Trace of momentum-space propagator

  • Thread starter RedX
  • Start date
  • #1
970
3
The integral:

[tex]\int \Pi_k d\phi_k e^{-\phi_i A_{ij} \phi_j} [/tex]

is a Gaussian and is equal to:

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

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 [tex]A^{-1}[/tex] would then be the propagator matrix.

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

For example if you take this expression:

[tex]e^{\int \int dx dy j(x) A^{-1}(x,y) j(y)} [/tex]

but change to momentum space then you get something like:

[tex]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)} [/tex]

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

Answers and Replies

  • #2
679
2
I'm pretty sure TrA (or Tr(lnA)) is not the same as [tex] \int \int dx dy j(x) A^{-1}(x,y) j(y)
[/tex] 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
[tex]{\omega ^2} = {k^2} + {m^2}[/tex], then the trace is just the sum of all eigenvalues:
[tex]\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}[/tex]
 
  • #3
970
3
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:

[tex]x^TAy= x^T(F^{T}DF)y=(Fx)^TD(Fy)
[/tex]
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:


[tex]
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)}
[/tex]

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.
 
  • #4
970
3
Actually, [tex]j^{\dagger}(k)=j(-k) [/tex] is the most general condition, so maybe it ought to be:

[tex]

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)}
[/tex]

since the Fourier transform that diagonalizes [tex]A^{-1}[/tex] should be unitary and not orthogonal, and hence the dagger on j is required.
 
  • #5
679
2
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.
 

Related Threads on Trace of momentum-space propagator

  • Last Post
Replies
4
Views
368
Replies
0
Views
1K
Replies
6
Views
265
  • Last Post
Replies
9
Views
14K
Replies
2
Views
5K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
8
Views
9K
Top