Questions tagged [hilbert-space]

This tag is for questions relating to Hilbert Space, a vector space equipped with an inner product, an operation that allows defining lengths and angles, and the space is complete. It arises naturally and frequently in mathematics and physics, typically as infinite-dimensional function spaces having the property that it is complete. Applies also to pre-Hilbert spaces, rigged Hilbert spaces, and spaces with negative norm or zero-norm states.

A Hilbert space, named by von Neumann after David Hilbert, is an inner product space that is complete with respect to the norm defined by the inner product.

This notion was invented by J. von Neumann in his famous book about the mathematical foundation of Quantum Theory to formalise the space of states of a quantum system.

The theory of Hilbert spaces that Hilbert, Riesz, von Neumann and others developed has not only greatly enriched the world of mathematics but has proven extremely useful in the development of scientific theories, particularly . For example, the ability to treat functions as vectors in a Hilbert space, as permitted by Hilbert space theory, has enabled quantum physicists to solve difficult differential and integral equations by using mere algebra. What is more, the theory and notation of Hilbert space has become so ingrained in the world of quantum mechanics that it is commonly used to describe many interesting phenomena, including the EPR paradox (entanglement), quantum teleportation, and quantum telecloning.

References:
$1.~~$ "A BRIEF INTRODUCTION TO HILBERT SPACE AND QUANTUM LOGIC"
$2.~~$ "Hilbert Space Quantum Mechanics"
$3.~~$ "Hilbert Space Operators in Quantum Physics"??
$4.~~$ "Hilbert spaces"
$5.~~$ "Hilbert space"

5160 questions
127
votes
10 answers

How is a quantum superposition different from a mixed state?

According to Wikipedia, if a system has $50\%$ chance to be in state $\left|\psi_1\right>$ and $50\%$ to be in state $\left|\psi_2\right>$, then this is a mixed state. Now, consider the state…
83
votes
5 answers

Understanding the Bloch sphere

It is usually said that the points on the surface of the Bloch sphere represent the pure states of a single 2-level quantum system. A pure state being of the form: $$ |\psi\rangle = a |0\rangle+b |1\rangle $$ And typically the north and south poles…
61
votes
4 answers

Hilbert space of harmonic oscillator: Countable vs uncountable?

Hm, this just occurred to me while answering another question: If I write the Hamiltonian for a harmonic oscillator as $$H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$ then wouldn't one set of possible basis states be the set of…
56
votes
1 answer

Differences between pure/mixed/entangled/separable/superposed states

I am currently trying to establish a clear picture of pure/mixed/entangled/separable/superposed states. In the following I will always assume a basis of $|1\rangle$ and $|0\rangle$ for my quantum systems. This is what I have so far: superposed: A…
55
votes
1 answer

What is the difference between general measurement and projective measurement?

Nielsen and Chuang mention in Quantum Computation and Information that there are two kinds of measurement : general and projective ( and also POVM but that's not what I'm worried about ). General Measurements Quantum measurements are described by a…
51
votes
2 answers

Rigged Hilbert space and QM

Are there any comprehensive texts that discuss QM using the notion of rigged Hilbert spaces? It would be nice if there were a text that went through the standard QM examples using this structure.
51
votes
6 answers

Why do we need infinite-dimensional Hilbert spaces in physics?

I am looking for a simple way to understand why do we need infinite-dimensional Hilbert spaces in physics, and when exactly do they become neccessary: in classical, quantum, or relativistic quantum physics (i.e. when particles can be created and…
Dee
  • 854
46
votes
7 answers

Why are Only Real Things Measurable?

Why can't we measure imaginary numbers? I mean, we can take the projection of a complex wave to be the "viewable" part, so why are imaginary numbers given this immeasurable descriptor? Namely with operators in quantum mechanics, why must measurable…
46
votes
3 answers

Not all self-adjoint operators are observables?

The WP article on the density matrix has this remark: It is now generally accepted that the description of quantum mechanics in which all self-adjoint operators represent observables is untenable.[17][18] The first footnote is to the appendix in…
46
votes
3 answers

What is the issue with interactions in QFT?

I've started studying QFT this year and in trying to find a more rigorous approach to the subject I ended up find out lots of people saying that "there is no way known yet to make QFT rigorous when there are interactions". As for the textbook…
Gold
  • 38,087
  • 19
  • 112
  • 289
46
votes
5 answers

Density matrix formalism

The density matrix $\hat{\rho}$ is often introduced in textbooks as a mathematical convenience that allows us to describe quantum systems in which there is some level of missing information. $\hat{\rho} = \sum_{i=1}^N c_i \rvert\psi_i\rangle \langle…
45
votes
3 answers

A "Hermitian" operator with imaginary eigenvalues

Let $${\bf H}=\hat{x}^3\hat{p}+\hat{p}\hat{x}^3$$ where $\hat{p}=-id/dx$. Clearly ${\bf H}^{\dagger}={\bf H}$, because ${\bf H}={\bf T} + {\bf T}^{\dagger}$, where ${\bf T}=\hat{x}^3\hat{p}$. In this sense ${\bf H}$ is "formally" self-adjoint. It…
45
votes
9 answers

Why ket and bra notation?

So, I've been trying to teach myself about quantum computing, and I found a great YouTube series called Quantum Computing for the Determined. However. Why do we use ket/bra notation? Normal vector notation is much clearer (okay, clearer because I've…
45
votes
2 answers

Physical interpretation of different selfadjoint extensions

Given a symmetric (densely defined) operator in a Hilbert space, there might be quite a lot of selfadjoint extensions to it. This might be the case for a Schrödinger operator with a "bad" potential. There is a "smallest" one (Friedrichs) and a…
45
votes
3 answers

What's wrong with this derivation that $i\hbar = 0$?

Let $\hat{x} = x$ and $\hat{p} = -i \hbar \frac {\partial} {\partial x}$ be the position and momentum operators, respectively, and $|\psi_p\rangle$ be the eigenfunction of $\hat{p}$ and therefore $$\hat{p} |\psi_p\rangle = p |\psi_p\rangle,$$ where…
1
2 3
99 100