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I am currently trying to learn Quantum Field Theory through David Tong's notes which only talk about canonical quantisation for the scalar field and Dirac spinor field.

In Chapter 2, the propagator for a real scalar field is defined as (and I believe this definition is standard since it is also the one used by Peskin and Schroeder):

$$ D(x-y) = \langle 0 | \phi(x)\phi(y) | 0 \rangle $$

However, in Chapter 5, when we are quantising the Dirac spinor field, the propagator, $S_{\alpha\beta}$, is defined as

$$ i S _ { \alpha \beta } = \left\{ \psi _ { \alpha } ( x ) , \bar { \psi } _ { \beta } ( y ) \right\} $$

This definition is also used in Ben Allanach's Lectures (notes taken by Dexter, see page 83).

However, this looks rather strange since the analogous thing to define would be something like $$ S_{\alpha\beta} = \langle 0 | \psi_\alpha(x)\bar{\psi}_\beta(y) | 0 \rangle $$

which should not be the anti-commutator.

Further down the Chapter, the Feynman propagator is defined in an analogous way to the Feynman propagator for the real scalar field as

$$ S _ { F } ( x - y ) = \langle 0 | T \psi ( x ) \bar { \psi } ( y ) | 0 \rangle \equiv \left\{ \begin{array} { c c } { \langle 0 | \psi ( x ) \bar { \psi } ( y ) | 0 \rangle } & { x ^ { 0 } > y ^ { 0 } } \\ { \langle 0 | - \bar { \psi } ( y ) \psi ( x ) | 0 \rangle } & { y ^ { 0 } > x ^ { 0 } } \end{array} \right. $$

Hence, if $\langle 0 | \psi_\alpha(x)\bar{\psi}_\beta(y) | 0 \rangle$ is not a propagator, then does it represent anything meaningful? And why is it not the correct propagator definition?

PS: This link might be helpful but does not discuss what the object $\langle 0 | \psi_\alpha(x)\bar{\psi}_\beta(y) | 0 \rangle$ might potentially be and why it does not work as the propagator definition.

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    $\begingroup$ I don't know why Tong call the anticommutator a "propagator." It's surely not standard language. I've always called it the "(anti) commutator function" $\endgroup$
    – mike stone
    Commented Jan 5, 2020 at 17:14
  • $\begingroup$ I think that we need compare complex scalar field and Dirac fermion $\endgroup$
    – Nikita
    Commented Jan 5, 2020 at 17:49
  • $\begingroup$ @mikestone That would have been a much better name for it! $\endgroup$
    – alfred
    Commented Jan 6, 2020 at 21:10

2 Answers 2

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Notation (and concepts) in QFT tends to be very subtle.

Propagators

A propagator is defined as the probability amplitude of annihilate at event $y$ a particle created at event $x$.

Since a particle is created out from thin air, and the set up return to "nothing" when the particle is annihilated, its definition always requires an IN and OUT state! In particular, something sandwiched by the vacuum, e.g.

$$\left< \Omega|\phi(y) \bar{\phi}(x) | \Omega \right>.$$

In the same foot, the for a fermion the propagator should be something like,

$$\left< \Omega|\psi(y) \bar{\psi}(x) | \Omega \right>.$$

However, there are several types of propagators one might define: retarded, advanced, Feynman, etc. But, the bottom line is that all of these are sandwich by the same IN and OUT state.

(micro) Causality

Causality ensures that axioms from special relativity are valid.

Being that fundamental, causality should be preserved regardless of the IN and OUT states!

Causality is measured by the (anti)commutator of the fields, i.e. $[\phi(y),\bar{\phi}(x)]$ for bosons and $\{\psi(y),\bar{\psi}(x)\}$ for fermions.

$$[\phi(y),\bar{\phi}(x)] = 0 \quad \text{for }(x - y) \text{ spacelike}$$ $$\{\psi(y),\bar{\psi}(x)\} = 0 \quad \text{for }(x - y) \text{ spacelike}$$

Why the confusion?

Love and marriage, love and marriage
They go together like a horse and carriage
Dad was told by mother
You can't have one without the other

--- Frank Sinatra

Because they are related! I recommend an old fashion classic by Wolfgang Pauli "Selected Topics in Field Quantization".

The micro causality condition is not imposed as stated above, but usually only the weaker conditions below are required:

$$\left<\Omega|[\phi(y),\bar{\phi}(x)]|\Omega\right> = 0 \quad \text{for }(x - y) \text{ spacelike}$$ $$\left<\Omega|\{\psi(y),\bar{\psi}(x)\}|\Omega\right> = 0 \quad \text{for }(x - y) \text{ spacelike}$$

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For Dirac spinor field:

$$ i S _ { \alpha \beta }(x-y) = \left\{ \psi _ { \alpha } ( x ) , \bar { \psi } _ { \beta } ( y ) \right\} = \langle 0 | \psi_\alpha(x)\bar{\psi}_\beta(y) | 0 \rangle + \langle 0 | \bar{\psi}_\beta(y)\psi_\alpha(x) | 0 \rangle $$

$$ (\gamma^\mu\partial_\mu + m)\langle 0 | \psi_\alpha(x)\bar{\psi}_\beta(y) | 0 \rangle =0 $$

For complex scalar field $\phi$:

$$ i\Delta(x-y) = [\phi(x),\bar{\phi}(y) ] = \langle 0 | \phi(x)\bar{\phi}(y) | 0 \rangle - \langle 0 | \bar{\phi}(y)\phi(x) | 0 \rangle $$

$$ (\partial^2 + m^2)\langle 0 | \phi(x)\bar{\phi}(y) | 0 \rangle =0 $$

So there is some analogy between complex scalar and Dirac field. In Feynman propogator you need to add $\theta$-functions.

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  • $\begingroup$ Thanks for your answer! However, if I recall correctly, $\Delta(x-y)$ is not the propagator for complex scalar field $\phi$ because the propagator is denoted by $D(x-y)$ (correct me if I am wrong). But in this case $S_{\alpha\beta}$ is defined as the propagator which will break the analogy. $\endgroup$
    – alfred
    Commented Jan 6, 2020 at 21:09
  • $\begingroup$ How do you define propogator for complex scalar field? $\endgroup$
    – Nikita
    Commented Jan 6, 2020 at 22:06
  • $\begingroup$ I have not come across such a definition in my notes. I only know how to define the Feynman propagator for complex scalar field but obviously that is a different object $\endgroup$
    – alfred
    Commented Jan 8, 2020 at 11:59

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