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This is kind of a follow-up of my previous question. I'm investigating the following infinite series of nested two-dimensional integrals

$$\sigma(t,t^\prime) = 1 - \int_{t^\prime}^t\mathrm dt_1 \int_{t^\prime}^{t_1}\mathrm dt_2 \, \Delta(t_1,t_2) + \int_{t^\prime}^t\mathrm dt_1 \int_{t^\prime}^{t_1}\mathrm dt_2 \int_{t^\prime}^{t_2}\mathrm dt_3 \int_{t^\prime}^{t_3}\mathrm dt_4 \, \Delta(t_1,t_2) \, \Delta(t_3,t_4) + \cdots \;,$$

with $\Delta(t_1,t_2)$ a complex function obeying the relation

$$\Delta(t_1,t_2) = \Delta^*(t_2,t_1) \;.$$

I'm looking for a representation of $\sigma(t,t^\prime)$ that can be calculated in completeness (the above infinite sum obviously isn't very useful in a concrete application). Treating $t^\prime$ as constant and taking the derivative of $\sigma(t,t^\prime)$ with respect to $t$ I obtain

$$\frac{\mathrm d \sigma(t,t^\prime)}{\mathrm d t} = - \int_{t^\prime}^t\mathrm dt_2 \, \Delta(t,t_2) \, \sigma(t_2,t^\prime) \;,$$

which after re-integrating and using the boundary condition $\sigma(t^\prime,t^\prime) = 1$ yields the recursion relation

$$\sigma(t,t^\prime) = 1 - \int_{t^\prime}^t\mathrm dt_1 \int_{t^\prime}^{t_1}\mathrm dt_2 \, \Delta(t_1,t_2) \, \sigma(t_2,t^\prime) \;.$$

However, I don't know how to solve either of these two equations for $\sigma(t,t^\prime)$.

In view of my previous question my first guess was to inspect

$$\sigma_1(t,t^\prime) = \exp \left[ -\int_{t^\prime}^t\mathrm dt_1 \int_{t^\prime}^{t_1}\mathrm dt_2 \, \Delta(t_1,t_2) \right] \;.$$

Obviously $\sigma_1(t^\prime,t^\prime) = 1$, but taking the derivative of $\sigma_1(t,t^\prime)$ with respect to $t$ at constant $t^\prime$ yields

$$\frac{\mathrm d \sigma_1(t,t^\prime)}{\mathrm d t} = - \int_{t^\prime}^t\mathrm dt_2 \, \Delta(t,t_2) \, \sigma(t,t^\prime) \;,$$

which is different from the derivative of the original series $\sigma(t,t^\prime)$ (see above). Still, to first order in $\Delta(t_1,t_2)$ the derivatives are equal and thus

$$\sigma(t,t^\prime) = \sigma_1(t,t^\prime) + \exp\left[\mathcal{O}\left(\Delta^2 \right) \right] \;,$$

which is actually quite acceptable to me, since in my case $\Delta(t_1,t_2)$ is a small perturbation. Nevertheless, I'm wondering if one can obtain a better solution to the problem.


Edit 1

I discovered the following propagator property of $\sigma(t,t^\prime)$

$$\sigma(t,t_1) \sigma(t_1,t^\prime) \stackrel{!}{=} \sigma(t,t^\prime) \;,$$

which follows from the fact that $\sigma(t,t^\prime)$ represents the interaction part of a Keldysh Green's function, i.e.

$$G(t,t^\prime) = G_0(t,t^\prime) \sigma(t,t^\prime) \;,$$

with $G_0(t,t^\prime)$ and $G(t,t^\prime)$ obeying the propagator relation

$$iG_{(0)}(t,t_1) iG_{(0)}(t_1,t^\prime) = iG_{(0)}(t,t^\prime)\;.$$


Edit 2

As pointed out by Didier Piau the above propagator property cannot hold, since

$$\frac{\mathrm d}{\mathrm d t} \left( \sigma(t,\bar t) \sigma(\bar t,t^\prime) \right) = - \int_{\bar t}^t\mathrm dt_2 \, \Delta(t,t_2) \, \sigma(t_2,\bar t) \sigma(\bar t, t^\prime) \;"="\; - \int_{\bar t}^t\mathrm dt_2 \, \Delta(t,t_2) \, \sigma(t_2, t^\prime) \;,$$

which is obviously different from $\frac{\mathrm d \sigma(t,t^\prime)}{\mathrm d t}$ (see above).


Edit 3

I verified that the above propagator property does in fact only hold for the unperturbed Green's functions $G_0(t,t^\prime)$, but in general not for the full ones $G(t,t^\prime)$ and thus not for $\sigma(t,t^\prime)$.

share|improve this question
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Assume that the relation $\sigma(t,t_1) \sigma(t_1,t^\prime)=\sigma(t,t^\prime)$ holds for every $t\ge t_1\ge t^\prime$, derive it with respect to $t$ and use the formula of your post for the derivatives of $\sigma(\ ,t_1)$ and $\sigma(\ ,t^\prime)$. Then the functions $\sigma(\ ,t_1) \sigma(t_1,t^\prime)$ and $\sigma(\ ,t^\prime)$ have different derivatives, which is absurd. Hence the propagator property mentioned in your edit cannot hold. –  Did Apr 20 '11 at 14:37
    
@Didier Piau Thanks for the input. Yes, I see. You're perfectly right. Something must have gone wrong. As far as I know the propagator property is fundamental to Keldysh and ordinary Green's functions, thus it must hold. I'm going to check this. –  hennes Apr 20 '11 at 15:59
    
@Didier Piau I checked back with my supervisor and it looks like I was wrong from the start. That propagator property is valid for the unperturbed Green's functions $G_0$, but in general not for the full ones $G$. Thank you for bringing this up. –  hennes Apr 22 '11 at 7:20
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