Feynman Propagator from the Generating Functional

1. Evaluating the Generating Functional via "Completing the Square"

The path integral for the free scalar field in the presence of a source $J(x)$ is:
$Z_0[J]=\int \mathcal{D}\varphi \exp \left(i\int d^{4}x\left[\frac{1}{2}\varphi (-{\partial }^2-m^2)\varphi +J\varphi \right]\right)$
The expression in the exponent is quadratic in the field $\varphi$, which makes this a Gaussian functional integral. You'll recall from your study of ordinary Gaussian integrals that an integral of the form $\exp (-\frac{1}{2}a{x}^2+bx)$ can be solved by "completing the square" in the exponent. We will use the exact same trick here, but for functions instead of variables.

The core idea is to shift the integration variable (the field $\varphi$) by a specific field configuration, ${\varphi }_{cl}$, that extremizes the action. This shift eliminates the term linear in the new integration variable.

The following notes provide a very clear and "slick" derivation of this.

Let's summarize the key steps of the derivation you just read:

1. The Goal: We want to evaluate $Z_0[J]$. The exponent is $\int d^{4}x\left[\frac{1}{2}\varphi (x)(-{\partial }^2-m^2)\varphi (x)+J(x)\varphi (x)\right]$.
2. The Shift: We define a new field ${\varphi }^{\prime }$ by shifting $\varphi$ by a carefully chosen amount:
   ${\varphi }^{\prime }(x)=\varphi (x)-{\varphi }_{cl}(x)$
   where ${\varphi }_{cl}(x)$ is chosen to make the exponent quadratic in ${\varphi }^{\prime }$ with no linear term. This leads to ${\varphi }_{cl}$ satisfying the classical equation of motion with source $J$.
3. The Propagator Emerges: The required shift involves the inverse of the Klein-Gordon operator. We define this inverse as the Feynman propagator, ${\Delta }_F$. In the notes you read, it was called $D_F$, and its defining relation is (using our conventions):
   $({\partial }_x^2+m^2-iϵ){\Delta }_F(x-y)=-{\delta }^{(4)}(x-y)$
   The $iϵ$ term, which we'll discuss shortly, is a small negative imaginary part added to the mass squared, which formally ensures the path integral converges.
4. The Result: After shifting the integration variable from $\varphi$ to ${\varphi }^{\prime }$, the path integral separates into two parts: a piece that depends only on $J$ and ${\Delta }_F$, and the original path integral with no source, $Z_0[0]$. This gives the final result:
   $Z_0[J]=Z_0[0]\exp \left(-\frac{i}{2}\iint d^{4}x{d}^{4}yJ(x){\Delta }_F(x-y)J(y)\right)$
   This is a remarkable result. The path integral formalism has automatically identified the Green's function for the Klein-Gordon equation—the Feynman propagator—as the fundamental object governing the dynamics of the free theory.

2. The Propagator in Momentum Space and the $iϵ$ Prescription

We have now formally derived the propagator ${\Delta }_F(x-y)$ as the inverse of the operator $-({\partial }_x^2+m^2-iϵ)$. To get an explicit formula for it, it's easiest to work in momentum space, where differential operators become algebraic expressions.

We can solve for the Fourier transform, ${\tilde{\Delta }}_F(k)$, by transforming the defining equation:
$({\partial }_x^2+m^2){\Delta }_F(x)=-{\delta }^{(4)}(x)$
Applying a Fourier transform $\int d^{4}x{e}^{ikx}$ to both sides, we use the facts that ${\partial }_{\mu }\to -i{k}_{\mu }$ (so ${\partial }^2\to -k^2$) and that the Fourier transform of a delta function is 1. This yields:
$(-k^2+m^2){\tilde{\Delta }}_F(k)=-1$
Solving for ${\tilde{\Delta }}_F(k)$ gives:
${\tilde{\Delta }}_F(k)=\frac{1}{k^2-m^2}=\frac{1}{k_0^2-{\mathbf{k}}^2-m^2}$
However, there is a problem. The inverse Fourier transform to get back to ${\Delta }_F(x)$ involves an integral over $d^{4}k$, including an integral over the energy component $k_0$. The denominator has poles where $k_0=\pm \sqrt{{\mathbf{k}}^2+m^2}=\pm E_p$, which lie directly on the path of integration. The integral is therefore ill-defined.

This is where the $iϵ$ prescription, which arose naturally as a convergence factor in the path integral, becomes essential. It tells us how to handle these poles. With the prescription, the propagator is:
${\tilde{\Delta }}_F(k)=\frac{1}{k^2-m^2+iϵ}$
This small imaginary term shifts the poles slightly off the real $k_0$ axis. The video below explains exactly how this works and why this specific shift is the correct one to produce the time-ordered propagator.

The video provides a beautiful physical argument for the $iϵ$ prescription. By demanding that the path integral is well-defined (which requires a rotation to Euclidean time, or "Wick rotation"), the contour of integration is uniquely fixed. This fixed contour is equivalent to shifting the poles as prescribed by $iϵ$.

To summarize, the Feynman propagator in momentum space, including the factor of $i$ that aligns with the conventions from canonical quantization (where $⟨T\varphi \varphi ⟩=i{\Delta }_F$), is:
$i{\tilde{\Delta }}_F(k)=\frac{i}{k^2-m^2+iϵ}$

Conclusion

In this lesson, we have solidified our understanding of the path integral for the free scalar field. We rigorously derived the generating functional $Z_0[J]$ and, in the process, derived the Feynman propagator as the Green's function of the Klein-Gordon operator.

Key Takeaways:

• The Gaussian functional integral for the free scalar field can be solved exactly by completing the square, analogous to finite-dimensional Gaussian integrals.
• This procedure naturally identifies the Feynman propagator ${\Delta }_F(x-y)$ as the kernel in the exponent of $Z_0[J]$, and fundamentally, as the inverse of the Klein-Gordon operator.
• In momentum space, the propagator is ${\tilde{\Delta }}_F(k)=\frac{1}{k^2-m^2+iϵ}$.
• The $iϵ$ prescription is not an ad-hoc rule, but a necessary consequence of ensuring the path integral is well-defined. It shifts the poles of the propagator in the complex energy plane, providing an unambiguous rule for integration that correctly yields the time-ordered two-point function.

Preview of the Next Lesson:
Now that we have derived the propagator from $Z_0[J]$, what about higher-order correlation functions like the four-point function? In the next lesson, we will see how to compute these by taking more functional derivatives. This will lead us to a powerful result known as Wick's Theorem, which will emerge as a simple consequence of differentiating a Gaussian function. This will be our final step before we use the path integral to tackle interacting theories.