# Wick contraction and propagator confusion

I am having trouble understanding the how the Wick contraction leads to the Feynman propagator for scalar fields. The Feynman propagator can be written as

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

and the Wick contraction of the two fields can be written as

$$\overline{\phi(x) \phi(y)}= \begin{cases} [\phi(x), \phi(y)], x^0>y^0 \\ [\phi(y), \phi(x)], y^0>x^0 \end{cases}.$$

In Peskin and Schörder's book and in another book I have they say that this contraction is equal to the Feynman propagator, how can that be? How can a commutator of two field operators be a propagator? Where did the expectation value go? In the book they show

$$T(\phi(x) \phi(y)) = \, :\phi(x) \phi(y): + \, \overline{\phi(x) \phi(y)}$$

which if we plug into the vacuum braket should result in the propagator, but again the book does not seem to do that.

Note: I used overline since I didn't figure out how to make the square bracket above the expression that seems to be the standard notation for the contraction.

• Mar 10, 2018 at 15:37

From the well-known definition of Feynman propagator that you have put above, we can re-write it as $$D_F (x-y)=\begin{cases} \langle 0|\phi(x), \phi(y)| 0 \rangle,\; x^0>y^0 \\ \langle 0|\phi(y), \phi(x) | 0\rangle,\; y^0>x^0 \end{cases}.\tag{1}$$ Let's decomposed the scalar field to a creation and anihilation parts, respectively $$\phi(x)=\phi_+(x)+\phi_-(x)=\frac 1 {\sqrt{(2\pi)^3}}\int \frac{d^3k}{\sqrt{2 \omega_k}}\bigg( \hat a^{\dagger}_{\vec k} e^{-ik\cdot x} +\hat a_{\vec k} e^{ik\cdot x} \bigg)$$ (where $k \cdot x = -k_0 x^0+ \vec k \cdot \vec x$)
Therefore we have $$\phi(x)\phi(y)=\phi_+(x)\phi_+(y)+\phi_+(x)\phi_-(y)+\phi_-(x)\phi_+(y)+\phi_-(x)\phi_-(y)$$ and then $$\boxed{\langle 0|\phi(x) \phi(y)| 0 \rangle=\langle 0|\phi_-(x) \phi_+(y)| 0 \rangle}\tag{2}$$ Since \begin{eqnarray} \langle 0| \phi_+ (x) &\sim& \langle 0| \hat a^\dagger_{\vec k}=0\;,\\ \phi_-(y)|0 \rangle &\sim& \hat a_{\vec k}| 0 \rangle =0 \;. \end{eqnarray} Similarly, we have $$\boxed{\langle 0|\phi(y) \phi(x)| 0 \rangle=\langle 0|\phi_-(y), \phi_+(x)| 0 \rangle} \tag{3}$$
The important trick is an observation that \begin{eqnarray} \langle 0|\phi_-(x) \phi_+(y)| 0 \rangle &=& [\phi_-(x), \phi_+(y)] \;\;\; \big(\equiv i \Delta^{(+)}(x-y)\big)\;,\\ \langle 0|\phi_-(y) \phi_+(x)| 0 \rangle &=& [\phi_-(y), \phi_+(x)] \;\;\; \big(\equiv i \Delta^{(+)}(y-x)\big)\;\tag{4} \end{eqnarray} These follow from the properties of creation and anihilation operators $$\boxed{{ \langle 0| \hat a_{\vec k}\hat a^\dagger_{\vec k'} |0\rangle=\langle \vec k| \vec k' \rangle =\delta^{(3)}(\vec k- \vec k')=[\hat a_{\vec k},\hat a^\dagger_{\vec k'}] }}$$
So if as you said $D_F (x-y)=\overline{\phi(x)\phi(y)}$, from (1),(2),(3), and (4) the correct definition of the Wick contraction would be $$\overline{\phi(x) \phi(y)}= \begin{cases} [\phi_-(x), \phi_+(y)], x^0>y^0 \\ [\phi_-(y), \phi_+(x)], y^0>x^0 \end{cases}.\tag{5}$$ or $\overline{\phi(x) \phi(y)}=\theta(x^0-y^0)i\Delta^{(+)}(x-y)+\theta(y^0-x^0)i\Delta^{(+)}(y-x)\;,$ where the step function is defined by $$\theta(x^0-y^0)=\begin{cases} 1,\; x^0>y^0 \\ 0,\; y^0>x^0. \end{cases}$$
By these details we can proof your last equation as the following \begin{eqnarray} T \phi(x)\phi(y) &=& \theta(x^0-y^0)\phi(x)\phi(y) +\theta(y^0-x^0)\phi(y)\phi(x)\;,\\ &=&\theta(x^0-y^0)\big[ \phi_+(x)\phi_+(y)+\phi_+(x)\phi_-(y)+\phi_-(x)\phi_+(y)+\phi_-(x)\phi_-(y) \big]\\ &&+\theta(y^0-x^0)\big[ \phi_+(y)\phi_+(x)+\phi_+(y)\phi_-(x)+\phi_-(y)\phi_+(x)+\phi_-(y)\phi_-(x) \big]\;.\tag 6 \end{eqnarray} Then we will make used of $$[\phi_+(x),\phi_+(y)]=[\phi_-(x),\phi_-(y)]=0$$ or $$\phi_+(x)\phi_+(y)=\phi_+(y)\phi_+(x)$$ $$\phi_-(x)\phi_-(y)=\phi_-(y)\phi_-(x)\tag 7$$ in addition $$\boxed{\theta(x^0-y^0)+\theta(y^0-x^0)=1} \tag 8$$