Tag Info

New answers tagged

1

Perhaps you are starting by the wrong end. Your concern seems to be related in the first term with the totally misleading notation of integrals in quantum mechanics, and this is more related with the spectral theorem than with distributions itself. Distributions only appear in Quantum mechanics when certain operators has empty spectrum in the usual Hilber ...


3

$$D(\hat{X}_i):= \left\{ \psi \in L^2(K, d^nx)\:\left|\: \int_K x_i^2|\psi(x)|^2 d^nx< \right.+\infty \right\} = L^2(K, d^nx)$$ where the last identity holds true if $K$ is bounded (in particular compact) because $x_i^2$ is bounded as well thereon (in this case the operator is bounded, too). Moreover $$D(\hat{J}^2):= \left\{\psi \in L^2(\mathbb S^2, ...


3

I) OP's question (v2) seems to be essentially a question of mathematical precision vs. the way physicists express themselves concisely when speaking of "infinitesimals" without becoming too technical by introducing epsilons and deltas and what not. See also this related Phys.SE post. Perhaps the easiest and most elementary way to make sense of the ...


0

Answer to question one : The Principle of Quantum Mechanics by R. Shankar page 149 reads "Barring a few exceptions, the schrodinger equation is always solved in a particular basis. Although all basis are equal mathematically, some are more equal that others. First of all, since H = H(X,P) the X and P basis recommend themselves....The choice between the two ...


0

You have to clarify the term "basis" in infinite dimensions. By definition, finite-dimensional spaces have a finite basis (so that, for instance, $\mathbb{R}^n\approx\mathbb{R}^m$ iff $n=m$). But for infinite-dimensional spaces, there are various types of bases (assuming you can take infinite sums). A natural one (using finite sums) is the Hamel basis, but ...


5

First, the term "basis spaces" isn't standard in quantum physics but let us assume that we understand what the sentences approximately mean. Second, the momentum is continuous (not quantized) if the position space is noncompact (infinite). The momentum only becomes quantized if the position space is compact (or periodic), and indeed, it's been ...


2

An observable is a self-adjoint operator $\mathcal{O}$ on the Hilbert space of states $\mathcal{H}$. The spectral theorem tells us that such an operator has an orthonormal basis of eigenvectors in $\mathcal{H}$, if it is compact. If it is not compact, we have to "enlarge" the Hilbert space to something called rigged Hilbert space or Gelfand tripel. A ...


0

Unfortunately not every operator, which is used to get an observable gives you enough elements to make a base of the Hilbert space. As far as I know every eigenstate is orthogonal to each other. But the “length” could be arbitrary. Mostly used eigenstates in quantum mechanics are normalized for better usage.


1

You can think about "solving X equations in Y unknowns". When $X>Y$, you generally expect infinite solutions, when $X=Y$ you generally expect a unique solution, when $X<Y$ you generally expect no solutions. This kind of statement is not always mathematically rigorous but you can usually argue it rigorously in specific circumstances. Pick an energy $W$ ...


6

Dirac being opaque and hard to follow? Well I never... In Chapter 10 Dirac argues on physical grounds that the eigenkets of an observable must form a complete set. His argument goes that say we have an observable with eigenstates $|\varepsilon\rangle$ and some general state $|P\rangle$. Then I can write $|P\rangle$ as \begin{equation}|P\rangle = \sum ...


0

IANAFD but I'll stick my neck out and say this: resolving the Clay problem one way or another won't cause people doing CFD to lose any more sleep than they already do. First of all, Jean Leray proved the existence of weak solutions to Navier Stokes in $R^3$ way back in the 1930s, and that is pretty much what matters for the task of getting numerical ...


1

I wondered about this one myself a while back. I'm not absolutely positive about this but it is definitely in the ballpark. Here's what I know for the background: I believe the first paper on exotic spheres in physics was by Witten [Commun. Math. Phys., 100, 197–229 (1985)] and centered around the idea that exotic spheres can be interpreted as ...


1

Yes! In fact, one of my (few) own research experiences was closely connected to random matrix theory! I did a small project under supervision of Enrico Pajer from Princeton University last January. The work I did was based on earlier work of his, which used random matrix theory in the context of inflationary theory, using a model where many fields ...


4

Brian Hall "Quantum Theory for Mathematicians" is a recent nice book that presents the basics of QM with mathematical rigor, as suggested by the title. It covers a fair amount of topics, and seems suitable for an undergraduate level. The short book of Mackey "Mathematical foundations of Quantum Mechanics" is also a very nice book on the axiomatization of QM, ...


1

Comments to the question (v2): The canonical expansion of the two-fermion wave function seems more related to the canonical form of antisymmetric real matrices in the framework of vector spaces and linear algebra. The Darboux' theorem in the framework of manifolds and differential geometry (which is a surprisingly potent result) is overkill for the ...


7

The phase space is a symplectic manifold, so any manifold $\mathcal{M}$ that admits a closed nondegenerate 2-form is a possible phase space. Now, what is necessary (or sufficent) for admitting such a form? First, as you mention, $\mathcal{M}$ must be even-dimensional. Second, $\mathcal{M}$ must be orientable. Why? Because orientability is equivalent to ...


10

The bible for the mathematical formulation of classical Mechanics, namely Foundations of Mechanics by Abraham and Marsden, defines a hamiltonian system as a triple $(M, \omega, X_H)$ where $(M, \omega)$ is a symplectic manifold, and $X_H$ is the Hamiltonian vector field corresponding to a hamiltonian function $H:M\to\mathbb R$. Now, are there typically any ...


0

This type of problem can be conveniently treated using the complex Laplace transform. For a function $f(t)$ with $t\geqslant 0$ it is defined as \begin{equation*} \hat{f}(z)=\int_{0}^{\infty }dt\exp [izt]f(t),\;Imz>0. \end{equation*} Setting $z=\omega +i\eta $ ($\eta >0$ but arbitrary otherwise) we have, with $\theta (t)$ the Heaviside step function ...


7

A Green's function is nothing but the (generally distributional) integral kernel of the inverse of a given operator. The point is that the operator $$A:= \left[ i\hbar\frac{\partial}{\partial t} - \hat{H}(\mathbf{r})\right]$$ does not admit a unique inverse. Conversely, $$A_{\pm\eta}:=\left[ i\hbar\frac{\partial}{\partial t} - \hat{H}(\mathbf{r}) \pm ...


0

I know complex analysis and the residuum theorem. This kind of thinks are introduce to evaluate some integrals of the form $\int\limits_{-\infty}^{+\infty}$ with some function which have a pole in $0$ and then You add an infinitesimal to evaluate an integral using residuum theorem. My question is why can't You just go through zero (where the pole is) make a ...


2

This is an optimal control problem, so I will use the rules of optimal control. First, we represent the state space equations. Also we take the total mass as a state and amount of fuel burnt as the input control. So we have: \begin{cases} \tag{1} \dot{x}_1=x_2 \\ \dot{x}_2=\frac{\eta \theta-k(x_1)x_2^2}{x_3}-g \\ \dot{x}_3=-\theta \end{cases} with these ...


0

I thought this question was interesting and I didn't want to do any proper work this afternoon so I made a simple model to find out what would happen. My matlab code is at the end of the question. So far I've tested three cases and considered changing the initial thrust and adding a linear increase in the thrust for each case. The thrust is given as a ...


1

The names of these creatures are a true mess and there are mainly two independent notation schemes: the mathematical and the physical one. Let $P \to M$ be a $G$-principal bundle. Then $G$ is called the structure group by mathematicians and the gauge group by physicians The (infinite-dimensional) group of automorphism of $P$, or equivalently the group of ...


1

This is not quite a complete answer, but more of an overly large comment on terminology. The definitions from nLab don't agree with Giovanni Giachetta, Luigi Mangiarotti, Gennadi Sardanashvily: Advanced Classical Field Theory, which I'll summarize briefly: The authors call the group $G$ of a (principal) $G$-bundle the structure group. This is standard ...


1

I admit I am a bit confused by your terminology, but here is how I learned it: Let $P$ be a $G$-principal bundle and $\Sigma$ a spacetime. gauge group: The fibers of the $G$-principal bundle over the spacetime, i.e. the group $G$. (Local) group of gauge transformations: The group of diffeomorphisms $t : P \rightarrow P$, which are fiber-preserving and ...


2

For 1D potentials, the sequence of bound state energy eigenvalues $E_n$ cannot grow faster than what happens in the case of an infinite well, i.e. $E_n$ cannot grow faster than $n^2$.


0

I think you need to add a postulate to (3) in order to obtain (1). This postulate it's actually assumed in every picture of the dinamic and it is that the evolution operator is a one group parameter. It is such a natural assumption that it comes no harm in taking it for true. Even in classical hamiltonian mechanics the flow in the coordinate space form a ...


4

The result can be proved in a general way using, well, math. In particular the theory of semigroups of linear operators on Banach spaces (I know that seems advanced and maybe not physical, but it is an elegant way of proving the result you seek ;-) ). Define the Banach space $\mathscr{L}^1(\mathscr{H})$ of trace class operators over a separable Hilbert ...


1

Just calculate (3) for $\hat{A}_H=\hat{U}_H$. With $\hat{U}_H(t)=\hat{U}^{\dagger}(t)\hat{U}(t)\hat{U}(t)=\hat{U}(t)$ and $\hat{H}_H=\hat{H}$ (as you said) you get: $\frac{d\hat{U}_H}{dt}=\frac{\partial \hat{U}}{\partial t}+\frac{1}{i\hbar}\underbrace{[\hat{H},\hat{U}]}_{=0 \text{(as you ...


4

CAUTION - ANSWER INCOMPLETE There is a gap in my argument (see the send); it relies on the claim that \begin{align} - \hat O^\dagger \hat A = \hat A\hat O \end{align} for all hermitian $\hat A$ implies $\hat O = 0$ which may not be true. Please comment if you know how to prove this or know of a counterexample. Update. Actually the claim above is ...


1

I) One mathematical problem is that the function $$\tag{1} f(q)~:=~ \frac{q\sin(rq)}{q^2+u^2}, \qquad q,r,u~>~0, $$ is not integrable $f\notin {\cal L}^{1}(\mathbb{R}_{+})$, because the integral over the absolute value of the integrand is infinite: $$\tag{2} \int_{\mathbb{R}_{+}} \! dq~|f(q)| ~=~\infty.$$ However it is still possible to define the ...


0

I can not quickly follow the detailed reasoning in your paper. But, usually, it is possible to use perturbation theory to calculate the derivatives of many quantities in quantum mechanics. The first derivative of an the energy of a bound state with respect to a change in the parameters of quantum mechanics is, in fact, very, very simple, just the ...


1

If you read any of the papers on 4d Yang-Mills referenced in the article you quote -- e.g., [3] by Balaban or [29] by Magnen, Seneor, & Rivasseau -- you'll discover that they are concerned with Yang-Mills on a 4-torus. This is standard in the subject, since no one wants to think about the boundary conditions on a cube.


0

A tricky question, really. Apart from the fact that your $\lvert e,p\rangle$ vector does not belong to $L^2$ (hence you cannot take scalar products of it), I don't see any other flaw. That, in my opinion, means you have a nice argument to prove the following mathematical statement: Let $\mathscr{H}$ be a separable Hilbert space, $0\neq z\in\mathbb{C}$. ...


8

As is customary in such question, I will point to this paper, which excellently discusses the problems the Dirac formalism has. Now, in your concrete example, the problem lies in the energy/momentum states $| p_0 \rangle$ themselves, which are non-normalizable, since the wave function associated is the Fourier transform of $\delta(p-p_0)$, which means that ...


10

You've got things slightly backwards. In constructive QFT, one almost always starts in Euclidean spacetime -- where it is 'easy' to define the path integral -- and then analytically continues to get correlation functions on Minkowski spacetime. (The Osterwalder-Schrader Theorem tells you when this analytic continuation 'works', meaning when the resulting ...


2

(add my comments as an answer) Both quantum and classical Yang-Mills theory can manifest various phenomena (e.g non-linear and others) which have not been studied fully. For example soliton solutions, asymptotic freedom, stability, confinement etc.. By the work of t'Hooft et al. Yang-Mills-type gauge theories are renormalizable Yang-Mills (quantum and ...



Top 50 recent answers are included