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Suppose I am studying a field theory at finite temperature or some black hole formation scenario from boundary theory perspective in the sense of AdS/CFT. How is it possible to gain information about them from say looking at the two point functions (propagators) of operators in field theory? I mean will there be some special pole structures etc. in that Green's function? Is there a generic such behavior? Can you suggest me some references?

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I don't have a complete answer but here's something to give you a rough idea. We consider a Gaussian scalar field with one-point Hamiltonian $\Omega$, i.e. Hamiltonian of this field is given by $$H = : {1 \over 2} \int {\rm d}^d {\mathbf x} \left( \pi(\mathbf x)^2 + \phi(\mathbf x) \Omega^2\phi(\mathbf x)\right): = \int {\tilde {\rm d} \mathbf p} E(\mathbf p) a^{\dagger}(\mathbf p)a(\mathbf p)$$ where we introduced $\tilde {\rm d} \mathbf p = {{\rm d} \mathbf p \over (2\pi)^d 2E(\mathbf p)} .$

The Euclidean two-point correlation function of this field is given by $$\left<{\mathbf x} \right| G_E(t, t') \left|{\mathbf x'} \right> = \int {{\rm d}^{d+1} p \over {2 \pi}^{d+1}} {e^{i p_0(t-t') + {\mathbf p} \cdot ({\mathbf x -\mathbf x'})} \over p_0^2 + E(\mathbf p)^2}$$

while for the thermal Green function we have

$$ \left<{\mathbf x} \right| G_{\beta}(t, t') \left|{\mathbf x'} \right> = {1 \over \beta} \sum_{\omega_n = {2 \pi n / \beta}} \int {{\rm d}^{d} p \over {2 \pi}^{d}} {e^{i \omega_n(t-t') + {\mathbf p} \cdot ({\mathbf x -\mathbf x'})} \over \omega_n^2 + E(\mathbf p)^2}. $$

This can be summed and separated into the contributions from the ground state and from the excited states $$\left<{\mathbf x} \right| G_{\beta}(t, t') \left|{\mathbf x'} \right> = \left<{\mathbf x} \right| G_E(t, t') \left|{\mathbf x'} \right> + \int {{\rm d}^{d} p \over {2 \pi}^{d}} {1 \over E(\mathbf p)} {\cosh E(\mathbf p) (t - t') \over e^{\beta E(\mathbf p)} - 1}.$$

Two general observations that can be seen from this:

  1. the thermal Green function has a pole at $\beta E = 2 \pi n$. This comes from the fact that we have compactified time (and so temperature) to a circle.
  2. in the limit $\beta \to \infty$ the contribution from excited states dies off and we are left with the vacuum Green function.
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Marek, please could you give us a reference for a derivation of the second equation. I'd like to see the connection between $\omega_n$ and $E(p)$ more clearly. Your statement (1) seems at first sight problematic, because the one-point Hamiltonian has a continuous spectrum, containing the time-like component of every 4-vector on the forward mass-shell. –  Peter Morgan Jul 22 '11 at 12:02
    
@Peter: you're right, I've skipped many details and what I've written isn't quite correct. I'll fix it later. –  Marek Jul 22 '11 at 12:18
    
@Peter: I've edited it a little, does it make some sense now? Also, as for the derivation of that equation, it can be obtained e.g. from the discretization of time, passing to the frequency space where the evolution operator with time-independent Hamiltonian in each time-step can be computed directly and finally taking a continuum limit. It's a bit long to be written out here in full and I don't have any reference (except from my notes from a course where we covered this). I suppose there must be a shorter derivation too but I am not aware of it. –  Marek Jul 22 '11 at 15:17
    
Please check the role of $\mathbf{x}'$ for typos. At two places, there should be an $i$, and then there is the last term in the very last eq. –  Qmechanic Jul 23 '11 at 8:43
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It's a little old now, but try Hiroomi Umezawa, "Advanced Field Theory; Micro, Macro, and Thermal Physics", AIP, 1993.

In a slightly different notation from that used by Marek, the two point function changes from $$\left<0\right|\phi(x)\phi(x+y)\left|0\right>=\hbar\int 2\pi\delta(k^2-m^2)\theta(k_0)\mathrm{e}^{-ik\cdot y}\frac{\mathrm{d}^4k}{(2\pi)^4},$$ the free Klein-Gordon vacuum state case, to the corresponding thermal state case, $$\omega_\mathsf{T}\Bigl[\phi(x)\phi(x+y)\Bigr]=\hbar\int 2\pi\delta(k^2-m^2)\theta(k_0)\coth\left[ \!\frac{\hbar k_0}{2\mathsf{k_BT}}\!\right]\mathrm{e}^{-ik\cdot y}\frac{\mathrm{d}^4k}{(2\pi)^4}.$$ On the basis of this presentation, the thermal free field case is no more than a different measure on the forward-light cone, which is, inevitably, not Lorentz invariant (the $k_0$ in the $\coth$ factor picks out a preferred frame). The deformation is no less smooth than the vacuum 2-point function. Additionally, although the above equations don't show it, the thermal field state is still Gaussian, like the vacuum state, so the whole structure of the thermal state over the free Klein-Gordon field is completely specified by the 2-point function.

One could also construct higher-order deformations of the mass-shell measure, by adding extra factors of $\coth\left[ \!\frac{\hbar k'_0}{2\mathsf{k_BT'}}\!\right]$, or otherwise, quite possibly corresponding to different time-like directions and temperatures.

A presentation of the Klein-Gordon field that is enough to reconstruct the above can be found in my "A succinct presentation of the quantized Klein–Gordon field, and a similar quantum presentation of the classical Klein–Gordon random field", quant-ph/0411156, Phys. Lett. A 338, 8-12(2005), although mostly I'm grinding quite a different axe in that paper.

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So, I guess the poles of the Green's function gives the energy of the field..right? Related to that, if we have a branch cut in the Green's function, that will possibly refer to the continuum energy state of the fields in question..right? –  user1349 Jul 23 '11 at 0:22
    
@user1349, Not sure what you mean by "energy of the field"? The poles of the two Green's functions above are all on the forward mass shell; that is, they are energy-momentum 4-vectors, not just an energy. The e-m of the vacuum state is 0 (assuming we subtract the zero-point energy). The expected value of the e-m of other states can be any value in the forward light-cone that we choose, by taking suitable superpositions and mixtures of states. I find energy to be a somewhat treacherous concept when dealing with thermal states. –  Peter Morgan Jul 23 '11 at 10:48
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