In quantum mechanics, an open system interacting with the environment is described by mixed state, which is represented by the density operator $\rho(t)$ acting on a finite dimensional Hilbert space $\mathscr{H}_n$. By the spectral theorem, $\rho(t)$ can be decomposed as $\rho(t)=\sum_{k=1}^n\omega_k(t)|\phi_k(t)\rangle\langle\phi_k(t)| $, where $\omega_k(t)$ are the eigenvalues, $|\phi_k(t)\rangle$ are the eigenvectors and $\langle\phi_k(t)|$ are the corresponding vectors in the dual space of $\mathscr{H}_n$. For simplicity, here only non-degenerate case is considered. I am now trying to assign a geometric phase for the mixed state represented by $\rho(t)=\sum_{k=1}^n\omega_k(t)|\phi_k(t)\rangle\langle\phi_k(t)| $. The geometric phase is defined as $$\sum_{k=1}^n\int_0^t \omega_k(t')\frac{d\,\gamma_k(t')}{d\,t'}d\,t'$$, where $\gamma_k(t)=\arg\langle\phi_k(0)|\phi_k(t)\rangle+i\int_0^t\langle\phi_k(t')|\dot{\phi}_k(t')\rangle d\,t'$ are the geometric phases for eigenvectors $|\phi_k(t)\rangle$. Here we assume that $|\phi_k(t)\rangle$ are not orthogonal to $|\phi_k(0)\rangle$ for any time t and $\gamma_k(t)$ are smooth functions of time t. Note that $\gamma_k(t)$ are put in differentiation of time t. There is no ambiguity caused by multiple values of $\gamma_k(t)$. Thus the above geometric phase is well-defined. Question: How to test in experiment the geometric phase for mixed states defined above?