• 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 ...

  Why do we have to double the degrees of freedom in the Keldysh-Schwinger technique?

+ 4 like - 0 dislike

The Keldysh-Schwinger (a.k.a. in-in) formalism avoids reference to the final state when doing QFT calculations by 'closing the time contour': we first go forward in the time direction, but then back again. This allows one to do non-equilibrium calculations, which is useful in a range of different settings (e.g. cosmology, condensed matter theory).

In reviews (e.g. http://arxiv.org/abs/0901.3586v3.pdf) it is usually stated without further explanation that one then has to consider two sets of fields, one 'living' on the forward and one on the backward time branch. This makes some intuitive sense, but I cannot make it precise; I would like a more rigorous argument as to why this is true. Can anyone provide me with one?

asked Apr 4, 2014 in Theoretical Physics by Danu (175 points) [ revision history ]
edited Oct 11, 2015 by Danu

2 Answers

+ 5 like - 0 dislike

The formal reason is that one needs to recover the wave functional at any fixed time, so at whichever time one starts one needs to return to it and (for scattering) have covered all other times. This leads to the doubled contour. http://arxiv.org/pdf/hep-th/9504073.pdf

In terms of the resulting Kadanoff-Baym equations, one needs the doubled contour to access all relevant time correlations. There is also a classical counterpart.

With proper care, the formalism is also equivalent to the thermofield approach, where the doubling comes from the fact that statistical mechanics works with density matrices rather than wave functions, and the number of indices required to describe density matrices is twice the number of indices required for wave functions.

answered Apr 12, 2014 by Arnold Neumaier (15,787 points) [ revision history ]
edited Apr 13, 2014 by Arnold Neumaier
+ 3 like - 0 dislike

Formally, I see it as emerging from the trace necessary to compute observables in quantum mechanics,

$$ \langle O(t) \rangle = \text{Tr}\left[\rho O(t) \right] \, .$$

I use Heisenberg's picture, $O(t) = \text{e}^{\text{i} t H} O \, \text{e}^{-\text{i} t H}$ and $\hbar = 1$. The path integral representation of the time evolution operator gives

$$\left. \langle x \right| \text{e}^{\text{i} t H} \left| x \right. \rangle = \int \text{D}x \, \text{e}^{i S} \, ,$$

with $S = \int_0^t \text{d}t [...]$ an action that contains a simple time integral from $t=0$ to $t=t$. In thermal equilibrium, you can set $\text{i}t = -\beta$ and compute observables from

$$ \langle O \rangle = \text{Tr}\left[\text{e}^{-\beta H} O \right] \, .$$

There is only one time path. Out-of-equilibrium however the two terms $ \text{e}^{\pm \text{i} t H}$ must be taken into account. Each leads to the same action. Only the direction of time changes. The full expression for the time dependent observable is

\begin{align*} \langle O(t) \rangle  & = \int \text{d}x \, \text{d}x'\,   \text{d}x'' \,   \text{d}x''' \,  \\ & \qquad \times \left. < x \right| \rho \left| x' > \right.  \left. < x' \right| \text{e}^{\text{i} t H} \left| x'' > \right. \left. < x'' \right| O \left| x''' > \right. \, \left. < x''' \right| \text{e}^{-\text{i} t H} \left| x > \right. \, ,\\ &   = \int Dx \, \text{e}^{\text{i} \left[ S_{+} + S_{-}\right]} \, \left. \, \left. < x_-(0) \right| \rho \left| x_+(0) > \right. \, < x_+(t) \right| O \left| x_-(t) \right. > \, ,\end{align*}

where $S_+ = \int_0^t \text{d}t[...]$, $S_- = \int_t^0 \text{d}t[...]$ and $x_{\pm}(t)$ are the paths on the forward and backwards paths respectively. The integrals over the boundaries, $x=x(0)$, $x'=x(0)$ $x''=x(t)$ and $x'''=x(t)$ have been absorbed into the functional integral, $\int Dx$.

This is a very quick answer. Tell me if you have trouble filling in the gaps. The generalisation from quantum mechanics to quantum field theory is straightforward.

answered Oct 5, 2015 by Steven Mathey (350 points) [ revision history ]
edited Oct 12, 2015 by Steven Mathey

Please spell out the way the action appears in the nonequilibrium case.

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