# Free propagator from generating functional in momentum space

I am reading the text by Daniel Amit, "Field Theory, the Renormalization Group, and Critical Phenomena" and trying to apply what is explained in chapters 3 and 4 to a real-life problem.

However, I don't understand pretty well how the propagator and Feynmann rules are obtained in momentum space. For example, let's focus in the free propagator. Let's say I have

$$Z_0[J] = \exp \left ( \frac{1}{2}\int dqd\omega J(-q,-\omega) G_0(q,\omega) J(q,\omega) \right )$$

then I can obtain any correlation function as

$$G_0(k_1,\ldots,k_n) = \left .\frac{\delta^n Z_0}{\delta J(-k_1)\ldots\delta J(-k_n)} \right |_{J=0}$$

Ok, so I want to explicitly obtain the free propagator. This is given by:

$$G_0(k,\omega)=\langle\phi(k) \phi(-k) \rangle = \left .\frac{\delta^2 Z_0}{\delta J(k)\delta J(-k)} \right |_{J=0}$$

What I obtain after the functional integration is:

$$G_0(k,\omega) = \frac{\delta}{\delta J(k)} \left \{ \exp(\ldots)\cdot \frac{1}{2} \int dq \left[ \delta(q+k)G_0(q)J(-q) + \delta(-q+k)G_0(q)J(q) \right] \right \} = \frac{\delta}{\delta J(k)} \left \{ \exp(\ldots)\cdot \frac{1}{2} \left[J(k)G_0(-k) + J(k)G_0(k) \right] \right \} = \exp(\ldots)\cdot \left \{ (\ldots) + \frac{1}{2} \delta(0)\left[G_0(-k) + G_0(k) \right] \right \}$$

When evaluated in $J=0$, all the things marked with $(\ldots)$ vanish, and then I have simply $(G_0(-k) + G_0(k))/2$. It is tempting to say that $G_0(-k)=G_0(k)$ and then I recover $G_0(k,\omega)$.

However, when I go to my real case, I realize this is false. In particular, I have

$$G_0(k)=\frac{1}{Dk^2-a-i\omega}$$

so when $\omega$ changes to $-\omega$ the propagator is different. Then, I am afraid that my derivation above is not correct.

In addition to that, my real-life model has two independent fields, so the generating functional is:

$$Z_0[\vec J] = \exp \left ( \frac{1}{2}\int dqd\omega \vec J(-q,-\omega) \hat G_0(q,\omega) \vec J(q,\omega) \right ) \tag{1}$$

with $\vec J = (J, \bar J)$ and

$$\hat G_0 = \left(\begin{array}{cc} 0 & G_0(k,\omega)\\ G_0(-k,-\omega) & 0 \end{array}\right).$$

In principle, since I am interested also in $G_0(k)$, I think the formula I have to use is the same as in one variable (I mean, deriving with respect to $J(-k)J(k)$, and not with respect to $\bar J(k) J(-k)$ or whatever else), but I am not sure. Amit notation for this case is not very friendly.

Okay, given all the above, I want to know:

1) If my computation of $G_0(k,\omega)$ with a single variable is ok (i.e. if I can sum $(G_0(-k) + G_0(k))/2 = G(k)$). And, more importantly, why is correct (or why not).

2) Why then in my case $G_0(-k)$ and $G_0(k)$ are not equal?

3) How do I compute the free propagator with several variables?

EDIT: For my problem, the Lagrangian (with sources) is given by:

$$\mathcal{L} = \bar{\phi}\left(\partial_{t}-D\nabla^{2}-a \right)\phi + \frac{\Gamma}{2}\left[ \bar{\phi}\phi^{2}-\bar{\phi}^{2}\phi \right] + J\phi + \bar J \bar\phi$$

For more detail, this is the correspondent field theoretic formulation of the Langevin equation of the contact process. In momentum space, the free action reads:

$$S_0 = \int d_{k\omega}\bar{\phi}(-k)\left(-i\omega+Dk^{2}-a\right)\phi(k)+ J(-k)\phi(k) + \bar J(-k) \bar\phi(k)$$

with $d_{k\omega}\equiv d\vec{k}d\omega/\left(2\pi\right)^{d+1}$. Defining the vector $\vec J$ and the matrix $\hat G_0(k,\omega)$, it is possible to write the free propagator as equation (1). From that point I should be able to compute correlation functions. Field $\bar \phi$ is not observable, so I think that the correlations of interests are $\langle \phi(k)\phi(-k) \rangle$, but I am not pretty sure of this.

I'm following this article, section 3.5: https://arxiv.org/pdf/cond-mat/0001070.pdf . I would like to reproduce the computations and get the results to learn how to apply the theory given in Amit.

• To obtain his functional, Amit uses that fact that $G_0$ is invariant under $\omega\to-\omega$., i.e. $G_0(k)=G_0(-k)$. That's why it doesn't work in you case. – Adam Oct 22 '17 at 15:15
• So I guess that having a preferred direction in time changes the formulas in some particular way. Do how know how? – VictorSeven Oct 22 '17 at 17:25
• Best thing is to compute the generating functional for a free non relativistic theory ;-) – Adam Oct 22 '17 at 17:34
• I have the generating functional for the free theory (I think), it's $Z_0[J]$. If you could write a complete answer, I would be grateful. I must admit that I am lost with this. – VictorSeven Oct 23 '17 at 8:36
• Could you give more details : lagrangian with source, to know what is coupled to what, and what correlation function in terms of the field $G_0$ is supposed to be. Are these fermions or bosons ? – Adam Oct 23 '17 at 8:54

A direct calculation of the generating functional shows that ($K=(k,\omega)$) $$\log Z_0[\vec J]=\int_{K}\bar J(K)G_0(K) J(-K),$$ which is indeed equal to Eq. (1) of the OP.
The only non-zero correlation function is the one that couples $\bar \phi$ and $\phi$ (since two derivatives with respect to $J$ or $\bar J$ would be zero), $$\langle\phi(Q)\bar \phi(P)\rangle=\frac{\delta^2\log Z_0}{\delta J(-Q)\delta\bar J(-P)},$$ which gives $$\langle\phi(Q)\bar \phi(P)\rangle=\delta(P+Q)G_0(Q)=\delta(P+Q)G_0(-P).$$
• Very nice, thank you. Also, I suspect that for the diagrams I need to connect $\bar \phi$ fields with the $\phi$ ones. – VictorSeven Oct 23 '17 at 17:05