1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

From the well-known definition of Feynman propagator that you have put above, we can re-write it as \begin{equation} 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} \end{equation} 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 $$\boxed{\theta(x^0-y^0)+\theta(y^0-x^0)=1} \tag 8$$

Back to (6), with these tools, we then have \begin{eqnarray} T\phi(x)\phi(y) &=& \phi_+(x) \phi_+(y) + \phi_-(x) \phi_-(y)\\ &&+\theta(x^0-y^0)\big[ \phi_+(x)\phi_-(y)+\phi_-(x)\phi_+(y) \big]\\ &&+\theta(y^0-x^0)\big[ \phi_+(y)\phi_-(x)+\phi_-(y)\phi_+(x) \big]\;,\\ &=& \phi_+(x) \phi_+(y) + \phi_-(x) \phi_-(y)\\ &&+\theta(x^0-y^0)\big[ \boxed{\boxed{ \phi_+(x)\phi_-(y)}}+\phi_-(x)\phi_+(y) -\phi_+(y) \phi_-(x)+\boxed{\phi_+(y) \phi_-(x)}\big]\\ &&+\theta(y^0-x^0)\big[ \boxed{ \phi_+(y)\phi_-(x)}+\phi_-(y)\phi_+(x)-\phi_+(x) \phi_-(y)+\boxed{\boxed{ \phi_+(x) \phi_-(y)}}\big]\;,\\ &=&\phi_+(x) \phi_+(y) + \phi_-(x) \phi_-(y)+ \boxed{\boxed{ \phi_+(x)\phi_-(y)}}+ \boxed{\phi_+(y)\phi_-(x)}\\ &&+\theta(x^0-y^0)[\phi_-(x),\phi_+(y) ]\\ &&+\theta(y^0-x^0)[\phi_-(y),\phi_+(x) ]\;,\\ &=& :\phi(x)\phi(y):+ \overline{\phi(x)\phi(y)}\;. \end{eqnarray} As we want.

$\endgroup$

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.