Spinors ground state energy via the Variational Method

Setup

We have a spinor evolving according to the Hamiltonian $H=-\omega {\sigma }^Z-\lambda {\sigma }^Y=\left(\begin{array}{cc}-\omega & -\lambda \\ -\lambda & \omega \end{array}\right)$, where

• $\omega ,\lambda \in \mathbb{R}$
• ${\sigma }^Z,{\sigma }^Y$ are the Pauli Matrixes.

Goal

We want to find the Ground state Energy $E_0$ using the Variational Method.

Solution

We parameterise our Quantum states as $|{\psi }_{\theta }⟩=\left(\begin{array}{c}\sin (\theta )\\ \cos (\theta )\end{array}\right)$. This parameterisation assumes that there is no relative phase of the spinor, but is already normalised. As such the expected energy is described by

$$\begin{array}{rl}{E}_{\theta }& =⟨{\psi }_{\theta }|H|{\psi }_{\theta }⟩\\ & ={\left(\begin{array}{c}\sin (\theta )\\ \cos (\theta )\end{array}\right)}^T\left(\begin{array}{cc}-\omega & -\lambda \\ -\lambda & \omega \end{array}\right)\left(\begin{array}{c}\sin (\theta )\\ \cos (\theta )\end{array}\right)\\ & ={\left(\begin{array}{c}\sin (\theta )\\ \cos (\theta )\end{array}\right)}^T\left(\begin{array}{c}-\omega \sin (\theta )-\lambda \cos (\theta )\\ -\lambda \sin (\theta )+\omega \cos (\theta )\end{array}\right)\\ & =-\omega {\sin }^2(\theta )-\lambda \cos (\theta )\sin (\theta )-\lambda \cos (\theta )\sin (\theta )+\omega {\cos }^2(\theta )\\ & =\omega [{\cos }^2(\theta )-{\sin }^2(\theta )]-\lambda [2\cos (\theta )\sin (\theta )]\\ & =\omega \cos (2\theta )-\lambda \sin (2\theta )\end{array}$$

In order to bound the Ground state Energy, we want to find $\underset{\theta }{\min }\{E_{\theta }\}$:

$$\begin{array}{rl}0& =\frac{\partial }{\partial \theta }{E}_{\theta }\\ & =\frac{\partial }{\partial \theta }(\omega \cos (2\theta )-\lambda \sin (2\theta ))\\ & =-2\omega \sin (2\theta )-2\lambda \cos (2\theta )\\ & \Downarrow \\ \theta & =\pm \frac{1}{2}\arctan \left(\frac{\lambda }{\omega }\right)\end{array}$$

We get that the minimum Energy is given by:

$$\begin{array}{rl}{E}_{\theta }{|}_{\theta =\pm \frac{1}{2}\arctan \left(\frac{\lambda }{\omega }\right)}& =\omega \cos (2\theta )-\lambda \sin (2\theta ){|}_{\theta =\pm \frac{1}{2}\arctan \left(\frac{\lambda }{\omega }\right)}\\ & =\omega \cos (2(\pm \frac{1}{2}\arctan \left(\frac{\lambda }{\omega }))\right)-\lambda \sin (2(\pm \frac{1}{2}\arctan \left(\frac{\lambda }{\omega }))\right)\\ & =\omega \cos (\pm \arctan \left(\frac{\lambda }{\omega })\right)-\lambda \sin (\pm \arctan \left(\frac{\lambda }{\omega })\right)\\ & =\pm \omega \cos (\arctan \left(\frac{\lambda }{\omega })\right)\pm \lambda \sin (\arctan \left(\frac{\lambda }{\omega })\right)\\ & =\pm \omega \left(\frac{\omega }{\sqrt{{\lambda }^2+{\omega }^2}}\right)\pm \lambda \left(\frac{\lambda }{\sqrt{{\lambda }^2+{\omega }^2}}\right)\\ & =\pm \frac{{\omega }^2+{\lambda }^2}{\sqrt{{\lambda }^2+{\omega }^2}}\\ & =\pm \sqrt{{\lambda }^2+{\omega }^2}\end{array}$$

As such we get the extremes of the Energy in this model, of which the minimum is $-\sqrt{{\lambda }^2+{\omega }^2}$.