t) = \int\psi^*(\mathbf r,t) q\mathbf r\psi(\mathbf r,t) d^3\mathbf r $$ The time evolution of $\psi(\mathbf r,t)$ is determined by the Hamiltonian $\hat{H}$. The most simple way to find approximate value of $\boldsymbol{\mu}(t)$ is to expand $\psi$ into eigenfunctions of $\hat{H}$ which depend on time as $e^{-i\frac{E_n t}{\hbar}}$. There will be many terms. Some are products of an eigenfunction with itself and contribution of these vanishes. Some are products of two different eigenfunctions. These latter terms depend on time as $e^{-i\frac{E_n-E_m}{\hbar}}$ and make $\boldsymbol{\mu}$ oscillate at the frequency $(E_m-E_n)/\hbar$. Schroedinger explained the Ritz combination principle this way, without any quantum jumps or discrete allowed states; $\psi$ changes continuously in time. Imperfection of this theory is that the function oscillates indefinitely and is not damped down; in other words, this theory does not account for spontaneous emission.
The water droplets that create a rainbow are not emitting the light that you see in a rainbow; if they were, you would see a glowing cloud of consistent color, not a rainbow. The rainbow is formed by sunlight refracting and reflecting through water droplets in the air; the water refracts through the "front" of the drop, reflects off the "back," and refracts again on the way back out. The refractions are what separate the colors, since different wavelengths of light refract to different degrees. If you used devices capable of imaging in other wavelengths of light, you'd see further bands of "color" beyond the red and violet sides of the rainbow, resulting from the infrared/ultraviolet (and other wavelengths beyond those) radiation in the sunlight.
So in short, the full-spectrum appearance of the rainbow is due to the fact that the source of the light (the sun) is a thermal blackbody and emits a blackbody spectrum.
Best Answer
$H_2$ contains 2 electrons in the same ground-state orbital; by Pauli exclusion, one must be spin-up and the other must be spin-down. The 21cm line is generated in a normal hydrogen atom when an electron's spin flips from being aligned with the proton to being anti-aligned with the proton. In $H_2$, an electron's spin cannot flip because it would then be occupying the same state as the other electron. Even if it could, the differing shape of the orbital would produce a line at a significantly different wavelength than 21cm.
The two protons, since they're not in the same orbital, can exhibit a similar type of transition as the 21cm line (i.e. proton spin flips from aligned to anti-aligned), but since the protons are highly localized, massive, and far apart, any spin-spin coupling between them is insignificant. Indeed, when measured, this transition has a frequency of just 72 kHz, corresponding to a wavelength of over 4 km.