Time evolution of an electron in an homogeneous magnetic field: get rid of $S_y$ in the exponential

Question

Context

We have a particle of spin $1/2$ and of magnetic moment $\vec{M} = \gamma\vec{S}$.

At time $t=0$, the state of the system is $$ |\psi(t=0) \rangle = |+\rangle_z $$

We let the system evolve under the influence of a magnetic field $B_0$ parallel to $Oy$.

We are asked to find the state of the system at time $t$.

Attempt

The energy of interaction between the $B$ field and the spin is $$\hat{H} = E_m \equiv -\vec{\mu} \cdot \vec{B} = - \gamma \vec{S} \cdot B_0 \hat{y} = - \gamma B_0 y S_y = \omega_0 S_y$$

where $\gamma$ denotes the gyromagnetic ratio and $S_y$ is the spin operator in the $y$ direction.

We can write the time-dependance of $\psi$ by applying the time-evolution operator on $|\psi(t=0)\rangle$, $$ \begin{align} |\psi(t)\rangle &= e^{-iHt/\hbar}|\psi(t=0)\rangle \\ &= e^{-i \omega_0 S_y t/\hbar}|+\rangle_z \end{align} $$

Now I would like to express $|+\rangle_z$ in the $\{|+\rangle_y, |-\rangle_y\}$ basis.

We can express $S_y$ in term of the Pauli matrix $\sigma_y$, $$ S_y = \frac{\hbar}{2} \underbrace{\begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}}_{\sigma_y} $$

Finding the eigenvalues of $S_y$, $$ det(S_y - \lambda I) = 0 \implies \lambda_{1,2} = \pm \frac{\hbar}{2} $$

Finding the eigenvectors, $$ (S_y - \lambda_{1,2}I)(\chi) = \vec{0} $$

• $\lambda_1 = \hbar/2$ $$ \frac{\hbar}{2} \begin{pmatrix}-1 & -i \\ i & -1\end{pmatrix} \begin{pmatrix}\chi_{_+} \\ \chi_{_-}\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix} \implies \begin{cases} -\chi_{_+} - i\chi_{_-} = 0 \\ i\chi_{_+} - \chi_{_{-}} = 0 \end{cases} $$

We find the normalised eigenvector $$ \chi_{_+} = \frac{1}{\sqrt{2}} \begin{pmatrix}1 \\ i\end{pmatrix} $$

We can write $$ |+\rangle_y = \frac{1}{\sqrt{2}}\left(|+\rangle_z + i|-\rangle_z \right) $$

• $\lambda_1 = -\hbar/2$ $$ \frac{\hbar}{2} \begin{pmatrix}1 & -i \\ i & 1\end{pmatrix} \begin{pmatrix}\chi_{_+} \\ \chi_{_-}\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix} \implies \begin{cases} \chi_{_+} - i\chi_{_-} = 0 \\ i\chi_{_+} + \chi_{_{-}} = 0 \end{cases} $$

We find the normalised eigenvector $$ \chi_{_-} = \frac{1}{\sqrt{2}} \begin{pmatrix}1 \\ -i\end{pmatrix} $$

We can write \begin{equation} |-\rangle_y = \frac{1}{\sqrt{2}}\left(|+\rangle_z - i|-\rangle_z \right) \end{equation}

Inverting the $|+\rangle_y$ and $|-\rangle_y$ relations we get $$ \begin{cases} |+\rangle_z = \frac{1}{\sqrt{2}}\left(|+\rangle_y + i|-\rangle_y \right) \\ |-\rangle_z = \frac{1}{\sqrt{2}}\left(|+\rangle_y - i|-\rangle_y \right) \end{cases} $$

Therefore, \begin{align} |\psi(t)\rangle &= e^{-i \omega_0 S_y t/\hbar}|+\rangle_z \\ &= e^{-i \omega_0 S_y t/\hbar} \frac{1}{\sqrt{2}}\left(|+\rangle_y + i|-\rangle_y \right) \end{align}

Now my question is how can i get rid of $S_y$ knowing that $S_y =\frac{\hbar}{2}\begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}$, I mean in class we saw that $$ |\psi(t)\rangle = e^{-i \frac{\omega_0}{2} t} \frac{1}{\sqrt{2}}\left(|+\rangle_y + i|-\rangle_y \right) $$ But I don't see how do we get that.

Answer 1

In the basis of eigenstates of $S_y$, the matrix representation of $S_y$ is diagonal and of the form \begin{align} S_y\to \frac{\hbar}{2}\left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right) \end{align} so that the exponential of $S_y$ is a diagonal matrix \begin{align} e^{-i \omega_0 t S_y/\hbar} \frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \\ i \end{array}\right)= \frac{1}{\sqrt{2}}\left(\begin{array}{c} e^{i\omega_0 t/2} \\ i e^{-i\omega_0 t/2}\end{array}\right) = \frac{1}{\sqrt{2}}e^{i\omega_0 t/2} \vert +\rangle_y +\frac{i}{\sqrt{2}}e^{-i\omega_0 t/2}\vert -\rangle_y\, . \end{align} This is all done in the basis of eigenstates of $S_y$. There remains to convert to the eigenstates of $S_z$.