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I know that population difference matters in NMR for sensitivity, that the splitting is dependent upon the gyromagnetic ratio of the nucleus, and that the population of the spin states is almost identical with a slight bias towards the lower energy spin state; but I don't really understand why both the magnitude of the splitting and the population difference matter in terms of sensitivity. For instance, DNP causes a massive splitting of energy levels, but why does that matter? I can't seem to find any good sources on the rationale that goes beyond a comparison of an electron's gyromagnetic ratio to that of a proton. Thanks!

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The magnitude of the splitting (let's call it $\Delta E$) leads to a larger equilibrium population difference, via the Boltzmann distribution. To be precise, we have that

$$\frac{p_\text{upper}}{p_\text{lower}} = \exp\left(-\frac{\Delta E}{kT}\right),$$

where $p_\text{upper}$ ($p_\text{lower}$) is the population of the upper (lower) state. If you further impose the condition that the populations must sum to $1$:

$$p_\text{upper} + p_\text{lower} = 1,$$

and use the approximation that $\exp(x) = 1 + x$ (which is valid when $|x| < 1$, and is definitely true of $\Delta E/kT$), then you can show that the population difference, $p_\text{lower} - p_\text{upper}$, is proportional to $\Delta E$. (See also: https://chemistry.stackexchange.com/a/133818/16683)

So, the remaining question is why a larger population difference means a larger signal.

To start off, let's look at what the $z$-angular momentum of this sample is. For a $\ce{^1H}$ nucleus (a proton), the 'lower' state is $\alpha$ (spin-up), and the 'upper' state is $\beta$ (spin-down). The $\alpha$ state has $z$-angular momentum of $+\hbar/2$, and the $\beta$ state $-\hbar/2$. So, the net angular momentum is

$$I_z = \frac{\hbar}{2}Np_\alpha - \frac{\hbar}{2}Np_\beta \propto \frac{\hbar}{2}\Delta E,$$

where $N$ is the total number of molecules, and $p_\alpha$ and $p_\beta$ are the populations of the $\alpha$ and $\beta$ states.

In an NMR experiment, you don't measure this angular momentum directly: what you measure is the voltage induced in a coil by the net magnetisation of the chemical sample. The larger this magnetisation, the stronger the signal. The magnetisation, in turn, is directly proportional to the angular momentum:*

$$M = \gamma I,$$

where $\gamma$ is a physical constant (the gyromagnetic ratio of the nucleus being studied). And the angular momentum is proportional to the population difference, which is proportional to the energy difference.

Much of this is covered in standard magnetic resonance textbooks such as Keeler's Understanding NMR Spectroscopy and Levitt's Spin Dynamics, which I'd strongly suggest reading.

DNP causes a massive splitting of energy levels

That's not true; DNP doesn't change the energy levels. Hyperpolarisation techniques (including DNP) seek to establish a non-equilibrium population difference which is much larger than the Boltzmann population difference.


* The magnetisation that is detected must lie along the $xy$-plane. The NMR experiment must rotate this magnetisation from the $z$-axis into the $xy$-plane. However, this doesn't affect the conclusions above, as the rotation cannot change the amount of magnetisation available. So the initial $z$-magnetisation still determines how much signal is observed. (For the very nitpicky, I'm ignoring non-unitary transformations.)

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  • $\begingroup$ In addition to your clear answer I would add that that in absorbing electromagnetic radiation of any type a difference in population between two levels is essential, otherwise no net absorption occurs, and the bigger the population difference the bigger the chance of absorption. In NMR the large static magnetic field splits the nuclear energy into two levels (in the case of protons) with a small population difference as described above. The bigger the mag field the bigger $\Delta E$ and the bigger the nmr signal. $\endgroup$
    – porphyrin
    Oct 20, 2022 at 8:02

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