- #1
geoduck
- 258
- 2
In Srednicki's QFT textbook, equation 9.6, he writes:
[tex]Z[J]=e^{i \int d^4x \, \mathcal L_I\left(\frac{1}{i}\frac{\delta}{\delta J(x)} \right)} \int \, \mathcal D\phi e^{i \int d^4x \, \mathcal [\mathcal L_0+J\phi]} \propto
e^{i \int d^4x \, \mathcal L_I\left(\frac{1}{i}\frac{\delta}{\delta J(x)} \right)} Z_0[J][/tex]
and said that we have to use "proportional to" rather than equals because the ε-trick does not give the correct overall normalization.
I don't quite understand this. Aside from physics, aren't the two sides equal mathematically? If you were to discretize the path integral of [itex] Z_0[J] [/itex], and take the limit of infinite number of time slices, then you would get [itex] Z_0[J]=e^{i\int \int J(y)\Delta(x-y) J(x)} [/itex]. If you then wanted to calculate
[tex]\int \, \mathcal D\phi e^{i \int d^4x \, \mathcal [\mathcal L_0+\mathcal L_I+J\phi]} [/tex]
then it would be equal mathematically to [itex]e^{i \int d^4x \, \mathcal L_I\left(\frac{1}{i}\frac{\delta}{\delta J(x)} \right)} Z_0[J] [/itex] would it not?
[tex]Z[J]=e^{i \int d^4x \, \mathcal L_I\left(\frac{1}{i}\frac{\delta}{\delta J(x)} \right)} \int \, \mathcal D\phi e^{i \int d^4x \, \mathcal [\mathcal L_0+J\phi]} \propto
e^{i \int d^4x \, \mathcal L_I\left(\frac{1}{i}\frac{\delta}{\delta J(x)} \right)} Z_0[J][/tex]
and said that we have to use "proportional to" rather than equals because the ε-trick does not give the correct overall normalization.
I don't quite understand this. Aside from physics, aren't the two sides equal mathematically? If you were to discretize the path integral of [itex] Z_0[J] [/itex], and take the limit of infinite number of time slices, then you would get [itex] Z_0[J]=e^{i\int \int J(y)\Delta(x-y) J(x)} [/itex]. If you then wanted to calculate
[tex]\int \, \mathcal D\phi e^{i \int d^4x \, \mathcal [\mathcal L_0+\mathcal L_I+J\phi]} [/tex]
then it would be equal mathematically to [itex]e^{i \int d^4x \, \mathcal L_I\left(\frac{1}{i}\frac{\delta}{\delta J(x)} \right)} Z_0[J] [/itex] would it not?