Normal ordering and path integrals

Question

What is the manifestation of normal ordering for creation/annihilation operators in the path-integral formalism?

Answer 1

This argument follows Srednicki's Quantum Field Theory, which is an excellent textbook.

Computing the matrix element for a transition between an initial and final state in a scalar interacting quantum field theory makes use of the Lehmann-Symanzik-Zimmermann (LSZ) reduction formula:

$$\langle f | i \rangle = i^{n + n'} \int d^4x_1 e^{i k_1x_1}(-\partial_1^2 + m^2)...d^4x_1' e^{i k_1'x_1'}(-\partial_{1'}^2 + m^2)...\times \langle 0|T \phi(x_1)...\phi(x_1')...|0 \rangle,$$ where $T$ is the time ordering operator.

Within the derivation of this formula, there is an important step where we are allowed to insert the $T$ operator into our expectation value because the operators are already in time-order, so the addition of $T$ alters nothing: $$\langle f | i \rangle = \langle 0 |a_{1'}(+ \infty)a_{2'}(+ \infty)a_{1}^{\dagger}(- \infty)a_{2}^{\dagger}(- \infty)|0 \rangle = \langle 0 |T a_{1'}(+ \infty)a_{2'}(+ \infty)a_{1}^{\dagger}(- \infty)a_{2}^{\dagger}(- \infty)|0 \rangle.$$

This is critical to the next step, where we plug in the definitions: $$a^{\dagger}(- \infty) = a^{\dagger}(+ \infty) + i \int d^3k f(\vec{k})\int d^4x e^{i k x}(-\partial^2 + m^2)\phi(x)$$

$$a(- \infty) = a(- \infty) + i \int d^3k f(\vec{k})\int d^4x e^{-i k x}(-\partial^2 + m^2)\phi(x)$$

After plugging these definitions in we'll have 16 terms, all but one with at least one $a$ or $a^{\dagger}$ operator. The $T$ operator will move these operators such that all the $a^{\dagger}(+ \infty)$'s annihilate $\langle0|$ and the $a(- \infty)$'s annihilate $|0\rangle$.

Thus, we're left with only one term because the other 15 were annihilated. Taking $f_{1,2}(\vec{k}) = \delta^3(\vec{k}-\vec{k}_{1,2})$, we finally arrive at the complete LSZ reduction formula:

$$\langle f | i \rangle = i^{n + n'} \int d^4x_1 e^{i k_1x_1}(-\partial_1^2 + m^2)...d^4x_1' e^{i k_1'x_1'}(-\partial_{1'}^2 + m^2)...\times \langle 0|T \phi(x_1)...\phi(x_1')...|0 \rangle.$$

Therefore, we conclude that $T$ is critical to being able to cancel all the $a$ and $a^{\dagger}$ terms and being able to write $\langle f|i\rangle$ in terms of a time-ordered vacuum expectation value of field operators.

This expectation value $\langle 0|T \phi(x_1)...\phi(x_1')...|0 \rangle$ can then be explicitly constructed as a perturbative expansion with the Feynman path integral and plugged into the LSZ formula.