# Tag Info

31

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 ...

28

The correlation function you wrote is a completely general correlation of two quantities, $$\langle f(X) g(Y)\rangle$$ You just use the symbol $x'$ for $Y$ and the symbol $x+x'$ for $X$. If the environment - the vacuum or the material - is translationally invariant, it means that its properties don't depend on overall translations. So if you change $X$ and ...

18

a very intuitive example for correlation functions can be seen in laser speckle metrology. If you shine light on a surface which is rough compared to the wavelength, the resulting reflected signal will be somehow random. This can also be stated as that you cannot say from one point of a signal how a neighbouring one looks like - they are uncorrelated. Such ...

17

This is just a property of Gaussian averaging analogous to the finite dimensional case: $\langle e^{ix} \rangle=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty} e^{ix}e^{-\frac{x^2}{2\sigma^2}}=e^{-\frac{\sigma^2}{2}}= e^{-\frac{\langle x^2 \rangle}{2}}$ The field can be decomposed into its independent Gaussian modes and integrated for each mode ...

17

Excellent question, Kostya. Lubos already gave a detailed answer using general arguments in the language of QFT. In astrophysics and cosmology, however, there is another, and very simple, reason why we use the correlation functions all the time. It turns out that the mean value of the function $f(\vec{x})$, denoted $\langle f(\vec{x})\rangle$, can often not ...

10

For a given quantum system, the kernel of the path integral is, in fact, the kernel of an integral transform as you explicitly write down. It is the transform that governs time evolution of the system as is manifest in your first equation. For this reason, it is often referred to as the propagator of a given system. For example, for a single, ...

10

First, the term "propagator" is usually defined as the Green's function of the first type, not the second type, i.e. as a solution to the diffential equation $\hat L G = \delta$. At any rate, those definitions are ultimately equivalent – when the details are correctly written down – because the Green's function defined as the correlator in the second ...

9

In axiomatic approaches to quantum field theory, the basic field operators are usually realized as operator-valued distributions. That's what Wightman fields are: operator-valued distributions satisfying the Wightman Axioms. Wightman functions are the correlation functions of Wightman fields, nothing more. There's a nice theorem that says if you have a ...

8

@ArnoldNeumaier and @dushya have both pointed out correct solutions, but I want to elaborate a bit. The easy approach is the one dushya suggested. (You can also do what Arnold Neumaier suggests: First define the time-ordered product of operator-valued distributions, and then take expectation values.) Begin by recalling how Wightman functions are defined. ...

6

David Bar Moshe's derivation is of course right. Let me offer you a Taylor-expansion-based alternative proof: $$\left\langle e^{ix} \right \rangle = \left\langle \sum_{n=1}^\infty \frac{(ix)^n}{n!} \right \rangle = \left\langle \sum_{k=1}^\infty \frac{(ix)^{2k}}{(2k)!} \right \rangle$$ Here, I just used that by some odd-ness, the odd powers have a ...

5

For simplicity, let's restrict the discussion to that of a single particle moving in one dimension. Path integrals can be performed in much broader contexts like quantum field theory, but I think that would conceptually obscure the issue at this point. Let $H$ denote the (time-independent) quantum hamiltonian. Then the time evolution on the system is ...

5

The goal is to find the single particle propagator in the presence of interactions. This propogator will be the sum of all diagrams which have two external vertices. This sum of diagrams would be difficult to compute, but it turns out it easy to write this big sum of diagrams in terms of a sum of a smaller set of diagrams: the set of "one particle ...

5


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