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New answers tagged mathematical-physics

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A good discussion of regularity properties of the Wiener measure is in section II.5 of the book "Functional Integration and Quantum Physics" by Barry Simon. He gives the proofs of most of the relevant theorems except the borderline case $\alpha=\frac{1}{2}$, namely Levy's Theorem showing that $\frac{|x(\tau_1)-x(\tau_2)|}{\sqrt|\tau_1-\tau_2|}$ diverges ...

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Think of this not as an extremely rigorous way of solving the differential equation, but rather as using your intuition to guess a solution. Often when you are given a differential equation, the solution is not at all obvious, and perhaps the equation isn't even solvable analytically. Instead of giving up, though, sometimes you can identify a parameter ...

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You are essentially correct. If you start on the real line with the Schrödinger equation $$\left(-\frac12 \frac{\partial^2}{\partial x^2} + V(x) \right ) \psi(x) = E\,\psi(x),$$ then at every point $x_0$ where $V(x)$ is analytic you are guaranteed that $\psi(x)$ will be analytic in a neighbourhood of that point. However, if $V(x)$ is not analytic then you ...

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Comments to the question (v2): To be specific, let us assume that the underlying 2D manifold is the Riemann sphere $S^2\cong \mathbb{C}\cup\{\infty\}$. The group of globally defined conformal transformations is the 6-dimensional group $PSL(2,\mathbb{C})$ of Moebius transformations. Mathematically speaking, one should consider the groupoid of locally ...

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There is a mathematical point that can be made, and in my opinion is related to a deeper understanding of what it means to solve a (partial) differential equation. I will try to keep things simple, and consider only linear models. Suppose that you have a space $X$ with some properties, for example it has a topology. We suppose that the state of our system ...

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QFTs in spacetime of dimension $<4$ have their use in real applications - 3D theories to quantum surfaces, and 2D theories to quantum wires. There much of the exceptional behavior of lower-dimensional QFTs can be observed. A famous example is the fractional Hall effect with anyonic (rather than Bose or Fermi) statistics.

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Comment to the question (v1): No, such decomposition is in general not unique. E.g. the unit element ${\bf 1}_{2\times 2}\in SU(2)$ can be written with parameters $b\in 4\pi\mathbb{Z}$ and $a=0=c$.

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You may indeed identify the generators in the way you did. However, the Lie algebras and Lie groups are different because – as quickly said by Qmechanic – you must use different reality conditions for the coefficients. A general matrix in the $SU(2)$ group is written as $$M = \exp[ i( \alpha J_+ + \bar\alpha J_- + \gamma J_0 )]$$ where $\alpha\in ... 7 The ladder operators do belong to the real Lie algebra $$\quad su(1,1)~\cong~ so(2,1)~\cong~sl(2,\mathbb{R}),$$ but they do not belong to the real Lie algebra $$su(2)~\cong~ so(3).$$ All the above real Lie algebras have complexifications isomorphic to$sl(2,\mathbb{C})$. 3 I) Well,$r=0$is the boundary of a$d$-dimensional spherical coordinate patch, i.e. an artifact of our choice of coordinate system, but$r=0$does not correspond to a physical boundary per se, other than what we can deduce from the TISE. The mantra is that a boundary condition (for finite$r$) should only be imposed if it is a consequence of the TISE. See ... 2$\renewcommand{ket}[1]{|#1\rangle}$As others have pointed out, you can go whole hog and solve the characteristic equation$\text{det}(\hat{Q} - \lambda \hat{I})=0$and find repeated solutions for$\lambda$. However, there is a simpler, more physically intuitive way to hunt for degeneracy: look for symmetry. When an operator$\hat{Q}$has a symmetry there ... 1 As Phoenix87 said, you need to solve the asociated eigenvalue problem.$\lambda$is an eigenvalue if and only if$\hat{Q}|\Psi>=\lambda|\Psi>$. If you have a matrix expression for the operator$\hat{Q}$, then the usual way to solve the problem is writing the above equation as Phoenix87 did, writing everything in the left side: ... 3 Assuming that your operator has a spectrum consisting of isolated points you can look for all the independent solutions of the eigenvalue equation $$(Q-\lambda I)\xi = 0$$ Let these solutions generate a vector space$V_\lambda$and then compute the dimension of$V_\lambda$. If it is greater than 1 then the eigenvalue$\lambda$is degenerate. 2 The tools and techniques for constructing QFTs are the same whatever the dimension. However you cannot prove conjectures which are false, no matter how powerful your tools are. The issue with 3+1 is that the class of models for which the conjecture "yeah it can be constructed" is in all likelihood true is much more narrow than in 1+1 and 2+1. Basically the ... 1 Well, consider a Lax pair for the Harmonic Oscillator; $$L = \begin{pmatrix} p & \omega q \\ \omega q & -p \\ \end{pmatrix}, \quad M=\frac{\omega}{2} \begin{pmatrix} 0 & -1 \\ 1& 0 \\ \end{pmatrix}$$ Since the Hamiltonian is $$H(q, p)=\frac{1}{2}(p^{2}+\omega^{2}q^{2})$$ It is eay to check that the Lax ... 2 We are not able, at least for the moment, to define in a rigorous mathematical fashion a meaningful interacting QFT in$3+1$dimensions that is coherent with the perturbative theory utilized by physicists (in more precise words, that satisfies the Wightman axioms). On the contrary, some rigorous interacting QFTs can be defined in lower (spatial) dimensions. ... 0 There is a way to understand Wick's Theorem as a particular instance of a very general connection between differentiation and combinatorics. At the bottom of it is just Leibniz's product rule. Given variables$x_1,\ldots,x_d$, the basic identity is: $$\frac{\partial}{\partial x_{i_1}}\ldots\frac{\partial}{\partial x_{i_n}}\ x_{j_1}\ldots x_{j_n}= ... 1 Unitarity of the time-evolution operator is exactly the point: Stone's theorem (see e.g. Reed, Simon: Theorems VIII.7, VIII.8) tells us If A is self-adjoint, the spectral theorem holds. This gives us a functional calculus which makes it possible to define U(t) = e^{itA} in the first place. A such defined U(t) is a strongly continuous unitary group. ... 7 In QM, a real valued observable A is mathematically represented by a projector valued measure over \mathbb{R}, P^{(A)}, i.e., if E is a Borel subset of the real line, then P^{(A)}(E) is a projector representing the proposition "the outcome of measuring A falls in E". In principle, that's all you need for representing, mathematically, ... 0 The momentum is a covector because it is a gradient, and gradients are always covariant. It does what it says on the tin. However, you are right that this is a subtle point and it's not particularly clear at first sight. For a lagrangian of the form L=T-V with V independent of \dot q, the canonical momentum is given by$$ p=\frac{\partial L}{\partial ... 0 I think there are mainly two reasons for motivating the introduction of$p$-adic models in physics. They could exist in nature. They provide insightful toy models for physical phenomena. There is a vast physical literature on the subject which cite Reason 1 as justification, this is sometimes called the Vladimirov Hypothesis. Namely, we do not know the ... 0 After thinking about Nick P's answer and re-reading the relevant chapter of Sussman's Structure and Interpretation of Classical Mechanics, I came up with the following elaboration of Nick's argument. It's not water-tight, but it convinced me, and perhaps it will help someone else. I will use Sussman's unorthodox but precise notation. The first step (and ... 1 For simplicity consider the 1-d case, with$\psi =\sqrt{n} e^{2i\phi}$, then $$i \psi_t =\frac{i}{2} \frac{\dot{n}}{\sqrt{n}} e^{2i\phi} -\sqrt{n} e^{2i\phi} 2\dot{\phi}.$$ Similarly$$\frac{\partial H}{\partial \psi^*} = \frac{\partial H}{\partial n}\frac{\partial n}{\partial \psi^*} + \frac{\partial H}{\partial \phi}\frac{\partial \phi}{\partial ... 3 In principle, yes,$\varphi$is a coordinate system, in that it tags every member of$\mathcal S$with an appropriate$n$-tuple of real numbers. If you know a set of coordinates$(c_1,\ldots,c_n)$, you can use$\varphi^{-1}$to find the corresponding point in$\mathcal S$. In practice, this will not be useful at all, mostly because the amount of real ... 0 No discrete symmetries are not infinitesimal and the derivation of Noether's theorem requires an infinitesimal symmetry to work as such.. Noether theorem is only valid for infinitesimal symmetries. However if you have a discrete symmetry which can expressed as an exponential form of some infinitesimal symmetry then you do have a corresponding conserved ... 0 Let me attempt an answer, based on the comments from Timaeus and ACuriousMind.. Answer: No, there is nothing that forces us to view non-relativistic quantum mechanics as a$U(1)\$ gauge theory. It's an elegant, equivalent way of looking at some situations like the Aharanov-Bohm effect. In more detail: We could treat the Aharanov-Bohm effect in the usual, ...

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