NMR Splitting for Adjacent, Equivalent Hydrogens?

Question

Why don't chemically equivalent hydrogens on adjacent carbons split the ¹H NMR signal? There appears to be an answer here, but I'm having trouble understanding it because I cannot rationalize what happens with the magnetic field.

Let's say we're analyzing acetylene. When we subject the molecule to an external B-field, both (equivalent) protons have 2 possible spin states. For each proton, I would expect that the other proton's magnetic moment may either augment or oppose B₀, causing a doublet. When we look at the signal for the other proton, it's the same story--I would expect 2 peaks identical to the 2 for the first proton (effectively doubling the integration).

Where have I gone wrong in extending spin-spin coupling to identical hydrogens? Why does the magnetic moment of adjacent, equivalent protons not affect the transition energy?

Edit: I asked my organic chemistry professor this question, and his response was that there does exist spin-spin coupling of identical protons, but the coupling constant is equal to 0. The reasoning involved quantum mechanics that went over my head.

Answer 1

There are two ways to look at this.

a) All transitions you can see are between a state where the spins are parallel and a state where they are antiparallel.

$aa <--> ab$ or $ab <--> bb$

Their distances are identical, as are the states $ab$ and $ba$, so, as porphyrin said in his comment, you only see one line.

b) If the two protons were chemically different, you would see two doublets.

If the difference in CS were small, you'd see the "roof effect": An AB spin system. The two inner lines come closer and grow at the cost of the outer lines.

But there is no difference in CS, and so the outer lines vanish, and the two inner lines coalesce into one singlet.