# Do the αβ and βα states of homonuclear AX system have same energy?

I was studying nuclear Overhauser effect and came across this diagram$$^1$$:

This diagram shows the energy levels of four possible nuclear spin states of homonuclear AX system. The white/grey boxes refer to population deficit/excess, respectively, w.r.t the population at each of the two states at the centre (α$$_\mathrm{A}$$β$$_\mathrm{X}$$, β$$_\mathrm{A}$$α$$_\mathrm{X}$$ ). This is the population distribution at thermal equilibrium.

It is stated clearly in the text that α$$_\mathrm{A}$$β$$_\mathrm{X}$$ and β$$_\mathrm{A}$$α$$_\mathrm{X}$$ have same energy, for homonuclear AX system.

My question is, how can the α$$_\mathrm{A}$$β$$_\mathrm{X}$$ and β$$_\mathrm{A}$$α$$_\mathrm{X}$$ have same energy?

The energy of each state is described by:

$$E(m_A, m_X) = -hv_Am_A -hv_Xm_X + hJm_Am_X$$

For the two states, third term at RHS are common, but the Larmor frequencies of A and X are not the same despite being homonuclear (already called 'AX', chemical shift very different, Larmor frequency also differ), so the states should have different energy. Did I understand something wrong?

1. Peter Atkins, Julio de Paula, James Keeler. Physical Chemistry (11th Edition). OUP. 2018. Page 516.

Or, 8th Edition: page 542; 9th Edition: Page 548

• Yes it is confusing and rather strange, and does not make sense. The energy $\alpha\beta;\; \beta\alpha$ levels are definitely different both with and without spin-spin coupling and these are shown correctly in fig15.12. In specialist texts such as Levitt 'Spin Dynamics' they are shown correctly when NOE's are explained, see fig 16.36 in that book. Jan 31 at 16:14

the Larmor frequencies of A and X are not the same despite being homonuclear (already called 'AX', chemical shift very different, Larmor frequency also differ), so the states should have different energy

Yes, that is precisely correct, so the two states $$|\alpha\beta\rangle$$ and $$|\beta\alpha\rangle$$ do not have the same energy. More formally, the Hamiltonian for the system (ignoring spin-spin coupling) is

$$\hat{H} = -\omega_I\hat{I}_{\!z} - \omega_S\hat{S}_z$$

where $$\omega_I$$ and $$\omega_S$$ are the Larmor frequencies of spin $$I$$ and $$S$$, and $$\hat{I}_{\!z}$$ and $$\hat{S}_z$$ are theie corresponding spin angular momentum operators, with the following properties:

\begin{align} \hat{I}_{\!z} |\alpha\rangle &= \frac{\hbar}{2}|\alpha\rangle \\ \hat{I}_{\!z} |\beta\rangle &= -\frac{\hbar}{2}|\alpha\rangle \end{align}

The energies of the states $$|\alpha\alpha\rangle$$, $$|\alpha\beta\rangle$$, $$|\beta\alpha\rangle$$, and $$|\beta\beta\rangle$$ can be evaluated fairly simply using

$$E = \langle \Psi | \hat{H} | \Psi \rangle$$

or in this case, since all of them are eigenstates of the Hamiltonian (without coupling), then

$$\hat{H}\Psi = E\Psi$$

and it suffices to read off the eigenvalue from the right-hand side. Proceeding in this manner, we have

\begin{align} \hat{H}|\alpha\beta\rangle &= -\omega_I \hat{I}_{\!z}|\alpha\beta\rangle - \omega_S \hat{S}_z|\alpha\beta\rangle \\ &= -\omega_I (\hbar/2)|\alpha\beta\rangle - \omega_S (-\hbar/2)|\alpha\beta\rangle \\ &= [\hbar(\omega_S - \omega_I)/2] |\alpha\beta\rangle \\ \Longrightarrow E_{\alpha\beta} &= \hbar(\omega_S - \omega_I)/2 \end{align}

Likewise,

\begin{align} \hat{H}|\beta\alpha\rangle &= [\hbar(\omega_I - \omega_S)/2] |\beta\alpha\rangle \\ \Longrightarrow E_{\beta\alpha} &= \hbar(\omega_I - \omega_S)/2 \end{align}

and in general, since $$\omega_I \neq \omega_S$$, these energies are different. The introduction of spin–spin coupling does not affect these conclusions.

One question remains, which is: why does Atkins depict the states as having the same energy, despite them not being equal? One possible answer is laziness, but if you read their writings, Atkins (and Keeler) are quite particular about notation and technical correctness, so I feel like this is not the primary reason, or maybe I am more inclined to give the benefit of the doubt here.

A more convincing answer would be that the difference between the energies $$E_{\alpha\beta}$$ and $$E_{\beta\alpha}$$ is very small:

$$E_{\alpha\beta} - E_{\beta\alpha} = \hbar(\omega_S - \omega_I)$$

The difference in Larmor frequencies between spins, i.e. $$\omega_S - \omega_I$$, is generally on the order of Hz to kHz. On the other hand, the other energy differences on the graph are far larger:

$$E_{\beta\beta} - E_{\beta\alpha} = \hbar\omega_S$$

The Larmor frequency itself $$\omega_S$$ is on the order of MHz. So the energy difference between $$\alpha\beta$$ and $$\beta\alpha$$ is pretty negligible compared to the other energy differences shown in the diagram.