On Peskin and Schroeder's QFT book, page 284, the book derived two-point correlation functions in terms of function integrals. $$\left\langle\Omega\left|T \phi_H\left(x_1\right) \phi_H\left(x_2\right)\right| \Omega\right\rangle=$$ $$\lim _{T \rightarrow \infty(1-i \epsilon)} \frac{\int \mathcal{D} \phi \phi\left(x_1\right) \phi\left(x_2\right) \exp \left[i \int_{-T}^T d^4 x \mathcal{L}\right]}{\int \mathcal{D} \phi \exp \left[i \int_{-T}^T d^4 x \mathcal{L}\right]} ,\tag{9.18} $$

Then, in the following several pages, the book calculate the numerator and denominator separately, using discrete Fourier series, and then take the continue limit.

Finally, the book obtain the calculation result of eq.(9.18) in eq.(9.27): $$ \left\langle 0\left|T \phi\left(x_1\right) \phi\left(x_2\right)\right| 0\right\rangle=$$ $$\int \frac{d^4 k}{(2 \pi)^4} \frac{i e^{-i k \cdot\left(x_1-x_2\right)}}{k^2-m^2+i \epsilon}=D_F\left(x_1-x_2\right). \tag{9.27}$$ I am troubled for the L.H.S of (9.27), why $\left\langle\Omega\left|T \phi_H\left(x_1\right) \phi_H\left(x_2\right)\right| \Omega\right\rangle$ became $\left\langle 0\left|T \phi\left(x_1\right) \phi\left(x_2\right)\right| 0\right\rangle$?

Also, the $\phi$ in eq.(9.27) maybe in interaction picture, which is different with Heisenberg picture $\phi_H$ in general case.

A parallel analysis is also in Peskin and Schroeder's QFT book, on page 87, in eq.(4.31) $$ \langle\Omega|T\{\phi(x) \phi(y)\}| \Omega\rangle=\lim _{T \rightarrow \infty(1-i \epsilon)} \frac{\left\langle 0\left|T\left\{\phi_I(x) \phi_I(y) \exp \left[-i \int_{-T}^T d t H_I(t)\right]\right\}\right| 0\right\rangle}{\left\langle 0\left|T\left\{\exp \left[-i \int_{-T}^T d t H_I(t)\right]\right\}\right| 0\right\rangle} .\tag{4.31}$$

And also we only considered the numerator of (4.31) in later analysis, why?


1 Answer 1


Perhaps it would be more symmetric if we write the denominator $\langle\Omega|\Omega\rangle=1$ on the left-hand sides explicitly. The denominators on the right-hand sides are important, and cannot in principle be omitted in later analysis.

  1. If we start with eq. (4.31), here the denominator on the right-hand side is crucial, since when we apply the theorem of Gell-Mann and Low a non-trivial factor is cancelled. Here $|\Omega\rangle$ and $|0\rangle$ denote the vacuum in the interaction and the free theory, respectively.

  2. Similarly, in eq. (9.18) the denominator is a convenient way to absorb an over-all normalization factor in the path integral measure.

  3. Eq. (9.27) refers to a free theory, so the Heisenberg and interaction pictures coincide.

  • $\begingroup$ Thank you very much, in free theory, $\left\langle\Omega\left|T \phi_H\left(x_1\right) \phi_H\left(x_2\right)\right| \Omega\right\rangle$ and $\left\langle 0\left|T \phi\left(x_1\right) \phi\left(x_2\right)\right| 0\right\rangle$ really same. $\endgroup$
    – Daren
    Commented Sep 26, 2022 at 10:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.