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

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

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.

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 $$S_z$$.