A Physical interpretation of correlator

Frank Castle

Consider the 2-point correlator of a real scalar field $\hat{\phi}(t,\mathbf{x})$, $$\langle\hat{\phi}(t,\mathbf{x})\hat{\phi}(t,\mathbf{y})\rangle$$ How does one interpret this quantity physically? Is it quantifying the probability amplitude for a particle to be created at space-time point $(t,\mathbf{x})$ and then propagate to, and be annihilated at space-time point $(t,\mathbf{y})$?! Or does it quantify whether the field amplitudes at two different points are related or not (i.e. whether the values of the field operator at the two different space-time points are correlated or not)? Or is there another explanation?

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A. Neumaier

Science Advisor
It has no simple probabilistic interpretation. it is a vacuum expectation value of an operator.

The physical meaning is, like for all space or time correlators, in terms of linear response theory. This is treated in any book on nonequilibrium statistical mechanics. An example are the Green-Kubo relations.

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"Don't panic!"

it is a vacuum expectation value of an operator.
How does one interpret the vacuum expectation value of a product of field operators though? In introductory texts that I've read in QFT, the quantity $$\langle\hat{\phi}(t,\mathbf{x})\hat{\phi}(t,\mathbf{y})\rangle := \langle 0\rvert T\lbrace\hat{\phi}(t,\mathbf{x})\hat{\phi}(t,\mathbf{y})\rbrace\lvert 0\rangle$$ is described as quantifying the probability amplitude for a particle to propagate from one space-time point to another. Wick's theorem can be used to express n-point correlators in terms of products of two-point correlators, so I assume that the interpretation of the two-point correlator can then be extended to interpretating the meaning of an n-point correlator?!)

Why are these vacuum expectation values of products of field operators (at different space-time points) referred to as correlators? The Wikipedia article states that "The correlation function can be interpreted physically as the amplitude for propagation of a particle or excitation between y and x." This seems a little vague - I assume that by amplitude the article is referring to probability amplitude, but I may be wrong here?!

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A. Neumaier

Science Advisor
Why are these vacuum expectation values of products of field operators (at different space-time points) referred to as correlators?
Because of the analogy to statistics where for Gaussian noise variables $\eta_j$ with zero mean and unit variances, $\langle\eta_j\eta_k\rangle$ is a correlation.
For free fields at equal time and with discrete space the analogy is quite close. For example, there is a ''Wick theorem'' for higher order expectation values of Gaussians long predating Wick 1948 (Isserlis 1918).
The correlation function can be interpreted physically as the amplitude
Don't buy everything Wikipedia says. It cannot be a probability amplitude since unlike for the latter the absolute values can be arbitrarily large.

"Don't panic!"

For free fields at equal time and with discrete space the analogy is quite close
In this case is the correlation between the field values at two different spatial points?

Don't buy everything Wikipedia says. It cannot be a probability amplitude since unlike for the latter the absolute values can be arbitrarily large.
This leaves me confused though, as reading through Peskin and Schroeder, they give a similar interpretation. Zee's introduction to QFT also describe it as "the amplitude for a disturbance in the field to propagate from a point x to a point y".
I get what your saying, since the expectation value has a pole at $p^{2}=m^{2}$, but it confuses me why such week known introductory texts describe it in this manner, which when first read seemed intuitively pleasing, but now feel disconcerted by.

bolbteppa

If it helps, Hatfield's QFT section 3.5 he gives the following probabilistic interpretation for a charged scalar field:

"Now consider the propagation of charge in the charge scalar field theory. The state corresponding to a particle of charge $+1$ at $x$ is $\phi^*(x)|0>$, while the state corresponding to the particle at $x'$ is $\phi^*(x')|0>$. Thus, the quantum mechanical amplitude to transport the charge from $x$ to $x'$ is $$<0|\phi(x')\phi^*(x)|0>.$$
Apparently, we can interpret the propagation of charge as the creation of a particle of $+1$ charge, an $a$ particle, out of the vacuum at $x$, the transport from $x$ to $x'$, and the reabsorption of the $a$ particle into the vacuum at $x'$. Since we can't absorb the particle before it is created, this process only makes sense if $t' \geq t$.

This is not the total amplitude for propagation of $+1$ charge. The total amplitude is the sum of all of the amplitudes of different processes that give equivalent physical results. In the process above, the charge at $x$ was increased by one unit, while the charge at $x'$ was lowered by one unit. We can accomplish the same thing by creating a particle of $-1$ charge, a $b$ particle, at $x'$ and transporting it to $x$, then destroying it. Since $\phi$ creates $b$ particles and $\phi^*$ destroys them, the amplitude for this process is $$<0|\phi^*(x)\phi(x')|0>.$$
As before, we can't destroy a particle before it is created, so this process only makes sense if $t \geq t'$.
The total amplitude, $G(x',x)$, for propagation is the sum of the two amplitudes, $$G(x',x') = \theta(t'-t)<0|\phi(x')\phi^*(x)|0> + \theta(t-t')<0|\phi^*(x)\phi(x')|0>."$$

A. Neumaier

Science Advisor
In this case is the correlation between the field values at two different spatial points?
yes.

This leaves me confused though, as reading through Peskin and Schroeder, they give a similar interpretation. Zee's introduction to QFT also describe it as "the amplitude for a disturbance in the field to propagate from a point x to a point y".
I get what your saying, since the expectation value has a pole at $p^{2}=m^{2}$, but it confuses me why such week known introductory texts describe it in this manner, which when first read seemed intuitively pleasing, but now feel disconcerted by.
Often ''amplitude'' is used just as a buzzword without a particular meaning, loosely suggesting an interpretation in Feynmans picture of particles traveling all possible paths.

This is a fictitious picture without any reality content. It serves to give a pseudo-intuition for abstract mathematical entities that helps beginners to swallow the formalism but later leads to stomach ache, as you experience it now.

As always in QFT, what counts is only the formulas, not the words, diagrams, and stories often associated with them. (see https://www.physicsforums.com/insights/misconceptions-virtual-particles/ for some loosely related stories.)

"Don't panic!"

If it helps, Hatfield's QFT section 3.5 he gives the following probabilistic interpretation for a charged scalar field:

"Now consider the propagation of charge in the charge scalar field theory. The state corresponding to a particle of charge $+1$ at $x$ is $\phi^*(x)|0>$, while the state corresponding to the particle at $x'$ is $\phi^*(x')|0>$. Thus, the quantum mechanical amplitude to transport the charge from $x$ to $x'$ is $$<0|\phi(x')\phi^*(x)|0>.$$
Apparently, we can interpret the propagation of charge as the creation of a particle of $+1$ charge, an $a$ particle, out of the vacuum at $x$, the transport from $x$ to $x'$, and the reabsorption of the $a$ particle into the vacuum at $x'$. Since we can't absorb the particle before it is created, this process only makes sense if $t' \geq t$.

This is not the total amplitude for propagation of $+1$ charge. The total amplitude is the sum of all of the amplitudes of different processes that give equivalent physical results. In the process above, the charge at $x$ was increased by one unit, while the charge at $x'$ was lowered by one unit. We can accomplish the same thing by creating a particle of $-1$ charge, a $b$ particle, at $x'$ and transporting it to $x$, then destroying it. Since $\phi$ creates $b$ particles and $\phi^*$ destroys them, the amplitude for this process is $$<0|\phi^*(x)\phi(x')|0>.$$
As before, we can't destroy a particle before it is created, so this process only makes sense if $t \geq t'$.
The total amplitude, $G(x',x)$, for propagation is the sum of the two amplitudes, $$G(x',x') = \theta(t'-t)<0|\phi(x')\phi^*(x)|0> + \theta(t-t')<0|\phi^*(x)\phi(x')|0>."$$
This is what I originally understood it as. I assumed that $\langle 0\rvert T\lbrace\hat{\phi} (t,\mathbf{x}) \hat{\phi} (t',\mathbf{y}) \rbrace\lvert 0\rangle$ describes a transition amplitude for a particle to propagate from a space-time point $(t,\mathbf{x})$ to the space-time point $(t',\mathbf{y})$?! (Apologies, in my previous posts, I forgot to express the field operator at two different points in time as well as space).

Science Advisor

"Don't panic!"

Often ''amplitude'' is used just as a buzzword without a particular meaning, loosely suggesting an interpretation in Feynmans picture of particles traveling all possible paths.

This is a fictitious picture without any reality content. It serves to give a pseudo-intuition for abstract mathematical entities that helps beginners to swallow the formalism but later leads to stomach ache, as you experience it now.

As always in QFT, what counts is only the formulas, not the words, diagrams, and stories often associated with them. (see https://www.physicsforums.com/insights/misconceptions-virtual-particles/ for some loosely related stories.)
I wish this wasn't done, it makes it much harder after one becomes a bit more familiar with the mathematics to actually understand the physics taking place.
How should one physically interpret the quantity $\langle 0\rvert T\lbrace\hat{\phi} (t,\mathbf{x}) \hat{\phi} (t',\mathbf{y}) \rbrace\lvert 0\rangle$ then?

A. Neumaier

Science Advisor
How should one physically interpret the quantity $\langle 0\rvert T\lbrace\hat{\phi} (t,\mathbf{x}) \hat{\phi} (t',\mathbf{y}) \rbrace\lvert 0\rangle$ then?
All sorts of correlators are most naturally interpreted in a 4D Fourier transform with respect to the difference of all arguments, where they have a meaningful spectral (4D momentum) interpretation. To understand what they really mean you need to turn to nonequilibrium statistical mechanics.

A. Neumaier

Science Advisor
we can interpret the propagation of charge as the creation of a particle
This only works when $\phi$ is a free field, and then it is clearly a fictitious process since free fields do not create anything.

In an interacting theory, $\phi(x)|0\rangle$ is not a single-particle state - particles have only asymptotic meaning.

"Don't panic!"

All sorts of correlators are most naturally interpreted in a 4D Fourier transform with respect to the difference of all arguments, where they have a meaningful spectral (4D momentum) interpretation. To understand what they really mean you need to turn to nonequilibrium statistical mechanics.
Would you be able to suggest any good texts on nonequilibrium statistical mechanics.

In an interacting theory, ϕ(x)|0⟩ϕ(x)|0⟩\phi(x)|0\rangle is not a single-particle state - particles have only asymptotic meaning.
In canonical quantisation QFT (in the interaction picture), doesn't the S-matrix element of a $2\rightarrow 2$ scattering $$\langle \mathbf{p}_{3},\mathbf{p}_{4}\rvert Te^{-i\int d^{4}x\;\mathcal{L}_{int}(x)}\lvert \mathbf{p}_{1}, \mathbf{p}_{2}\rangle$$ describe the scattering amplitude between asymptotically free particle states $\lvert \mathbf{p}_{1}, \mathbf{p}_{2}\rangle$ and $\lvert \mathbf{p}_{3}, \mathbf{p}_{4}\rangle$, as $t\rightarrow -\infty$ and $t\rightarrow\infty$, respectively.
Is this a pseudo-intuition as in the previous case we've been discussing?!

A. Neumaier

Science Advisor
Would you be able to suggest any good texts on nonequilibrium statistical mechanics.
Initially Reichl, A modern course in statistical physics; this explains linear response theory and its relation to Green's functions and correlations in simpler, nonrelativistic classical and quantum situations. Later Calzetta and Hu, Nonequilibrium quantum field theory would be a good choice.
In canonical quantisation QFT (in the interaction picture), doesn't the S-matrix element of a $2\rightarrow 2$ scattering $$\langle \mathbf{p}_{3},\mathbf{p}_{4}\rvert Te^{-i\int d^{4}x\;\mathcal{L}_{int}(x)}\lvert \mathbf{p}_{1}, \mathbf{p}_{2}\rangle$$ describe the scattering amplitude between asymptotically free particle states $\lvert \mathbf{p}_{1}, \mathbf{p}_{2}\rangle$ and $\lvert \mathbf{p}_{3}, \mathbf{p}_{4}\rangle$, as $t\rightarrow -\infty$ and $t\rightarrow\infty$, respectively.
Is this a pseudo-intuition as in the previous case we've been discussing?!
S-matrix elements have a clear physical meaning. But one gets probability amplitudes only after integrating over a range of momenta corresponding to a normalized wave packet. The S-matrix elements themselves could perhaps be described to be ''probability density amplitudes''. But I think there is nothing gained by replacing the well-understood notion of an S-matrix element by such an expression.

Also, the time-ordered expression you wrote for the S-matrix makes sense only after renormalization (and only in the absence of bound states); so expanding it into free low order correlators doesn't give a good interpretation of free time-ordered correlators.

"Don't panic!"

Initially Reichl, A modern course in statistical physics; this explains linear response theory and its relation to Green's functions and correlations in simpler, nonrelativistic classical and quantum situations. Later Calzetta and Hu, Nonequilibrium quantum field theory would be a good choice.
Thanks, I'll take a look.

S-matrix elements have a clear physical meaning. But one gets probability amplitudes only after integrating over a range of momenta corresponding to a normalized wave packet. The S-matrix elements themselves could perhaps be described to be ''probability density amplitudes''. But I think there is nothing gained by replacing the well-understood notion of an S-matrix element by such an expression.

Also, the time-ordered expression you wrote for the S-matrix makes sense only after renormalization (and only in the absence of bound states); so expanding it into free low order correlators doesn't give a good interpretation of free time-ordered correlators.
Fair enough. Thanks for the information.

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