Hot answers tagged correlation-function
16
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 ...
11
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 ...
8
Cutology
The math books aren't good for this, you need seat-of-the-pants intuition. The quick and dirty physicist answer is that a branch cut is best thought of as a continuum of poles densely spread over a line. You reproduce branch cuts by integrating poles spread over an interval, for example with a constant "residue density" (this is not the standard ...
7
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 ...
6
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 ...
5
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 ...
5
@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. ...
5
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 ...
4
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 ...
4
Your first answer is the correct one. The temporal auto-correlation function (or the time correlation function) of a fluctuating quantity $A(t)$ is
$$C_{AA}(\tau) = \langle A(t) A(t+\tau)\rangle$$
This is a measure of how correlated $A(t)$ is to its value at another time $t+\tau$.
The Green-Kubo formula that connects a response function (the electrical ...
4
Okay, so I gather from the link that $G^{(k)}$ in your notation refers to the correlation between field values at $2k$ points, with $\varepsilon^+$ inserted at half of them and $\varepsilon^-$ inserted at the other half.
This concept of an $n$-point correlation function is very similar to the $n$th moment of a random variable or statistical distribution. ...
4
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 ...
3
Computation of the $S_z$ probability distribution for each of the
manifolds of equal entanglement:
Remark: Notations and references from Kuś and Žyczkowski are used.
Case 1: The separable case: The state vector is parametrized as (equation: 24)
$w = \begin{bmatrix} \cos \alpha \cos \beta e^{i \chi_1},& \cos \alpha \sin\beta e^{i \chi_2}, &\sin ...
3
Edited to add the second part
Edited again, for part 3 and 4
$\newcommand\ket[1]{\left|#1\right>} \newcommand\bra[1]{\left<#1\right|} $
1. Absence of Quantum Loophole
You can easily see that there is no "quantum loophole" in your argument by writing explicitly any pure separable state. With your notations, we have :
$$
...
2
Assuming that:
by "a perfect DFT solution" you mean the set of Kohn-Sham orbitals for the exact (presently unknown) denistiy functional that existance of which Hohenberg and Kohn have proven;
you are asking whether the individual energies of these "perfect" Kohn-Sham orbitals can be used as estimates of exact ionization and/or excitation energies,
the ...
2
The time-ordering operator is well-defined rigorously through the Epstein-Glaser approach of distribution splitting. You can read about this in Scharf's book on QED, or http://arxiv.org/abs/arXiv:0906.1952. See also the summary in http://de.wikipedia.org/wiki/FQFT
2
$X$ isn't a Hermitian operator, as you've discerned, and $\langle 0 | X |0\rangle$ isn't the expectation value of an observable.
It's traditional to call the quantum mechanical quantity $\langle 0| T(\phi(t_2) \phi(t_1) )| 0 \rangle$ a "correlation function", but this is actually an abuse of terminology. It is actually an amplitude! Or, if you prefer, ...
2
How about that way of looking at it.
Starting from path-integral definitions:
$$\langle\phi_1\phi_2\rangle = \frac{\int [D\phi_1][D\phi_2] \phi_1\phi_2e^{iS[\phi_1,\phi_2]}}{\int [D\phi_1][D\phi_2]e^{iS[\phi_1,\phi_2]}}$$
$$\langle\phi_1\rangle = \frac{\int [D\phi_1][D\phi_2] \phi_1e^{iS[\phi_1,\phi_2]}}{\int [D\phi_1][D\phi_2] e^{iS[\phi_1,\phi_2]}},\quad ...
2
Consider non-relativistic quantum mechanics of a point particle in 1 dimension with the classical Lagrangian
$$\tag{A} L~:=~\frac{m}{2}\dot{x}^2-V(x).$$
Let $\Delta t:=t_f-t_i$ and $\Delta x:=x_f-x_i$. OP's question concerns the following properties (B) and (C) of the kernel $K(x_f,t_f;x_i,t_i)$:
$$\tag{B} K(x_f,t_f;x_i,t_i) ...
2
Mathematically, the expression
$$C(r) = \langle m(0)m(r) \rangle - \langle m \rangle^2$$
does not support the two relations. Without knowing how $C(r)$ changes with $r$, all this expression does is define a correlation. Without knowing that the correlation decreases with $r$ and without defining a cut-off for what we consider to be small, this relationship ...
1
Here we are just rephrasing user1504's correct answer using slightly different words.
Off-shell correlation functions
$$\tag{1}\langle 0| T\varphi(x_1)\ldots \varphi(x_m)\varphi(y_1)\ldots \varphi(y_n) | 0 \rangle$$
are related via the LSZ reduction formula to on-shell $S$-matrix elements
$$\tag{2}\langle p_1, \ldots p_n~{\rm out} | p_1, \ldots p_m~ ...
1
But, if we get the correlation functions, then, can't we just employ the LSZ formula to find the scattering amplitudes, and so we need to calculate fewer diagrams?
I think that you are missing the perturbation theory in your picture. -- If you know the correlation functions, then you get the amplitudes by LSZ. But in order to get the correlation ...
1
You define the density auto-correlation function as
$$S_{\rho\rho} = \langle \delta \rho(\mathbf{x}_1) \delta \rho(\mathbf{x}_2)\rangle$$
where $\delta \rho(\mathbf{x}) = \rho(\mathbf{x}) - \langle \rho(\mathbf{x}) \rangle$ is deviation from the local mean value.
The Fourier transform of $S_{\rho\rho}$ is related to the structure-factor
$$S(\mathbf{q}) = ...
1
I don't have a complete answer but here's something to give you a rough idea.
We consider a Gaussian scalar field with one-point Hamiltonian $\Omega$, i.e. Hamiltonian of this field is given by
$$H = : {1 \over 2} \int {\rm d}^d {\mathbf x} \left( \pi(\mathbf x)^2 + \phi(\mathbf x) \Omega^2\phi(\mathbf x)\right): = \int {\tilde {\rm d} \mathbf p} E(\mathbf ...
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