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Results tagged with hilbert-space
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user 326183
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.
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Some questions on extraction of quantum probabilities from vectors and Hilbert space?
In the framework of textbook quantum mechanics, states are abstract vectors $\lvert \psi \rangle$ of a Hilbert space $\mathcal{H}$. More technically, a state is an equivalence class of vectors defined …
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4
votes
3
answers
886
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Is the zero vector necessary to do quantum mechanics?
Textbook quantum mechanics describes systems as Hilbert spaces $\mathcal{H}$, states as unit vectors $\psi \in \mathcal{H}$, and observables as operators $O: \mathcal{H} \to \mathcal{H}$. Ultimately, …
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1
answer
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What are the distinct mathematical formalisms of quantum mechanics?
Consider the physical theory called non-relativistic quantum mechanics. What are the distinct mathematical formalisms for this physical theory? That is, different mathematical frameworks for this phys …
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2
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answers
45
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Simple way to see if $n$-qubit Hamiltonian has no interaction terms?
Consider a system of $n$-qubits. The Hilbert space is $\mathcal{H} \cong \mathbb{C}^2 \otimes ... \otimes \mathbb{C}^2$. Fix each tensor factor $\mathbb{C}^2$ to have the $\sigma_z$ vector space basis …
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The abstract state of a particle
At a high level, a state of a quantum system is a vector $\lvert \psi \rangle$ of a complex vector space $\mathcal{H}$. If you recall the definition of a vector space, a vector need not be a column of …
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3
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Is maximal entanglement basis independent?
Let $\mathcal{H}$ with dimension $\dim \mathcal{H} = 2^n$ where $n \in \mathbb{N}$ be a Hilbert space describing your system of interest. This could be a Hilbert space for a system of $n$ qubits, for …
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2
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95
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Is Sakurai's derivation of the Lippmann-Schwinger equation correct?
I am using Sakurai's Modern Quantum Mechanics 3rd ed. The following is from the beginning of chapter 6.
The defining equation for the $T$-matrix is
$$\langle \vec{k}' \lvert U_I(t, t_0) \lvert \vec{k} …
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2
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Does Sakurai's definition of $S$-matrix assume a particular type of scattering?
I am using Sakurai's Modern Quantum Mechanics 3ed.
In chapter 6, Sakurai defines the $T$-matrix via the equation
$$\langle \vec{k}' \lvert U_I(t, t_0) \lvert \vec{k} \rangle = \delta_{k'k} - \frac{i}{ …
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2
votes
0
answers
93
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Computing Fubini-Study expectation values over $\mathbb{C}P^n$
In finite-dimensional textbook quantum mechanics, we postulate that states of our system are rays in a Hilbert space $\mathcal{H}$ with dimension $\dim{\mathcal{H}} = n+1$ where $n \in \mathbb{N}$, in …
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How to find the expectation value of momentum operator?
I am going to rewrite your formulas because the placement of some of your quantities is personally confusing.
$$\left \langle \hat{p} \right \rangle = \int_{-\infty}^{\infty}dx \ \psi^{*}(x)\left(-i\h …
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Derivative of $c(t)$ in Adiabatic Approximation
You are incorrectly assuming that the instantaneous eigenstates $\lvert n; t\rangle$ satisfy the Schrödinger equation. Instead, they are merely defined by the equation
$$H(t)\lvert n;t\rangle = E_n(t) …
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5
votes
2
answers
654
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What is the Majorana stellar representation?
One can geometrically visualize spin-1/2 states using the Bloch sphere. A natural question then is: "Can one geometrically visualize spin-$s$ states using a similar object to the Bloch sphere?" It see …
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8
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Accepted
What is the Majorana stellar representation?
The Majorana stellar representation is a way to geometrically visualize pure spin-s states. In essence, the Majorana stellar representation 1) establishes a bijection between states of Hilbert space a …
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1
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Time evolution of operators in the Heisenberg picture
The crux of the matter is this, the Schrödinger and Heisenberg pictures are both valid because they both yield the same observable results. That is to say, the amplitude
$$\langle \psi\lvert U^\dagger …
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vote
Accepted
How does Weinberg definition of particle states from standard momentum work?
I believe 3. is literally a definition. That is, while the unitary representation of an arbitrary homogenous Lorentz transformation $U(\Lambda)$ acts as:
$$U(\Lambda) \Psi_{\rho, \sigma} = \sum_{\sigm …
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