Continuous-wave nuclear magnetic resonance

For constant drive, set the three rotating frame derivatives to zero. Substitution first gives $u=\Delta T_{2}v$, then $w=M_0-{\omega }_1{T}_{1}v$, henceThe absorption is proportional to ${\omega }_{1}v$. With $s\ll 1$, its frequency dependence is a Cauchy distribution profile, often called a Lorentzian. With appreciable $s$, the response broadens and $w=M_0/(1+s)$ on resonance: nuclear magnetic resonance saturation limits the response. The short-pulse amplitude in the previous example is not a weak continuous drive; these are different operating regimes of the same equations.

Drive reduces the longitudinal population imbalance when excitation competes with recovery. The dimensionless $s$ in the steady Bloch equations measures that competition.

For weak drive in this homogeneous model, half maximum occurs at $|\Delta |=1/T_2$. The absorption full width at half maximum is thereforeThe example gives $2.653\mathrm{Hz}$ or $0.06230\mu \mathrm{T}$. At finite drive the model width is multiplied by $\sqrt{1+s}$. Field inhomogeneity and unresolved components can also broaden an actual line, so its width alone does not determine pure dephasing.

The same width appears by an original Fourier transform calculation. For $a=1/T_2$ and a free homogeneous signal demodulated exactly at resonance,This connects exponential memory to a Lorentzian absorptive line. It is a mathematical connection, not evidence that the 1946 teams performed pulsed Fourier transform spectroscopy. Instrumental signal bandwidth is a separate choice: narrowing a receiver band can reject thermal electrical noise, but too narrow a band distorts a changing signal. A sample linewidth and a receiver bandwidth are not interchangeable.