Chapter 1: Introduction to NMR

Atoms are the basic composition units in chemical and biological activities. Each atom consists of a nucleus with one or more surrounding electrons. Molecules are formed by bonding valence electrons of different atoms. The arrangements of valence electrons in these bonds determine the chemical properties of the molecule. For example, the same carbon atom in a CO₂ molecule behaves differently to that in a CO molecule. The variations of the bonding between valence electrons also affect NMR signals, manifested as shifts in NMR signals, called chemical shifts. This correlation is used to derive the protein secondary structure, or electronic and magnetic properties in various insulators, conductors, and superconductors.

The diameter of an atom, for example, a helium atom, is roughly 1 Å (10⁻¹⁰ m). The nucleus occupies only a tiny spot at the center, with a diameter of around 1 fm (10⁻¹⁵ m). Outside the nucleus, the electrons are moving about at high speed. The ratio of the electrons’ speed to the speed of light is proportional to the charge of the nucleus and the fine-structure constant (1/137). However, the location of the electrons cannot be precisely determined at any specific time if the velocity is measured precisely. Instead, quantum mechanics uses the wave function φ(r) to represent the probabilistic distribution of their locations, which is often described as the electron cloud. Their most probable locations around the nucleus resemble the orbits of the planets in the solar system, so they are called atomic orbitals.

Let's make an analogy: if an atom is a Boeing 747, then the nucleus is about the size of a sesame seed, or a speckle on your face. If the atom is the Earth, the nucleus will be about 60 m in diameter. However, in contrast to the size difference, more than 99.94% of the atomic mass is associated with the nucleus. The properties of the nucleus do not change during chemical or biological reactions. However, the nucleus has its own underlying structure, being composed of protons and neutrons. Protons, neutrons, and many nuclei have the quantum mechanical property called “spin.” In the presence of an external magnetic field, these particles precess in the direction of the magnetic field. The ratio between the precession frequency, the so-called Larmor frequency, and the magnetic field, is a parameter named gyromagnetic ratio γ, with units rad/(T s), or γ/2π with units Hz/T. For example, γ is 42.5774 MHz/T for ¹H, 10.7084 MHz/T for ¹³C, and −4.316 MHz/T for ¹⁵N. The sign here indicates the precession sense, with + for clockwise and − for counterclockwise when viewed from the top, of the direction pointed by the spin.

So what is the consequence of this nucleus precession?

We can model the nucleus as a charged particle q in the classical picture. When a charged particle q adopts a uniform circular motion with a radius of r and period T, it produces a current

where v is the speed of the particle. According to electromagnetic theory, this circular motion of the charged particle creates a loop current, which carries a magnetic moment $μ⇀$⁠,

Here, $S⇀$ is the vector, with its magnitude equal to the area of the loop, and its direction described by the right-hand rule following the direction of the loop current, which is represented by unit vector $e⇀z$⁠. Meanwhile, such a gyration motion (rotation motion) also can be quantified by an angular momentum $J⇀$⁠:

The ratio between $J⇀$ and $μ⇀$ is the gyromagnetic ratio γ:

So, this classical picture shows that the gyromagnetic ratio is related to the charge and mass of the particle. Taking this knowledge, think about γ of various elementary particles, for example, electrons, protons, ¹³C, and ¹⁵N. What kind of estimation can we get?

Of course, classical mechanics is not the most accurate theoretical framework to describe microscopic particles such as nuclei or electrons. We need quantum mechanics. For an isolated (noninteracting) charged particle, the Hamiltonian H is

where $p⇀$ is the momentum operator: $p⇀=−iℏ∇⇀=−iℏ(e⇀x∂∂x+e⇀y∂∂y+e⇀z∂∂z)$⁠, $A⇀$ is the vector potential associated with the magnetic field, $H⇀=B⇀=∇⇀×A⇀$⁠, and ϕ is the scalar field associated with the electric field, $E⇀=−∇ϕ−∂A⇀∂t$⁠.

Here, we assume the particle exists in a vacuum, where the magnetization associated with the medium vacuum is 0, so magnetic induction or magnetic field density $B⇀$ comes entirely from the magnetic field strength $H⇀$⁠.

We want to emphasize that, in general, $B⇀≠H⇀$⁠. So, $B⇀=μ0(H⇀+M⇀)=(1+χm)μ0H⇀=μμ0μ0H⇀=μrμ0H⇀$⁠, where μ₀ is the magnetic permeability of free space, equal to 4π × 10⁻⁷ N A⁻², μ_r is the relative permeability, and χ_m (dimensionless) is the magnetic susceptibility.

What does magnetic permeability mean? Conceptually, it describes the ability of a medium to support the formation of a magnetic field within itself. According to this property, materials can be categorized as diamagnetic, paramagnetic, antiferromagnetic, or ferromagnetic.

How will magnetic permeability impact our NMR experiment? What are diamagnetism, paramagnetism, antiferromagnetism, and ferromagnetism?

A diamagnetic material does not exhibit a permanent magnetization macroscopically in the absence of an external magnetic field. However, the application of an external magnetic field induces an opposing internal magnetic field inside the material. Correspondingly, it has a negative χ_m, or μ_r < 1. The magnitude of the induced field is very small in general, and is created by the preferential precession of the electronic orbitals around the external magnetic field, which produces a very small magnetic field opposing the external magnetic field. The details are given in the chemical shift shielding calculation in Sec. 5.4. Most biological materials, such as proteins, are diamagnetic. Superconductors can be considered as perfect diamagnetic materials that under appropriate conditions can expel completely the external magnetic field.

A paramagnetic material also does not exhibit any permanent magnetization macroscopically. However, in the presence of an external magnetic field, an internal field is induced by the orbital motions of electrons, aligning along the external magnetic field. It has a positive χ_m, or μ_r > 1. The induced field is directly proportional to the external magnetic field. Its magnitude is normally larger than that of the diamagnetic field, but still very weak.

Ferromagnetism is created by the overall consistent alignment of the magnetic dipole moment and also the orbital angular momentum of unpaired electrons. These unpaired electrons tend to align their spins in parallel due to the exchange interactions, because of the Pauli exclusion principle. According to the Pauli principle, if two identical fermion particles have the same spin orientation, they cannot be physically located in the same orbital state, which in return decreases the electrostatic energy of the electrons compared to the antiparallel alignment that can assume the same orbital. This exchange interaction is normally much stronger (thousands of times) than the magnetic dipole–dipole interactions, which would have spins aligned in antiparallel. At temperatures higher than the so-called Curie temperature, thermal motions will randomize the orientation and kill the alignment.

Using the vector potential, and assuming all fields have no time dependence,

This expression is equivalent to saying that we have an external magnetic field H₀ along $e⇀z$ in direction $H⇀=H0e⇀z$⁠. The corresponding x, y, and z components of the vector potential are $Ax=−12H0y$⁠, $Ay=12H0x$⁠, and A_z = 0. Hence, the Hamiltonian H can be rewritten as

where the subscript i represents the x, y, and z components. Rearranging the equation, we get

Recall the definition of angular momentum in Eq. (1.3):

We can see that

Hence, we can rewrite this as

with

The second term describes the interaction of angular momentum with external magnetic field, which is called the Zeeman effect or Zeeman interaction:

Here, ω₀ is the Larmor frequency. By comparing to H_b, the gyromagnetic ratio γ is

Note that Eq. (1.9) has an extra factor of c in the denominator, compared to Eq. (1.4). This is because we used the formula in cgs units in the quantum mechanical derivation above, while SI and cgs units assume the same format as for the classical mechanical derivation. To convert to SI units, all charges have to be multiplied by a factor of c. Then you can see the quantum mechanical derivation of gyromagnetic ratio is the same as the previous classical derivation. With more rigorous derivation, beyond the scope of this book, the gyromagnetic ratio is

where g_e for electrons is about 2, as can be measured by the one-electron cyclotron.

Let us inspect the third term H_c in Eq. (1.8). If we assume $ω0=qH02mc$⁠, then the third term can be rewritten as

Let us compare the magnitude of H_b and H_c in Eq. (1.8). We can treat $x2+y2=23r2$⁠, where r is the radius of the electron orbital. In hydrogen, this can be simplified to the Bohr radius. The Bohr radius is the most probable distance between the electron and nucleus in a hydrogen atom at its ground state. We will replace the angular momentum operator J_z in H_b by $ℏ$⁠. Then, it is trivial to show that $HbHc<10−4$⁠.

In summary, the above quantum mechanical derivation also shows that the motion of a charged particle creates an angular momentum, which then couples with the external magnetic field, similar to that predicted in the classical picture. The coefficient resembles the gyromagnetic ratio, except it is missing the correct g-factor obtained in more rigorous derivations.

In 1905, Paul Langevin presented an explanation for the diamagnetism due to the orbital motions of electrons just like the second term:¹ 

Here, we have used the relation 1 T = 1 kg/(C s). Compared to the true gyromagnetic ratio value for the electron, this is smaller by a factor of 2, due to the missing g-factor. Let us pretend that we still do not know the g-factor and proceed to the computation of the proton Larmor frequency in a 14 T magnetic field. This will give us

To lower the energy, the angular momentum associated with the orbital motion needs to adopt the parallel alignment along H₀. This means that the magnetic moment associated with the current loop of the electron should then be antiparallel to H₀, and is diamagnetic. We can easily compute the current from the circulating frequency according to Eq. (1.1):

The magnetic moment associated with the current can then be calculated according to Eq. (1.2):

Let us use the parameters of the electron in the hydrogen atom. The orbital radius is r = a₀ = 5.29 × 10⁻¹¹ m, so $x2+y2=23r2$⁠:

The total diamagnetization due to the protons in per volume water is

where n is the density of proton per volume of water. Inserting the magnetic permeability in free space μ₀ = 1.257 × 10⁻⁶ T · m/A, and we have

Recall that the direction of the field opposes the external field, so there should be a negative sign: M = −7.73438 × 10⁻⁷ T. Of course, a perfect alignment of all protons in the water is assumed in the above derivation, which is obviously an overestimation. In reality, the alignment should obey the Boltzmann distribution. According to the definition of magnetic susceptibility, we can estimate the per volume magnetic susceptibility of protons in water is

This rough estimation is about 20 times larger than the true value of water's susceptibility, which is −9.051 × 10⁻⁹ (m³/kg), computed by the molecular orbitals of the valence electrons in water between protons and oxygen. But you get the idea of how to use simple known constants to estimate physical parameters of interest.

In addition to the angular momentum associated with orbital motions, elementary particles can also possess another kind of angular momentum called the spin angular momentum. The spin momentum is normally represented by symbol I or S in quantum mechanics. This is also quantized, as proved experimentally in 1922 by Stern and Gerlach. The projection of spin and orbital angular momentum to a given direction is labeled by I_z. As shown by the Stern–Gerlach experiment,^2–4  I_z is not a continuous value for elementary particles, but can only assume a set of discrete values between −I, −I + 1, −I + 2, …, I − 2, I − 1, I, where I is the major spin momentum value. For electrons, ¹H, ¹³C, and ¹⁵N, as the major spin quantum number is 1/2, the quantized I_z can only be either −1/2 or 1/2.

Just like the orbital angular momentum, elementary particles with nonzero spins will exhibit a dipolar magnetic moment, μ, associated with the spin: μ = γI, where γ is the gyromagnetic ratio. The only difference is, the particle does not have to carry any charge, nor is any physical orbital motion needed to produce the spin angular momentum and the magnetic moment. Nonetheless, similar to angular momentum, you can imagine these spins are vectors, each assuming a specific direction with quantized values between −γI and γI. Without an external magnetic field, these spins and the magnetic moments exhibit no preferred orientation, and, as they are completely randomized, there is no bulk magnetization when summed together.

When an external magnetic field is applied, the spins and the magnetic moment will assume a preferred direction along the magnetic field direction, with more aligned along the magnetic field. In solid state NMR (ssNMR) experiments, we normally need a considerable amount of samples. For example, ∼1 µmol sample is needed to produce a reasonable NMR signal-to-noise ratio (SNR) in ssNMR. Why is that? This is because not all magnetic moments will align. Let us do a simple estimation, and assume that we have N identical nuclei with spin I, and that they only interact with the external magnetic field H₀, without interactions between each other. The Hamiltonian H of the system is just the Zeeman term:

The coupling with magnetic field lifts degeneracy, and causes shifting of energy levels dependent on the spin quantum states, from $−γH0Iℏ$ to $γH0Iℏ$⁠. The total magnetization is the sum of the magnetic moments of all the particles. According to thermodynamics and statistics, the distribution of their spin orientation is neither entirely random nor completely aligned, but follows the Boltzmann distribution. For a single spin,

where the partition function $Z=∑m=−IIe−EmkT$ and $Em=−mγH0ℏ$⁠. For a system of N noninteracting spins, the total energy should be

and the total system magnetic moment is

We can write a specific state function of the system as

where m_i represents the spin alignment state I_z of particle i, with quantized value between −I and I. This uses the so–called the bra-ket notation to represent the system wave function in quantum mechanics. We will give a detailed introduction in Sec. 4.1. If quantum mechanics is new to you, you can treat it as a convenient way to label the system according to its distinct spin orientation.

A particle would like to align along the magnetic field (assume γ is positive) for lower energy. As will be shown in the introduction of density matrix in Sec. 4.2, for those who are not familiar with quantum mechanics, the observed magnetization M is the trace evaluation:

Since all particles are identical, we can evaluate the contribution from jth particle, and total magnetic moment M of the system would be N-fold of this value, as shown below:

Here, we separate the jth particle's contribution from the rest in the weight factor of the energy term. Obviously, the summation over all possible states is now decomposed into two parts: the term $exp(∑i≠j⁡miγH0ℏkT)$ will sum over all possible states with possible values of other particles i ≠ j, then it will multiply with the sum over all possible states of the jth value with the term $mjexp(mjγH0ℏkT)$⁠. These two summations are independent of each other, as we have a system of noninteracting independent particles. Therefore, the common sum of the term $exp(∑i≠j⁡miγH0ℏkT)$ on both numerator and denominator will cancel with each other, and we have

When we do the evaluation of the trace, note that we need to count all possible combinations of all particles. We can evaluate the trace in the numerator over all possible states of the jth particle, and then it will be repeated over all possible states of other particles. This means that for each m_i, it will be repeated 2I + 1 times first over all its possible quantized values. Subsequently, the results of this sum over the states of the jth particle will be repeated (2I + 1)^N−1 times for the combination of the possible states of the rest of the particles. Let us do the sum over the jth particle first:

Here we need to apply some approximation using the conditions in our normal experiment: say the magnetic field H₀ is 20 T, the temperature is 300 K,

In comparison with the denominator:

So $mγH0ℏkT∼0.0001≪1$⁠. Hence, we can justify the high temperature approximation:

When performing the summation of the index m_j over all m_j states from −I to I, $mγH0ℏkT$ and m_j will both come out to be zero. The summation of the unit in the denominator will also be summed over all possible values, and be repeated 2I + 1 times, so we will just have

The term $Ijz2$ contains the square of I_jz. Recall in quantum mechanics that $Ix2+Iy2+Iz2=I2=$ $I(I+1),$ and $Ix2=Iy2=Iz2$⁠, so $Ijz2=13I(I+1)$⁠:

So, the magnetization of N noninteracting particles will be just $Nγ2ℏ2I(I+1)3kTH0$ in the presence of an external magnetic field. This is, in fact, Curie's law. The part in front of H₀ is essentially the susceptibility χ. We can apply this estimation to the protons in water, although water is diamagnetic. Treat N as the number of protons per unit volume, and note that there are two protons per water molecule:

Remember the energy of a magnetic field is $E=12B⋅HdV$⁠, so 1 J = 1 T A/m · m³ = 1 T A m². Therefore,

Of course, this rough estimation is not very accurate for the water susceptibility, because water is diamagnetic, not paramagnetic. The per volume diamagnetic susceptibility of water is 7 × 10⁻⁷⁵  in cgs units, and to convert to SI units we have to multiply by 4π. A rigorous derivation of the susceptibility needs to account for the dominant effect from the orbital motions of electrons, not the magnetic moments of nuclei. But the above estimation does show how weak even the magnetization of protons in water can be.

In normal NMR experiments, resonance signals are proportional to the sample's magnetization. To detect resonance signals, various methods can be used. For example, in the magnetic resonance force spectroscopy, the resonance is detected by light signals associated with the displacement resulted from the force on the ferromagnetic tip of the cantilever in the force-detected magnetic resonance spectroscopy. The detection of single electronic spin was demonstrated by this means, which is probably the most sensitive method.⁶  However, the most popular approach is still the detection of the electromotive force (EMF) by a solenoid. The signals in a solenoid associated with NMR are described by Hoult and Richards in their 1976 paper:⁷ 

Here, ω₀ is the Larmor frequency, I is the current in the coil, V is the sample volume, M is the magnetization per volume of the sample, and B₁ is the transverse magnetic field created in the solenoid. We will skip the derivation of Eq. (1.14). Readers can refer to the original paper if interested.

What does this expression tell us about our NMR experiment sensitivity?

First of all, nuclei with higher Larmor frequency or signals in a higher magnetic field will produce stronger NMR signals due to the ω₀ factor in Eq. (1.14). That is why we prefer carbon detection to nitrogen detection. With the capability of faster and faster magic angle spinning (MAS), now proton-detected ssNMR for proteins is one hopeful direction.⁸  This also explains the drive for high magnetic fields for NMR spectroscopy, given other parameters are identical. Secondly, higher B₁ fields give more sensitivity, that can usually be achieved by using a solenoid with a smaller diameter. Finally, more sample materials would also improve our NMR signal. In the same publication, Hoult and Richards also gave an expression for B₁:

Interested readers can show the derivation of Eq. (1.15) using basic electromagnetism and calculus. Here, a represents the radius of the coil, and z is the distance of the round of the coil to the evaluated point. Therefore,

We can use this expression to estimate the voltage induced by a $π2$ pulse in our NMR experiments. First, let us estimate the B₁ field of a single-turn coil with a = 1.6 mm, z = 2 mm:

Here, we can estimate how strong the current should be to produce a $π2$ pulse for the proton channel with t_p = 2 μs, which is quite common in ssNMR experiments:

Compared with the $B1I$ expression in Eq. (1.16), we see the current necessary to generate such a field in the solenoid coil is ∼60.6 A. This is a very high current. Now we understand why we need hundreds to thousands of watts power in our transmitter amplifier.

If we want a $π2$ pulse with t_p = 4 μs on carbon channel, what would be the B₁ field? And how much current is needed?

So, even larger currents are needed for lower channel with a decent $π2$ pulse.

The sample volume of a 3.2 mm MAS rotor is about 30 µl. Recall that we estimated χ_H = 1.60576 ×10⁻⁷A/(T m) for protons in water. According to Eq. (1.14), in a 14 T magnetic field, the total magnetization of a 3.2 mm rotor filled with water would be

The signal detected by the solenoid is the voltage signal in volts. To see that, we need to do some unit conversion:

So, you can see the voltage induced in the coil is in the nV range; very small. Of course, here we assumed the solenoid consists of a single-turn coil during the B field estimation. In reality, a 3.2 mm probe coil may have multiple turns. But the point is proved that the NMR signal is very tiny. However, this is not the complete story. What really matters in experiment is the SNR. So, what is the noise during our NMR experiments?

Here, we briefly review some basic concepts about common noise sources. In a RF electronic circuit, there are two kinds of noise: shot noise and Johnson noise. Flicker noise, which is inversely proportional to the frequency, is not significant in NMR circuits.

The first candidate is the shot noise.⁹  This arises from the fluctuation of discreteness/quantization of charge carriers in the DC current. It is independent of temperature and frequency, and described by the Poisson distribution. Hence, at normal current magnitude that consists of a large number of charge carriers, it is insignificant compared to Johnson noise.

The dominant noise contribution in NMR experiments is therefore Johnson noise.¹⁰  This is also called Johnson–Nyquist noise, Nyquist noise, or thermal noise. It arises from the thermal motion of electrons or other charge carriers in the electrical conductors at equilibrium. It exists with or without the application of voltage. In an ideal resistor, the power spectral density of Johnson noise is nearly uniform over all frequencies, the so-called white noise. When the bandwidth is limited, Johnson noise can be approximated by a Gaussian distribution. Its voltage variance per hertz bandwidth $vn2¯$ can be described by

where k is the Boltzmann constant, T is the temperature in kelvins, and R is the resistance in ohms. This noise dominates the SNR in the NMR experiments detected by solenoid coils. Hence, to enhance SNR in a cryo-probe, the coil temperature is controlled at 4 K to reduce the contribution of the Johnson noise in the detection coil. Theoretically, by going from 300 K to 4 K, the voltage/frequency due to Johnson noise decreases by $300/4=75$ times. Of course, the actual S/N enhancement of normal cryo-probe is much more modest, roughly about a factor of 4, as the entire detection circuit encompasses more than the detection coil. At room temperature,

Equation (1.19) indicates that Johnson noise is proportional to both the resistance and the bandwidth, so we should limit the detection bandwidth to a reasonable value to help improve the SNR. For example, assuming that we are detecting a 10 ppm proton spectrum on a 600 MHz spectrometer with a bandwidth of 10 000 Hz, and the resistance of the probe coil is 1 Ω, the standard deviation of voltage due to the Johnson noise from the probe coil will be: 0.13 × 100 = 13 nV. This is about the same level of magnitude as the signal as estimated earlier. It demonstrates the low sensitivity weakness of NMR.

What should we do?

The answer is to signal average. Why does signal average enhance S/N, and how? Suppose we acquire a 1D cross polarization spectrum with four scans, with SNR = 2. We want to get a spectrum with SNR = 4, how many scans should we go?

The randomness factor associated with Johnson noise can be described by the normal distribution in statistics. According to the Central Limit Theorem (CLT) of normal distribution, the variance of N independent random variables will be decreased N-fold. Hence, when you accumulate N scans of NMR spectra, your signal will increase N-fold, since the signal is coherent. Meanwhile, the variance of Johnson noise will be $Nvn2¯$⁠. The corresponding standard deviation for noise amplitude will be $Nvn2¯$⁠. Hence, now we have

which shows that the SNR increases with $N$⁠, not N.

In addition to the signal average, NMR also exploits the characteristics of the RLC resonance circuit for signal excitation and sensitive detection, which will be introduced in Chapter 2.