# Tag Info

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 ...

13

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 ...

9

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 ...

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 ...

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

@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

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 ...

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 ...

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

The primary utility in introducing the generating functional is in using it to compute correlation functions of the given quantum field theory. Let's restrict the discussion to that of a theory of a single, real scalar field on Minkowski space, and let $x_1, \dots, x_n$ denote spacetime points. Of central importance are time-ordered vacuum expectation ...

3


3

If you can calculate vacuum-to-vacuum transition amplitudes, you can calculate S-matrix elements, because the two are related by the LSZ reduction formula. The LSZ will in any case chop off the propagators for external lines that the generating functional inserts, so you will end up only needing to compute amputated diagrams.

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

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: $$\langle\phi_1\phi_2\rangle = \frac{\int [D\phi_1][D\phi_2] \phi_1\phi_2e^{iS[\phi_1,\phi_2]}}{\int [D\phi_1][D\phi_2]e^{iS[\phi_1,\phi_2]}}$$ $$\langle\phi_1\rangle = \frac{\int [D\phi_1][D\phi_2] \phi_1e^{iS[\phi_1,\phi_2]}}{\int [D\phi_1][D\phi_2] e^{iS[\phi_1,\phi_2]}},\quad ... 2 Mathematically, the expression$$C(r) = \langle m(0)m(r) \rangle - \langle m \rangle^2$$does not support the two relations. Without knowing how C(r) changes with r, all this expression does is define a correlation. Without knowing that the correlation decreases with r and without defining a cut-off for what we consider to be small, this relationship ... 2 Let W be any d-dimensional pseudo-Riemannian manifold with metric h_{ab} and let S(W) be any action defined on W. The energy-momentum tensor is usually defined by (up to a scalar multiple anyways; it seems in your case the definition is modified by a factor of -4\pi):$$ T_{ab}:=\frac{1}{\sqrt{|h|}}\frac{\delta S}{\delta h^{ab}}, $$where ... 2 To obtain the classical (in the sense of non quantum mechanical) potential due to a pair of charged particles, says, two electrons, you may calculate the scattering amplitude and compare it to$$\langle p'|iT|p \rangle=-i2\pi\tilde{V}( \textbf{q} )\delta^{3}(E_{p'}-E_{p}), \hspace{2cm} \bf{q}=\bf{p'}-\bf{p'}  this is equation $(4.123)$ in Peskin's & ...

2

You need to refine your intuition a little bit. If you bring two operators together, you indeed get singular behaviour, which is taken into account by the OPE. The unit operator is the most singular, and the higher the scaling dimension of an operator $\phi$ in the $O_1 \times O_2$ OPE, the less singular its contribution will be. (In your case, the ...

2

Here is my proposal : First, I am using some notations : $n'_F(\omega) =n_F(-\omega+ E_F)$ I supposed here that $n_F(\omega) = \large \frac{1}{e^{\beta \omega} + 1}$, so that $n'_F(\omega) = \large \frac{1}{e^{\beta (-\omega+E_F)} + 1}$. Finally, I define $C'_{12}(t) = \pi C_{12}(t)$. And we suppose $t >0$. With these notations and hypothesis, we have ...

2

I have just found out that it is fundamentally wrong to approximate the fermi function with chebyshev polynomials. Representation of the Fermi function on the real axis by Chebychev polynomials might be okay, but the representation of the Fermi function in the complex plane and especially close to the poles of the Fermi function is certainly very poor (no ...

2

Hint $\int_0^t \int_0^{t_1} dt_1 dt_2 \, a(t_1) a(t_2) = \frac{1}{2!} \int_0^t\int_0^t dt_1 dt_2\, \mathcal{T}\{ a(t_1) a(t_2) \}$ and so forth. You can see this by noting that the (square) integration region in the second integral can be split up into two triangular integration regions like in the first integral. This is one way to define the ...

Only top voted, non community-wiki answers of a minimum length are eligible