I am quoting Clayden et al 2nd edition:

1. The sample is irridiated with a short pulse of radio-frequency energy. This disturbs the equilibrium balance between the two energy levels: some nuclei absorb the energy and are promoted to a higher energy level.

2. When the pulse finishes, the radiation given out as the nuclei fall back down to the lower energy level is detected using what is basically a sophisticated radio receiver.

Now, my confusion is about the equilibrium of nuclei in higher & lower energy level , after the introduction of radio wave.

Are the nuclei in higher level steady i.e. not radiating and coming back to the lower level constantly?

Or, are they radiating and constantly re-energized by the radio wave till the radio wave is on(still keeping a equilibrium constant between nuclei in lower and higher energy level)?

• Do you understand the fundamental way in which modern Fourier transform NMR works generally, or is the whole concept new to you? Jun 24 '17 at 16:42
• It's a new concept for me @NotEvans Jun 25 '17 at 5:46

I'll try to be brief. Suppose a spin exists in the state

$$\require{begingroup} \begingroup \newcommand{ket}{\left|#1\right>} \ket{\psi} = c_\alpha\ket{\alpha} + c_\beta\ket{\beta}$$

where the coefficients $c_\alpha, c_\beta$ are time-dependent. When a $90^\circ$ RF pulse, with strength $B_1$, is applied about the $x$-axis, the Hamiltonian of the system becomes (in the rotating frame)

$$\hat{H} = \omega_1 \hat{I}_{\!x}; \quad \omega_1 = -\frac{B_1}{\gamma}$$

If you plug this into the time-dependent Schrodinger equation

$$\mathrm{i}\hbar \frac{\mathrm{d}\ket{\psi}}{\mathrm{d}t} = \hat{H}\ket{\psi}$$

you will get two coupled differential equations (the proof is detailed in most QM texts, as far as I know)

\begin{align} \frac{\mathrm dc_\alpha}{\mathrm{d}t} &= \frac{1}{2}\mathrm{i}\omega_1c_\beta \\ \frac{\mathrm dc_\beta}{\mathrm{d}t} &= \frac{1}{2}\mathrm{i}\omega_1c_\alpha \end{align}

These can be solved using standard methods to get

\begin{align} c_\alpha(t) &= c_\alpha(0)\cos\left(\frac{\omega_1 t}{2}\right) - \mathrm{i}c_\beta(0)\sin\left(\frac{\omega_1 t}{2}\right) \\ c_\beta(t) &= c_\beta(0)\cos\left(\frac{\omega_1 t}{2}\right) - \mathrm{i}c_\alpha(0)\sin\left(\frac{\omega_1 t}{2}\right) \end{align}

For a $90^\circ$ pulse, the flip angle $\omega_1 t = \pi/2$, and this simplifies to

\begin{align} c_\alpha(t) &= \frac{1}{\sqrt{2}}(c_\alpha(0) - \mathrm{i}c_\beta(0)) \\ c_\beta(t) &= \frac{1}{\sqrt{2}}(c_\beta(0) - \mathrm{i}c_\alpha(0)) \end{align}

Now let's talk about a spin that happens to be initially 100% up spin (which is the "lower energy state" for a nucleus with positive gyromagnetic ratio), such that

$$\ket{\psi(0)} = \ket{\alpha}; \quad c_\alpha(0) = 1, c_\beta(0) = 0.$$

After applying a $90^\circ$ pulse, the wavefunction will be changed into

$$\ket{\psi(t)} = \frac{1}{\sqrt{2}}\ket{\alpha} - \frac{\mathrm{i}}{\sqrt{2}}\ket{\beta}$$

i.e. equal amounts of up and down character, in the sense that the mod-squares of the coefficients are the same. This is the so-called "excitation".

Likewise one can verify that if $\ket{\psi(0)} = \ket{\beta}$, then the final wavefunction after the RF pulse will be

$$\ket{\psi(t)} = -\frac{\mathrm{i}}{\sqrt{2}}\ket{\alpha} + \frac{1}{\sqrt{2}}\ket{\beta}$$

i.e. equal amounts of up and down character, as well. So if a spin is initially in the "higher-energy state" $\ket{\beta}$, the pulse will actually cause it to return towards the lower-energy state!

In fact there is no stipulation that the spins have to be initially either up or down. This misconception is very widely propagated in introductory accounts of NMR. In truth the initial coefficients $c_\alpha, c_\beta$ can be anything. There is no restriction on them such that they can only be $(1,0)$ or $(0,1)$. In general the spins in a sample will have all sorts of values of $c_\alpha$ and $c_\beta$. Some of them will be leaning more towards $\alpha$ than they do towards $\beta$, and vice versa. This is a direct consequence of the small energy gap between the two levels - there isn't a very large driving force for the spins to fall into the lower level.

Nevertheless, we can say that because of the external magnetic field, in an ensemble of spins (i.e. many spins), the values of $|c_\alpha|^2$ tend to be larger than the values of $|c_\beta|^2$. Formally, we would write

$$\overline{|c_\alpha|^2} > \overline{|c_\beta|^2}$$

where the bar indicates an ensemble average. These two quantities are called the populations of states $\alpha$ and $\beta$. If you are interested, the idea of an ensemble average is accounted for very neatly by density matrices.

If you work through the relevant maths, you would find that after the $90^\circ$ pulse, you would get

$$\overline{|c_\alpha|^2} = \overline{|c_\beta|^2}$$

i.e. equal populations of the two spin states. Therefore, there is a net conversion of the spins in the system, from $\ket{\alpha}$ to $\ket{\beta}$, and this is the "excitation" that textbooks like to speak of.

However, we can only speak of the populations in terms of an ensemble average. The so-called "excitation" only means that as a whole, the system loses $\alpha$-character ($\overline{|c_\alpha|^2}$ decreases) and gains $\beta$-character ($\overline{|c_\beta|^2}$ increases).

That tells us nothing about each individual spin. As shown in the previous section, it is perfectly possible to have an individual spin lose $\beta$-character ($c_\beta$ decreases) and gain $\alpha$-character ($c_\alpha$ increases).

As an addendum: after this excitation, the main form by which $\overline{|c_\alpha|^2}$ and $\overline{|c_\beta|^2}$ return to their equilibrium values is not by spontaneous emission of a photon (the rate of spontaneous emission is proportional to $\nu^3$ and is hence incredibly small) but rather spin-lattice relaxation, which is a non-radiative process.

The nuclei absorb the rf radiation and the spin population in the upper level increases at the expense of that in the lower level. There is now a chance that the upper levels will radiate away their extra energy and so return to equilibrium. However, in an nmr experiment the excitation pulses used in most if not all nmr machines are very short, just a few microseconds (and so spectrally wide) but the natural radiative lifetime of the excited levels is some seconds. Thus it is possible to excite the nuclear spins and build up a relatively large population before any significant decay occurs.

(This is one of the reasons why nmr works so well considering the tiny population difference that is produced because the energy levels are so close in energy and so all levels are significantly thermally populated. The long lifetime of upper levels means that it is possible to coherently build up a significant macroscopic magnetisation in the sample and this what is detected.)