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The phase space has points $x^{i}\in V_{2}$. The standard inner product is $x^{i}\delta_{ij}x^{j}$ with $\delta_{ij}$ an invariant tensor under the action of the isometry group SO(2). If $R^{i}_{j}$ are the group matrices, $$\delta_{ij}=[R^{-T}]_{i}^{\ k}[R^{-T}]_{j}^{\ l}\delta_{kl} \ .$$ Set up a one-parameter subroup ...

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The other answers are correct. I would like to add to them with an example. Take a spring, with spring constank $k$, with a mass, $m$, at one end and fixed to a large immovable object at the other. Let the only force acting on the mass be due to the spring and the difference from the equilibrium position to be $x$, which can be positive and negative. This ...

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This is a usual term about solving differential equations. By "analytic" (or mathematical analysis), we mean finding an algebraic expression like $y=f(t)$ which satisfy the desired differential equation. But sometimes we solve the equation only at some special points. The latter method is called "numerical". Since the Newton's law (and other principal ...

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A multi-body problem consisting of $N$ objects requires $N$ coupled differential equations that need to be solved simultaneously (if you want to find the objects trajectories in time with known initial positions) When you solve $x + 2y = 3 \text{and} x + y = 2$, this is what is known as mathematical analysis. The exact solution can be found: $$x = 1, y = ... 0 This is to add up a little more detail to the discussion. Water as it exists in the form of H_3O^+ and OH^- ions are bound together by "Van der waals forces" which is "the sum of the attractive or repulsive forces between molecules other than those due to covalent bonds, or the electrostatic interaction of ions with one another, with neutral molecules, ... 3 A Community Wiki answer to make some other people's comments permanent and tie some loose ends up. To add to Mark Mitchison's Answer, the reason that the prevailing shape is the one that minimises surface energy as he states is that, in the case of water, the liquid's total energy is an (almost) constant offset (the potential and kinetic energy of the ... 1 Another way to look at it is the following. The main force on the molecules will come from other water molecules and be due to cohesion. The system will try to minimize it's energy and bond the molecules together as much as possible. This means minimizing the surface which results in a sphere. 4 The droplet wants to minimise its surface energy. This energy is proportional to its surface area. So the equilibrium shape is that which minimises the surface area for fixed volume (the bulk density is fixed by the temperature and pressure). 2 Then, there is the case that such an operator is defined on the full interval I assume that by "full interval" you mean the whole real line. First question: Do we then need any boundary conditions? Yes, as noted by Sam Bader, boundary conditions are part of the Hamiltonian. In my physics lecture we used so-called Born von Karmann boundary ... 1 The Lagrangian for GR is$$ L \propto \int R \sqrt{-g} \, d^4 x $$where R is the Ricci scalar$$ R = R^\mu_\mu = R^{\mu \nu}_{\mu \nu} $$So, this is a scalar which is related linearly to all the components of the Riemann tensor, and is a second-order differential of the metric g of the form$$ R \sim g \partial^2 g + (\partial g)^2 $$This is ... 0 They are not analogous. R_{abcd} is just Riemann Tensor and R_{abcd}R^{abcd} is Riemann Tensor squared. Mathematically they must be squared since having a Single term Riemann Tensor / Ricci Tensor in gravitational action doesn't make sense. Physically speaking they are modification of Einstein Hilbert action. They are curvature not field one shall ... 3 You are correct in your interpretation that Weisner's method is geometric in nature: it is a method for finding generating functions for special functions using representation theory of Lie groups and Lie algebras. And as you know, Lie groups play an enormous role in modern geometry, on several different levels. Lie groups are smooth differentiable ... 1 It's good that you're considering questions like this; I find that this type of questions really forces a student to a deeper understanding of the math involved. Do we then need any boundary conditions? Yes, boundary conditions should be considered as part of the definition of the Hamiltonian and its domain. Different boundary conditions can result in ... 4 any eigenfunction to this Schrödinger operator is automatically periodic with the potential's period, is this true? No!! The eigenfunctions are Bloch waves \psi(x) = u(x)e^{ikx}, where u is periodic (with the period of the lattice). But the product \psi is not periodic (with the period of the lattice) unless k=0. I put up an example on Wikipedia ... 2 Restricting ourselves to just vector spaces without any extra structure, the theorem is true. One way to see this is to note that any member f of the dual space is uniquely defined by the value it returns acting on the basis \{\psi_n\}, say f(\psi_n) = z_n for complex numbers z_n. Then V^* is isomorphic to \mathbb{C}^\mathbb{N}, the set of ... 9 There are two concepts of duality for vector spaces. One is the algebraic dual that is the set of all linear maps. Precisely, given a vector space V over a field \mathbb{K}, the algebraic dual V_{alg}^* is the set of all linear functions \phi:V\to \mathbb{K}. This is a subset of \mathbb{K}^V, the set of all functions from V to \mathbb{K}. The ... 3 Your "imaginary eigenvalues" don't work, because the eigenfunctions are no eigenfunctions. They do not lie in L^2, as you seem to be aware of. So, let's deal with the Laplacian itself: -\Delta=-\frac{d^2}{dx^2}. What I want to do is, I want to calculate the Fourier transform of this operator, because the Fourier transform diagonalizes -\Delta, as we ... 4 First off, if k =: i\kappa is imaginary, the eigenvalue (“energy”) is -\kappa^2, i.e. real but negative:$$-\frac{d^2}{dx^2} e^{ikx} = -\frac{d^2}{dx^2} e^{-\kappa x} = -\kappa^2 e^{-\kappa x}.$$Physically, that is an evanescent wave in one direction, but grows without bound in the other, so if your space is all of x\in\mathbb R, it is not a valid ... 1 As mentioned by Qmechanic, the answer to your questions is no. However, assuming space-time is oriented, we have the following: For any pseudo-Riemannian metric g, there exists a normalized time-like one-form h^0 and a Riemannian metric g^R so that$$ g = 2h^0\otimes h^0 - g^R $$This yields a locally Euclidean topology compatible with the manifold ... 2 The spectral theorem only holds for normal operators. Self-adjoint operators are normal, symmetric ones not necessarily so. In physicist-speak, we want the generalized eigenvectors to from a 'complete basis' of the Hilbert space. For example, the generalized eigenvectors of the momentum operator in position representation are plane waves, and even though ... 2 I'm pretty sure there exists an answer for this here already, but I can't find it (it's always about unboundedness). For the Hamiltonian, the answer is basically given by Stone's theorem on one-parameter unitary groups. There is a one-to-one correspondence between self-adjoint operators and strongly continuous one-parameter families of unitaries. Why is ... 1 Consider an operator A on a Hilbert space \cal H, say L^2(\mathbb R) in order to deal with QM of a particle on a real axis without spin. Let D(A) \subset \cal H be the domain of A. The spectrum \sigma(A)\subset \mathbb C of A is defined as the union of the following three pairwise disjoint subsets \sigma_p(A), \sigma_c(A), \sigma_c(A). ... 0 the Dirac Point on Graphene is protected by hidden symmetry. And it is explained very well in the paper arXiv:1406.3800. It is not that easy to understand the hidden symmetry. Personally speaking, I thought it is combination of inversion, time reversal and reflection symmetry, though the hidden symmetry in that paper has a totally different form with my ... 3 Any Hilbert space \mathcal{H} with the notion of unitary time evolution also possesses the notion of Hamiltonian. If \mathcal{U}(t) : \mathcal{H} \to \mathcal{H} is the time evolution operator for every t \in \mathbb{R}, then it forms a one-parameter Lie subgroup of the Lie group of unitary operators, which is generated by some distinct element H ... 4 No, I don't see why that should be the case. The notion of a Hilbertspace underlying a quantum-mechanical system is quite independant of the postulate that time evolution is generated by a Hamiltonian. The notion of a vectorspace enters QM, because fundamentaly QM should be a linear theory and thus allow for arbitrary superpositions. The more wonderous ... 2 The Dirac operator, as we know it, is D_\mu\gamma^\mu with D as the gauge covariant derivative. Using the Fujikawa method of deriving the Adler-Bell-Jackiw or chiral anomaly, one finds that the anomaly of the chiral current is given by$$ \partial_\mu \langle (j^5)^\mu\rangle = 2 \mathrm{i}A(x)$$where$$ A(x) = \int \sum_n \psi_n^\dagger \gamma^5 ...

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Ok, there are a lot of points here. 1) First of all, an operator in Hilbert spaces is not defined only by its action (e.g. the operation of derivation for the momentum), but also by the so-called domain of definition, i.e. the subspace of vectors of the Hilbert space where it can act. Unbounded operators are not defined for every vector of the Hilbert ...

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If you have different Hilbert spaces, you cannot say it is the same operator on them, since operators are defined on the Hilbert space. The momentum operator is a tricky one for many systems, and rigor requires the discussion of concepts like rigged Hilbert spaces. A nice introductory discussion of this is "Mathematical surprises and Dirac's formalism in ...

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Quantum dynamics is commonly known to be generated by self-adjoint operators. Therefore in order to properly define the dynamics of a system it is necessary to introduce a suitable self-adjoint Hamiltonian operator. In quantum field theories, this task is extremely difficult, because the formal operators that emerge quantizing a classical field theory ...

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Advanced Classical Field Theory (2009) by Giachetta, Mangiarotti, Sardanashvily remarks on p. 248: A non-compact world manifold admits a Dirac spinor structure if and only if it is parallelizable. For a compact world manifold $X$, its Euler characteristic and the second Stiefel-Whitney class $w_2$ must be zero, and its first Pontryagin number ...

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$\nabla\times\vec{A}$ is the solenoidal component of the vector field: it is the divergenceless component. A good way to intuitively visualise the Helmholtz theorem is to think in Fourier space, so that all fields become their Fourier transforms. In this visualisation, $\nabla\times \vec{E}$ is the component of the Fourier transform that is at right angles ...

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$\vec{A}$ is the vector potential. $\vec{E}$ is equal to $-\nabla\phi$ only in the electrostatic case. If the electric field varies with time, we have $$\nabla\times\vec{E} = -\frac{\partial\vec{B}}{\partial t}$$ Since $\nabla\cdot\vec{B}$ is always zero, $\vec{B}$ is written as a curl of another field, called the vector ...

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As showed by Solovay here, in a non-separable Hilbert space $H$ there may be probability measures that cannot be written, for any $M$ closed subspace of $H$, as $\mu (M)=\mathrm{Tr}[\rho \mathbb{1}_M]$, for some positive self-adjoint trace class $\rho$ with trace 1 (density matrix). Here $\mathbb{1}_M$ denotes the orthogonal projection on $M$. [The proof of ...

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