• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Definition of geometric phase for mixed states and test in experiment

+ 2 like - 0 dislike
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.

Your answer

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):
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:
Then 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).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights