I was reading Sakurai's "Modern Quantum Mechanics" and he states that if the Hamiltonian $H$ is time-dependent and the $H$'s commute at different times, then the time evolution is given by the relation $$u(t,t_0) = \text{exp}\left[-\left(\frac{i}{\hbar}\right)\int_{t_0'}^{t}dt'H(t')\right]$$ I was thinking about having the following magnetic field $$ \begin{equation} \textbf{B} = B_0 \cos(\omega t) \hat{z} \end{equation} $$ with $$ \Psi(0) = \frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ 1 \end{pmatrix} $$ i.e, the eigenstate of $S_x$ with eigenvalue $\hbar/2$. The Hamiltonian will be $$ H = -\gamma B_0 \cos(\omega t) S_z = -\frac{\gamma B_0 \hbar}{2} \cos(\omega t) \sigma_z $$

The time evolution now will be $$ u(t,0) = \text{exp}\left[\left(\frac{i}{\hbar}\right)\int_{0}^{t} \frac{\gamma B_0 \hbar}{2} \cos(\omega t') \sigma_z dt'\right] \rightarrow u(t,0) = \text{exp} \left[\left(\frac{i\gamma B_0 \hbar}{2\omega} \sin(\omega t)\right) \sigma_z\right] $$ and from here I can get the time-dependent wavefunction.

My question is, is it possible to get expressions for the polar angle $\theta (t)$ and the azimuthial $\phi(t)$ from this notation?


1 Answer 1


If I am not mistaken, you can do something like this:

The Pauli-Heisenberg equation should be $$i\hbar \, \frac{d\,\psi}{dt} \,=\, -\,\frac{\gamma B_0\hbar}{2} \,\Big( \cos(\omega\, t)\,\sigma_z \,\Big)\, \psi$$ where $$\psi \, : \, \mathbb{R} \, \to \, \mathbb{C}^2$$ and as you have already observed, the evolution operator is $$U \, : \, \mathbb{R} \, \to \, \text{US}(2)$$ is $$U(t) \, =\, \text{exp}\left(- i \, \frac{\gamma B_0}{2\omega}\,\sin(\omega t)\right)\sigma_z$$ Hence, given initial spinor $$\psi_0 \, =\, \begin{bmatrix} \psi_{0,1}\\ \psi_{0,2} \end{bmatrix}\, \in \, \mathbb{C}^2$$ after time $t$, the spinor evolves into the spinor $$\psi \, =\, \psi(t) \, =\, U(t)\,\psi_0 \,=\, \text{exp}\left(- i \, \frac{\gamma B_0}{2\omega}\,\sin(\omega t)\right)\sigma_z \, \psi_0$$ In general, given a spinor $\psi = \begin{bmatrix} \psi_{1}\\ \psi_{2} \end{bmatrix} \in \mathbb{C}^2$ such that $\psi^*\psi = |\psi_1|^2 + |\psi_2|^2 = 1$, we can associate to it a unique $\text{SU}(2)$ matrix $$\Psi \,=\, \begin{bmatrix} \psi_{1} & - \bar{\psi}_2\\ \psi_{2} & \bar{\psi}_1\end{bmatrix} \, \in \, \text{SU}(2)$$ Then, this spinor matrix decomposes uniquely into $$\Psi \,=\, \cos\Big(\frac{\mu}{2}\Big) \, + \, i\,\sin\Big(\frac{\mu}{2}\Big)\,\big(\vec{u}\cdot \vec{\sigma}\big) $$ where $\vec{u} \,=\, u_x \,\vec{i} + u_y \, \vec{j} + u_z \, \vec{k} \, \in \mathbb{R}^3$ is the 3D unit vector along the axis of rotation of the spinor frame $\Psi$, the angle $\mu$ is the angle of 3D rotation that the spinor frame $\Psi$ defines, and $\vec{\sigma} = \sigma_x \, \vec{i} + \sigma_y \, \vec{j} + \sigma_z \, \vec{k}$ with $\sigma_x, \, \sigma_y, \, \sigma_z$ being the three standard Pauli matrices. In other words, $$\vec{u}\cdot \vec{\sigma} \,=\, u_x \,\sigma_x + u_y \, \sigma_y + u_z \, \sigma_z$$ Form this representation, you can take only the first column of the spinor frame and its decomposition, yielding $$\begin{bmatrix} \psi_1\\ \psi_2 \end{bmatrix} \, =\, \begin{bmatrix} \cos\Big(\frac{\mu}{2}\Big) \\ 0 \end{bmatrix}\, + \, i \sin\Big(\frac{\mu}{2}\Big) \begin{bmatrix} u_z \\ u_x + i\,u_y \end{bmatrix} $$ or componentwise \begin{align} \psi_1 \,&=\,\cos\Big(\frac{\mu}{2}\Big) + i\, \sin\Big(\frac{\mu}{2}\Big)\, u_z \\ \psi_2 \,&=\, i\,\sin\Big(\frac{\mu}{2}\Big) \big( u_x + i\,u_y\big)\, =\, \sin\Big(\frac{\mu}{2}\Big) \big( -u_y + i\,u_x\big) \end{align} Therefore, \begin{align} \sin\Big(\frac{\mu}{2}\Big)\, u_z \,&=\, \text{Im}(\psi_1)\\ \sin\Big(\frac{\mu}{2}\Big) \, u_x \,&=\, \text{Im}(\psi_2)\\ \sin\Big(\frac{\mu}{2}\Big) \, u_y \,&=\, -\text{Re}(\psi_2) \end{align} And from here, you can obtain the coordinates of the unit vector $\vec{u}$ along the axis of rotation \begin{align} u_z \,&=\, \frac{\text{Im}(\psi_1)}{\sqrt{\, |\psi_2|^2 + |\text{Im}(\psi_1)|^2\,}\,}\\ u_x \,&=\, \frac{\text{Im}(\psi_2)}{{\sqrt{\, |\psi_2|^2 + |\text{Im}(\psi_1)|^2\,}\,}}\\ u_y \,&=\, \frac{-\,\text{Re}(\psi_2)}{{\sqrt{\, |\psi_2|^2 + |\text{Im}(\psi_1)|^2\,}\,}} \end{align} In your case, you have an explicit egienspinor as initial condition $\psi_0$, so componentwise, the time evolution is \begin{align} \psi_1 \,&=\, \frac{1}{\sqrt{2}} \, \text{exp}\left(-i\frac{\gamma B_0}{2\omega} \sin(\omega t)\right)\\ \psi_2 \,&=\, -\,\frac{1}{\sqrt{2}} \, \text{exp}\left(-i\frac{\gamma B_0}{2\omega} \sin(\omega t)\right)\\ \end{align} Extract the necessary imaginary and real components \begin{align} \sqrt{2}\,\text{Im}(\psi_2) \, &= \,\sin\left(\frac{\gamma B_0}{2\omega} \,\sin(\omega t)\right)\\ -\,\sqrt{2}\,\text{Re}(\psi_2) \, &= \,\cos\left(\frac{\gamma B_0}{2\omega} \,\sin(\omega t)\right)\\ \sqrt{2}\,\text{Im}(\psi_1) \, &= \, -\, \sin\left(\frac{\gamma B_0}{2\omega} \,\sin(\omega t)\right)\\ \end{align} so, if I have not made calculation mistakes (maybe double check), the evolution of the spinor's rotation axis is given by \begin{align} u_x \,&=\, \frac{\sin\left(\frac{\gamma B_0}{2\omega} \,\sin(\omega t)\right)}{\,\sqrt{\,1 \,+\, \sin^2\left(\frac{\gamma B_0}{2\omega} \,\sin(\omega t)\right)\,}\,}\\ u_y \,&=\, \frac{\cos\left(\frac{\gamma B_0}{2\omega} \,\sin(\omega t)\right)}{\,\sqrt{\,1 \,+\, \sin^2\left(\frac{\gamma B_0}{2\omega} \,\sin(\omega t)\right)\,}\,}\\ u_z \,&=\, \frac{-\,\sin\left(\frac{\gamma B_0}{2\omega} \,\sin(\omega t)\right)}{\,\sqrt{\,1 \,+\, \sin^2\left(\frac{\gamma B_0}{2\omega} \,\sin(\omega t)\right)\,}\,}\\ \end{align}

From these last expressions, you can see that the azimuthal and polar angles are \begin{align} \phi(t) \, &=\,\frac{\pi}{2} - \frac{\gamma B_0}{2\omega} \,\sin(\omega t)\\ \theta(t) \, &=\, \arccos\left(\, \frac{-\,\sin\left(\frac{\gamma B_0}{2\omega} \,\sin(\omega t)\right)}{\,\sqrt{\,1 \,+\, \sin^2\left(\frac{\gamma B_0}{2\omega} \,\sin(\omega t)\right)\,}\,} \,\right) \end{align}


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