17

This is a good question, and the answer is rather subtle, and I think a physicist and a mathematician would answer it differently. Mathematically, a Hilbert space is just any complete inner product space (where the word "complete" takes a little bit of work to define rigorously, so I won't bother). But when a physicist talks about "the Hilbert space of a ...


17

A quantum state includes the information about a particle's position. Two particles with the same quantum numbers at different locations are in different states, so are allowed by the exclusion principle.


11

The signs are important for fixing an out of order machine. Define the states $|\pm\rangle$ as: $$|\pm\rangle = \frac{1}{\sqrt{2}}\left[\left |\text{Working}\right\rangle\pm \left |\text{Down}\right\rangle\right]$$ And we define the observable $O$ as: $$O = |+\rangle\langle + |\ - \ |-\rangle\, \langle -|$$ Suppose then that coffee machine is out of ...


10

Superposition is often claimed, particularly in popular physics texts, to be "an object that is in two states at once", but this is in fact quite misleading. Superposition states, in fact, go beyond this property. Say I have a box, and I claim that it contains a cat that's both dead and alive, with the proviso that when you actually look at it you will ...


9

The time dependent Schroedinger equation looks like this: $$ i\hbar \frac{\partial \Psi}{\partial t} = H \Psi = \left ( -\frac{\hbar^2 }{2 m}\frac{\partial^2}{\partial x^2} + V(x,t) \right ) \Psi(x,t) ,$$ you attempt a solution via separation of variables: $\Psi(x,t) = \psi(x) T(t)$, plug it in. If the potential $V$ is time independent such that $V(x,t) = ...


9

The statement by yuggib is correct. To put it in perspective, I'll start with a completely general formulation, and then I'll show how vector-states and density operators fit into that picture. I won't try to be mathematically rigorous here, but I'll try to give an overview with enough keywords and references to enable further study. State = normalized ...


9

Two complex topological vector spaces $X$ and $Y$ are said to be in duality if there is a sesquilinear map $$b:X\times Y\to \mathbb{C}\; .$$ The idea is that, given such map, a dual action of $X$ on $Y$ and viceversa is defined: for any $x\in X$, $$b(x,\cdot): Y\to \mathbb{C}$$ is a linear functional. There are two natural spaces in duality with a given ...


8

The most general description of a quantum system is given by a density matrix $\rho$. It has dimensions of $N \times N$, where $N$ is the number of degrees of freedom of the system: 2 for a 2 level quantum system (qubit), 3 for 3-level etc. But often we deal with the systems that have infinite number of degrees of freedom. Such systems are quantum harmonic ...


8

A Gaussian state is a ground or thermal state of a (bosonic or fermionic) Hamiltonian which is quadratic in the creation and annihiliation operators. Those states are fully characterized by expectation values of quadratic operators, and thus $4N^2$ parameters for $N$ fermions or bosons.


8

To start, the kets are vectors, which means if we want an explicit realization of them, we would need to write them with respect to some basis. The first basis most people see is the position basis, where the basis kets are the states of definite position. Then, an arbitrary state $|\psi\rangle$ can be written as $$ |\psi\rangle = \int_{-\infty}^{\infty}dx \...


7

This is all just a result of sloppy language on the part of people describing quantum mechanics. The state $$ \left\lvert \Psi \right\rangle = \frac{1}{\sqrt{2}} \left( \left\lvert \uparrow \right\rangle + \left\lvert \downarrow \right\rangle\right) \tag{1}$$ is a superposition of the two orthogonal states $\left\lvert \uparrow \right\rangle$ and $\left\...


7

The second view is strictly broader than the first, and it is used to describe mixed states as well as the pure states described by the first view. There are several different situations that require the use of mixed states, but the simplest to understand is that of a (classically) probabilistic source for the states. In this paradigm, you have a ...


7

The solution to this is quite simple: you calculated the expectation value $\langle x \rangle$ at a particular moment in time, call it $t=0$. In order to calculate it at an arbitrary moment in time, introduce time development to the system in one of two ways: i) Schrödinger picture. States develop according to $$ i \hbar \frac{\text{d}|\psi\rangle}{\text{...


7

Restricting to finite-dimensional Hilbert spaces, you have a complete basis of orthonormal eigenkets if and only if the operator is normal, meaning that it commutes with its adjoint, $[A, A^\dagger] = 0$. In particular, Hermitian operators are normal since they are equal to their adjoints, anti-Hermitian operators are normal, and unitary operators are ...


6

Let me first answer your questions: Q1: No, the partial transposition acts on one part of the subsystem. To make this more clear, let me give a real definition of the criterion: Let $\mathcal{H}_1,\mathcal{H}_2$ be two Hilbert spaces, then the partial transpose on the second system is defined via $$ ^{PT}: \mathcal{B}(\mathcal{H}_1\otimes \mathcal{H}_2)\...


6

The GNS construction generates a representation of the $C^*$ algebra over a Hilbert space. We know that if we start from a pure state, we reach an irreducible representation, which describes an elementary quantum system. If we start from another state, the representation describes another quantum system. Some of these representations are unitarily equivalent,...


6

$\newcommand{\bs}[1]{\boldsymbol{#1}}\newcommand{\on}[1]{\operatorname{#1}}$Let $f:\mathbb R^{d^2-1}\to\mathscr B(\mathscr H)$ be the mapping from points in $\mathbb R^{d^2-1}$ to bounded operators on the Hilbert space $\mathscr H$, defined by $$f(\bs b)\equiv \frac{1}{d}\left(I+\sum_{\mu=1}^{d^2-1}b_\mu F_\mu\right).$$ It can be easily checked that, for all ...


6

You can't do $\langle \mathbf{r} | + \rangle$, because you'd be mixing two different Hilbert spaces. The states $|+\rangle$ and $|-\rangle$ are a basis for the Hilbert space of a two-state spin system, which is two-dimensional. Meanwhile, the states $|\mathbf{r}\rangle$ are basis for the Hilbert space of a spinless particle moving on $\mathbb{R}^3$ (or your ...


6

There is actually a Hilbert space where mixed states (pure and non-pure) are represented by unit vectors up to phases. This Hilbert space is not the same Hilbert space where pure states are represented by all unit vectors (up to phases). However it includes the set of pure states (the vectors of the initial Hilbert space up to multiplication with scalars) ...


5

The fact that every mixed state $\rho$ acting on a finite dimensional Hilbert spaces can be viewed as the reduced state of some pure state $|\psi\rangle$ on a bigger Hilbert space is known as purification, see this Wikipedia page, where also the algorithm is given. In OP's case of $$\rho~\in~\mathcal{B}(\mathbb{C}^n\otimes \mathbb{C}^n),$$ one may choose a ...


5

In Quantum Mechanics, a physical state is always normalized by definition. So, in fact, you have to normalize your linearly dependent vectors, obtaining the same normalized vector, and, in fact, both represent the same physical state. Norm is not important. Note however that scalars in QM are complex numbers, so your two vectors could be $|{\psi_1} \rangle$ ...


5

Intuitively, a pure state is a state that "lies entirely inside" the Hilbert space, i.e. the system it describes is not entangled with anything outside of the Hilbert space. For example, for two spins entangled in the singlet configuration, the state of the two spins together is a pure state, because the two spins are not entangled with anything else. But ...


5

The formula for the expectation value $\langle A\rangle=\langle\psi|\hat{A}|\psi\rangle$ is given for the normalized states $\langle\psi|\psi\rangle=1$. You can generalize it as \begin{equation} \langle A\rangle=\frac{\langle\psi|\hat{A}|\psi\rangle}{\langle\psi|\psi\rangle} \end{equation} Of course this expression would still be ill-defined for $|x\rangle$ ...


5

Your prescription is incorrect. The state you've written down is the (external) direct sum of the two separate vector spaces, but independent quantum particles are described by states in the tensor product of their individual state spaces. Thus, if you write each of the states as $|\psi_1\rangle = a|0\rangle + b|1\rangle$ and $|\psi_2\rangle = \alpha|0\...


5

Start with a pure qubit and a mixed qubit, then trace out the mixed qubit. The overall state goes from mixed to pure.


5

A GHZ in $M=2$ would be $|00\rangle+|11\rangle$. Sure you could still call it a GHZ, but it is not very useful because this state already has a name: it's a Bell state. A standard reference to understand entanglement and how it is measured is Gühne and Toth 2008.


5

First of all, your problem is a special version of a more general problem, namely finding the minimum number of states which minimize the entanglement of formation, this is, given a state $\rho$ on AB$\equiv \mathbb C^D\otimes \mathbb C^{D'}$, find the decomposition $$ \rho = \sum_{i=1}^m p_i |\psi_i\rangle\langle\psi_i| $$ which minimizes $\sum_i p_i E(|\...


5

Although there are many perspectives on this, largely united by the (correct) notion that Hilbert spaces allow for geometric tools to be applied, I'd like to present another overlooked perspective: Hilbert spaces are the tool of choice in QM because they can uniquely mimic probability spaces. In quantum mechanics, we are given an object that entirely ...


5

The vertical bar means nothing by itself. The notation $|\psi\rangle$ is called a “ket” and indicates a vector in a Hilbert space, representing some quantum state. The corresponding notation $\langle\psi|$ is called a “bra” and denotes the Hermitian conjugate. Scalar products between two vectors are written as $\langle\phi|\psi\rangle$, with only one ...


4

It seems that OP's question arises because he assumes that a state $|\psi\rangle$ is normalized $\langle\psi |\psi\rangle=1$ at all stages of developing the quantum mechanical language. Let $H$ be a Hilbert space. Note that the set $$\{|\psi\rangle\in H \mid \langle\psi |\psi\rangle=1\}$$ of normalized states is not a vector space, and therefore not a ...


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