I'm fairly new to the subject of quantum field theory (QFT), and I'm having trouble intuitively grasping what a n-point correlation function physically describes. For example, consider the 2-point correlation function between a (real) scalar field $\hat{\phi}(x)$ and itself at two different space-time points $x$ and $y$, i.e. $$\langle\hat{\phi}(x)\hat{\phi}(y)\rangle :=\langle 0\rvert T\lbrace\hat{\phi}(x)\hat{\phi}(y)\rbrace\lvert 0\rangle\tag{1}$$ where $T$ time-orders the fields.

Does this quantify the correlation between the values of the field at $x=(t,\mathbf{x})$ and $y=(t',\mathbf{y})$ (i.e. how much the values of the field at different space-time points covary, in the sense that, if the field $\hat{\phi}$ is excited at time $t$ at some spatial point $\mathbf{x}$, then this will influence the "behaviour" of the field at later time $t'$ at some spatial point $\mathbf{y}$)? Is this why it is referred to as a correlation function?

Furthermore, does one interpret $(1)$ as physically describing the amplitude of propagation of a $\phi$-particle from $x$ to $y$ (in the sense that a correlation of excitations of the field at two points $x$ and $y$ can be interpreted as a "ripple" in the field propagating from $x$ to $y$)?


2 Answers 2


Yes, in scalar field theory, $\langle 0 | T\{\phi(y) \phi(x)\} | 0 \rangle$ is the amplitude for a particle to propagate from $x$ to $y$. There are caveats to this, because not all QFTs admit particle interpretations, but for massive scalar fields with at most moderately strong interactions, it's correct. Applying the operator $\phi({\bf x},t)$ to the vacuum $|0\rangle$ puts the QFT into the state $|\delta_{\bf x},t \rangle$, where there's a single particle whose wave function at time $t$ is the delta-function supported at ${\bf x}$. If $x$ comes later than $y$, the number $\langle 0 | \phi({\bf x},t)\phi({\bf y},t') | 0 \rangle$ is just the inner product of $| \delta_{\bf x},t \rangle$ with $| \delta_{\bf y},t' \rangle$.

However, the function $f(x,y) = \langle 0 | T\{\phi(y) \phi(x)\} | 0 \rangle$ is not actually a correlation function in the standard statistical sense. It can't be; it's not even real-valued. However, it is a close cousin of an honest-to-goodness correlation function.

If make the substitution $t=-i\tau$, you'll turn the action $$iS = i\int dtd{\bf x} \{\phi(x)\Box\phi(x) - V(\phi(x))\}$$ of scalar field theory on $\mathbb{R}^{d,1}$ into an energy function $$-E(\phi) = -\int d\tau d{\bf x} \{\phi(x)\Delta\phi(x) + V(\phi(x))\}$$ which is defined on scalar fields living on $\mathbb{R}^{d+1}$. Likewise, the oscillating Feynman integral $\int \mathcal{D}\phi e^{iS(\phi)}$ becomes a Gibbs measure $\int \mathcal{D}\phi e^{-E(\phi)}$.

The Gibbs measure is a probability measure on the set of classical scalar fields on $\mathbb{R}^{d+1}$. It has correlation functions $g(({\bf x}, \tau),({\bf y},\tau')) = E[\phi({\bf x}, \tau)\phi({\bf y},\tau')]$. These correlation functions have the property that they may be analytically continued to complex values of $\tau$ having the form $\tau = e^{i\theta}t$ with $\theta \in [0,\pi/2]$. If we take $\tau$ as far as we can, setting it equal to $i t$, we obtain the Minkowski-signature "correlation functions" $f(x,y) = g(({\bf x},it),({\bf y},it'))$.

So $f$ isn't really a correlation function, but it's the boundary value of the analytic continuation of a correlation function. But that takes a long time to say, so the terminology gets abused.

  • $\begingroup$ By "amplitude", is it meant to be a transition amplitude? Also, what is the associated correlation function describing? Is it simply a measure of how the field values at separate points covary? $\endgroup$
    – user35305
    Aug 11, 2016 at 19:39
  • $\begingroup$ 1) Yes. It's the inner product of two states, so an amplitude or a transition amplitude, or whatever you wish to call it. 2) Yes. The associated correlation function is doing exactly what you say: it's measuring how the values of the stochastic fields in the associated statistical field theory covary as we move around the points where we sample the field values. $\endgroup$
    – user1504
    Aug 11, 2016 at 19:58
  • $\begingroup$ So the use of the phrase "correlation function" in QFT is a slight abuse of terminology, but is consistent in the sense that you describe in your answer?! Does this analogy hold for n-point correlation functions? $\endgroup$
    – user35305
    Aug 11, 2016 at 20:12
  • $\begingroup$ Yes, it's also true for n-point functions (of composite operators, even). It's not guaranteed that a Euclidean statistical theory exists for every theory, though. The existence of spinors depends on spacetime signature, for example, as does the gauge invariance of the Chern-Simons action. $\endgroup$
    – user1504
    Aug 11, 2016 at 20:59
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    $\begingroup$ Please back up the statement that the propagator is the amplitude to propagate from x to y. This quantity does not transform like a spacial probability density. From what I have read this is only a heuristic understanding, not actually true if taken literally. $\endgroup$ Oct 30, 2019 at 9:49

No, $⟨0|T{ϕ(y)ϕ(x)}|0⟩$ is NOT the probability amplitude for a particle to propagate from $x$ to $y$, even for a free scalar field. It seems to be a common false belief that it is. There is one obvious reason and one deep reason why it cannot be.

The obvious reason is that the square of this value, which is supposed to be the probability density, does not integrate to 1 (see wikipedia):

$⟨0|T{ϕ(y)ϕ(x)}|0⟩=$ $$G_F(x,y) =\lim_{\epsilon\to 0}\frac{1}{(2\pi)^4}\int d^4p\frac{e^{-ip(x-y)}}{p^2-m^2+i\epsilon} =\begin{cases} -\frac{1}{4\pi}\delta(s)+\frac{m}{8\pi\sqrt{s}}H_1^{(1)}(m\sqrt{s}) & s\ge 0\\ -\frac{im}{4\pi^2\sqrt{-s}}K_1(m\sqrt{-s}) & s<0 \end{cases}$$ where $s:=(x^0-y^0)^2-(\vec{x}-\vec{y})^2$, hence $\int dy_1 dy_2 dy_3\,|⟨0|T{ϕ(y)ϕ(x)}|0⟩|^2$ is infinite (if makes sense at all). Interpretation as a 'relative probability amplitude' does not fix that because the most of 'probability to propagate from $x$ to $y$' would be anyway concentrated on the cone $s=0$ due to $\delta(s)$-term.

A more deep reason is that 'probability' does not make any sense until a probability space, i.e. the set of possible outcomes of an experiment, is introduced. The least we have to assume is that these outcomes are mutually exclusive. In nonrelativistic quantum mechanics this set is an orthonormal basis in a Hilbert space (or a bit more general object such as the set of delta-functions supported at different spatial points $\mathbf{y}$). Then the inner product with the basis elements is interpreted as the transition probability density. But for the free quantum field, delta-functions supported at different spatial points $\mathbf{y}$ and the same time $t$ are not orthogonal anymore (in no sense): $⟨0|T{ϕ(\mathbf{y},t)ϕ(\mathbf{x},t)}|0⟩\ne 0$ by the above expression. In other words, one can find the particle at different points simultaneously, making just impossible to speak of the probability density to find a particle at a point. Hence, the inner product of $|δ_x,t⟩$ with $|δ_y,t'⟩$ cannot be interpreted as probability density (keeping aside the question what does this notation at all mean in the Fock space), quite opposite to the answer by user1504.

Finally, one cannot introduce any experiment to measure the quantity under name 'the probability for a free particle to propagate from $x$ to $y$', because the precision of measurement of the particle coordinates cannot be better than the particle Compton wavelength.

Beware: having absolutely no understanding of the free quantum field, writing just to get a chance to be corrected by those who have. In particular, would be grateful for explanation what is meaning of $⟨0|T{ϕ(y)ϕ(x)}|0⟩$, if it is not a probability amplitude. Being stuck with this question for quite a long time.

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    $\begingroup$ @Thomas Fritsch Thanks for typesetting the equations! $\endgroup$ Oct 26, 2019 at 5:11
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    $\begingroup$ Please tell me the meaning of correlation function, as I am confused between the matrix element of propagator and correlation function. How are they different? Do they both represent transition amplitude? $\endgroup$
    – sawan kt
    May 16, 2021 at 12:57
  • $\begingroup$ @Shine kk You could try to have a look at the following discussion: physics.stackexchange.com/questions/20797/… $\endgroup$ May 17, 2021 at 13:09

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