- #1

Silicon-Based

- 51

- 1

- Homework Statement
- Calculate the Berry connection

- Relevant Equations
- $$

A_R = i\langle \psi | \nabla_R| \psi \rangle

$$

This isn't technically a homework problem, but I'm trying to check my understanding of the geometric phase by explicitly calculating the Berry connection for a simple 2x2 Hamiltonian that is not a textbook example of a spin-1/2 particle in a three dimensional magnetic field solved via a Bloch sphere representation and my results don't make sense. Suppose I have the Hamiltonian

$$

H = a\sigma_x+c\sigma_z =

\begin{pmatrix}

c & a\\

a & -c

\end{pmatrix}

$$ To simplify the calculations slightly I switch to polar coordinates using $$a=r\cos(\theta)$$ and $$c=r\sin(\theta).$$ The eigenvalues of the Hamiltonian are

$$

E_{\pm}=\pm r

$$ and the eigenvector associated with the negative eigenvalue becomes

$$

|-\rangle = (r^2-2r\sin(\theta)+1)^{-1/2} (\sin(\theta)-r,\cos(\theta))^T

$$ It is evident from the expression for the above eigenvector that the Berry connection for this eigenstate, which for the $r$-component is defined as

$$

A_r = i\langle - | \frac{\partial}{\partial r} |- \rangle,

$$ will be purely imaginary. This is obviously wrong as the Berry phase, which is the integral over the Berry connection, must be real, but I'm confused as to why my calculations are incorrect. In the textbook examples of a spin-1/2 particle, there is a convenient phase factor of $e^{i\phi}$ present in the eigenstates, which results in a real value for the $\phi$-component of the Berry connection upon differentiation, unlike in the example above.

$$

H = a\sigma_x+c\sigma_z =

\begin{pmatrix}

c & a\\

a & -c

\end{pmatrix}

$$ To simplify the calculations slightly I switch to polar coordinates using $$a=r\cos(\theta)$$ and $$c=r\sin(\theta).$$ The eigenvalues of the Hamiltonian are

$$

E_{\pm}=\pm r

$$ and the eigenvector associated with the negative eigenvalue becomes

$$

|-\rangle = (r^2-2r\sin(\theta)+1)^{-1/2} (\sin(\theta)-r,\cos(\theta))^T

$$ It is evident from the expression for the above eigenvector that the Berry connection for this eigenstate, which for the $r$-component is defined as

$$

A_r = i\langle - | \frac{\partial}{\partial r} |- \rangle,

$$ will be purely imaginary. This is obviously wrong as the Berry phase, which is the integral over the Berry connection, must be real, but I'm confused as to why my calculations are incorrect. In the textbook examples of a spin-1/2 particle, there is a convenient phase factor of $e^{i\phi}$ present in the eigenstates, which results in a real value for the $\phi$-component of the Berry connection upon differentiation, unlike in the example above.