Trace of momentum-space propagator

[RedX]

The integral:

$$\int \Pi_k d\phi_k e^{-\phi_i A_{ij} \phi_j}$$

is a Gaussian and is equal to:

$$(\pi)^{n/2}\sqrt{det(A^{-1})}= (\pi)^{n/2} e^{\frac{1}{2}Tr ln A^{-1}}$$

Now usually A is a diagonal matrix that represents the Lagrangian (so that the sum over i and j collapses to a sum just over i, and this sum is converted to an integral for continuum), and $$A^{-1}$$ would then be the propagator matrix.

The book I'm reading says that the trace of $$A^{-1}$$ is best evaluated in the momentum representation where it is diagonal. But how is $$A^{-1}$$ diagonal in this representation?

For example if you take this expression:

$$e^{\int \int dx dy j(x) A^{-1}(x,y) j(y)}$$

but change to momentum space then you get something like:

$$e^{\int dk j(-k) A^{-1}(k) j(k)}=e^{\int \int dk dq j(q) A^{-1}(k) \delta(k+q) j(k)}$$

Doesn't this suggest that $$A^{-1}$$ is not diagonal in momentum space? If it were diagonal, then there would be a $$\delta(k-q)$$ and not $$\delta(k+q)$$ on the RHS.

 

[kof9595995]

I'm pretty sure TrA (or Tr(lnA)) is not the same as $$\int \int dx dy j(x) A^{-1}(x,y) j(y)$$ because this expression depend on j(x) you choose.
I think your book means usually A's eigenvectors are monochromatic waves, for example if A is the Klein-Gordon operator, then A's eigenvectors are expi(kx-wt) with eigenvalues
$${\omega ^2} = {k^2} + {m^2}$$, then the trace is just the sum of all eigenvalues:
$$\begin{array}{l} TrA = \int {dk(} {k^2} + {m^2}) \\ Tr\ln {A^{ - 1}} = \int {dk} \ln \frac{1}{{{k^2} + {m^2}}} \\ \end{array}$$

 

[RedX]

A Fourier transform F is a type of linear transform, so if F diagonalizes a matrix A, then for any two vectors x and y:

$$x^TAy= x^T(F^{T}DF)y=(Fx)^TD(Fy)$$
for a diagonal matrix D.

So if you allow j(x) and j(y) to be your vectors x and y except in continuum, then for the Fourier transforms F[j(x)]=j(k), I think the continuum analog should be:


$$e^{\int \int dx dy j(x) A^{-1}(x,y) j(y)}=e^{\int \int dk dq j(q) D(k,q) j(k)}$$

But D(k,q) is not diagonal since it has a delta(k+q) and not a delta(k-q).

I would like to make the claim that maybe j(k)=j(-k) if you require j(x) to be real, but that's not necessarily true.

 

[RedX]

Actually, $$j^{\dagger}(k)=j(-k)$$ is the most general condition, so maybe it ought to be:

$$e^{\int \int dx dy j(x) A^{-1}(x,y) j(y)}=e^{\int \int dk dq j^{\dagger}(q) D(k,q) j(k)}= e^{\int \int dk dq j^{\dagger}(q) D(k)\delta(k+q) j(k)} =e^{\int dk j^{\dagger}(-k) D(k) j(k)}=e^{\int dk j(k) D(k) j(k)} =e^{\int \int dq dk j(q) D(k)\delta(k-q) j(k)}$$

since the Fourier transform that diagonalizes $$A^{-1}$$ should be unitary and not orthogonal, and hence the dagger on j is required.

 

[kof9595995]

I‘m not saying your math is wrong, but it''s just not Tr A, because Tr A should be a fixed value independent of j(x). And by "momentum representation" my understanding is to express A in terms of monochromatic plane waves, which are the eigen solutions of A, not simply change the integration measure to ,momentum space.