CLASSICAL DESCRIPTION OF NMR SPECTROSCOPY

• Suggested Reading: Sanders & Hunter Chapter 1

An NMR spectrum of a sample is fundamentally no different than any other form of spectroscopy. A photon of energy is absorbed that promotes a nuclear spin from the ground state to the excited state. As with other forms of spectroscopy, the intensity of the absorption is proportional to the population difference between the ground and the excited state. This population difference depends on the molar concentration of the sample and thus NMR can be used as quantitative technique. The energy difference between the ground and the excited state depends on the local environment of the nuclear spins, thus NMR can provide information on the local environment.

In addition to the above, NMR spectroscopy provides several unique forms of information. The first is the ability to obtain information on the structure of bio-polymers. This structural information can be obtained in two ways, dipole-dipole interactions between nuclear spins and spin-spin coupling between spins. Thus NMR can be used to obtain structures of biological molecules in solution. These structures are similar in fidility to medium-resolution X-ray derived structures. Finally, NMR spectroscopy can provide information on the dynamical behaviour of atoms over a wide range of time scales. This is perhaps the most unique (and useful) aspect of NMR.

Interaction of Nuclear Spins with the Magnetic Field

To perform NMR spectroscopy it is necessary to generate at least two different energy levels. These energy levels arise from the interaction of the nuclear magnetic dipole moment with an external magnetic field. The generation of a nuclear magnetic dipole moment requires the existence of spin angular momentum. All nuclei with an odd mass number (e.g. $^{1}$H, $^{13}$C, $^{15}$N) have spin angular momentum. All nuclei with an even mass number and an odd charge (e.g. $^{2}$H, $^{14}N$) also have spin angular momentum. Spin angular momentum is a quantum mechanical property of matter and has no analog in classical physics. Angular momentum is quantized, represented by a quantum number I, and has units of $\hbar$. Nuclear spin angular momentum is quantized (as is all angular momentum) and restricted to integral or half integral values. The z component of the angular momentum, $\vec{\mu}_z$, is restricted to values of I, I-1, ..., -I+1, -I. For a spin 1/2 nuclei (such as protons) only spin values of +1/2 and -1/2 will be found for the z component of the angular momentum.

The magnetic moment of a nuclear spin, $\vec{\mu}$, is proportional to the spin angular momentum, $\hbar \vec{I}$ by a factor, $\gamma$:

$$\vec \mu_n = \gamma_n \hbar I$$ (1)

The units of $\gamma$ are:

$$[ \gamma_n ]= radians\ sec^{{-1}}\ gauss^{{-1}}$$ (2)

The magnitude of $\gamma$ can be estimated using a classical model. Nuclei are charged and if they possess spin angular momentum then they behave like little current loops. Current loops will produce a magnetic field (the basis of electromagnets). In a similar manner charged particles with spin angular momentum also produce a magnetic field. This field is called a magnetic dipole in analogy to an electric dipole (which is a result in the separation of unlike charges). This argument gives $\vec{\mu}$ in more fundamental parameters:

$$\vec \mu_n = g_n \beta_n I$$ (3)

Where $\beta_n$ is the nuclear magneton and is equal to:

$$\beta_n = { e \bar h \over 2 M c }$$ (4)

$g_n$ is a fudge factor to make reality agree with theory.

The magnitude of $\gamma$ depends on the type of nuclei (protons, carbon etc.). Values of $\gamma$ for some commonly observed nuclei are:

Nuclei	$\gamma$
$^1$H	26,753
$^{19}$F	25,179
$^{13}$C	6,728
$^{15}$N	-2,712

Energy of Interaction

The existence of two (or more) values of angular momentum imply an energy difference for the different states in the presence of a magnetic field. The relative energies of each state can be found from classical electromagnetic theory. Consider a magnetic dipole in a static magnetic field, $\vec{B}$:

The energy required to change the angle, $\theta$, is:

$$E = \int \Gamma d \theta\ =\ \int ( \vec \mu \times \vec B )... ...eta ) d \theta = \vert \mu \vert\vert B \vert \cos ( \theta )$$ (5)

The torque, $\Gamma$ arises as the static field attempts to align the magnetic dipole of the nucleus. The energy of interaction between the field and the dipole is therefore given by the dot product of the two vectors:

$$H \ = \ - \vec \mu \ \cdot \ \vec B$$ (6)

Note that this Hamiltonian is classical in form, thus to a large extent we can describe the behavior of nuclear spins without resorting to quantum mechanics.

In a typical NMR spectrometer there is a single static (time independent) magnetic field. The direction of this field defines the co-ordinate axis system and the z direction is in the direction of the static field.

$$\vec B = \hat k B$$ (7)

$$H = - u_z B_z$$ (8)

$$= - \gamma \hbar B m_z$$ (9)

$\mu_z$ is the spin angular momentum along the z axis and $m_z$ is the quantum number for the z component of the angular momentum. Using Eq. 9 it is possible to draw an energy diagram for the system as a function of magnetic field. For a spin 1/2 particle $m_z$ is either +1/2 or -1/2 therefore the energy as a function of the field for spin 1/2 nuclei is (this is a diagram for a nuclear spin with a neg. $\gamma$, such as $^{15}$N.

The existence of two energy levels ($\alpha$ and $\beta$) implies that at thermal equilibrium there will be a population difference between the levels. The energies of the two levels will be:

$$E_{\alpha} = - { \gamma \bar h B \over 2 }$$ (10)

$$E_{\beta} = + { \gamma \bar h B \over 2 }$$ (11)

The energy difference between the two states is:

$$\Delta E = E_{\alpha} - E_{\beta} = \gamma \bar h B$$ (12)

or,

$$\omega = \gamma B$$ (13)

The energy of an NMR transition is quite low, in fact, the wavelength of light required to excited transitions is in the radio frequency range. This has two important consequences:

• The population difference between the ground and excited state is very small, on the order of 1 part in 10$^6$.
• The lifetime of the excited state can be quite long, on the order of $msec$ to $sec$

The consequence of having a small population differences is that NMR is a relatively insensitive experimental technique. Typically, protein concentrations on the order of 1-2 mM are required. However, in some cases concentrations in the range of 50 $\mu$M have be used.

Having a long lifetime for the excited state provides two benefits. First, the resonance lines are relativly narrow, thus the intrinsic resolution of NMR spectroscopy is quite high. For example, NMR linewidths can be on the order of 1 Hz, while IR and UV-Vis spectral lines are $\gg$ 1 KHz. The long lifetime of the excited state also allows experimental manupulation of these states. This is quite unique to NMR spectroscopy and is the basis of multi-dimensional NMR experiments.

Equation of Motion of Nuclear Spins

All NMR experiments involve the excitation, manipulation, and detection of excited nuclear spin states. The simplist NMR experiment is a one-pulse experiment:

In this experiment the nuclear spins are excited by a short burst of light (radiofrequency) and the resultant excited states produce a magnetic field which is detected (the Free induction decay). The subsequent Fourier transformation of the free induction decay gives the normal NMR spectrum.

During this simple experiment the motion of the spins in a magnetic field can be described to a satisfactory level by the use of classical mechanics. In more complicated experiments it will be necessary to utilize some quantum mechanics to describe the evolution of the nuclear.

The evolution of the nuclear spins during the one-pulse experiment can be treated as three discrete time intervals:

• Prior to the excitation pulse (Static $B_o$ field, thermal equilibrium)
• During the excitation pulse (Static $B_o$ field plus excitation field, $B_1$)
• During Detection (Static $B_o$ field)

Since the nuclear spins possess angular momentum the effect of applying an external field ($B_o$ and $B_1$) to the spins is to generate a torque on the spin. This torque will cause a change in angular momentum as described by the following equation:

$${dJ \over dt }= \mu \times B$$ (14)

Using $\vec{\mu} = \gamma \vec{J}$ we can write:

$$\frac{d \mu}{dt }= \gamma \mu \times \ B$$ (15)

This equation could be solved by standard methods, but it is much more instructive to introduce a rotating frame of reference before attempting its solution. In a rotating reference frame, the basis vectors change their direction according to:

$${d x_i \over dt }= \Omega \ \times \ x_i$$ (16)

This is just the cartesian frame rotating with a speed $\vert\Omega\vert$ around some axis in the direction of $\Omega$. For the case of the magnetic moment, $\mu$:

$${d\mu \over dt }= \sum x_i {du_i \over dt }+ \sum u_i {dx_i \over dt}$$ (17)

$$={ \delta u \over \delta t} + \sum u_i \Omega \ \times \ x_i$$ (18)

$$= { \delta u \over \delta t} + \Omega \ \times \ u$$ (19)

$\delta u \over \delta t$ is the change of with respect to time in the rotating frame. Therefore:

$${ \delta u \over \delta t} = \gamma u \ \times \ (B + {\Omega \over \gamma })$$ (20)

This equation is of the same form as d/dt, but with the addition of a fictitious field $\Omega / \gamma$. What happens when $B+\Omega/\gamma=0$? In this case neither the direction or the amplitude of changes, it becomes a stationary vector in the rotating frame. Under these conditions we find:

$$\Omega = \ - \ \gamma B$$ (21)

This frequency is called the "Larmor" frequency and is the rate at which precesses around $\vec{B}$ in the laboratory frame. This frequency is also the same as the resonance frequency, or absorption energy of the nuclear transition.

Equation 20 describes the motion of a single nuclear spin in the rotating co-ordinate frame under the influence of the static magnetic field. Therefore, prior to the excitation pulse each individual spin precesses about the static field. If there are a large number of spins in the sample it is possible to define a net magnetic moment for the sample:

$$M_i = \sum u_i$$ (22)

In the presence of the static field the z-component of each magnetic dipole will add to give a detectable bulk magnetization. This is non-zero because there is a slight difference in the population of spins pointing one way versus spins point another (given by the Boltzmann distribution). The net magnetization along z is:

$$M_z = M_o$$ (23)

In contrast, the distribution of the magnetic dipoles in the x-y plane is random. In other words, there is no relationship between the transverse (x-y) magnetization of one spin to another. The transverse magnetization is termed to be incoherent. Consequently, there is no bulk magnetization at thermal equilibrium:

$$M_x = M_y = 0$$ (24)