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

+ 4 like - 0 dislike
407 views

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 (12,425 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.

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