- #1

RedX

- 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.

[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.

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