# Differentiating Propagator, Green's function, Correlation function, etc

For the following quantities respectively, could someone write down the common definitions, their meaning, the field of study in which one would typically find these under their actual name, and most foremost the associated abuse of language as well as difference and correlation (no pun intended):

Maybe including side notes regarding the distinction between Covariance, Covariance function and Cross-Covariance, the pair correlation function for different observables, relations to the autocorrelation function, the $n$-point function, the Schwinger function, the relation to transition amplitudes, retardation and related adjectives for Greens functions and/or propagators, the Heat-Kernel and its seemingly privileged position, the spectral density, spectra and the resolvent.

Edit: I'd still like to hear about the "Correlation fuction interpretation" of the quantum field theoretical framework. Can transition amplitudes be seen as a sort of auto-correlation? Like... such that the QFT dynamics at hand just determine the structure of the temporal and spatial overlaps?

• The propagator, the two-point correlation function, and the two-point Green's function are all synonymous. They are used primarily in quantum mechanics, and quantum field theory. They represent the probability of preparing a one particle state at $\vec{x}$ and then finding the particle at $\vec{y}$. – kηives Feb 10 '12 at 14:08
• A more explicit relation between the Kernel and Green's function is given here. And as a comment to josh's answer, unicity (kernel or Green's function) of course depend on the equation at stake, but for the wave equation, it is both a condition on the boundary and at initial time. As tparker points out, the difference kernel/Green has to do with whether one considers the homogeneous or inhomogeneous equation. – Noix07 May 23 '19 at 15:02

## 3 Answers

The main distinction you want to make is between the Green function and the kernel. (I prefer the terminology "Green function" without the 's. Imagine a different name, say, Feynman. People would definitely say the Feynman function, not the Feynman's function. But I digress...)

Start with a differential operator, call it $L$. E.g., in the case of Laplace's equation, then $L$ is the Laplacian $L = \nabla^2$. Then, the Green function of $L$ is the solution of the inhomogenous differential equation $$L_x G(x, x^\prime) = \delta(x - x^\prime)\,.$$ We'll talk about its boundary conditions later on. The kernel is a solution of the homogeneous equation $$L_x K(x, x^\prime) = 0\,,$$ subject to a Dirichlet boundary condition $\lim_{x \rightarrow x^\prime}K(x,x^\prime) = \delta (x-x^\prime)$, or Neumann boundary condition $\lim_{x \rightarrow x^\prime} \partial K(x,x^\prime) = \delta(x-x^\prime)$.

So, how do we use them? The Green function solves linear differential equations with driving terms. $L_x u(x) = \rho(x)$ is solved by $$u(x) = \int G(x,x^\prime)\rho(x^\prime)dx^\prime\,.$$ Whichever boundary conditions we what to impose on the solution $u$ specify the boundary conditions we impose on $G$. For example, a retarded Green function propagates influence strictly forward in time, so that $G(x,x^\prime) = 0$ whenever $x^0 < x^{\prime\,0}$. (The 0 here denotes the time coordinate.) One would use this if the boundary condition on $u$ was that $u(x) = 0$ far in the past, before the source term $\rho$ "turns on."

The kernel solves boundary value problems. Say we're solving the equation $L_x u(x) = 0$ on a manifold $M$, and specify $u$ on the boundary $\partial M$ to be $v$. Then, $$u(x) = \int_{\partial M} K(x,x^\prime)v(x^\prime)dx^\prime\,.$$ In this case, we're using the kernel with Dirichlet boundary conditions.

For example, the heat kernel is the kernel of the heat equation, in which $$L = \frac{\partial}{\partial t} - \nabla_{R^d}^2\,.$$ We can see that $$K(x,t; x^\prime, t^\prime) = \frac{1}{[4\pi (t-t^\prime)]^{d/2}}\,e^{-|x-x^\prime|^2/4(t-t^\prime)},$$ solves $L_{x,t} K(x,t;x^\prime,t^\prime) = 0$ and moreover satisfies $$\lim_{t \rightarrow t^\prime} \, K(x,t;x^\prime,t^\prime) = \delta^{(d)}(x-x^\prime)\,.$$ (We must be careful to consider only $t > t^\prime$ and hence also take a directional limit.) Say you're given some shape $v(x)$ at time $0$ and want to "melt" is according to the heat equation. Then later on, this shape has become $$u(x,t) = \int_{R^d} K(x,t;x^\prime,0)v(x^\prime)d^dx^\prime\,.$$ So in this case, the boundary was the time-slice at $t^\prime = 0$.

Now for the rest of them. Propagator is sometimes used to mean Green function, sometimes used to mean kernel. The Klein-Gordon propagator is a Green function, because it satisfies $L_x D(x,x^\prime) = \delta(x-x^\prime)$ for $L_x = \partial_x^2 + m^2$. The boundary conditions specify the difference between the retarded, advanced and Feynman propagators. (See? Not Feynman's propagator) In the case of a Klein-Gordon field, the retarded propagator is defined as $$D_R(x,x^\prime) = \Theta(x^0 - x^{\prime\,0})\,\langle0| \varphi(x) \varphi(x^\prime) |0\rangle\,$$ where $\Theta(x) = 1$ for $x > 0$ and $= 0$ otherwise. The Wightman function is defined as $$W(x,x^\prime) = \langle0| \varphi(x) \varphi(x^\prime) |0\rangle\,,$$ i.e. without the time ordering constraint. But guess what? It solves $L_x W(x,x^\prime) = 0$. It's a kernel. The difference is that $\Theta$ out front, which becomes a Dirac $\delta$ upon taking one time derivative. If one uses the kernel with Neumann boundary conditions on a time-slice boundary, the relationship $$G_R(x,x^\prime) = \Theta(x^0 - x^{\prime\,0}) K(x,x^\prime)$$ is general.

In quantum mechanics, the evolution operator $$U(x,t; x^\prime, t^\prime) = \langle x | e^{-i (t-t^\prime) \hat{H}} | x^\prime \rangle$$ is a kernel. It solves the Schroedinger equation and equals $\delta(x - x^\prime)$ for $t = t^\prime$. People sometimes call it the propagator. It can also be written in path integral form.

Linear response and impulse response functions are Green functions.

These are all two-point correlation functions. "Two-point" because they're all functions of two points in space(time). In quantum field theory, statistical field theory, etc. one can also consider correlation functions with more field insertions/random variables. That's where the real work begins!

• Very nice answer. I wonder why when you introduce the Kernel, the $lim$ is taken to be w.r.t. the same arguments $x$ and $x'$ as the delta function, but later, you only use times. Also, in statistical mechanics, is the correlation function (which depend on the correlation length and which specify how macroscopic the effect are) a Green(s) function? I don't see any differential equations there. That's generally the problem I think, that I read the name Greens function, where there are no Differential equations and deltas around. Lastly, what about the functions characterizing susceptibilities? – Nikolaj-K Feb 11 '12 at 23:46
• In how many dimensions you take the limit (i.e. just time or time and space) is sort of a matter of terminology, due to the fact that the $\delta$ function is zero everywhere except one point. For the limit of the heat kernel, for example, all I'm getting at is that if the two time coordinates approach one another and the spatial points are not equal, the result vanishes. But if they are equal and then the time coordinates are made to approach, you get a quantity that behaves like a $d$-dimensional $\delta$ function. – josh Feb 12 '12 at 17:46
• To see how quantities like $W(x,x^\prime)=\langle0|\varphi(x)\varphi(x^\prime)|0\rangle$ satisfy the right differential equations and boundary conditions, read about Schwinger-Dyson Equations in QFT. And don't forget that when you canonically quantize a Klein-Gordon field, the canonical momentum $\pi = \partial_t\varphi$ and so $[\varphi(x,t),\partial_t\varphi(x^\prime,t)] = i\hbar\delta(x-x^\prime)$. This will matter in getting the right boundary conditions on the time-slice boundary. – josh Feb 12 '12 at 17:50
• @josh "If one uses the kernel with Neumann boundary conditions on a time-slice boundary"... Shouldn't this be Dirichlet boundary? – Jahan Claes Jun 29 '17 at 0:03
• Could you give an example of a distinct Green's function and kernel for a single linear differential operator? – tparker Nov 10 '18 at 3:21

It has been many years since you asked this question. I assume that over time you have compiled meaning definitions and distinctions for the other terms in your list. However, there are terms not defined by @josh's response (A response which I have relied on multiple times, thank you for posting it @josh). Personally, my background is in Lattice QCD, which is both a quantum field theory and statistical field theory. So I have also had to sit down and organize the meanings of all these terms. Here's what I have come up with during my PhD program.

-----------------The Short and Sweet----------------------------------

• The problem is that a lot of people are confused about this and so OFTEN times people just define their own lingo. If you assume the free field and linear response limit, then propagators and linear response functions are the same. When you include some non-linear interaction term these things become murky. To be facetious, everything is the same if you don't want to think too hard about it, hence why there's so much confusion.
• Regardless, the propagator, Green function, Wightman function, and linear response function can ALWAYS be understood as 2pt-correlation functions (shown below).
• By definition the wightman functions are just correlation functions. Nothing special other than they are the building blocks for the time-ordered (Feynman) propagator, here $$\Theta$$ is the heavy-side function. (Peskin, Zee, Zuber, Huang) $$\Delta^{(+)} = \left< \phi(x) \phi(y) \right>$$ $$\Delta^{(-)} = \left< \phi(y) \phi(x) \right>$$ $$G_F = \Theta(x^0-y^0) \Delta^{(+)} - \Theta(y^0-x^0) \Delta^{(-)} = \left< \mathcal{T} \phi(x) \phi(y) \right>$$
• The propagator, is the causal (a.k.a. retarded) Green function. The propagator is understood as the quantum transition amplitude (Le Bellac, Wiki). There is difference in convention, but it can be defined (Peskin v.s. Hong Lectures & Wiki respectively) $$G_R = \Theta(x^0-y^0) \left< [\phi(x), \phi(y)] \right> = \Theta(x^0-y^0) \left( \Delta^{(+)} - \Delta^{(-)} \right)$$ $$G_R = \Theta(x^0-y^0) \left< \phi(x), \phi(y) \right>$$
• The linear response function is equivalent to the Retarded green function a.k.a. Propagator (shown below).

------------------Linear Response Functions are 2pt correlation functions------------

I'll start with Kubo formulae. This derivation follows Tong "Kinetic Theory", Gale $$\&$$ Kapusta. Assume we have some system at equilibrium and apply a small perturbation to it. This looks like an equilibrium Hamiltonian $$H_0$$ and the perturbation $$V_I$$, $$H(t) = H_0 + V_I(t)$$ For this example, let it be that we have applied an electric field to a wire. Then the linear response function will end up being the conductivity. We write the interaction potential as some source term, $$\phi$$ (time dependent, external, c-valued, scalar field) multiplied by an an observable, $$J$$ like, $$V_I(t) = \phi(t) J(t)$$

Now consider the expectation value of the observable, $$J(t)$$ after perturbation $$V_I(t)$$ is applied. $$\left< J(t) \right> = \left< U^{-1}(t,t_0) J(t) U(t,t_0) \right>_{eq}$$ Where by the Schwinger-Dyson series (https://en.wikipedia.org/wiki/Dyson_series) we have that $$U^{-1}(t,t_0) = \mathcal{T}\exp(- i \int_{t_0}^t dt' V_I(t'))$$, which to linear order gives: $$\left< J(t) \right> \approx \left< \left(1 + i \int_{t_0}^t dt' V_I(t') \right) J(t) \left(1 - i \int_{t_0}^t dt' V_I(t') \right) \right>_{eq}$$ We can expand this expectation value by distribution property and dropping the non-linear term $$\propto \left( \int_{t_0}^t dt' V_I(t') \right)^2$$. We are left with, $$\left< J(t) \right> \approx \left< J(t) \right>_{eq} + \left< i \int_{t_0}^t dt' V_I(t') J(t) - i \int_{t_0}^t dt' J(t) V_I(t') \right>_{eq}$$ $$\left< J(t) \right> \approx \left< J(t) \right>_{eq} + i \left< \int_{t_0}^t dt' [ V_I(t'), J(t) ] \right>_{eq}$$ Insert definition of $$V_I$$ from above and subtract equilibrium value of observable $$\delta \left< J(t) \right> \approx i \int_{t_0}^t dt' \phi(t') \left< [ J(t'), J(t) ] \right>_{eq}$$ Let the source be turned on infinitely long ago ($$t_0 \rightarrow -\infty$$) and insert heavy-side function ($$t \rightarrow \infty$$). $$\delta \left< J(t) \right> \approx i \int_{-\infty}^{\infty} dt' \Theta(t-t') \phi(t') \left< [ J(t'), J(t) ] \right>_{eq}$$ We can group terms to define the linear response function, $$\chi$$. Where due to time translation invariance, $$i \Theta(t-t') \left< [ J(t'), J(t) ] \right>_{eq} = \chi (t',t) = \chi (t' - t)$$ Thus we arrive at our final expression. $$\delta \left< J(t) \right> \approx \int_{-\infty}^{\infty} dt' \phi(t') \chi (t'- t)$$ We see here that the linear response function is equivalent to a 2pt correlation function. It is also the retarded green function, a.k.a. propagator

We can also generalize, to when the observable in the expectation value and the observable in the observable in the Hamiltonian aren't the same observable. The observable being measured isn't the observable coupled to the source term. For example, $$\left< \mathcal{O}_i(t) \right> \approx \left< \mathcal{O}_i(t_0) \right>_0 + i \int dt' \phi(t') \left< [ \mathcal{O}_j(t'), \mathcal{O}_i(t_0) ] \right>$$ Then you are calculating a cross-correlation function.

----------------The Propagators are 2pt correlation function----------------

The path integral formulation of quantum mechanics and the generating functional will show us that the propagator is a 2pt-correlation function.

Starting from the path-integral formulation of Quantum mechanics transition amplitude (https://en.wikipedia.org/wiki/Path_integral_formulation#Path_integral_formula) we add a source term, $$\int d^4x J[x]\phi[x]$$, to our action $$S_E$$ as we see in (https://en.wikipedia.org/wiki/Partition_function_(quantum_field_theory)). To arrive at the the generating functional $$\mathcal{Z}[J] = \int D_{\phi} e^{-S_E[\phi] + i\int d^4x J[x]\phi[x])}$$ Exactly as in our linear response discussion, our source term is a field $$\phi$$, with an observable $$J$$.

Note that by Wick rotation $$S_E$$ is equivalent to the Hamiltonian http://www.math.ucr.edu/home/baez/classical/spring_garett.pdf) So that $$\mathcal{Z}$$ is a generalized partition function. Therefore, a generating functional is a generalized case of both the partition function and the quantum transition amplitude. As a partition function, then the generating functional is also characteristic function from probability theory whose argument is a set of stochastic variables (the quantum fields $$\phi[x]$$). The variables distribution is defined by Gibbs measure. This can be expressed as: $$\mathcal{Z}[J] = \int D\mu\{x\} e^{ i\int d^4x J[x]\phi[x]}= \mathbb{E}\left[ \exp[i\int d^4x J[x]\phi[x] ]\right]$$ $$D\mu\{x\} = D_{\phi} \frac{e^{-S_E[\phi]}}{\mathcal{Z}[0]}$$

A $$\#$$pt-correlation function (shortened to $$\#$$pt-function or correlation function) can be expressed via functional derivatives of the generating functional. $$\left< \prod_k \phi[x_k] \right> = (-i)^n\frac{1}{\mathcal{Z}[0]}\frac{\partial^n\mathcal{Z}}{\prod_k \partial J[x_k]}|_{J=0}$$ Then, by definition, the $$n$$-point function are the $$n^{th}$$ moments of the Gibbs measure.

We can see by definition that the transition amplitude (i.e. propagator) is the 2nd moment of the Gibbs measure. Thus, it is also a 2pt function

Furthermore, the Kahlen-Lehmann spectral representation of the 2pt-function, shows that a free particle's 2pt-function is equivalent to its propagator. Otherwise the 2pt-correlation function is the convolution of the free particle propagator with the particles spectral density (https://en.wikipedia.org/wiki/Spectral_density), which by the Wiener-Khinhcin theorem is equal to a auto-correlation function.

---------------A note about the connection between propagators and linear response functions------------

We could have short cut all these derivations and simply done a Volterra expansion (like a Taylor expansion but with convolutions instead of derivatives - https://en.wikipedia.org/wiki/Volterra_series#Continuous_time). To linear order the Volterra expansion is... you guessed it! $$\left< J(t) \right> \approx \left< J(t) \right>_{eq} + \int_{t_0}^t dt' \phi(t') \chi (t'- t)$$ Note that we have truncated our non-linear Volterra expansion at linear order so we choose to have a linear system for which Green function approaches could solve. To beat a dead horse: It says on the wiki page for green functions "If the operator is translation invariant then the Green's function can be taken to be a convolution operator. In this case, the Green's function is the same as the impulse response of linear time-invariant system theory."

Furthermore, the source term, $$\phi(t)$$ in my perturbation, $$V_I(t)$$, is equivalent to the "driving force" that @josh refers to as $$\rho$$. From this Volterra series vantage point you can see how our answers are connected.

If you want to consider non-linear interactions, then you can't truncate your Voltarre series at first order and your response kernels become non-linear. The whole system is no longer solvable with green function approaches.

---------------CITATIONS---------------------------

https://ocw.mit.edu/courses/physics/8-324-relativistic-quantum-field-theory-ii-fall-2010/lecture-notes/MIT8_324F10_Lecture7.pdf

David Tong "Kinetic Theory lecture notes"

Gale Kapusta "Finite Temperature F.T."

Le Bellac "Thermal F.T."

Peskin $$\&$$ Schroder "Intro to Q.F.T."

Huang "Operators to Path Integral"

Zee "Q.F.T. in a Nutshell"

Itzykson Zuber "Intro to Q.F.T."

josh's answer is good, but I think there are two points that require clarification.

First, his sentence defining the kernel makes no sense, because as written the dummy limit variable appears on both sides of the equation. In this context, we need to distinguish between a single "time-type" dependent variable $$t$$ and the other "space-type" dependent variables $${\bf x}$$, which are treated inequivalently. (I'm not using the terms "timelike" or "spacelike" to avoid confusion with special relativity, as this distinction can apply whether or not the PDE is Lorentz invariant.)

The correct statement is "The kernel is a solution of the homogeneous equation $$L_{{\bf x}, t}\, K({\bf x}, t; {\bf x}', t') = 0$$, subject to a Dirichlet boundary condition [in time] $$K({\bf x}, t; {\bf x}', t) = \delta^d({\bf x} - {\bf x}')$$ or a Neumann boundary condition $$\partial_t K({\bf x}, t; {\bf x}', t) = \delta^d({\bf x} - {\bf x}')$$, where $$d$$ is the number of spatial dimensions."

Also, I think it's misleading to bold the word "linear" only when discussing the Green's function, because that seems to imply that the linearity is important for distinguishing the Green's function and the kernel. In fact, the kernel is also used to solve linear differential equations. I would say the primary difference between their use cases is that the Green's function is used to solve inhomogeneous differential equations, and the kernel is used to solve homogeneous boundary value problems. (For inhomogeneous boundary value problems, the idea of the kernel is effectively subsumed into the process of choosing the Green's function to get the boundary conditions right.)