NMR: Why do "identical" nuclei not couple with each other?

[Glyphicon]

TL;DR

Spin–spin coupling is encountered in the majority of NMR spectra. However, it does only occur when the two interacting nuclei are non-"identical". But why is the coupling not occuring between "identical" nuclei?

Hello there,

Spin–spin coupling is encountered in the majority of NMR spectra. However, it does only occur when the two interacting nuclei are non-"identical". But why is the coupling not occurring between "identical" nuclei?

I have been wondering about this because I could not think of a reason for this behavior. I went on to google my question. I found this thread:
https://www.physicsforums.com/threads/nmr-why-do-identical-protons-not-experience-splitting.799129/
This is what brought me to this forum.
However, I am afraid the final answer, which was considered to be the solution, cannot be correct.
Unfortunately the thread is closed. This is why I open a new one.

NMR-Signals in general arise from the transition between the energetically favoured α spin (singlet state) and the energetically disfavoured β spin (triplett state). (Nevertheless, the energy difference between them is very slight and the difference in occupancy is close to zero.)

However, the spin of a nucleus featuring a bonding interaction with another nucleus is energetically favoured when their spins are anti-parallel (as a result of Fermi contact interaction and Pauli excltion principle). Consequently, the energy the energy difference between, the αα state (both nuclei feature an α spin) and the βα state, is different to, the energy differenc between the βα state and the ββ state. These two different energy differences result in one duplet of signals.

This is the summary of my understanding of spin–spin coupling.
Now I would like to express what I would expect – based on my "knowledge" – to happen to two "identical" nuclei.
As example molecule I will take formaldehyde (H2CO, two hydrogen (1H nuclei) bond to one carbon which features a double bond to an oxygen)
As the protons have one nucleus in between, it is energetically favoured when their spins are parallel (and the carbon nucleus in between is antiparallel to both of the hydrogen nuclei). Consequently the αα energy state (α H-1 and α H-2) and the ββ (β H-1 and β H-2) are lowered and the transitions αα→αβ and βα→ββ are energetically different. (As the hydrogen nuclei are "identical", the αα→αβ transition is identical to the αα→βα transition and the βα→ββ is identical to the αβ→ββ. Nevertheless, there are two different transitions.) These two different transions correspond to a douplett of signals. Well, it does not. But why?

So now my question: Where am I wrong? How does the (identical) chemical environment interfer with spin–spin coupling?

I do not think that the answer of DrDu in the thread from 2015 (see below) is correct because no transition would mean no signal at all. Moreover, the "splitting" is the result of two transitions which are (slightly) different in energy.

The point is that total spin becomes a good quantum number if both nuclei are identical. So there is no transition between singlet and triplet states, but only within the triplet, which are equidistant. So the splitting is still there, only the transitions are missing.

If you could help me with my question, I would appreciate it a lot.

Kind regards,
Glyphicon

 

[mjc123]

Your understanding of NMR seems to be lacking at some points. The α state is not a singlet and the β state is not a triplet. The hydrogen nucleus with I = 1/2 is a doublet, with states m_I = +1/2 and -1/2, between which the NMR transition occurs.

You can't talk of the carbon nucleus being antiparallel; if it is a ¹²C nucleus, as it usually is, it has no magnetic moment and is neither parallel nor antiparallel.

Most important is @DrDu's point, which you seem to have misunderstood. When the two nuclei are identical, you can't distinguish the spin states of the individual nuclei. You have to deal with the total spin of the system. Thus the "states" αβ and βα are not acceptable solutions because they don't conform to the symmetry of the system; you need to take symmetric and antisymmetric linear combinations (αβ + βα) and (αβ - βα) [ignoring normalisation constants]. What you end up with is states with m_I = +1, 0, 0 and -1. These can be divided into a triplet with I = 1 and m_I = +1, 0 and -1; and a singlet with I = 0 and m_I = 0. Transitions between singlet and triplet are forbidden (this doesn't mean there are no transitions, which your confusion about singlet and triplet above may have led you to think), but transitions within the triplet between m_I = +1 and 0, or 0 and -1, are allowed, and these transitions have the same energy.

 

[TeethWhitener]

One might very well ask why identical electrons don't split each other in XPS. Same reason: indistinguishability.

 

[Glyphicon]

Thank you a lot for your detailed explanation, mjc123.

I had already suspected that the model itself may be the flaw in here.