- #1
MadMax
- 98
- 0
Einstein summation convention employed throughout
We want to calculate
\hbar \ln \int D x_i \exp[\frac{1}{32 \pi^3} \int ds \int d^3 r x_i(-is,r) M_{ij}(s,r) x_j(is,r)]
The answer is
\hbar \int \frac{ds}{2\pi} \ln \det[M_{ij}\delta^3(r-r')]
I know that
\int d^3 x_i e^{\frac{1}{2}x_i B_{ij} x_j} = \sqrt{\frac{(2\pi)^n}{\det B_{ij}}}
and that standard logarithmic properties will be used. Also the \delta^3(r-r') means that a Fourier transform involving that delta function will be employed at some point.
Beyond that I'm at a complete loss as to how to continue. One question is why we don't need to employ a Fourier transform involving a \delta(s-s')]. Any help would be much appreciated.
We want to calculate
\hbar \ln \int D x_i \exp[\frac{1}{32 \pi^3} \int ds \int d^3 r x_i(-is,r) M_{ij}(s,r) x_j(is,r)]
The answer is
\hbar \int \frac{ds}{2\pi} \ln \det[M_{ij}\delta^3(r-r')]
I know that
\int d^3 x_i e^{\frac{1}{2}x_i B_{ij} x_j} = \sqrt{\frac{(2\pi)^n}{\det B_{ij}}}
and that standard logarithmic properties will be used. Also the \delta^3(r-r') means that a Fourier transform involving that delta function will be employed at some point.
Beyond that I'm at a complete loss as to how to continue. One question is why we don't need to employ a Fourier transform involving a \delta(s-s')]. Any help would be much appreciated.
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