Spin $\frac{1}{2}$ in Magnetic Field

This case is, for example, extremely important for NMR (nuclear magnetic resonance). Even here the Hamiltonian ${\mathcal H}(t)$ is in general not exactly soluble, its form is

$${\mathcal H}(t) \equiv {\bf B}(t) \mbox{\boldmath$\sigma$} \equiv B_x(t) \hat{\sigma}_{x} + B_y(t) \hat{\sigma}_{y}+ B_z(t) \hat{\sigma}_{z}$$

$$\equiv \left( \begin{matrix}B_z(t) & B_{\Vert}^*(t)\\ B_{\Vert}(t) & - B_z(t)\end{matrix}\right),\quad B_{\Vert}(t)\equiv B_x(t) + i B_y(t),$$ (2.1)

where the Pauli-matrices are defined as

$$\hat{\sigma}_{x} \equiv \begin{pmatrix}0 & 1 1 & 0 \end{pmatrix},\quad \hat{\sigma}_{y... ...hat{\sigma}_{z}\equiv \left( \begin{matrix}1 & 0 0 & -1 \end{matrix}\right).$$ (2.2)

Why is that so difficult? Let us write the Schrödinger equation

$$i\partial_t \vert\Psi(t)\rangle = {\mathcal H}(t)\vert\Psi(t)\rangle,\quad \vert\Psi(t)\rangle \equiv \begin{pmatrix}\psi_1(t) \psi_2(t)\end{pmatrix}$$

$$\leadsto i\frac{d}{dt}\psi_1(t) = B_z(t) \psi_1(t) + B_{\Vert}^*(t) \psi_2(t)$$

$$i\frac{d}{dt}\psi_2(t) = B_{\Vert}(t) \psi_1(t) - B_z(t) \psi_2(t).$$ (2.3)

We assume ${ B_{\Vert}}\ne 0$ and write (omit the $t$-dependence for a moment)

$$\psi_1 = \frac{i\dot{\psi}_2 + B_z \psi_2}{ B_{\Vert}}$$ (2.4)

$$i \ddot{\psi}_2 = \dot{B_{\Vert}} \psi_1 + B_{\Vert}\dot{\psi_1} - \dot{B_z} \psi_2 - B_{z}\dot{\psi_2}$$

$$= \frac{\dot{B_{\Vert}}}{B_{\Vert}} [ i\dot{\psi}_2 + B_z \psi_2] -... ...\Vert} [B_z \psi_1 + B_{\Vert}^* \psi_2] - \dot{B_z} \psi_2 - B_{z}\dot{\psi_2}$$

$$= \frac{\dot{B_{\Vert}}}{B_{\Vert}} [ i\dot{\psi}_2 + B_z \psi_2] -... ...z \psi_2] -iB_{\Vert} B_{\Vert}^* \psi_2 - \dot{B_z} \psi_2 - B_{z}\dot{\psi_2}$$

$$= i\frac{\dot{B_{\Vert}}}{B_{\Vert}} \dot{\psi}_2 + \left[ \frac{\d... ...}{B_{\Vert}} B_z -i B_z^2 -i \vert B_{\Vert}\vert^2 - \dot{B}_z \right] \psi_2.$$ (2.5)

This is a second order ODE with time-dependent coefficients, which in general is not solvable in terms of known functions (it can of course be solved numerically quite easily).