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Let us consider a system in state $\rho$ with an internal hamiltonian $H_0$ on which we apply a cyclic, unitary evolution

$H_t = H_0 + V(t)$

Where $V(t)$ is a time dependent external potential for some duration $\tau$.

Assuming $H_0 = \mid 1 \rangle \langle 1 \mid$ we parametrize the total Hamiltonian with control functions $v^x_t$,$v^y_t$ and $v^z_t$ for the pauli operators $\sigma_x$,$\sigma_y$ and $\sigma_z$.

Additionally the eigenvalues of the external potential $V(t)$ are defined as $\lambda^{\pm} = \pm \sqrt{v_x^2 + v_y^2+v_z^2}$.

Can anyone help me to understand this? I especially have a hard time understanding what $v^i_t$ are, how they relate to pauli operators and why they define the eigenvalues of $V(t)$

Thanks in advance!

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I guess the external potential has been parametrized with the functions $\vec v(t)$. It is a common choice to do so on bidimensional hamiltonian, because the pauli matrices are a real base of the $2\times 2$ hermitian matrices.

To be more explicit, I guess that your potential is

$\hat V(t) = \vec v(t) \cdot \vec \sigma = v_x(t) \sigma_x + v_y(t) \sigma_y + v_z(t) \sigma_z = \left(\begin{array}{cc}v_z & v_x - i v_y \\ v_x + i v_y & -v_z\end{array}\right)$

If you compute the eigenvalues of that matrix you will find out that they are exactly $\lambda^\pm$ that you defined earlier.

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  • $\begingroup$ Thank you! I was able to do my calculations with it and everything worked out perfectly! $\endgroup$ – Benjamin Jabl Dec 30 '18 at 12:55

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