Qu'est-ce que cela signifie que deux qubits soient enchevêtrés?

J'ai fait une sorte de recherche en ligne sur les qubits et les facteurs les rendant infâmes, c'est-à-dire permettre aux qubits de contenir 1 et 0 en même temps et une autre est que les qubits peuvent être enchevêtrés d'une manière ou d'une autre afin qu'ils puissent avoir des données connexes en eux, peu importe la distance ils le sont (même aux côtés opposés des galaxies).

En lisant à ce sujet sur Wikipedia, j'ai vu une équation qui est encore difficile à comprendre pour moi. Voici le lien vers Wikipedia .

Des questions:

1. Comment sont-ils enchevêtrés en premier lieu?

2. Comment relient-ils leurs données?

Pour un exemple simple, supposons que vous ayez deux qubits dans des états définis et | 0 ⟩ . L'état combiné du système est | 0 ⟩ ⊗ | 0 ⟩ ou | 00 ⟩ en sténographie.$|0\rangle$$|0\rangle$$|0\rangle\otimes |0\rangle$$|00\rangle$

Ensuite, si nous appliquons les opérateurs suivants aux qubits (l'image est coupée de la page wiki de codage superdense ), l'état résultant est un état intriqué, l'un des états de la cloche .

Tout d'abord dans l'image, nous avons la porte hadamard agissant sur le premier qubit, qui dans une forme plus longue est sorte qu'elle est l'opérateur d'identité sur le deuxième qubit.$H\otimes I$

La matrice de hadamard ressemble à où la base est ordonnée{| 0⟩,| 1⟩}.

$$H=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$$ $\{|0\rangle,|1\rangle\}$

Donc, après que l'opérateur hadamard a agi, l'État est maintenant

$$(H\otimes I)(|0\rangle\otimes|0\rangle)=H|0\rangle\otimes I |0\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\otimes (|0\rangle)=\frac{1}{\sqrt{2}}(|00\rangle+|10\rangle)$$

La partie suivante du circuit est une porte non commandée, qui n'agit sur le deuxième qubit que si le premier qubit est un .$1$

Vous pouvez représenter comme | 0 ⟩ ⟨ 0 | ⊗ I + | 1 ⟩ ⟨ 1 | ⊗ X , où | 0 ⟩ ⟨ 0 | est un opérateur de projection sur le bit 0 , ou sous forme matricielle ( 1 0 0 0 ) . De même | 1 ⟩ ⟨ 1 | est ( 0 0 0 1 ) .$CNOT$$|0\rangle\langle0|\otimes I+|1\rangle\langle1|\otimes X$$|0\rangle\langle0|$$0$$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$|1\rangle\langle1|$$\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$

L' opérateur est l'opérateur de basculement de bits représenté par ( 0 1 1 0 ) .$X$$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

Globalement, la matrice est ( 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0$CNOT$$\begin{pmatrix} 1 & 0 &0 & 0 \\ 0 & 1 &0 & 0 \\ 0 & 0 &0 & 1 \\0 & 0 &1 & 0 \\\end{pmatrix}$

Lorsque nous appliquons le nous pouvons soit utiliser la multiplication matricielle en écrivant notre état comme vecteur ( $CNOT$, ou nous pouvons simplement utiliser la forme du produit tensoriel.$\begin{pmatrix}\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \\0 \end{pmatrix}$

$$CNOT (\frac{1}{\sqrt{2}}(|00\rangle+|10\rangle))=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$$

On voit que pour la première partie de l'Etat le premier bit est 0 , de sorte que le second bit est laissé seul; la deuxième partie de l'état | 10 ⟩ le premier bit est 1 , de sorte que le second bit est lancée d' 0 à 1 .$|00\rangle$$0$$|10\rangle$$1$$0$$1$

Notre état final est

$$\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$$ which is one of the four Bell states which are maximally entangled states.

To see what it means for them to be entangled, notice that if you were to measure the state of the first qubit say, if you found out that it was a $0$ it immediately tells you the second qubit also has to be a $0$, because thats our only possibility.

Compare to this state for instance:

$$\frac{1}{2}(|00\rangle+|01\rangle+|10\rangle+|11\rangle).$$

If you measure that the first qubit is a zero, then the state collapses to $\frac{1}{\sqrt{2}}(|00\rangle+|01\rangle)$, where there is still a 50-50 chance the second qubit is a $0$ or a $1$.

Hopefully this gives an idea how states can be entangled. If you want to know a particular example, like entangling photons or electrons etc, then you would have to look into how certain gates can be implemented, but still you might write the mathematics the same way, the $0$ and $1$ might represent different things in different physical situations.

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Update 1: Mini Guide to QM/QC/Dirac notation

Usually there's a standard computational (ortho-normal) basis for a single qubit which is $\{|0\rangle,|1\rangle\}$, say $\mathcal{H}=\operatorname{span}\{|0\rangle,|1\rangle\}$ is the vector space.

$|0\rangle$$\begin{pmatrix} 1\\ 0 \end{pmatrix}$$|1\rangle$$\begin{pmatrix} 0\\ 1 \end{pmatrix}$$X$$\sigma_x$$|0\rangle\mapsto |1\rangle$$|1\rangle \mapsto |0\rangle$, can be written as $\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$, the first column of the matrix is the image of the first basis vector and so on.

When you have multiple say $n$-qubits they should belong to the space $\mathcal{H}^{\otimes n}:=\overbrace{\mathcal{H}\otimes\mathcal{H}\otimes\cdots\otimes \mathcal{H}}^{n-times}$. A basis for this space is labelled by strings of zeros and ones, e.g. $|0\rangle\otimes|1\rangle\otimes |1\rangle\otimes\ldots \otimes|0\rangle$, which is usually abbreviated for simplicity as $|011\ldots0\rangle$.

A simple example for two qubits, the basis for $\mathcal{H}^{\otimes 2}=\mathcal{H}\otimes \mathcal{H}$, is $\{|0\rangle\otimes|0\rangle,|0\rangle\otimes|1\rangle,|1\rangle\otimes|0\rangle,|1\rangle\otimes|1\rangle\}$ or in the shorthand $\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\}$.

There's different ways to order this basis in order to use matrices, but one natural one is to order the strings as if they are numbers in binary so as above. For example for $3$ qubits you could order the basis as

$$\{|000\rangle,|001\rangle,|010\rangle,|011\rangle,|100\rangle,|101\rangle,|110\rangle,|111\rangle\}.$$

The reason why this can be useful is that it corresponds with the Kronecker product for the matrices of the operators. For instance, first looking at the basis vectors:

$$|0\rangle\otimes |0\rangle=\begin{pmatrix} 1\\ 0 \end{pmatrix}\otimes \begin{pmatrix} 1\\ 0 \end{pmatrix}:=\begin{pmatrix} 1\cdot\begin{pmatrix} 1\\ 0 \end{pmatrix} \\ 0\cdot\begin{pmatrix} 1\\ 0 \end{pmatrix} \end{pmatrix}=\begin{pmatrix} 1\\ 0\\0\\0 \end{pmatrix}$$

and

$$|0\rangle\otimes |1\rangle=\begin{pmatrix} 1\\ 0 \end{pmatrix}\otimes \begin{pmatrix} 0\\ 1 \end{pmatrix}:=\begin{pmatrix} 1\cdot\begin{pmatrix} 0\\ 1 \end{pmatrix} \\ 0\cdot\begin{pmatrix} 1\\ 0 \end{pmatrix} \end{pmatrix}=\begin{pmatrix} 0\\ 1\\0\\0 \end{pmatrix}$$

and similarly

$$|1\rangle\otimes |0\rangle=\begin{pmatrix} 0\\ 0\\1\\0 \end{pmatrix},\quad |1\rangle\otimes |1\rangle=\begin{pmatrix} 0\\ 0\\0\\1 \end{pmatrix}$$

If you have an operator e.g. $X_1X_2:=X\otimes X$ which acts on two qubits and we order the basis as above we can take the kronecker product of the matrices to find the matrix in this basis:

$$X_1X_2=X\otimes X=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\otimes \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0\cdot\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} & 1\cdot\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\ 1\cdot\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} & 0\cdot \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \end{pmatrix}=\begin{pmatrix} 0 &0&0&1\\ 0 &0&1&0\\0 &1&0&0\\1 &0&0&0\\ \end{pmatrix}$$

If we look at the example of $CNOT$ above given as $|0\rangle\langle0|\otimes I+|1\rangle\langle1|\otimes X$.$^*$ This can be computed in matrix form as $\begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}\otimes \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}+\begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}\otimes\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$, which you can check is the $CNOT$ matrix above.

It's worthwhile getting used to using the shorthands and the tensor products rather than converting everything to matrix representation since the computational space grows as $2^n$ for $n$-qubits, which means for three cubits you have $8\times 8$ matrices, $4$-qubits you have $16\times 16$ matrices and it quickly becomes less than practical to convert to matrix form.

Aside$^*$: There are a few common ways to use dirac notation, to represent vectors like $|0\rangle$; dual vectors e.g. $\langle 0|$, inner product $\langle 0|1\rangle$ between the vectors $|0\rangle$ and $|1\rangle$; operators on the space like $X=|0\rangle\langle1|+|1\rangle\langle0|$.

An operator like $P_0=|0\rangle\langle0|$ is a projection operator is a (orthogonal) projection operator because it satisfies $P^2=P$ and $P^\dagger=P$.

Although the linked wikipedia article is trying to use entanglement as a distinguishing feature from classical physics, I think one can start to get some understanding about entanglement by looking at classical stuff, where our intuition works a little better...

Imagine you have a random number generator that, each time, spits out a number 0,1,2 or 3. Usually you'd make these equally probability, but we can assign any probability to each outcome that we want. For example, let's give 1 and 2 each with probability 1/2, and never give 0 or 3. So, each time the random number generator picks something, it gives 1 or 2, and you don't know in advance what it's going to be. Now, let's write these numbers in binary, 1 as 01 and 2 as 10. Then, we give each bit to a different person, say Alice and Bob. Now, when the random number generator picks a value, either 01 or 10, Alice has one part, and Bob has the other. So, Alice can look at her bit, and whatever value she gets, she knows that Bob has the opposite value. We say these bits are perfectly anti-correlated.

Entanglement works much the same way. For example, you might have a quantum state

$$\left|\psi\right\rangle=\frac{1}{\sqrt{2}}\left(\left|01\right\rangle-\left|10\right\rangle\right)$$ where Alice holds one qubit of $\left|\psi\right\rangle$, and Bob holds the other. Whatever single-qubit projective measurement Alice chooses to make, she'll get an answer 0 or 1. If Bob makes the same measurement on his qubit, he always gets the opposite answer. This includes measuring in the Z-basis, which reproduces the classical case.

The difference comes from the fact that this holds true for every possible measurement basis, and for that to be the case, the measurement outcome must be unpredictable, and that's where it differs from the classical case (you may like to read up about Bell tests, specifically the CHSH test). In the classical random number example I described at the start, once the random number generator has picked something, there's no reason why it can't be copied. Somebody else would be able to know what answer both Alice and Bob would get. However, in the quantum version, the answers that Alice and Bob get do not exist is advance, and therefore nobody else can know them. If somebody did know them, the two answers would not be perfectly anti-correlated. This is the basis of Quantum Key Distribution as it basically describes being able to detect the presence of an eavesdropper.

Something further that may help in trying to understand entanglement: mathematically, it’s no different to superposition, it’s just that, at some point, you separate the superposed parts over a great distance, and the fact that that is in some sense difficult to do means that making the separation provides you with a resource that you can do interesting things with. Really, entanglement is the resource of what one might call ‘distributed superposition’.

Entanglement is a quantum physical phenomenon, demonstrated in practical experiments, mathematically modeled in quantum mechanics. We can come up with several creative speculations of what it is (philosophically), but at the end of the day we just have to accept it and trust the math.

From a statistics point of view we can think of it as a complete correlation (1 or -1) between two random variables (the qubits). We may not know any these variables outcome beforehand, but once we measure one of them, due to the correlation, the other will be previsible. I recently wrote an article on how quantum entanglement is handled by a quantum computing simulator, wich you may find helpful as well.