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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 ...


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 ...


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 ...


4

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 ...


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 ...


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).


4

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 ...


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 ...


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 ...


3

There is no better definition than what Wikipedia offers - in general, a topological excitation is a (field) state, i.e. a localized quantity since fields depend on spacetime, whose integral is a topological invariant. One prime example are Yang-Mills theories in 4D, where the integral $\int \mathrm{Tr}(F\wedge F)$, as essentially the second Chern class of ...


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 $H$ ...


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 ...


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 ...


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 ...


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 usually see it in the reverse way, but it is a matter of taste. Hilbert spaces, in general, can have bases of arbitrarily high cardinality. The specific one used on QM is, by construction, isomorphic to the space L2, the space of square-integrable functions. From there you can show that this particular Hilbert space is separable, because it is a theorem ...


2

$\vec{A}$ 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 \begin{equation} \nabla\times\vec{E} = -\frac{\partial\vec{B}}{\partial t} \end{equation} Since $\nabla\cdot\vec{B}$ is always zero, $\vec{B}$ is written as a curl of another field, called the vector ...


2

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 ...


2

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 ...


2

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 ...


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 ...


2

When you introduce an auxiliary variable, such as a regularization parameter, at the end of the calculation you have to take the limit that sets the expression back to the original one. If you introduce multiple auxiliary variables, you have to do this for all of them. Otherwise you're just doing a different integral. In this case specifically, ...


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 ...


2

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 = ...


1

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 ...


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.


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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 ...


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 ...


1

$\nabla\times\vec{A}$ 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 ...


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 ...



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