# Tag Info

9

Let me take a stab at answering this (somewhat vague) question. You said you are interested in the analytic structure of QFT. But you also mentioned the RG, which is somewhat different. I will try to address the analytic structure of QFT and then emphasize that the renormalization group can be thought of as merely a trick to improve perturbation theory. ...

8

Ok so the way I see it, let's strip away the physics first. Then we're left with the math. So the question you're posing is, given only $\frac{d^nx(t)}{dt^n}|_{t=c} \forall n$, can we reconstruct the function $x(t)$ uniquely in some domain $(a,b)$ s.t. $c \in (a,b)$? Well the answer is no. Even if you impose a $C^\infty$ condition on $x(t)$, the example ...

8

Could this imply there is a formulation where that value comes naturally... This sentence implicitly assumes that analytic continuation is "unnatural". But the truth is the other way around: analytic continuation is one of the most natural mathematical procedures in physics. On the contrary, it's functions – especially functions of momenta or energy – ...

6

There are no exact solutions, only approximations and numerical solutions. Don't forget that orbiting black holes will radiate gravitational waves so any solution would have to include those and the corresponding decay of the orbit until the black holes coalesce.

6

This is a really interesting, but equally beguiling, question. Shock waves are discontinuities that develop in solutions of the wave equation. Phase transitions (of various kinds) are non-continuities in thermodynamics, but as thermodynamics is a study of aggregate quantitites, one might argue that the microscopic system is still continuous. However, the ...

5

Carefully following Feynman's procedure one actually finds: $$\langle x''|e^{-i\frac{(z''-z')}{\hbar}H}| x' \rangle$$ $$=\langle x'', z''|x', z'\rangle =\lim_{N\to \infty\: \epsilon \to 0} \left[\frac{m}{2\pi i \hbar \epsilon} \right]^{N/2}\int_{-\infty}^{+\infty}\cdots \int_{-\infty}^{+\infty} \left(\prod_{i=1}^{N-1} dx_i \right) ... 5 Any analytic function is defined everywhere on its Riemannian surface just by its values in an arbitrarily small neighborhood of a point. The expression to be ''analytically continued'' therefore just specifies which function is meant, but it has ''direct and natural'' values everywhere on its Riemann surface. Except that not all of these values can be ... 5 For a very recent authoritative review of the numerical approach, see Centrella et. al. http://arxiv.org/abs/1010.5260 For the alternate parameterized post Newtonian approach, see Living Reviews of Relativity http://relativity.livingreviews.org/Articles/subject.html and look for articles number 2007-2, 2006-4 and 2003-6. 4 The pole of Green's function is related to the spectrum of the particle which is propagating. One dimension for example$$\tilde{G}(\omega)= \frac{i}{\omega-(\epsilon+i\Gamma)}$$If pure real, G(t) is some oscillation function which shows that the particle is stable. If pure imaginary, G(t) has some exponential decay behavior which shows that the particle is ... 3 Let us for simplicity consider a bosonic string X:\Sigma\to M in the matter sector only (as opposed to the ghost sector). Here M is a target manifold. The worldsheet \Sigma is a Lorentzian manifold of real dimension 2. Locally in a neighborhood U\subseteq \Sigma of the worldsheet, we may work in a so-called conformal gauge, which means to choose a ... 3 Franz Pretorius has worked on this and developed animations. http://prl.aps.org/abstract/PRL/v95/i12/e121101 The field is numerical relativity. Matthew Choptuik also, I believe, has done work on this. 3 According to general relativity, a pair of massive bodies that orbit each other emits gravitational waves - for analogous reasons to the reasons why accelerating charges in electrodynamics emit electromagnetic waves. So there can't be any static solutions resembling binary stars or binary black holes. The solutions have to be non-static and a complicated ... 3 Some rules for the cross product are:$$ A \times (\beta B + \gamma C) = \beta (A \times B) + \gamma (A \times C) (e_x \times e_y) \cdot e_z = 1$$The equation$$ (B \times C)\cdot A = d$$is antisymmetric with respect to transpositions, meaning that if you take any two of those vectors and switch them, you multiply d by -1. These rules ... 3 Suppose that for all z in some open set Z of complex numbers containing z_0, the Hamiltonian H(z) is a compact perturbation of the self-adjoint H(z_0) depending analytically on z. Then, for every simple eigenvalue E_0 of H(z_0) and associated normalized eigenstate \psi_0, there exist a complex neighborhood N of z_0 and unique functions ... 2 To perform the analytic continuation of the step function, start with the second equation stated in the question. Since the step function is the derivative of the Dirac delta function, substitute the integration of the Dirac delta for the step function. Then perform the Fourier transformation on the integral. After simplifying this equation, it is equal to ... 2 A way of physically thinking about this is that a two body problem in general relativity does not generally have closed orbits. If one of the bodies is very large and the other a small satellite the problem is integrable. The periapsis (perihelion) advance of the small satellite is repeated with each orbit, which makes the problem integrable. If the two ... 2 One way to look at it is simply a mathematical trick that encodes the boundary conditions of the Schroedinger equation. An alternative and only slightly more intuitive view is the following. In order to obtain only outgoing solutions it is essential to assume that the scattering potential is slowly switched on adiabatically. Formally one can achieve this by ... 1 1) OP is basically wondering how Weinberg on the middle of p. 112 can extend the integration region from^1$${\cal J}^{\pm}_{\beta}~=~ \int_{m_{\alpha}}^{\infty} \!dE_{\alpha}\frac{e^{-iE_{\alpha}t}g(E_{\alpha})T_{\beta\alpha}^{\pm}} {E_{\alpha}-E_{\beta}\pm i0^{+}}$$to include the negative real axis$${\cal J}^{\pm}_{\beta}~=~ \int_{-\infty}^{\infty} ...

1

First, as I understand the Lippman-Schwinger equation, there are actually two different cases, denoted by the $\pm$'s in the standard form of the equation: $$| \psi^{(\pm)} \rangle = | \phi \rangle + \frac{1}{E - H_0 \pm i \epsilon} V |\psi^{(\pm)} \rangle. \$$ In this way, we're talking about not just the outgoing solution (the ...

1

Finite fields are essential for constructing MUBs (see this paper by Wootters and Fields). Also, dealing with quantum error correcting codes (for qubits) is essentially the same as doing linear algebra over GF(4) because of the way Pauli matrices behave (see arXiv:0904.2557 or arXiv:quant-ph/9608006).

1

The zeta function always has poles, depending on the number of dimensions you're working in. For $d=4$ you will generically have poles at $s=2$ and $s=1$ (but only there). The sum $\sum_n \omega_n^{-s}$ only converges if $\text{Re } s > 2$ (staying in $4d$ for simplicity). So in every case you'll encounter, you will need to use 'analytic continuation'. ...

1

What quantities supposed to be $C^\infty$? I don't know if it answers you question, but AFAIK smooth functions are nice and useful tool to describe many aspects of the physical world. However, I don't see why they should be considered as fundamental in any sense. When it comes to QFT, even there you often encounter Dirac delta (and you can't get rid of it ...

1

I will answer the question of why $\dot{\hat{e}_x} = \hat{e}_y \omega$ when the rotation is $\vec{\omega}=\hat{e}_z \omega$ At any instant the components of $e_x$ and $e_y$ are $$\hat{e}_{x}=\begin{bmatrix}\cos\theta & \sin\theta & 0\end{bmatrix}$$ $$\hat{e}_{y}=\begin{bmatrix}\mbox{-}\sin\theta & \cos\theta & 0\end{bmatrix}$$ where ...

1

The only two "somewhat nontrivial" facts that the trick relies upon is: a) the independence of de facto closed contour integrals of holomorphic functions on the paths in the complex space (I say "de facto closed" because infinite curves combined with functions that quickly, e.g. in a Gaussian way, decrease at infinity allows us to consider the curve closed ...

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