Lecture Notes for MATH4017 Quantum Field Theory — Correlation functions and Feynman propagator

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3.4 Correlation functions and Feynman propagator

Using the Hilbert space representation from the previous section, we can now determine and study the correlation functions of the Heisenberg picture field operators $\Phi (x)$. One defines the $n$-point correlation function, which is sometimes also called the $n$-point Wightman function, to be the vacuum expectation value

\begin{equation} W_n(x_1,\dots ,x_n) := \expect {0}{\Phi (x_1)\cdots \Phi (x_n)}{0} \end{equation}

of the product of $n\geq 1$ quantum fields at different points $x_1,\dots ,x_n\in \bbR ^d$ of Minkowski spacetime. Using the presentation of $\Phi (x)$ in terms of creation and annihilation operators from (3.20), the definition of the vacuum state (3.39) and the commutation relations (3.10), one can compute these correlation functions explicitly. For low $n$, one finds

\begin{flalign} W_1(x) &= \expect {0}{\Phi (x)}{0}=0\quad ,\\ W_2(x,y) &=\expect {0}{\Phi (x)\Phi (y)}{0}=\int _{\bbR ^{d-1}} \frac {1}{2\,\omega _{\mathbf {k}}}~e^{\ii \,k\,(x-y)}\, \frac {\dd \mathbf {k}}{(2\pi )^{d-1}}\quad ,\label {eqn:KGWightman2pt} \end{flalign} where we recognize again the Poincaré invariant measure. In particular, this implies that the $1$- and $2$-point functions are invariant under proper and orthochronous Poincaré transformations, i.e. $W^\prime _1(x^\prime ) = W_{1}(x)$ and $W^\prime _2(x^\prime ,y^\prime )= W_2(x,y)$. (The same will hold true for the higher $n$-point functions.) As an exercise, you can try to compute the $3$- and $4$-point correlation functions. In general, it turns out that the $n$-point function $W_n(x_1,\dots ,x_n)$ of a free QFT is determined completely by the $2$-point function $W_2(x,y)$, but we are not going to prove this now. Using again the trick in (3.26), we can also rewrite the $2$-point function as an integral over $k\in \bbR ^d$

\begin{flalign} W_{2}(x,y) = \int _{\bbR ^d} \delta (k^2+m^2)~\Theta (k^0)\,e^{\ii \,k\,(x-y)}\,\frac {\dd k}{(2\pi )^{d-1}}\quad , \end{flalign} from which Poincaré invariance is even better visible. Note that $W_2(x,y)$ satisfies the Klein-Gordon equation in both entries, i.e.

\begin{flalign} \label {eqn:2pointKGeqn} (-\partial _x^2 +m^2)W_{2}(x,y) = 0 = (-\partial _y^2 +m^2)W_{2}(x,y) \quad , \end{flalign} where the subscript indicates the coordinates along which one takes partial derivatives, and that the commutator function (3.22) is the following difference

\begin{flalign} \Delta (x-y) = W_2(x,y) - W_2(y,x)\quad , \end{flalign} which is of course clear from the definition $W_2(x,y) =\expect {0}{\Phi (x)\Phi (y)}{0}$ and the normalization condition $\braket {0}{0}$ of the vacuum state. In contrast to the causality of the commutator function (see (3.27)), the $2$-point function $W_2(x,y)$ does not vanish for spacelike separated point $x,y\in \bbR ^{d}$. This means that the vacuum state $\ket {0}\in \HH $ exhibits nonlocal features, which one may interpret as a quantum entanglement over spacelike separated distances. It is a common feature of QFT that the field operators behave local and causal, but the vacuum state is a nonlocal concept.

As we shall see in the next chapter, it turns out that in QFT a different type of $n$-point functions is more important, namely the so-called time-ordered $n$-point functions

\begin{equation} \label {eqn:TOnpointKG} G_n(x_1,\dots ,x_n) := \expect {0}{\TO \big (\Phi (x_1)\cdots \Phi (x_n)\big )}{0} \quad . \end{equation}

The symbol $\TO \big (\cdots \big )$ denotes the time-ordered product of field operators, which means that field operators that are located at later times are placed to the left of field operators that are located at earlier times. For example, the time-ordered product of two field operators is given by

\begin{flalign} \nn \TO \big (\Phi (x)\Phi (y)\big ) &= \begin{cases} \Phi (x)\,\Phi (y)&,~\text {if }x^0\geq y^0\quad ,\\ \Phi (y)\,\Phi (x)&,~\text {if }y^0\geq x^0\quad , \end {cases}\\[5pt] &=\Theta (x^0-y^0)\,\Phi (x)\,\Phi (y) +\Theta (y^0-x^0)\,\Phi (y)\,\Phi (x) \quad . \label {eqn:TO2tmp} \end{flalign} Note that the two cases agree for $x^0=y^0$, which is a consequence of the equal-time commutation relations (3.1). A general formula for $\TO \big (\Phi (x_1)\cdots \Phi (x_n)\big )$ can be obtained by noting that for any tuple $x_1,\dots ,x_n\in \bbR ^{d}$ of $n$-points in Minkowski spacetime there exists a permutation $\sigma \in S_n$, such that the time coordinates $t_a = x_a^0$ are ordered according to $x^0_{\sigma (1)}\geq x^0_{\sigma (2)}\geq \cdots \geq x^0_{\sigma (n)}$. Then the time-ordered product is given by

\begin{flalign} \label {eqn:TOproductn} \TO \big (\Phi (x_1)\cdots \Phi (x_n)\big ) = \Phi (x_{\sigma (1)})\cdots \Phi (x_{\sigma (n)})\quad . \end{flalign}

Remark 3.4. It is not immediately clear that the time-ordered product is invariant under proper and orthochronous Poincaré transformations because its definition uses a fixed choice of time coordinate $x=(t,\mathbf {x})$. A nice property of such Poincaré transformations, which is relatively easy to prove but not shown here, is that they can invert the time-ordering of two points $x,y\in \bbR ^d$, i.e. $x^0> y^0$ transforms to $x^{\prime 0}< y^{\prime 0}$, only if these points are spacelike separated, i.e. $(x-y)^2>0$. Due to causality (3.27), we know that any two field operators located at spacelike separated points commute with each other, hence the time-ordered products computed in the two different coordinate systems $x$ and $x^\prime $ coincide. This implies that $\TO $ is indeed invariant under proper and orthochronous Poincaré transformations.

Let us compute the first two examples of the time-ordered $n$-point functions. For the time-ordered $1$-point function, we find again zero

\begin{flalign} \expect {0}{\TO \big (\Phi (x)\big )}{0} = \expect {0}{\Phi (x)}{0} = W_1(x)= 0\quad . \end{flalign} The time-ordered $2$-point function, which is called the Feynman propagator $\Delta _F$, is more interesting and, using (3.54), it is given by

\begin{flalign} \nn \Delta _F(x-y):&=\expect {0}{\TO \big (\Phi (x)\,\Phi (y)\big )}{0} = \begin{cases} W_2(x,y) &,~\text {if }x^0 \geq y^0\quad ,\\ W_2(y,x) &,~\text {if }y^0 \geq x^0\quad , \end {cases}\\[5pt] &= \Theta (x^0-y^0)\,W_2(x,y) +\Theta (y^0-x^0)\,W_2(y,x)\quad , \label {eqn:FeynmanpropagatorKGcases} \end{flalign} where $W_2$ is the $2$-point correlation function. Note that in both cases the two arguments of $W_2$ are ordered such that the later spacetime point is to the left of the earlier one, which is of course precisely what time-ordering does. We will see later that all other time-ordered $n$-point functions can be determined from $\Delta _F(x-y)$ via Wick’s theorem, see Theorem 4.1 and Corollary 4.3. The formula above for the time-ordered $2$-point function is not very useful for calculations because it involves case distinctions. Using the concept of contour integration, there is a clever way to rewrite the Feynman propagator $\Delta _F(x-y)$ in terms of an integral formula over the relativistic Fourier momenta $k\in \bbR ^d$. Let me first state the result and then show you how to prove it.

Proposition 3.5. The time-ordered $2$-point function/Feynman propagator can be written as the limit $\epsilon \to 0$ of the Fourier integral

$\seteqnumber{0}{3.}{56}$
\begin{equation} \label {eqn:FeynmanpropagatorKGintegral} \Delta _F(x-y) = \expect {0}{\TO \big (\Phi (x)\,\Phi (y)\big )}{0} = \lim _{\epsilon \to 0} \int _{\bbR ^d} \frac {-\ii }{k^2 + m^2 -\ii \,\epsilon }~e^{\ii \,k\,(x-y)}\,\frac {\dd k}{(2\pi )^d} \quad , \end{equation}

where $\epsilon >0$ is a positive parameter whose role will become clear in the proof below.

Proof. Throughout the proof, we assume that $\epsilon >0$ is positive and sufficiently small so that terms of order $\epsilon ^2$ can be neglected. To compare the integral formula (3.57) with (3.56b), we carry out the integral over the $0$-component $k^0$ of the relativistic Fourier momentum. Splitting into time and space components, the denominator reads as
$\seteqnumber{0}{3.}{57}$
\begin{flalign} k^2 + m^2 -\ii \,\epsilon = - (k^0)^2 + \omega _{\mathbf {k}}^2 - \ii \,\epsilon = -\big (k^0 - \omega _{\mathbf {k}}+\ii \,\epsilon ^\prime \big ) ~\big (k^0 + \omega _{\mathbf {k}}-\ii \,\epsilon ^\prime \big ) \quad , \end{flalign} where $\epsilon ^\prime = \epsilon /(2\omega _{\mathbf {k}})>0$ and we recall that terms of order $\epsilon ^2$ are neglected. This means that the integrand in (3.57) has two first order poles in $k^0$ that are located as follows in the complex plane:

$(3.59) \{begin}{flalign} \label {eqn:KGFeynmanpoles} \parbox {7cm}{ \begin{tikzpicture}[scale=0.75] \draw [thick,->] (-4,0) -- (4,0); \draw (4.2,-0.4) node{$\mathrm {Re}(k^0)$}; \draw [thick,->] (0,-2) -- (0,2); \draw (-0.4,2.4) node{$\mathrm {Im}(k^0)$}; \draw [thick] (2,-0.15) -- (2,0.15); \draw (2,-0.4) node{$\omega _{\mathbf {k}}$}; \draw [thick] (-2,-0.15) -- (-2,0.15); \draw (-2,-0.4) node{$-\omega _{\mathbf {k}}$}; \draw [red,fill=red] (2,-1) circle (.5ex); \draw [red,fill=red] (-2,1) circle (.5ex); \end {tikzpicture} } \{end}{flalign}$

The idea is now to perform a contour integration in order to compute the $k^0$ integral. To figure out which contour to choose, note that the integrand in (3.57) also contains a complex exponential $e^{\ii \,k\,(x-y)} = e^{-\ii \,k^0\, (x^0-y^0) + \ii \,\mathbf {k}\,(\mathbf {x}-\mathbf {y})}$. Since $k^0 = \mathrm {Re}(k^0) + \ii \,\mathrm {Im}(k^0)$ was extended to a complex number, this yields a factor of the form
$\seteqnumber{0}{3.}{59}$
\begin{flalign} e^{-\ii \,k^0\, (x^0-y^0)} = e^{-\ii \,\,\mathrm {Re}(k^0)\,(x^0-y^0) + \mathrm {Im}(k^0)\,(x^0-y^0)}\quad . \end{flalign} This means that the integrand falls off in the imaginary direction $\mathrm {Im}(k^0)$ for the following cases: 1.) If $x^0 \geq y^0$, it falls off for $\mathrm {Im}(k^0)\to -\infty $. 2.) If $y^0 \geq x^0$, it falls off for $\mathrm {Im}(k^0)\to +\infty $. This justifies choosing (the infinite radius limit of) the following contours

$(3.61a) \{begin}{flalign} \parbox {7cm}{ \begin{tikzpicture}[scale=0.75] \draw [thick,->] (-4,0) -- (4,0); \draw (4.2,-0.4) node{$\mathrm {Re}(k^0)$}; \draw [thick,->] (0,-2) -- (0,2); \draw (-0.4,2.4) node{$\mathrm {Im}(k^0)$}; \draw [thick] (2,-0.15) -- (2,0.15); \draw (2,-0.4) node{$\omega _{\mathbf {k}}$}; \draw [thick] (-2,-0.15) -- (-2,0.15); \draw (-2,-0.4) node{$-\omega _{\mathbf {k}}$}; \draw [red,fill=red] (2,-1) circle (.5ex); \draw [red,fill=red] (-2,1) circle (.5ex); \draw [very thick, blue] (-3,0) -- (3,0); \draw [very thick, blue,->] (-3,0) -- (3,0) arc(0:-90:3); \draw [very thick, blue] (-3,0) -- (3,0) arc(0:-180:3) node[midway,below]{$C_-$}; \node [draw] at (5,2) {for $x^0\geq y^0$}; \end {tikzpicture} } \{end}{flalign}$

and

$(3.61b) \{begin}{flalign} \parbox {7cm}{ \begin{tikzpicture}[scale=0.75] \draw [thick,->] (-4,0) -- (4,0); \draw (4.2,-0.4) node{$\mathrm {Re}(k^0)$}; \draw [thick,->] (0,-2) -- (0,2); \draw (-0.4,2.4) node{$\mathrm {Im}(k^0)$}; \draw [thick] (2,-0.15) -- (2,0.15); \draw (2,-0.4) node{$\omega _{\mathbf {k}}$}; \draw [thick] (-2,-0.15) -- (-2,0.15); \draw (-2,-0.4) node{$-\omega _{\mathbf {k}}$}; \draw [red,fill=red] (2,-1) circle (.5ex); \draw [red,fill=red] (-2,1) circle (.5ex); \draw [very thick, blue] (-3,0) -- (3,0); \draw [very thick, blue,->] (-3,0) -- (3,0) arc(0:90:3); \draw [very thick, blue] (-3,0) -- (3,0) arc(0:180:3) node[midway,above]{$C_+$};; \node [draw] at (5,2) {for $y^0\geq x^0$}; \end {tikzpicture} } \{end}{flalign}$

Using Cauchy’s residue theorem, one then finds for the case $x^0\geq y^0$ that
$\seteqnumber{0}{3.}{61}$
\begin{flalign} \nn \int _{\bbR } \frac {-\ii }{k^2 + m^2 -\ii \,\epsilon }~e^{-\ii \,k^0\,(x^0-y^0)}\,\frac {\dd k^0}{2\pi } &= \oint _{C_-} \frac {\ii }{\big (k^0 - \omega _{\mathbf {k}}+\ii \,\epsilon ^\prime \big ) \,\big (k^0 + \omega _{\mathbf {k}}-\ii \,\epsilon ^\prime \big )}~e^{-\ii \,k^0\,(x^0-y^0)}\,\frac {\dd k^0}{2\pi }\\ \nn &= \frac {1}{2 (\omega _{\mathbf {k}}-\ii \,\epsilon ^\prime )}~e^{-\ii \,(\omega _\mathbf {k}-\ii \,\epsilon ^\prime )\,(x^0-y^0)}\\ &\stackrel {\epsilon \to 0}{\longrightarrow } \frac {1}{2 \omega _{\mathbf {k}}}~e^{-\ii \,\omega _\mathbf {k}\,(x^0-y^0)} \end{flalign} and similarly for the case $y^0\geq x^0$ that
$\seteqnumber{0}{3.}{62}$
\begin{flalign} \nn \int _{\bbR } \frac {-\ii }{k^2 + m^2 -\ii \,\epsilon }~e^{-\ii \,k^0\,(x^0-y^0)}\,\frac {\dd k^0}{2\pi } &= \oint _{C_+} \frac {\ii }{\big (k^0 - \omega _{\mathbf {k}}+\ii \,\epsilon ^\prime \big ) \,\big (k^0 + \omega _{\mathbf {k}}-\ii \,\epsilon ^\prime \big )}~e^{-\ii \,k^0\,(x^0-y^0)}\,\frac {\dd k^0}{2\pi }\\ \nn &= \frac {1}{2 (\omega _{\mathbf {k}}-\ii \,\epsilon ^\prime )}~e^{+\ii \,(\omega _\mathbf {k}-\ii \,\epsilon ^\prime )\,(x^0-y^0)}\\ &\stackrel {\epsilon \to 0}{\longrightarrow } \frac {1}{2 \omega _{\mathbf {k}}}~e^{+\ii \,\omega _\mathbf {k}\,(x^0-y^0)}= \frac {1}{2 \omega _{\mathbf {k}}}~e^{-\ii \,\omega _\mathbf {k}\,(y^0-x^0)} \quad . \end{flalign} Inserting these back into (3.57), and recalling the formula (3.49b) for the $2$-point correlation function $W_2$, then shows that (3.57) agrees with (3.56b). □

To conclude this section, let us note that, in contrast to the $2$-point correlation function (3.51), the Feynman propagator (3.57) is not a solution of the Klein-Gordon equation but rather a Green’s function, i.e.

\begin{flalign} (-\partial ^2_x +m^2)\Delta _F(x-y) = -\ii \,\delta (x-y) \quad . \end{flalign} What distinguishes the Feynman propagator from other choices of Green’s functions, e.g. the retarded/advanced one that is obtained by moving instead of (3.59) both poles to the lower/upper complex half-plane, is the following feature: As visualized by the contours (3.61) in the proof above, the Feynman propagator $\Delta _F(x-y)$ propagates the positive frequencies $\omega _\mathbf {k}$ forward in time $x^0\geq y^0$ and the negative frequencies $-\omega _{\mathbf {k}}$ backward in time $y^0\geq x^0$.

3.4 Correlation functions and Feynman propagator

Further reading