# How to explain spin-spin relaxation in MRI/NMR in terms of quantum mechanics?

I noticed two approaches to explanation of MRI/NMR. The most popular classical approach exploits the notion of magnetization vector and Larmor precession. Everything then is described as oscillations of this vector relative to static field direction. In quantum approach there are energy states splitting due to Zeeman effect. Photons of radio-frequency field make protons to jump between these states, and we measure how these states are populated in time.

Quantum approach is more accurate. Classical description is very clear illustration, but it doesn't explain what happens to individual proton.

Spin-spin relaxation is always explained in terms of magnetization vectors. Can you explain what it is in terms of quantum mechanics?

In general, interaction of proton magnetization states with neighboring atoms allows states transitions. If we can consider them as small perturbations then in terms of time-dependent perturbation theory the transition probability rate form state i to f is given, $$W_{fi} = \frac{2\pi}{\hbar} |<f|V|i>|^{2} \delta(E_{f}-E_{i})$$ To conserve energy if transition from $m=-1/2$ to $m=+1/2$ happens, it should be followed by another proton from $m=+1/2$ to $m=-1/2$. Then take any two state $a,b$ with b state has higher energy. Denote an upwards energy transition by $a-$ $\rightarrow$ $b+$ and downward transition by $b+$ $\rightarrow$ $a-$.

If $N_{+}$ is the initial number of spins with $m=+1/2$ then,

$$\frac{dN_{+}}{dt} = W_{b+a-}N_{-}n_{a}-W_{a-b+}N_{+}n_{b}$$

where $n_{b}$ is the initial number states with higher energy and $n_{a}$ lower energy, related by Boltzmann factor,

$$\frac{n_{a}}{n_{a}} = e^{-\hbar B_{0} \gamma /kT}$$

For our case probability transitions are equal, which is the case in $1^{st}$ or perturbation theory,

$$W_{b+a-} = W_{a-b+} = W$$

Total number of states is fixed, $$N = N_{+}+N_{-}$$ and writing, $$N_{\pm} = \frac{N \pm \Delta N}{2}$$ Now combining the previous equations, $$\frac{d \Delta N}{dt} = WN(n_{a}-n_{b})-W\Delta N(n_{a}-n_{b})$$ We want in the equilibrium $$\frac{d \Delta N}{dt} = 0$$ But this implies $$\Delta N_{0} = \frac{n_{a}-n_{b}}{n_{a}+n_{b}}N$$ Define $W(n_{a}+n_{b})$ as $\frac{1}{T_{1}}$ longitidunal relaxation time constant, then we get

$$\frac{d \Delta N}{dt} = \frac{\Delta N_{0} -\Delta N}{T_{1}}$$

Finally, take the average over the volume and you will get something similar to

$$\frac{d M_{z}}{dt} = \frac{M_{0}-M_{z}}{T_{1}}$$ which is the longitudunal component relaxation equation of spin.

• But $T_1$ is due to the spin-lattice relaxation, isn't it? The question concerns spin-spin relaxation. – abir Aug 3 '20 at 14:32