Tag Info

6

If $\cal H$ is a complex Hilbert space, and $A :D(A) \to \cal H$ is linear with $D(A)\subset \cal H$ dense subspace, there is a unique operator, the adjoint $A^\dagger$ of $A$ satisfying (this is its definition) $$\langle A^\dagger \psi| \phi \rangle = \langle \psi | A \phi \rangle\quad \forall \phi \in D(A)\:,\forall \psi \in D(A^\dagger)$$ with: ...

5

Quantization usually means the association of a Hilbert space to the classical phase space (in our case a Poisson manifold). However, in deformation quantization, this task is achieved indirectly, first through the construction of an associative $C^*$ algebra, in this case the deformed algebra of functions equipped with a star product which serves as the ...

5

General Remarks. In general, you cannot "derive" a representation of a given group $G$ on the objects you're considering, but there are some really standard definitions of certain group representations which are given special names like "scalar," "vector," and so on. However, given the representation of a Lie group $G$, this induces a representation of its ...

4

The eigenvalue equation $$\tag{1} \hat{x}\psi(x)~=~x_0\psi(x)$$ in the standard Schrödinger position representation $$\tag{2} \hat{x}~=~x, \qquad \hat{p}~=~-i\hbar\frac{\partial}{\partial x},$$ reads $$\tag{3} (x-x_0)\psi(x)~=~0,$$ which has general solution $$\tag{4} \psi(x) ~\propto~ \delta(x-x_0).$$

4

I cannot understand why you wrote that $\eta$ is the rapidity: It would be the rapidity if $L$ were the boost, but this is not the case because the internal sign in the RHS of the formula defining $L$ is wrong. Your $L$ is formally an angular momentum if you do not pay attention to the weird name of the variable $t$, time? Well, in addition to the ...

4

Suggestions: Study how to derive the Lie algebra $so(p,q)$ from the Lie group $SO(p,q)$, cf. e.g. this Phys.SE post. Work from now on at the level of Lie algebra (as opposed to Lie group). Show that $$\hat{J}^{\mu\nu}=\hat{x}^{\mu}\hat{p}^{\nu}-\hat{x}^{\nu}\hat{p}^{\mu}$$ are generators (of a representation) of the Lie algebra $so(p,q)$. In this way the ...

4

I do not know it this is an answer, since I am not sure to have understood your question. The structure of the equation is formally hyperbolic: $$\frac{\partial^2 \psi}{\partial t^2} - A\psi = S\quad (1)$$ where $\psi =(p,q)^t$. If $A$ were self-adjoint and non-negative (or non positive, changing a sign and inserting a further $i$ in front of $\sqrt{-A}$ as ...

4

You can prove a general formula $$\int\frac{d^dp_{\mathrm{E}}}{(2\pi)^d} \frac{(p_{\mathrm{E}}^2)^m}{(p_{\mathrm{E}}^2+\Delta)^n}= \frac{1}{(4\pi)^{d/2}}\frac{\Gamma(m+d/2)\Gamma(n-m-d/2)}{\Gamma(d/2)\Gamma(n)} \left(\frac{1}{\Delta}\right)^{n-m-d/2},\quad n>m+d/2$$ by using Gaussian integral and Euler integral of the first kind.

3

The answer is that $\{|x\rangle\}$ is not a basis of $L^2(\mathbb R)$ which admits only countable basis. The point is that objects like $|x\rangle$ are not vectors in $L^2(\mathbb R)$. To provide them with a rigorous mathematical meaning one should enlarge $L^2(\mathbb R)$ into an extended (non-Hilbert) vector space structure including Schwartz distributions ...

3

I think that Wolfram is arguing that the study of cellular automata and perhaps similar computational systems could serve as an organizational principle, providing a coherent framework to look at different problem (just like the more familiar frameworks provided by physics and chemistry). This explains the title of his new book, A new kind of Science (i.e. ...

3

I'll add a small disclaimer as well, I am a mathy with little to no physics background, so if any of the below needs expanding, feel free to ask! (Though I'll mention that I felt a lot more comfortable with this stuff when I first computed the induced actions I'll mention below and really got my hands dirty verifying everything.) To see that there is only ...

3

A linear operator $A: D(A) \to {\cal H}$ with $D(A) \subset {\cal H}$ a subspace and ${\cal H}$ a Hilbert space (a normed space could be enough), is said to be bounded if: $$\sup_{\psi \in D(A)\:, ||\psi|| \neq 0} \frac{||A\psi||}{||\psi||} < +\infty\:.$$ In this case the LHS is indicated by $||A||$ and it is called the norm of $A$. Notice that, ...

2

That particular sum will not be very well behaved, because you are summing many terms very close to one. Most of the mechanics with this type of sum are fairly easy; for example, you should definitely be able to do the conversion to relative coordinates (or the whole problem will be much too complicated). There is one tricky step, though, at the heart of the ...

2

The conformal group is defined for any spacetime you want. The conformal group of d-dimensional Euclidean space, which has isometry group SO(d), is SO(d+1,1). The conformal group of d+1 dimensional Minkowski space, whose isometry group is SO(d,1), is SO(d+1,2). The defining property of the conformal group is that its the set of transformations that leave the ...

2

Couder and Fort's experiment is based on a mathematical analogy between the Hilbert space of a particle moving in two dimensions, and the two surface of a vibrating oil bath, which interacts with an oil droplet bouncing on top of it. Naïvely, one might try to extend this analogy a two-particle system by having two oil droplets bouncing on a single ...

1

I) OP mentions in a comment that his sum should be normalized with a factor $\frac{1}{2N+1}$. Hence OP is considering $$S_N ~:=~\frac{1}{2N+1}\sum_{-N\leq k,\ell\leq N} e^{-\alpha (k+\ell)^2} ~\stackrel{(3)}{=}~ \frac{1}{2N+1}\sum_{n,m\in\mathbb{Z}}^{-2N\leq n\pm m\leq 2N} e^{-\alpha n^2}$$ $$\tag{1}~=~\sum_{-2N\leq n\leq 2N}\frac{2N-|n|+1}{2N+1} ... 1 This is more an extended comment than an answer. Given the principle of relativity (along with a handful of almost axiomatic assumptions), it is indeed possible to derive a general coordinate transformation that involves an invariant speed. Which leaves the conjecture that the invariant speed is the measured speed of light, c, a matter of empirical ... 1 This is much to do with the possible eigenvalues of the operators. Normal operators on a Hilbert space are closely analogous to complex numbers, with the adjoint taking the role of the conjugate; these relations are typically inherited directly to the operator's eigenvalues. Thus, if a linear operator L has an eigenfunction f with eigenvalue \lambda, ... 1 The special feature of contact geometry is the contact 1-form \lambda, which satisfies \lambda\wedge d\lambda\ne0 (let's restrict to 3-dimensions). In our Lagrangian mechanics example, \lambda = dq-vdt. You want this to pull-back to zero on the permissible'' curves in phase space -- these curves represent the motions of your system. For a more ... 1 As has been mentioned in the comments, it is an assumption that the QFT has a vacuum state which is annihilated by P^\mu. This is actually a very important point, since it is one of the crucial differences between flat space QFT and QFT in curved spacetime. This is explained in Wald's book QFT in Curved spacetime. Essentially in quantum theories, the ... 1 I) Let there be given a local action functional$$\tag{1} S[\phi]~=~\int_V \mathrm{d}^nx ~{\cal L}, $$with the Lagrangian density$$\tag{2} {\cal L}(\phi(x),\partial\phi(x),x).  [We leave it to the reader to extend to higher-derivative theories.] II) Assume that a variation of $S$ for arbitrary $x$-dependent infinitesimal $\epsilon(x)$ takes the ...

1

He is saying that if $A$ is the set of configurations of a binary star system, then the sum $\oplus: A \times A \to A$ can be not defined, because if $a_1$ is the configuration at some time and $a_2$ is the configuration some time later, then $a_1 \oplus a_2$ will not be a configuration of a binary star system, since it will have four stars. In other words, ...

1

Although the answer given by soliton is sufficient, I've found a way to explicitly evaluate this integral (in case anybody might be interested). Let us start from equation $(2)$ in the original message: I = \int\limits^1_0 \mathrm{d} x \int \frac{d^d p'_\text{E}}{(2 \pi)^d} \frac{1}{\left[p_\text{E}'^2 + q_\text{E}^2 x(1-x) + m^2 \right]^2} ...

1

My naive guess is that the position "basis" vectors are not in fact in this Hilbert space. My reasoning is that while you can write any function with a finite number of jump discontinuities in the basis $| n\rangle$, the position basis vectors $\delta(x - x_0) = |x\rangle$ cannot be written as such. Hence they are not in that Hilbert space.

1

Wolfram claims that the universe at its core is described by some Turing universal computations (it doesn't matter what specific form it takes, such as cellular automata, tag systems, Peano arithmetic etc)as all Turing universal computations are equivalent). He also claims that mathematical descriptions are a special case of general computations where you ...

Only top voted, non community-wiki answers of a minimum length are eligible