¿Qué significa que se enreden dos qubits?

He hecho algún tipo de investigación en línea sobre qubits y los factores que los hacen infames, es decir, permitir que los qubits mantengan 1 y 0 al mismo tiempo y otro es que los qubits pueden enredarse de alguna manera para que puedan tener datos relacionados en ellos, sin importar cuán lejos son (incluso en lados opuestos de las galaxias).

Mientras leía sobre esto en Wikipedia, he visto algunas ecuaciones que todavía me cuesta comprender. Aquí está el enlace a Wikipedia .

Preguntas:

1. ¿Cómo se enredan en primer lugar?

2. ¿Cómo relacionan sus datos?

Para un ejemplo simple, suponga que tiene dos qubits en estados definidos y | 0 ⟩ . El estado combinado del sistema es | 0 ⟩ ⊗ | 0 ⟩ o | 00 ⟩ en taquigrafía.$|0\rangle$$|0\rangle$$|0\rangle\otimes |0\rangle$$|00\rangle$

Luego, si aplicamos los siguientes operadores a los qubits (la imagen se corta de la página wiki de codificación superdensa ), el estado resultante es un estado enredado, uno de los estados de campana .

Primero en la imagen tenemos la puerta hadamard que actúa en el primer qubit, que en una forma más larga es para que sea el operador de identidad en el segundo qubit.$H\otimes I$

La matriz de hadamard se parece a donde se ordena la base{| 0⟩,| 1⟩}.

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

Entonces, después de que el operador hadamard actúa, el estado ahora

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

La siguiente parte del circuito es una puerta no controlada, que solo actúa en el segundo qubit si el primer qubit es un .$1$

Puede representar como | 0 ⟩ ⟨ 0 | ⊗ I + | 1 ⟩ ⟨ 1 | ⊗ X , donde | 0 ⟩ ⟨ 0 | es un operador de proyección en el bit 0 , o en forma de matriz ( 1 0 0 0 ) . De manera similar | 1 ⟩ ⟨ 1 | es ( 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}$

El operador es el operador de cambio de bit representado como ( 0 1 1 0 ) .$X$$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

En general, la matriz es ( 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}$

Cuando aplicamos podemos usar la multiplicación de matrices escribiendo nuestro estado como un vector ( $CNOT$, o simplemente podemos usar la forma del producto tensorial.$\begin{pmatrix}\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \\0 \end{pmatrix}$

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

Vemos que para la primera parte del estado el primer bit es 0 , por lo que el segundo bit se queda solo; la segunda parte del estado | 10 ⟩ el primer bit es 1 , por lo que el segundo bit se invierte a partir de 0 a 1 .$|00\rangle$$0$$|10\rangle$$1$$0$$1$

Nuestro estado final es que es uno de los cuatro estados de Bell que son máximamente estados entrelazados.

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

Para ver qué significa para ellos estar enredados, observe que si midiera el estado del primer qubit diga, si descubrió que era un , inmediatamente le dice que el segundo qubit también debe ser un 0 , porque Esa es nuestra única posibilidad.$0$$0$

Compare con este estado, por ejemplo:

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

Si mide que el primer qubit es cero, entonces el estado colapsa a , donde todavía hay una posibilidad de 50-50 el segundo qubit es un0o un1.$\frac{1}{\sqrt{2}}(|00\rangle+|01\rangle)$$0$$1$

Esperemos que esto dé una idea de cómo se pueden enredar los estados. Si desea conocer un ejemplo en particular, como enredar fotones o electrones, etc., entonces debería investigar cómo se pueden implementar ciertas puertas, pero aún así podría escribir las matemáticas de la misma manera, el y el 1 podrían representar cosas diferentes en Diferentes situaciones físicas.$0$$1$

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Actualización 1: Mini guía de notación QM / QC / Dirac

Por lo general, hay una base computacional estándar (orto-normal) para un solo qubit que es , por ejemplo H = lapso { | 0 ⟩ , | 1 ⟩ } es el espacio vectorial.$\{|0\rangle,|1\rangle\}$$\mathcal{H}=\operatorname{span}\{|0\rangle,|1\rangle\}$

En este ordenamiento de la base podemos identificar con ( 1 0 ) y | 1 ⟩ con ( 0 1 ) . Cualquier operador de qubit único se puede escribir en forma de matriz utilizando esta base. Por ejemplo, un operador de cambio de bit X (después de pauli- σ x ) que debería tomar | 0 ⟩ ↦ | 1 ⟩ y | 1 ⟩ ↦ | 0 ⟩ , se puede escribir como ( 0 1 1 $|0\rangle$$\begin{pmatrix} 1\\ 0 \end{pmatrix}$$|1\rangle$$\begin{pmatrix} 0\\ 1 \end{pmatrix}$$X$$\sigma_x$$|0\rangle\mapsto |1\rangle$$|1\rangle \mapsto |0\rangle$$\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 $|011\ldots0\rangle$.

$\mathcal{H}^{\otimes 2}=\mathcal{H}\otimes \mathcal{H}$$\{|0\rangle\otimes|0\rangle,|0\rangle\otimes|1\rangle,|1\rangle\otimes|0\rangle,|1\rangle\otimes|1\rangle\}$$\{|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.