关于传播子（特指费米子传播子）定义的依据及标量场相关疑问

Great question—let’s unpack these two key points from David Tong’s QFT notes, since they’re foundational and easy to get hung up on when you’re starting out!

关于对易子表达式中的隐含单位算子 #

You’re absolutely right to notice this detail! In quantum field theory (and quantum mechanics more broadly), we follow a standard convention: when an operator equation has a c-number (a regular complex number, like the difference $D(x-y)-D(y-x)$ here) on the right-hand side, it’s implicitly multiplied by the identity operator $\mathbb{I}$.

The reason we omit $\mathbb{I}$ is straightforward—it’s the multiplicative identity for operators, just like 1 is for regular numbers. Writing $[\phi(x),\phi(y)] = (D(x-y)-D(y-x))\mathbb{I}$ is technically precise, but since multiplying any operator by $\mathbb{I}$ leaves it unchanged, the community universally drops the $\mathbb{I}$ to keep notation clean. Tong’s notes follow this standard convention, so you can rest assured the equality holds with that implicit identity operator in place.

传播子定义背后的推理逻辑（标量场与费米子场） #

To understand why propagators are defined the way they are, let’s start with their core purpose: they describe the amplitude for a quantum to propagate from one spacetime point to another, or more broadly, the correlation between field values at two points.

实标量场传播子 #

The definition $D(x-y) = \langle 0|\phi(x)\phi(y)|0\rangle$ comes from focusing on vacuum expectation values (VEVs)—the vacuum is the simplest, most fundamental state in QFT, so using it to measure field correlations makes intuitive sense.

Breaking it down: $\phi(y)$ acts on the vacuum to create a scalar particle at point $y$, and $\phi(x)$ acts to annihilate that particle at point $x$. The VEV gives the amplitude for this process (or equivalently, for a particle to travel from $y$ to $x$). This definition also ties directly to causal behavior, as you saw with the commutator $[\phi(x),\phi(y)] = D(x-y)-D(y-x)$—the difference ensures that for spacelike-separated points, the commutator vanishes, preserving causality (no superluminal influence between measurements).

费米子传播子 #

For fermions, the propagator is typically defined as $S(x-y) = \langle 0|\psi(x)\bar{\psi}(y)|0\rangle$, where $\bar{\psi}(y) = \psi^{\dagger(y)\gamma}0$ is the Dirac adjoint. Here’s the reasoning behind this specific order:

• The Dirac field $\psi(x)$ splits into two parts: terms that annihilate positive-energy fermions and terms that create negative-energy fermions (which we interpret as antiparticles).
• The adjoint $\bar{\psi}(y)$ splits into terms that create positive-energy fermions and terms that annihilate negative-energy fermions.

When we take the VEV $\langle 0|\psi(x)\bar{\psi}(y)|0\rangle$, only the annihilation term from $\psi(x)$ and the creation term from $\bar{\psi}(y)$ will produce a non-zero result—this corresponds to the amplitude for a positive-energy fermion to propagate from $y$ to $x$. If we reversed the order to $\langle 0|\bar{\psi}(x)\psi(y)|0\rangle$, we’d get the amplitude for an antiparticle to travel from $y$ to $x$.

Additionally, this definition aligns with causality for fermions: the anticommutator ${\psi(x),\bar{\psi}(y)}$ (or related combinations) vanishes for spacelike separations, ensuring that measurements of fermion fields at causally disconnected points don’t interfere with each other.

Finally, propagators are critical for calculating scattering amplitudes via Feynman diagrams—each internal line in a diagram corresponds to a propagator, representing the quantum propagating between interaction vertices. This makes the VEV definition not just conceptually meaningful, but practically essential for doing QFT calculations.

内容的提问来源于stack exchange，提问作者Guilherme Tomishiyo