Nuclear magnetic resonance

Consider a spin 1/2 particle in a time varying magnetic field

\begin{align*} \vec B(t) = B_0 \hat z + B_1 (\cos \omega t \, \hat x - \sin\omega t \,\hat y). \end{align*}

The Hamiltonian is

\begin{align*} H(t) = -\gamma \vec B(t) \cdot \vec S = -\gamma \big(B_0 S_z - B_1 (\cos \omega t \, S_x - \sin\omega t\, S_y)\big) \end{align*}

Since $[H(t_1), H(t_2)] \ne 0$ , we can’t use the simplified form of time evolution. Usually we would have to apply the time-ordered exponential, but in this case we can use a special trick.

Notice that the circularly polarized component of the field rotates in the $xy$ plane. We can write that term of the Hamiltonian as the $x$ spin operator $S_x$ rotated about the $z$ axis

\begin{align*} H(t) = -\gamma B_0 S_z - \gamma B_1 \, e^{i\omega t S_z/\hbar} S_x e^{-i\omega t S_z/\hbar}. \end{align*}

We can just as well apply the rotation to the $S_z$ operator too, since rotating $S_z$ about the $z$ axis does nothing. This allows us to write the expression as a time-independent commutator

\begin{align*} H(t) = e^{i\omega t S_z / \hbar} \left[ -\gamma B_0 S_z - \gamma B_1 S_x \right] e^{-i\omega tS_z/\hbar}. \end{align*}

Now define a state $\ket{\psi_R,t} = e^{-i\omega t S_z/\hbar} \ket{\psi,t}$ in a rotating frame. Applying the Schrödinger equation to $\ket{\psi,t}$ , written in terms of $\ket{\psi_R,t}$

\begin{align*} i\hbar \frac{d }{d t} \left[ e^{i\omega t S_z/\hbar} \ket{\psi_R,t} \right] &= e^{i\omega tS_z/\hbar} \left[ -\gamma B_0 S_z - \gamma B_1 S_x \right] e^{-i\omega t S_z/\hbar} e^{i\omega t S_z/\hbar} \ket{\psi_R,t} \\ -\omega S_z e^{i\omega tS_z/\hbar} \ket{\psi_R,t} + i\hbar e^{i\omega tS_z/\hbar} \frac{d }{d t} \ket{\psi_R,t} &= e^{i\omega tS_z/\hbar} \left[ -\gamma B_0 S_z - \gamma B_1 S_x \right] \ket{\psi_R,t} \\ i\hbar \frac{d }{d t} \ket{\psi_R,t} &= \left[ (\omega - \gamma B_0) S_z - \gamma B_1 S_x \right] \ket{\psi_R,t}. \end{align*}

In the rotating frame, the spin processes around an effective magnetic field

\begin{align*} \vec B_\eff &= \left( B_0 - \frac{\omega}{\gamma} \right) \hat z + B_1 \hat x. \end{align*}

This gives us the solution

\begin{align*} \ket{\psi_R,t} &= \exp \left(i\gamma \vec B_\eff \cdot \vec S t/\hbar\right) \ket{\psi_R,0} \\ \ket{\psi,t} &= \exp \big(i\omega t S_z/\hbar\big) \exp \left(i\gamma \vec B_\eff \cdot \vec S t/\hbar\right) \ket{\psi,0}. \end{align*}

Now take $B_0 \gg B_1$ and set $\ket{\psi,0} = \ket{z+}$ . If $\omega$ is far from $\omega_0 = \gamma B_0$ (off resonance), then $\vec B_\eff \sim \hat z$ , and the state proceses around the $z$ axis. Since our state is $\ket{z+}$ it doesn’t change.

If $\omega = \omega_0 = \gamma B_0$ , then $\vec B_\eff = B_1 \hat x$ and the state in the rotating frame $\ket{\psi_R,t}$ processes around the $\hat x_R$ axis. After a time $\Delta t = \pi/2\gamma B_1$ it is $\ket{\psi_R,\Delta t} = \ket{y_R+}$

After this time $\Delta t$ , set $B_1 = 0$ with $B_0 = \omega_0/\gamma$ unchanged. Then $\ket{\psi_R,t} = e^{i\omega t S_z/\hbar} \ket{\psi_R,\Delta t}$ and the state continues to process around the $z$ axis.

In the lab frame, this looks like the state slowly rotating from $+z$ to the $xy$ plane, all while quickly rotating around the $z$ axis.