# Spin 1/2 systems


## Single Spin 1/2 System: State Space

The state space of a single spin 1/2 system is the simplest nontrivial quantum mechanical space possible, $$\cn^2$$. The obervables corresponding to the components of the angular momentum are the matrices $$\hat{s}_x=\frac{\hbar}{2} \smx,\hat{s}_y=\frac{\hbar}{2} \smy,\hat{s}_z=\frac{\hbar}{2} \smz,$$ represented here in the eigen-basis, $$\{\ket{+},\ket{-}\}$$, of the z-component of the momentum, $$\hat{s}_z$$. These angular momentum observables are related to the Pauli matrices $$\hat{\sigma}_x=\left( \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array} \right),\hat{\sigma}_y=\left( \begin{array}{cc} 0 & -i \\ i & 0\end{array} \right),\hat{\sigma}_z=\left( \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right).$$ A standard set of operators used in NMR are the dimensionless spin matrices $$\hat{I}_x=\frac{1}{2} \smx,\hat{I}_y=\frac{1}{2} \smy,\hat{I}_z=\frac{1}{2} \smz,$$ and we will also adopt them in this text. It is easy to check that all Pauli matrices have two eigen-values, +1 and -1, so measurements of the spin operators can only yield +1/2 or -1/2 as outcomes. For completeness we define the magnetization operators as $$\vec{\hat{M}}=\mu \vec{\hat{\sigma}}$$, where $$\mu$$ is the magnetic moment of the spin-1/2 particle.

## Single Spin 1/2 System: Zeeman Hamiltonian

The interaction of a magnetic moment, $$\vec{\mu}$$, with a magnetic field, $$\vec{B}_0$$, is described by the interaction Hamiltonian -$$\vec{M}\cdot\vec{B}_0$$. This translates to a quantum mechanical Hamiltonian $$\hat{H}(\vec{\hat{\sigma}})=-\mu \vec{\hat{\sigma}}\cdot \vec{B}_0.$$ If the field is parallel to the z axis we obtain the standard form of the Zeeman Hamiltonian $$\hat{H}(\vec{\hat{\sigma}})=-\mu B_0 \hat{\sigma}_z=\mu B_0 \left( \begin{array}{cc} -1 & 0 \\ 0 & 1\end{array} \right).$$ Using 3 from the previous section we find that the evolution operator for the spin-1/2 Zeeman Hamiltonian is $$\hat{U}(t)=e^{i(\mu B_0 \hat{\sigma}_z)/\hbar}=\left( \begin{array}{cc} e^{i\omega_0 t} & 0 \\ 0 & e^{-i\omega_0 t} \end{array} \right),$$ where we have defined the Larmor frequency $$\omega_0=(\mu B_0)/\hbar$$.

## Free precession

Since the Hamiltonian is proportional to $$\hat{I}_z$$ they have the same eigen-vectors, $$\ket{+}$$ and $$\ket{-}$$, that have only a trivial phase evolution. Once the spin is polarized to be in the $$\ket{+}$$ or $$\ket{-}$$ state it stays there and subsequent measurements of $$\hat{I}_z$$ always yield a z component of spin that is either +1/2 or -1/2 depending on which state the spin was initialy in. Let us examine the time dependence of the expectation value of the x and y component of the magnetization if the the system was initially in the $$\ket{+}$$ state \begin{eqnarray} \av{\hat{M}_x(t)}&=&\av{\mu \hat{\sigma}_x(t)}=\mu\ev{+}{\hat{U}^\dagger (t)\sx\hat{U}(t)}{+} \\ \nonumber &=&\mu (1 \; 0)\mat{e^{-i\omega_0 t}}{0}{0}{e^{i\omega_0 t}}\smx \mat{e^{i\omega_0 t}}{0}{0}{e^{-i\omega_0 t}}\col{1}{0} \\ \nonumber &=&\mu (e^{-i\omega_0 t} \; 0) \smx \col{e^{-i\omega_0 t}}{0}=(e^{-i\omega_0 t} \; 0) \col{0}{e^{-i\omega_0 t}}=0. \end{eqnarray} It is easy to show, in an analogous way, that $$\av{M_y(t)}=0$$ also. The eigenstates of \sx are $$\ket{+}_x=\frac{1}{\sqrt{2}}\col{1}{1} \;\;\; \mbox{and} \;\;\; \ket{-}_x=\frac{1}{\sqrt{2}}\col{1}{-1}$$ since $$\smx \frac{1}{\sqrt{2}}\col{1}{1}=\frac{1}{\sqrt{2}}\col{1}{1} \;\;,\;\; \smx \frac{1}{\sqrt{2}}\col{1}{-1}=-\frac{1}{\sqrt{2}}\col{1}{-1}.$$ Analogously to $$\ket{+}$$, if we measure $$\hat{I}_x$$ when the system is in the state $$\ket{+}_x$$ we will always measure +1/2 as the result - the spin is polarized along the positive x-axis. Let us now examine the evolution of the magnteization average when the initial state of the spin is $$\ket{+}_x$$, which is the case after a 90 degree rf pulse was applied to a spin polarized along the z axis \begin{eqnarray} \av{\hat{M}_z(t)}&=&\av{\mu \hat{\sigma}_z(t)}=\\ \nonumber &=&\mu (1 \; 1)\mat{e^{-i\omega_0 t}}{0}{0}{e^{i\omega_0 t}}\smz \mat{e^{i\omega_0 t}}{0}{0}{e^{-i\omega_0 t}}\col{1}{1} \\ \nonumber &=&\mu (1 \; 1)\smz\col{1}{1}=0. \end{eqnarray} As expected the average of the z magnetization is always zero. The situation is different for the x and y components \begin{eqnarray} \av{\hat{M}_x(t)}&=&\av{\mu \hat{\sigma}_x(t)}= \\ \nonumber &=&\frac{\mu}{2} (1 \; 1)\mat{e^{-i\omega_0 t}}{0}{0}{e^{i\omega_0 t}}\smx \mat{e^{i\omega_0 t}}{0}{0}{e^{-i\omega_0 t}}\col{1}{1} \\ \nonumber &=&\frac{\mu}{2} (e^{-i\omega_0 t} \; e^{i\omega_0 t} ) \smx \col{e^{i\omega_0 t}}{e^{-i\omega_0 t}}=\frac{\mu}{2}(e^{-i\omega_0 t} \; e^{i\omega_0 t}) \col{-e^{i\omega_0 t}}{e^{i\omega_0 t}}\\ \nonumber &=&\mu \cos(\omega_0 t). \end{eqnarray} Similarly we can show that $$\av{\hat{M}_y(t)}=\mu \sin(\omega_0 t).$$ The three just derived equations $$\av{\hat{M}_x(t)}=\mu \cos(\omega_0 t), \; \; \av{M_y(t)}=\mu \sin(\omega_0 t) \;\; \mbox{and} \;\; \av{M_z(t)}=0,$$ describe the quantum analog of the classical free precession. One has to be carefull in interpreting these equations since they deal with averages. To clarify this we examine the evolution of the $$\ket{+}_x$$ state of the system $$\ket{\psi (t)}=\hat{U}(t)\ket{+}_x=\mat{e^{i\omega_0 t}}{0}{0}{e^{-i\omega_0 t}} \frac{1}{\sqrt{2}} \col{1}{1}=\frac{1}{\sqrt{2}} \col{e^{i\omega_0 t}}{e^{-i\omega_0 t}}.$$ If we measure $$\hat{I}_x$$ only at times such that $$\omega_0 t=2\pi n+\Delta \phi$$, where $$n$$ is integer, the result will be $$+1/2$$ with probability $$p(+1/2,\ket{\psi (t)},\hat{I}_x)=\vert \frac{1}{\sqrt{2}}(1\;1)\frac{1}{\sqrt{2}}\col{e^{i(2\pi n +\Delta \phi)}}{e^{-i(2\pi n + \Delta \phi)}} \vert^2=\cos^2(\Delta \phi). \label{caveat}$$ If we measured $$\hat{I}_x$$ for each individual spin only at times for which $$\phi=\pi/4$$, half the time the outcome will be $$1/2$$ and half the time $$-1/2$$. On average the magnetization along the x axis will be zero as predicted by \ref{caveat}, however each measurement of individual spins would produce one of the two non-zero eigen-values. By contrast, in a classical measurement each individual spin measured at those times would have zero magnetization along the x direction.