Following the book "Condensed Matter Field Theory" from Altland (Chapter 5.1), the free propagator is defined as

$$ G_0(\mathbf{x}) \equiv \langle \phi (\mathbf{0}) \phi(\mathbf{x})\rangle_0 \equiv \frac{\int D\phi \, \mathrm{e}^{-S_0[\phi]} \phi (\mathbf{0}) \phi(\mathbf{x}) }{\int D\phi \, \mathrm{e}^{-S_0[\phi]}} \qquad (1) $$

with $$D\phi \equiv \prod_{\mathbf{x}_i} \mathrm{d}\phi(\mathbf{x}_i) \qquad (2) \text{,}$$ where the points $\mathbf{x}_i$ correspond to some finite ($d$-dimensional) lattice and $\phi(\mathbf{x})$ being real-valued integration-variables,

\begin{align} S_0[\phi] &= \frac{1}{2} \int \mathrm{d}^d x \, \left( (\partial \phi)^2 + r \phi^2\right) \qquad(2) \\ &= \frac{1}{2} \sum_{\mathbf{p}} \phi_{\mathbf{p}} (p^2 + r) \phi_{-\mathbf{p}} \qquad(3) \end{align}

where the integral extends over the finite lattice and the sum extends over all the allowed reciprocal lattice points.

Now, using the relation

$$ \frac{\int \mathrm{d} \mathbf{v}\, \mathrm{e}^{-\frac{1}{2} \mathbf{v}^{\text{T}} \mathbf{A} \mathbf{v}} v_m v_n}{\int \mathrm{d} \mathbf{v} \mathrm{e}^{-\frac{1}{2} \mathbf{v}^{\text{T}} \mathbf{A} \mathbf{v}}} = A_{mn}^{-1} $$

it should be easy to see that

$$ G_{0 \mathbf{p}} = \int \mathrm{d}^d x \, \mathrm{e}^{ \mathrm{i} \mathbf{p} \cdot \mathbf{x}} G_0(\mathbf{x}) \stackrel{\text{I}}{=} \sum_{\mathbf{p}^{\prime}} \langle \phi_{\mathbf{p}} \phi_{\mathbf{p}^{\prime}}\rangle_0 \stackrel{\text{II}}{=} \sum_{\mathbf{p}^{\prime}} \delta_{\mathbf{p} + \mathbf{p}^{\prime}, 0} \frac{1}{p^2 + r} \qquad \left( = \frac{1}{p^2 + r} \right) \text{,} $$ but I am completely lost with the equation signs I and II.

From the equation sign II, it seems like the expression $\langle \phi_{\mathbf{p}} \phi_{\mathbf{p}^{\prime}}\rangle_0$, is being interpreted as

$$ \langle \phi_{\mathbf{p}} \phi_{\mathbf{p}^{\prime}}\rangle_0 = \frac{\int D\phi \, \mathrm{e}^{-S_0[\phi]} \phi_{\mathbf{0}} \phi_{\mathbf{p}} }{\int D\phi \, \mathrm{e}^{-S_0[\phi]}} $$

with $$D\phi \equiv \prod_{\mathbf{p}_i} \mathrm{d}\phi_{\mathbf{p}_i}$$

and $S_0[\phi]$ similar as in eq. $(3)$. (Contrary to the definition in equations $(1)$ and $(2)$!?)

Please, can somebody explain the two equation signs I and II?

  • $\begingroup$ Hint: A standard method is to introduce external sources. $\endgroup$ – Qmechanic Dec 4 '19 at 19:32
  • $\begingroup$ Where I can find the derivation of $\frac{\int \mathrm{d} \mathbf{v}\, \mathrm{e}^{-\frac{1}{2} \mathbf{v}^{\text{T}} \mathbf{A} \mathbf{v}} v_m v_n}{\int \mathrm{d} \mathbf{v} \mathrm{e}^{-\frac{1}{2} \mathbf{v}^{\text{T}} \mathbf{A} \mathbf{v}}} = A_{mn}^{-1}$?, I tried looking at the book but couldn't find it, sorry for the unrelated question and thanks $\endgroup$ – Daniel D. Dec 5 '19 at 16:52
  • $\begingroup$ dandide: Chapter 3.2. ... I still have to think about the other answer... $\endgroup$ – Antihero Dec 7 '19 at 1:45
  • $\begingroup$ Found it, thank you, also not sure but I think if you write your variable as its fourier series and replace them you should get the identity (I) but doesn't looks easy, so $ \frac{\int D\phi_x e^{-S_0[\phi_x]}\phi_x\phi_{x'}}{\int D\phi_x e^{- S_0[\phi_x]}}=\frac{\int D(\sum e^{-2i\pi px}\phi_p) e^{-S_0[\sum e^{-2i\pi px}\phi_p]} (\sum e^{-2i\pi px}\phi_p) (\sum e^{-2i\pi p'x}\phi_{p'})}{\int D(\sum e^{-2i\pi px}\phi_p) e^{-S_0[\sum e^{-2i\pi px}\phi_p]}}=\int d^dx\, e^{-2i\pi(p-p')x}(\frac{\int D\phi_p e^{-S_0[\phi_p]}\phi_p \phi_{p'}}{\int D\phi_p e^{- S_0[\phi_p]}})$ $\endgroup$ – Daniel D. Dec 7 '19 at 23:39
  • $\begingroup$ and by (2) and (3) you already know that $S_0[\phi_x]=S_0[\sum e^{-2i\pi px}\phi_p]=S_0[\phi_p]$ so you don't need to do anything there and it seems that to conclude (II) you would need to show that $\frac{\int D\phi_p e^{-S_0[\phi_p]}\phi_p \phi_{p'}}{\int D\phi_p e^{- S_0[\phi_p]}}=\frac{1}{p^2+r}$ that as I understand is what you already believe $\endgroup$ – Daniel D. Dec 7 '19 at 23:39

Given your equation (3), it seems natural to begin with the propagator in momentum space. We have $$ \langle \phi_{\mathbf{p}} \phi_{\mathbf{p}'} \rangle_0 = \frac{\int \mathcal{D} \phi \, e^{- S_0[\phi]} \phi_{\mathbf{p}} \phi_{\mathbf{p}'}}{\int \mathcal{D} \phi \, e^{- S_0[\phi]}}. $$ In this case, it is worth being careful about how the integration measure of the fields works after Fourier-transforming to momentum space. See the discussion in this answer to see one solution: we can restrict our attention to taking $q_i > 0$ for one of the components of momentum, and then integrate over both the real and imaginary parts of $\phi_{\mathbf{q}}$. So $$ \mathcal{D} \phi = \prod_{q_i > 0} d \mathrm{Re}[\phi_{\mathbf{q}}]d \mathrm{Im}[\phi_{\mathbf{q}}]. $$ Perhaps this is what's tripping you up - there is a subtle difference between this and how one would interpret the integration measure in the last equation you give! Fundamentally, it comes from thinking about how the integration measure changes under the Fourier transformation, which is a linear transform after all.

Meanwhile, the action $S_0[\phi]$ is diagonal in $\mathbf{q}$-space, so we've really just reduced this to a product of Gaussian integrals. The denominator is $$ \int \mathcal{D} \phi \, e^{- S_0[\phi]} = \prod_{q_i > 0} \int d \mathrm{Re}[\phi_{\mathbf{q}}]d \mathrm{Im}[\phi_{\mathbf{q}}] \, \exp\left[ (q^2 + r) (\mathrm{Re}[\phi_{\mathbf{q}}]^2 + \mathrm{Im}[\phi_{\mathbf{q}}]^2) \right] = \prod_{q_i > 0} \frac{\pi}{q^2 + r}. $$ An identical set of manipulations shows $$ \int \mathcal{D} \phi \, e^{- S_0[\phi]} \phi_{\mathbf{p}} \phi_{\mathbf{p}'} = \delta_{\mathbf{p}+\mathbf{p}',0} \frac{\pi}{(p^2 + r)^2} \prod_{\mathbf{q} \neq \mathbf{p},q_i>0} \frac{\pi}{q^2 + r}. $$ So then taking the ratio we get $$ \langle \phi_{\mathbf{p}} \phi_{\mathbf{p}'} \rangle_0 = \frac{\delta_{\mathbf{p}+\mathbf{p}',0}}{p^2 + r} $$ directly. And of course a usual inverse Fourier transform gives your desired equalities.

  • $\begingroup$ Hi, could you please explain again why it is sufficient to restrict our attention to $q_i > 0$ for one of the momentum components? (And therefore (???) $\mathcal{D} \phi = \prod_{q_i > 0 } \mathrm{d} \operatorname{Re}\, \phi_q \mathrm{d} \operatorname{Im} \, \phi_q$?) $\endgroup$ – Antihero Feb 21 '20 at 15:15
  • 1
    $\begingroup$ @Antihero Since $\phi$ is real, the complex conjugate of the Fourier transformed field satisfies $\phi_{-\mathbf{q}} = \phi_{\mathbf{q}}^{\ast}$. If you sum over all complex values of $\phi_{\mathbf{q}}$ and all $\mathbf{q}$, you are double-counting the degrees of freedom. $\endgroup$ – Seth Whitsitt Feb 22 '20 at 22:55

It took me quite some time, but now I got it.

The "double counting" that Seth Whitsitt mentioned is the important catchphrase. The discrete Fourier-Transformation that transforms the integration variables(see e.g. Wikipedia) corresponds to an invertible square matrix.

Usually this matrix takes values from the complex numbers to the complex numbers. Instead one can also interpret this mapping as an invertible matrix from real numbers to complex numbers.

This, alongside with some calculations regarding the determinant of the matrix (it's one for proper prefactor conventions of the DFT), reproduces the results from the book.

Thank you all for your effort.


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