# Definition of geometric phase for mixed states and test in experiment

+ 2 like - 0 dislike
280 views
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?
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)
@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.
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsO$\varnothing$erflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.