64

I had this same problem, too. The trick with it is realizing that there's an important difference between Taylor series and Taylor approximations or polynomials, whose behavior is described by Taylor's theorem. Yes, very often I suspect a common mistake is that you first see Taylor polynomials and theorem, and then you get Taylor series and that becomes the ...


24

The Stone-Weierstrass theorem says that any continuous function on a compact interval is arbitrarily well approximated by polynomials. Thus, as long as we're only interested in explaining experimental results (and not in the exact solutions of theoretical models), series expansions are plenty good enough. That is, for whatever we'd like to describe, there is ...


13

Proposition. Let there be given two complex numbers $a,b\in \mathbb{C}$ such that ${\rm Re}(a)\geq 0$. In the case ${\rm Re}(a)=0$, we demand furthermore that ${\rm Im}(a)\neq 0$ and ${\rm Re}(b)=0$. The Gaussian integral is well-defined and is given by $$ \underbrace{\int_{\mathbb{R}}\!dx~ e^{-\frac{a}{2}x^2+bx}}_{=:~ I_{\mathbb{R}}(a,b)} ~=~\lim_{\begin{...


12

Let me expand a little more on what Craig Thone just said : Consider the energy/frequency-dependent Green function : $$ \tilde{G}(\omega)=\frac{1}{\omega-(a-\mathrm{i}b)} $$ with one single pole in $\omega=a-\mathrm{i}b$ (with $b>0$), which is Fourier transform of the time-dependent $G(t)$ Green function such as : $$ G(t)=\int\frac{\mathrm{d}\omega}{2\pi}...


11

First of all, one need to prove the following key integral formula: \begin{align*} \boxed{I = \int_{-\infty}^{+\infty} d x e^{ i a x^2} = \sqrt{\dfrac{i \pi}{a}} \qquad (a>0) } \end{align*} Usually one can pick up an analytic function $f(z)=e^{ i a z^2}$ and then performs the complex integral along the closed contour showed above. $$J = \oint d z e^{ i ...


9

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 – that ...


9

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

This is a somewhat broad question, because there are a number of different Green's functions in quantum physics. Perhaps the simplest one is the resolvent Green's function for a single-particle system. Its definition is $$G(\omega^{\pm})=\lim_{\delta \rightarrow 0^+}\left[ \omega \pm i\delta - H \right]^{-1}\equiv \frac{1}{\omega\pm i\delta - H},$$ where $H$ ...


8

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 ...


7

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) \exp\left\{\frac{i\...


7

I) L&L are referring to a linearized model where the TISE $$ \hbar^2\psi^{\prime\prime}(x) +P^2(x) \psi(x)~=~0,\qquad P(x)~:= \sqrt{2m (E-V(x))}~=~|P(x)| e^{i\theta(x)}, \tag{1}$$ becomes the Airy differential equation. Near a turning point $x_0$, we can approximate the square momentum $$P^2(x)~\approx~ \alpha (x\!-\!x_0), \qquad (P^2)^{\prime}(x)~\...


7

The Kramers-Kronig relations are the expression, in the Fourier frequency domain, of the fact that the linear susceptibility $\chi(\tau)$ is a causal function, i.e. that the dielectric response of the signal $f$ to a forcing $F$ has the form $$ f(t) = \int_0^\infty \chi(\tau) F(t-\tau) \mathrm d\tau = \int_{-\infty}^\infty \theta(\tau)\chi(\tau) F(t-\tau) \...


6

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} ...


6

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 ...


6

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 ...


6

Landau free energy is just an approximation to the real free energy in the thermodynamic limit. For that reason, Landau free energy can be analytic, while the real one is not. Let me show how the approximation works. As you may know, the Landau free energy is defined in the following way, assuming the Ising model: $$Z\left(h,T\right)=\sum_{\left\{ s_{i}\...


5

I) Preliminaries. First let us state explicitly the Sommerfeld expansion to all orders. To this end, let $$ \tag{1} B(x)~:=~\frac{x}{e^x-1}~=~\sum_{m=0}^{\infty}\frac{B_m}{m!}x^m~=~1-\frac{x}{2}+\frac{x^2}{12}-\frac{x^4}{720}+\frac{x^6}{30240}+{\cal O}(x^8)\qquad $$ be the generating function for the Bernoulli numbers$^1$. Let $$ \widehat{A}(x)~:=~\frac{...


5

Every regularization scheme is somewhat arbitrary. There are three popular regularization schemes when it comes to path integrals and their associated perturbative divergent integrals: time slicing, mode regularization, and dimensional regularization. Time slicing is the usual procedure used to derive the path integral, and it is the discretization of time ...


5

Heuristically, one approach to justifying Weinberg's application of Cauchy's formula is to treat the non-analytic integrand as the boundary behavior of a meromorphic function (kind of like Fourier series): perform all internal integrations until you are left with an integral over energy, solve the Cauchy-Riemann equations within the upper half plane (...


5

One example: Complex analysis is used heavily in the proofs of the CPT theorem and spin-statistics theorem in relativistic quantum field theory. The classic book Streater and Wightman, PCT, Spin and Statistics, and All That is filled with complex analysis, such as the "edge of the wedge" theorem described in Section 2-5. This example isn't just some ...


5

When we integrate the propagator with respect to $k^0$ (i.e. the energy), we encounter two poles: one at $\omega_{\mathbf{k}}=\sqrt{\mathbf{k}^2+m^2}$ and one at $-\omega_{\mathbf{k}}=-\sqrt{\mathbf{k}^2+m^2}$, where $\mathbf{k}$ is the 3-momentum. In order to regularize this integral, we move these poles slightly off of the real line by adding or ...


4

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 ...


4

In dimensional regularization, $d$ is a complex number, not a true dimension. The $d$-dimensional integrals of a rational function are defined for any complex $d$ with sufficiently negative real part (the threshold depending on the integrand), and therefore can be analytically continued to a (provably meromorphic) function for all $d$. For a concise, ...


4

The Wikipedia page is being sloppy. They mean that the free energy density is an analytic function of the mean-field order parameter, whereas at a thermal phase transition the free energy density is a non-analytic function of the temperature (or for a zero-temperature phase transition, of the external parameter being tuned across the transition). The order ...


4

Short answer: Air resistance pushes it to its stable configuration. If you dropped it in a vacuum, it wouldn't turn over. Long answer: As the shuttlecock falls, the air exerts an upward force on all parts of the shuttlecock. The net effect of all of these upward forces is a torque on the shuttlecock that depends on its orientation. There are two ...


4

The $i\epsilon$ is used as a prescription to tell you how to integrate in the complex $p^{0}$ plane. In a sense, it enforces the boundary conditions of your propagator, and uniquely fixes the contour over which you have to integrate to obtain the Green's function. To make this more precise, note the following identity, $$ \lim_{\epsilon\rightarrow 0^+}\...


4

If we know the value of $f$ at $t$, and we want to know the value of $f(t+\Delta t)$ for small $\Delta t$, then the most basic thing to do is to just assume that $f(t+\Delta t) = f(t)$. This is known as the "zeroth order approximation". In calculus, you learned about tangent lines. With a tangent line, instead of approximating the function with a fixed value,...


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