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Definition of geometric phase for mixed states and test in experiment

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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?
asked Feb 11, 2015 in Theoretical Physics by Guang-Le Du (10 points) [ revision history ]
edited Feb 11, 2015 by Guang-Le Du
I'm a bit puzzled by the question. Isn't one of the key features of density matrix that the relative phase information is erased? In your case, before calculating $\alpha_k(t)$, how do you measure $|\phi_k(t)\rangle$? I mean, if you redefine $|\phi_k(t)\rangle\to e^{i\beta_k(t)}|\phi_k(t)\rangle$ , the density matrix $\rho(t)$ stays the same, and this means $\alpha_k(t)$ cannot be uniquely defined if all you have is a $\rho(t)$.(note this ambiguity doesn't go away by just differentiating $\alpha_k(t)$ since $\beta_k$ is time-dependent)
You are right, @Yiyang. When I said the relative phase, the relative phase should be dependent of a certain "purification" and is surely not gauge invariant. Actually I want to know how to measure the geometric phase stated in the question after edition. If the relative phase of a certain "purification" can be measured like pure state case, then the geometric phase can be tested after the parallel transportation condition for each eigenvector is applied.
@Guang-LeDu, "purification" means the same thing as before edition?

@Yiyang. The well-known purification of $\rho(t)$ is of the form $|\psi(t)\rangle=\sum_{k=1}^n\sqrt{\omega_k(t)}|\phi_k(t)\rangle\otimes|a_k\rangle$. The relative phase of $|\psi(t)\rangle$ is given by $\arg\langle\psi(0)|\psi(t)\rangle=\arg(\sum_{k=1}^n\sqrt{\omega_k(0)\omega_k(t)}\langle\phi_k(0)|\phi_k(t)\rangle)$. Before edition, I wanted to know if it is possible to construct a pure state related to $\rho(t)$, like the well-known purification, giving the relative phase I mentioned. In this sense, the "purification". It may be a possible way to give an experimental scheme to test the geometric phase defined here.

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