Nuclear magnetic resonance (NMR)

Recall that the spin states of a particle of spin $S$ will have different energies in an external magnetic field ${\bf B}$. We saw this in the case of an electron, which is a spin $S=1/2$ particle. Earlier in the course, we noted that nuclei also have spins. In particular, recall that the proton $^1H$ is also a spin-1/2 particle. The magnetic moment of a proton is roughly three orders of magnitude smaller than that of an electron because magnetic moment is inversely proportional to the mass of the particle. Recall that the ``classical'' energy of a particle in a magnetic field ${\bf B}$ is

$$E = -{\bf\mu}\cdot {\bf B}$$

where ${\bf\mu}$ is the magnetic moment. For a proton, the magnetic moment is given by

$${\bf\mu} = {ge \over 2m_{\rm p}}{\bf S}$$

Here $g$ is a constant that has the value of roughly 5.6. It is known as the ``nuclear $g$-factor'' and is different for different nuclei. ${\bf S}$ is the spin vector. Thus, if ${\bf B}= (0,0,B)$, then the energy levels for a proton would be determined by the fact that $S_z = \pm \hbar/2$, and we find two energy levels

$$E_1 = -{ge\hbar \over 4m_{\rm p}}B\;\;\;\;\;\;\;\;E_2 = {ge\hbar \over 4m_{\rm p}}B$$

so that the energy difference is $\Delta E = E_2-E_1 = (ge\hbar/2m_{\rm p})B$. The constant $e\hbar/2m_{\rm p}$ is known as the nuclear magneton and is denoted $\beta$. Thus $\Delta E = g\beta B$.

Hence, a transition from $E_1$ to $E_2$ can be induced if a photon of frequency $\nu = g\beta B/h$ is used, and this is typically very low frequency in the radio wave part of the spectrum. In this way, the spin states of nuclei can be probed. However, NMR goes well beyond this: it is able to probe the local chemical environment of a nucleus, which makes it particularly useful for medical applications (as well as numerous others). Basically, when a magnetic field is applied to system, the nuclear spin states will be split. However, the electrons are also affected by the field. Since most electrons are in pairs with net spin of zero, the effect is not to split the spin states but rather to cause the electrons, which are charged particles to move in circular orbits. These circular orbits are like small current loops that produce a local magnetic field $B'$ that opposes the applied magnetic field and gives rise to a reduction in the field felt by the nuclei. This phenomenon is known as shielding. Expressing the opposing field $B'$ in terms of the applied field $B$, we have $B' = -\sigma B$, where $\sigma$ is a fraction $\sigma < 1$. $\sigma$ is known as the shielding constant. (In some cases, $\sigma$ is not just a simple constant but a matrix that mixes the components of the magnetic field. We will not consider such cases here.) Hence, the effective field felt by the nuclei is

$$B_{\rm eff} = B-B' = B(1-\sigma)$$

and the energy splitting $\Delta E$ for a proton, for example, would actually be $\Delta E = g\beta B_{\rm eff}$. The important thing about $\sigma$ is that it is very sensitive to the local chemical environment, i.e. the local electron density near the nucleus. Although we could now plot the spectrum as a function of the radio-frequency pulse $\nu$ needed to induce the transitions for the given magnetic field $B$ used to create the spin-state splitting. This turns out to be somewhat problematic because no two magnets will produce exactly the same external field $B$. Thus, different experiments based on different magnets will have a different set of peak frequencies, making comparison difficult.

The resolution of this problem is to report the transition frequency relative to that of some standard reference compound measured with the same magnet. The reference compound can be anything, but it should be a substance that 1) chemically nonreactive; 2) easy to remove from the sample; 3) should not produce spectral features that interfere with those of organic compounds, where using NMR is particularly useful. One example of such a compound is tetramethylsilane (TMS) Si(CH$_3$)$_4$. Let $\nu_{\rm ref}$ be the transition frequency associated with this reference compound. Then, we report our NMR spectra as a function of the chemical shift $\delta$ given by

$$\delta = {\nu - \nu_{\rm ref} \over \nu_{\rm ref}}\times 10^6$$

that is, we report $\delta$ in parts per million.

Another effect that creates sensitivity to the local chemical environment results from the fact that nuclear spins can couple to each other. If ${\bf S}_1$ and ${\bf S}_2$ are the spins of two nuclei, the energy of the coupling is of the form

$$E = -J{\bf S}_1\cdot{\bf S}_2$$

where $J$ is known as the spin-spin coupling constant or simply the ``J-coupling'' factor. $J$ will also be sensitive to the type of nuclei in the local chemical environment, and the energy level splittings will be affected by the strength of this coupling.