# Tag Info

2

The term to look for is Coulomb wave. These wavefunctions are well explained in the corresponding Wikipedia article. Depending on your mathematical background, you should be ready for a bit of a formula jolt, as these wavefunctions rely very intimately on the confluent hypergeometric function. If you want the short of it, then I can tell you that the ...

0

I think the point is that if you took the limits to infinity without doing anything else, you'd be implicitly redefining the matrix element in order to make the equations consistent. So he just calls the redefined matrix element $T_{ni}$ instead of $V_{ni}$. Later he solves for $T_{ni}$ in terms of $V_{nj}$ (see the section "Solving for the T matrix").

3

An intuitive answer to motivate the Stone theorem that ACuriousMind's answer cites. Take a quantum system. For now, let it be a finite dimensional one (the Stone theorem is needed to make the reasoning work in a separable, infinite dimensional Hilbert space). Let it evolve for time $t_1$. The evolved state is some linear operator $U(t_1)$ imparted on the ...

3

A very simple proof proceeds along the following lines. $$\Psi(t_1+\Delta t)=\Psi(t_1)+\frac{\partial \Psi}{\partial t} \Delta t$$ where we omitted higher order $\Delta t$ terms (will disappear when we let $\Delta t \to 0$ later). Applying this formula recursively, we get $$\Psi(t_1+2\Delta t)=\Psi(t_1)+2\frac{\partial \Psi (t_1)}{\partial t} \Delta ... 5 This rather detailed answer will potentially chart out the issues and assumptions that are needed for a straightforward idea. Understanding how it is done might be more useful than actually doing it. Assumption 0) The Hamiltonian is time independent. To be honest, I never find arguments about taking derivatives of operators very appealing. We will have a ... 2 An elementary way to proceed is as follows. Let's put an explicit factor of \epsilon in B. The problem is then to solve the following equation for the matrix X:$$A + \epsilon B = X^{2}$$We want to do this perturbatively, so we assume that X can be represented as:$$X = \sum_{n=0}^{\infty}\epsilon^{n}X^{(n)}$$We can then write:$$ ...

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The relation between $U=U(t_1,t_1+t)$ and $H$ is $$H={\partial U\over\partial t}\Big|_{t=0}$$

8

For time-independent Hamiltonians, $U(t) = \mathrm{e}^{\mathrm{i}Ht}$ follows from Stone's theorem on one-parameter unitary groups, since the Schrödinger equation $$H\psi = \mathrm{i}\partial_t \psi$$ is just the statement that $H$ is the infinitesimal generator of a one-parameter group parameterized by the time $t$. For time-dependent Hamiltonians $H(t) = ... 1 If the eigenvectors of$\hat H$don't depend explicitly on$t$(the eigenvalues can) then we can easily make sense of$\hat H~U = i \partial_t U$by working in the eigenbasis of$\hat H$where$H$is diagonal, because if$\hat H = \operatorname{diag}(\lambda_1, \lambda_2, \dots)$then$\exp(-i~\hat H~t) = \operatorname{diag}(e^{-i\lambda_1t}, ...

3

Hints: The square root function has a Taylor expansion around $a>0$ $$\tag{1} \sqrt{a+b}~=~\sum_{n=0}^{\infty} \begin{pmatrix}\frac{1}{2} \cr n\end{pmatrix}a^{\frac{1}{2}-n}b^n, \qquad |b| ~<~a.$$ One may show that a possible non-commutative generalization reads $$\sqrt{A+B}~=~\sqrt{A}+\sum_{n=1}^{\infty} \begin{pmatrix}\frac{1}{2} \cr ... 0 Some algebraic geometry with the main purpose of understanding D-branes in the context of mirror symmetry is reviewed in Paul Aspinwall's 'D-branes on Calabi-Yau manifolds' (http://arxiv.org/abs/hep-th/0403166). I have not begun reading it thoroughly myself, but it seems accessible to physicists, at least those with the basic math background for string ... 2 Some time ago I have asked a similar question; although I have accepted one of the answers, It did not satisfy my main interest concerning the physics of the case when the diagonal \Delta is not removed. Now I have some more information that I can share with you. Most of the authors give two reasons for the removal of the diagonal: If we include the ... 3 TL;DR: It is the wedge product \wedge and the exterior derivative/differential d (which squares to zero d^2=0) that give rise to Grassmann-odd elements and supersymmetry. More concretely, Ref. 1 first writes down a (non-relativistic) SUSY algebra {\cal A}$$\tag{10} \{Q_1,Q_1\}_+~=~2H~=~\{Q_2,Q_2\}_+, \qquad \{Q_1,Q_2\}_+~=~0, $$spanned by two ... 1 As said in the commentaries, the first one comes from the Dirac formalism. Simply put, it deals with quantum states as vectors \lvert a \rangle whose components contain the projections of the system into different eigenstates. For a system, the set of state eigenvectors \lvert a_i \rangle must be linearly independent \langle a_i \lvert a_j ... 3 The proof is given in great generality by the beautiful and powerful spectral theorem; in its functional calculus form. This theorem clarifies the meaning of your first expression, and specifies the functions for which you can write the second (that however are many). In addition, it applies to any self-adjoint operator, bounded or unbounded. Sadly, the ... 3 First of all you need the vectors |O_k\rangle to form an orthonormal system. Then to prove what you need you start with polynomials and verify that, e.g. O^2 has the desired form etc... For a continuous function f, assuming that O is a bounded operator you can then invoke the Stone-Weierstrass theorem. 0 I would recommend S.T. Yau's book on Mathematical Aspects of String Theory, following @Tomas Smith. There is also a two volume set based on lectures given at Princeton. The books can be found on Amazon at http://www.amazon.com/Quantum-Fields-Strings-Course-Mathematicians/dp/0821820125 and ... 2 Let \mathcal{M} be our spacetime. Then, a gauge theory is given by a connection form A on a principal bundle over it (that locally projects onto the spacetime in a way compatible with gauge transformations), which is the gauge potential. Maxwell's equations1 (in vacuum) are the equations of motion for the gauge field for the Yang-Mills action coupled to ... 1 Hints to the question (v2): First of all, one should realize the abuse of notation in eq. (6.69) of Ref. 1 where \phi is used in two meanings: both as an external parameter and as internal integration/dummy variable. It is more properly written as$$ \widehat {Z}[\phi]~=~\frac{{e}^{iS[\phi]}}{{\cal N}}, \qquad {\cal N}~:=~\int\!{\cal D}\phi~e^{iS[\phi]} ...

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Here I constructed perturbation-like approximants converging to the vacuum in $\phi^4_2g(x)$ (technically an interacting QFT, although not translation invariant, so Haag's theorem does not apply). There are no "infinities" in this case.

2

So, it looks like you're trying to answer a question like this: you've got a plate which is held at $T=0$ at $x = 0, x = L$ for all $y$, with $T= T_0(x)$ at $y=0$. What is $T(x, y)$ in this space $y \ge 0, 0\le x\le L?$ Then the answer is that on this strip we can solve Laplace's equation $\nabla^2T = 0$ with the functions $T(x, y) = A \exp(-\alpha y) ... 0 The Laplacian is a linear operator - it does't include any non linear functions of the derivatives, thus includes only linear combination of (second order) derivatives. For the laplacian in cartesian coordinates -$\nabla^2=(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}) $i.e.$ \nabla^2 ...

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The most recent textbook answer I could find would be from Choquet-Bruhat's 2009 book "General Relativity and the Einstein Equations". She writes on page 403: Remark A curvature singularity does not imply geodesic incompleteness. The geodesic flow depends only on the $C^{1,1}$ structure of the metric. Conversely, does geodesic incompleteness imply a ...

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Is there any extension of the theorems where the conclusion of the theorem is curvature blow-up rather than geodesic incompleteness? No. Remember that spacetime is an abstract 3+1 dimensional mathematical space where we plot motion through space against time, which is in essence motion through space inside a clock. Also remember that a geodesic is in ...

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