# What exactly are control functions (used for parametrization)?

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 $$V(t)$$

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.

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