I'd like to know which parts of the following is due to convention and which part has to be a certain way.

Let's assume a complex scalar field $\phi(x)$ with Lagrangian: $$ \mathcal L = \partial_\mu \phi^\dagger \partial^\mu \phi - m^2 \phi^\dagger \phi. \tag{1} $$ The solutions of the Euler--Lagrange equations are usually written as (I'll use the notation of Schwartz' QFT book, e.g., eqs. (9.14) and (9.15)): $$ \begin{align} \phi &= \int\frac{\text{d}^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left[ a_p \text{e}^{-\text{i} p\cdot x} + b_p^\dagger \text{e}^{\text{i} p\cdot x} \right] \tag{2} \\ \phi^\dagger &= \int\frac{\text{d}^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left[ a_p^\dagger \text{e}^{-\text{i} p\cdot x} + b_p \text{e}^{\text{i} p\cdot x} \right] \tag{3} \end{align} $$ Question 1: Is it correct that $\phi$ annihilates a particle state via $a_p$ as in $\phi |\text{particle}\rangle=|0\rangle$ and creates an antiparticle state using $b^\dagger$ as in $\phi|0\rangle = |\text{antiparticle}\rangle$? Is there even room for debate to define $\phi^\dagger$ as a particle-creation/antiparticle-annihilation operator?

Now for the propagator, I'd like to write down the propagation of a particle.

Question 2: Which of the following is the correct expression for the propagation of a particle?

$$ \begin{align} G_\text{particle}(x,y) &= \langle 0| \phi(x)\phi^\dagger(y) |0\rangle \tag{4} \\ G_\text{particle}(x,y) &= \langle 0| \phi(x)^\dagger\phi(y) |0\rangle \tag{5} \end{align} $$ I think it should be eq. (4), since we should first create a particle using $\phi^\dagger(y) |0\rangle$, which can then be annihilated by $\phi(x)$.

  • $\begingroup$ What do you mean with "room for debate" exactly? $\endgroup$
    – ohneVal
    Aug 15, 2021 at 14:14
  • $\begingroup$ @ohneVal Whether everyone agrees that phi annihilates a particle or whether one can do calculations where phi creates a particle. $\endgroup$
    – ersbygre1
    Aug 15, 2021 at 14:16

2 Answers 2


Question 1: Be careful when you say that $\phi(x)|\text{particle}\rangle = |0\rangle$. This is in general not true, as it will depend on the specific state. What is true is the fact that one can formally write

$$\phi(x) |{0}\rangle = \frac{1}{(2\pi)^\frac{3}{2}}\int \frac{d^3 p}{\sqrt{2\omega_p}}e^{i p \cdot x}\hat{b}_{p}^\dagger |{0}\rangle =|a(x)\rangle$$ and $$\phi^\dagger(x) |{0}\rangle = \frac{1}{(2\pi)^\frac{3}{2}}\int \frac{d^3 p}{\sqrt{2\omega_p}}e^{-i p \cdot x}\hat{a}_{p}^\dagger |{0}\rangle =|p(x)\rangle,$$

where by the kets $|p(x)\rangle$ and $|a(x)\rangle$ I mean the formal particle and anti-particle states that represents a particle/anti-particle centered at $x$. I say "formal" because in principle these states are singular and have ill defined energy, and other issues.

Question 2: The "propagator" for a particle state to go from $x$ to $y$ will be given by

$$G_p(x,y) = \langle p(x)|p(y)\rangle = \langle 0|\phi(x)\phi^\dagger(y)|0\rangle,$$

as you have anticipated. However, notice that the usually when one talks about the propagator, there is usually a time ordering operation acting on the fields.


The first thing to understand is that you should be looking at the creation and annihilation operators $a$ and $b$, not at the field in configuration space generally, that is why they receive such names, not the fields.

Answer 1: What creates and annihilates particles are the $a$ and $b$ their names are up to you to choose. By convention the underlying Hilbert is built upon the condition $a_p|0\rangle = 0$ (no dagger) for particles of species $a$, the rest follows from algebraic properties. You could choose to do it with daggers I guess, but nobody does this. (Also see the Historical note 9.1.1 from Schwartz book)

Answer 2: Sadly here the answer is more complicated than what you might expect. Propagators coincide with certain types of Green's functions. They exist with different boundary conditions and properties e.g. Feynman, Retarded, Advanced, etc. The boundary conditions determine where the ${}^\dagger$ goes. Section 6.2 of the book explains more for the case of the Feynman propagator which is neither of the ones you wrote but $$\langle 0 | T[\phi^\dagger(x)\phi(y)] | 0 \rangle,$$ where $T$ stands for the time ordering operator.

So the propagator is not to be confused with perhaps the number operator, which follows the structure you describe of creating and then destroying the same particle while "counting" it.

These other Green's functions are commonly used in quantum field theory at finite temperature for example.

  • 3
    $\begingroup$ In the case of a free complex scalar theory the quantity you wrote is identically vanishing: $\langle 0|\phi(x)\phi(y)|0\rangle = 0$. (Same is true for time ordering, of course). $\endgroup$
    – Rick
    Aug 15, 2021 at 14:36
  • $\begingroup$ Thanks for the correction $\endgroup$
    – ohneVal
    Aug 15, 2021 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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