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I am struggling to understand Green's functions, as used in Quantum Field Theory. One of my main problems is that the source I have been reading has a definition which is certainly correct, but involves enough integrals to obscure any intuition I initially had. Could someone explain what a Green function $G^{(n)}(x_1,x_2,\dots,x_n)$ intuitively represents (or provide a reference to a readable explanation)? In particular, I am being told that the Green function enjoys the invariance $$G^{(n)}(x_1+y,x_2+y,\dots,x_n+y) = G^{(n)}(x_1,x_2,\dots,x_n)$$ Supposedly, this is very simple and intuitive, but I can't see it (other than going a few levels down in the definitions, which in not very enlightning). Could someone please explain this property?

(my usual disclaimer: I am not a physicist, but a mathematician who occasionally comes into contact with physics. Please forgive my lack of physical insight.)

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    $\begingroup$ Would you like an explanation of "why" quantum field theories are constructed in such a way so that their Green's functions exhibit translation invariance, or "why" in the context of a particular QFT this can be shown to be true via computation? Or perhaps you are looking for a hybrid of these two, or something else entirely? $\endgroup$ – joshphysics Jul 8 '13 at 17:50
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Roughly speaking, a Green's function measures the correlation between fields at different points of spacetime. The reason we care about them in particle physics is that they are related to scattering amplitudes by the LSZ formula. A scattering amplitude is just the transition amplitude from an $n$-particle initial state to an $m$-particle final state.

To answer your specific question, a Green's function should respect the symmetries of the underlying theory. The easiest way to see this is by writing \begin{align*} \langle F(\phi)\rangle=\int d\phi F(\phi)\exp(iS[\phi]), \end{align*} where $S$ is the action. Then if it's possible to define a quantum mechanical measure that respects a symmetry of the action, then the Green's functions should be invariant under that symmetry. For example, if the action is translation-invariant, then the Green's functions are also.

One should be careful about the measure, since the classical symmetry might be anomalous, meaning that it can't be carried over to the quantum theory. This is impossible for translations, since it's easy to come up with regulators that preserve translation invariance (dim reg, Pauli Villars, zeta function etc. etc.).

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    $\begingroup$ I want to flag something here, which can be confusing for mathematicians trying to read the physics literature: Minkowski signature Green's functions do 'measure the correlation between field values' in a poetic sense of the word correlation. But they don't literally measure correlations in the sense of statistics. They're just matrix elements of composite operators. (Euclidean signature correlation functions, on the other hand, actually do measure the correlation between field values.) $\endgroup$ – user1504 Jul 8 '13 at 18:03
  • $\begingroup$ Yeah, but Minkowski correlators are just rotated Euclidean correlators, so in this sense the measure is inherited from Euclidean space. $\endgroup$ – Matthew Jul 8 '13 at 18:08
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    $\begingroup$ Sure. My point is simply that it can be confusing for mathematicians to be told that some number measures a correlation, and then to notice that that number has non-zero imaginary part. $\endgroup$ – user1504 Jul 8 '13 at 18:09
  • $\begingroup$ That's a good point. $\endgroup$ – Matthew Jul 8 '13 at 18:10
  • $\begingroup$ Strictly speaking they're kernels of integral transforms, or elements of linear operators that are the inverses of another linear operator representing our field equation, for example. That doesn't exactly correspond to a correlation in the statistical sense of the word but you can certainly construct one later. $\endgroup$ – Chay Paterson Jul 8 '13 at 19:18
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The intuition comes from Poincare-invariant scalar field theories, where a (Minkowski-signature) Green's function is the vacuum expectation value of a product of field operators:

$G(x_1,...,x_n) = \langle vac | \phi(x_1)...\phi(x_n)| vac\rangle.$

Here, I'm writing $\phi(x_i)$ for the observable which measures the value of a scalar field $\phi$ at a point $x_i \in \mathbb{R}^d$. (This is a mild abuse of notation. If we were being careful I would have to explain that fields are operator-valued distributions, and that the Green's function is a Schwarz kernel for the distribution in $n$ variables given by $f_1,...f_n\mapsto \langle vac | \phi(f_1)...\phi(f_n)|vac\rangle$.)

In such theories, the group of translations of $\mathbb{R}^d$ has a unitary representation $U$ on the Hilbert space where $|vac\rangle$ lives, and it acts by conjugation on operators on this Hilbert space. In particular, translation by $y \in \mathbb{R}^d$ sends $\phi(x_i)$ to $U^*(y)\phi(x_i)U(y) = \phi(x_i + y)$. These translation operators also leave the vacuum vector invariant. This is where the translation formula you asked about comes from. $\langle vac | \phi(x_1)...\phi(x_n)| vac\rangle \\= \langle vac | U^*(y)\phi(x_1)...\phi(x_n) U(y)| vac\rangle \\= \langle vac | U^*(y)\phi(x_1)U(y)U^*(y)...U(y)U^*(y)\phi(x_n)U(y)| vac\rangle \\= \langle vac | \phi(x_1+y)...\phi(x_n+y)| vac\rangle$.

In the first equality, I used translation invariance of the vacuum: $U(y)|vac\rangle = |vac\rangle$. In the second, I inserted $1 = U(y)U^*(y)$ between every pair of adjacent field operators.

Your source may be talking about Euclidean signature Green's functions. These are obtained by analytically continuing in the variables $x_i$.

Generally, Streater & Wightman's book is a nice source for this stuff. Also, David Kazhdan's contribution to the IAS QFT & Strings for Mathematicians book.

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  • $\begingroup$ I think it might be good to qualify this response a bit. Not all QFTs that one might dream up possess translation-invariant vacuum states. This is certainly true, however, for Poincare-invariant field theories for example. $\endgroup$ – joshphysics Jul 8 '13 at 18:29
  • $\begingroup$ @joshphysics That's a fair point. (Although OP is certainly reading about translation invariant Green's functions.) $\endgroup$ – user1504 Jul 8 '13 at 18:39

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