# AX Spin system ordering

I would like to ask a question about the ordering of the $$\ce{X}$$ spins on the right hand side of an $$\ce{AX2}$$ energy diagram.

I was given the diagram here below:

and asked to evaluate why this diagram results in a triplet. This is simple, since there are 4 possible energy transitions, but one of them is equivalent to the singlet transition, hence only 3 transitions would appear in the NMR spectrum.

Prior to this though, I was confused because I thought the ordering for the bottom 3 spins was incorrect. Taking $$\ce{\alpha}$$ as the lower energy spin, I thought the bottom 3 spins should read as:

\begin{align}&\beta \beta \\ \beta \alpha &~~~~ \alpha \beta \\ &\alpha \alpha \end{align}

and the energy levels would be drawn more appropriately. I attempted but failed to mention this to the lecturer during the presentation.

Am I correct in this instance?

• Neither of those diagrams make sense afai can tell. How can you have a transition that doesnt flip any spins? – Buck Thorn Nov 13 at 18:51
• @BuckThorn I questioned whether the order of the bottom 3 spins was incorrect but I did think about the selection rule and why that was not obeyed in this slide – vik1245 Nov 13 at 19:09
• @BuckThorn it is flipping the A spin. The A spin state is on the left. The X spin states (which don’t flip) are on the right. – orthocresol Nov 14 at 8:19
• @orthocresol Gotcha, I misread the diagram. – Buck Thorn Nov 14 at 8:21
• @BuckThorn that was a minor point, though. You are quite correct that the diagram is wrong in many ways. It is not a physically accurate description of coupling. The diagram implies that the splitting between the peaks in the triplet has a frequency on the order of an X spin flip, which is completely... well... off. – orthocresol Nov 14 at 8:25

There are a few things that seem to be wrong with the diagram you present, but since your focus is on the relative energies of the $$\alpha$$ and $$\beta$$ spin states I'll address only this. If you check a reliable source such as Cavanagh et al.s NMR textbook [ref 1] it is explained that
the state with $$m=+\frac{1}{2}$$ is referred to as the $$\alpha$$ state, and the state with $$m=-\frac{1}{2}$$ is referred to as the $$\beta$$ state. If $$\gamma$$ [the gyromagnetic ratio] is positive, then the $$\alpha$$ state has lower energy than the $$\beta$$ state.
Therefore for protons $$\alpha$$ is lower energy (since $$\gamma>0$$). If the nucleus is $$\ce{^{15}N}$$ ($$\gamma<0$$), $$\beta$$ becomes the lower energy state.