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## Hot answers tagged mathematical-physics

13

It is a distribution. The easiest, cleanest way to think of it is as a linear functional $\mathscr{H}\to\mathbb{R}$ on the Hilbert space $\mathscr{H}$ of functions $\mathbb{R}^N\to\mathbb{R}$ that are $\mathbf{L}^2(\mathbb{R}^N)$. Input a function $f\in\mathbf{L}^2(\mathbb{R}^N)$, and DiracDelta spits out $f(0)$. It's a manifestly linear operator. ...

9

There is no general rule. However there is a class of bounded self-adjoint operators whose spectrum is made of a bounded set of isolated points (proper eigenvalues) -- except for $0$ at most -- and the eigenspaces associated to these eigenvalues are finite dimensional. They are the so-called compact operators (this class includes classes of operators ...

6

Your question has, indeed, been beaten to a pulp in the 70 years of the formulation, and, as you suggested, the necessary conditions are not all independent, so parts are redundant. For a pure state real $f(x,p)$ the sufficient condition is straightforward, eqn (6) of Ref. 1: Given its Fourier transform (the cross-spectral density) must left-right" ...

5

There are two main problems in defining the logarithm, and are related to the fact that the creation/annihilation operators are unbounded and not self-adjoint, neither normal. The first problem is that there is not a functional calculus for non-normal operators; in addition an usual power series expansion is difficult to manage for unbounded operators. It ...

3

Separable Hilbert spaces are all isomorphic to one other, since they are all isomorphic to $\ell_2$, despite the dimensions (this holds even in the infinite dimensional case). No matter how you realise it, you may just choose one Hilbert space and its properties will exhaust all you need to perform the calculations and derive observable quantities (they do ...

3

There are many approaches. But first I want to make sure you know that when you have $n$ spin zero particles in a 3d space and have a wavefunction that the function is from $\mathbb R^{3n},$ i.e. from configuration space. But also I want you to know when someone says infinite dimensional Hilbert Space, they mean the size of a set of mutually orthonormal ...

3

In QM the position operator and the Hilbert space of a particle are defined contextually: The Hilbert space is $L^2(\mathbb R^3, d^3x)$ and the operator position along $x_k$ is defined, in that space, as $(X_k\psi)(x):= x_k \psi(x)$ with the obvious domain. You can adopt a more abstract viewpoint if you simultaneously define the momentum and the position ...

3

The delta function is an instruction to switch the order of limits. Here's how to think about it. Let's say we have an infinite sequence of functions $\delta_1$, $\delta_2$, ..., which satisfy the following properties: 1) $\int\delta_i(x)dx=1$ 2) The integral of $\delta_i$ over any patch not containing $0$ approaches $0$ as $i\rightarrow\infty$ Then you ...

2

In infinite dimensions, you can define the trace only for a special class of compact operators: the so-called trace-class operators. Given an Hilbert space $\mathscr{H}$, the space of trace class operators $\mathscr{I}_1(\mathscr{H})$ is a two-sided ideal of the bounded operators $\mathscr{L}(\mathscr{H})$. The two operations $\mathrm{Tr}$ and ...

2

As mentioned in the comments by Bubble, this is answered in Ground State Energy Calculations for the Quartic Anharmonic Oscillator, Robert Smith. Notes for Math 4901, University of Minnesota, Morris (2013). but as the document is not crawlable by the Wayback Machine I'll summarize it here. Smith considers hamiltonians of the form  H=-\frac12 ...

2

It is a mathematical theorem that self-adjoint operators in Hilbert space have a complete spectrum. Note that "self-adjoint" has a special mathematical meaning. Not every Hermitian symmetric operator is self-adjoint. For example, the 1D free Schrodinger Hamiltonian on an open interval without boundary conditions is not self-adjoint. The reason is that we ...

2

The "time-independent Schrödinger equation" is just an equation for the eigenvalues and eigenvectors of the Hamiltonian operator on the Hilbert space of states (typically $L^2(\mathbb{R}^3,\mathrm{d}x)$, the "space of wavefunctions") The spectral theorem tells us that the eigenvectors of any self-adjoint operator form a basis for the space the operator ...

1

This isn't too hard. Translations clearly are isometries: if $\vec a' = \vec a + \vec c$ and $\vec b' = \vec b + \vec c$ then $|\vec a' - \vec b'| = |\vec a - \vec b|.$ Consider any isometry $f$; consider the isometry $g(x) = f(x) - f(0)$ which preserves the origin. It's not too hard to see that this has to be linear and is therefore described by a matrix ...

1

Actually there are analogies or generalisations of results which reduce to Noether's theorems under usual cases and which do hold for discrete (and not necesarily discretised) symmetries (including CPT-like symmetries) For example see: Anthony C L Ashton (2008) Conservation Laws and Non-Lie Symmetries for Linear PDEs, Journal of Nonlinear Mathematical ...

1

I think Shankar's Principles of Quantum Mechanics contains decent exercises that promote understanding of the material, many of which are proofs and derivations rather than simply computations.

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