Time Development of a Spin ${1\over 2}$ State in a B field

Assume that we are in an arbitrary spin state $\chi(t=0)=\pmatrix{a\cr b}$ and we have chosen the z axis to be in the field direction. The upper component of the vector (a) is the amplitude to have spin up along the z direction, and the lower component (b) is the amplitude to have spin down. Because of our choice of axes, the spin up and spin down states are also the energy eigenstates with energy eigenvalues of $\mu_BB$ and $-\mu_BB$ respectively. We know that the energy eigenstates evolve with time quite simply (recall the separation of the Schroedinger equation where $T(t)=e^{-iEt/\hbar}$). So its simple to write down the time evolved state vector.

$$\bgroup\color{black}\chi(t)=\pmatrix{ae^{-i\mu_BBt/\hbar}\cr ... ...t/\hbar}} =\pmatrix{ae^{-i\omega t}\cr be^{i\omega t}} \egroup$$

where ${\omega=\mu_BB\over\hbar}$.

So let's say we start out in the state with spin up along the x axis, $\chi(0)=\pmatrix{{1\over\sqrt{2}}\cr {1\over\sqrt{2}}}$. We then have

$$\begin{eqnarray*} \chi(t) & = & \pmatrix{{1\over\sqrt{2}}e^{-i\omega Bt}\cr {1\o... ...+e^{-2i\omega Bt}\right) ={\hbar\over 2}\cos(2\mu_BBt/\hbar) \\ \end{eqnarray*}$$

So again the spin precesses around the magnetic field. Because $g=2$ the rate is twice as high as for $\ell=1$.

The remainder of this section is another, more complete, explanation of the time development. It is somewhat notationally challenged. Recall that, for any arbitrary system in Quantum Mechanics, a state vector $\psi(0)$ describing the system at some particular time $t=0$ will evolve under the action of the ``Time Evolution Operator'':

$$\bgroup\color{black}\psi(t)=e^{iH t/\hbar}\psi(0)\egroup$$

where H is the Hamiltonian operator (for more details, see example ``time evolution of $\ell=1$ system...''). For this system $H=\mu_B B \sigma_z$, so that the time evolution operator becomes $e^{i\mu_B B \sigma_z t/\hbar}$ and, using spinor notation, we write

$$\bgroup\color{black}\chi(t)=e^{i\mu_B B \sigma_z t/\hbar}\chi(0)\egroup$$

which in matrix notation becomes,

$$\bgroup\color{black}\left(\matrix{\chi_1(t) \cr \chi_2(t)}\ri... ...right]} \left(\matrix{\chi_1(0) \cr \chi_2(0)}\right ) \egroup$$

where $\omega = \mu_B B/\hbar$.

?? . . . Mathematical Note on exponentiation of matrices . . . ??

As usual, we know how the operator $\left(\matrix{1&0 \cr 0&-1}\right)$ in the exponent acts on its eigenstates (basis vectors), so we decompose the arbitrary state vector into a linear combination of eigenstates:

$$\bgroup\color{black}\left(\matrix{\chi_1(0) \cr \chi_2(0) }\r... ...ix{1 \cr 0}\right ) + a_2 \left(\matrix{0 \cr 1}\right )\egroup$$

so now we can write

$$\bgroup\color{black}\left(\matrix{\chi_1 (t) \cr \chi_2 (t)}\... ...0}\right ) + a_2 \left(\matrix{0 \cr 1}\right ) \right) \egroup$$

$$\bgroup\color{black} = a_1 e^{i\omega t} \left(\matrix{1 \cr ... ... ) + a_2 e^{-i\omega t} \left(\matrix{0 \cr 1}\right ).\egroup$$

Let's look at the expectation values of the three spin operators on an such an arbitrary state:

$$\bgroup\color{black}\langle S_z \rangle=\langle \chi(t)\vert S_z\vert\chi(t)\rangle\egroup$$

$$\bgroup\color{black} = \pmatrix{a^*_1 e^{-i\omega t} & a^*_2... ...-1} \pmatrix{ a_1 e^{i\omega t} \cr a_2 e^{-i\omega t} }\egroup$$

$$\bgroup\color{black} = {\hbar\over 2} \left( \vert a_1\vert^2 - \vert a_2\vert^2 \right) \egroup$$

so it appears that the expectation $\langle \chi(t)\vert S_z\vert\chi(t)\rangle$ will remain stationary in time for any arbitrary state. Meanwhile, substituting $S_x={\hbar\over 2}\left(\matrix{0&1\cr 1&0}\right)$ and $S_y={\hbar\over 2} \left(\matrix{0&-i \cr i&0}\right)$ in the above derivation yields

$$\bgroup\color{black}\langle S_x \rangle = {\hbar\over 2} \lef... ...1a^*_2 e^{2i\omega t} + a^*_1a_2 e^{-2i\omega t}\right] \egroup$$

$$\bgroup\color{black} \langle S_y \rangle = {i\hbar\over 2} \... ...1a^*_2 e^{2i\omega t} - a^*_1a_2 e^{-2i\omega t}\right] \egroup$$

To see that this is a rotation, absorb $a_1a^*_2$ into the exponent i.e., express $a_1a^*_2= r e^{i\delta}$ where r and $\delta$ are real constants. Then

$$\bgroup\color{black}a_1a^*_2 e^{2i\omega t} = r e^{i(2\omega t + \delta)}= {1\over 2} e^{i\alpha}\egroup$$

(the case $a_1=a_2\Rightarrow a_1=1/\sqrt{2}$ for normalization so that $r=1/2$). Thus,

$$\bgroup\color{black} \langle S_x \rangle = {\hbar\over 4} \l... ...cos{\alpha} = {\hbar\over 2} \cos{(2\omega t + \delta)} \egroup$$

$$\bgroup\color{black} \langle S_y \rangle = {i\hbar\over 4} \... ...\sin{\alpha} = {\hbar\over 2} \sin{(2\omega t + \delta)}\egroup$$

Hence the state precesses about the z-axis with angular frequency $2\omega$. Compare this to the $\ell=1$ case: here the state rotates twice as fast!