# Che cosa significa intrappolare due qubit?

14

Ho fatto una sorta di ricerca online sui qubit e sui fattori che li rendono famigerati, ovvero consentendo ai qubit di contenere 1 e 0 contemporaneamente e un altro è che i qubit possono essere intrecciati in qualche modo in modo che possano avere dati correlati in essi, non importa quanto lontano sono (anche ai lati opposti delle galassie).

Durante la lettura di questo su Wikipedia ho visto alcune equazioni che è ancora difficile da comprendere per me. Ecco il link a Wikipedia .

Domande:

1. Come sono intrappolati in primo luogo?

2. Come si collegano i loro dati?

2
Puoi prendere in considerazione il collegamento all'articolo Wikipedia / includere la formula nella tua domanda? Ciò renderà più facile per gli altri capire quale sia esattamente il tuo problema.
MEE - Ripristina Monica il

la risposta di snulty è una risposta di alta qualità alla domanda 1 in questo post, ma non è abbastanza per rispondere alla domanda del titolo. L'entanglement è un concetto sottile non completamente riducibile a "due sistemi perfettamente correlati". La risposta di DaftWullie va un po 'oltre nel cercare di spiegare perché l'entanglement non è solo correlazioni perfette. Le parole chiave per le ricerche future sono le disuguaglianze di Bell e questo eccellente aper di Mermin web.pdx.edu/~pmoeck/lectures/Mermin%20longer.pdf
Andrea

Risposte:

15

Per un semplice esempio supponiamo di avere due qubit in stati definiti $|0⟩$$|0\rangle$ e $|0⟩$$|0\rangle$ . Lo stato combinato del sistema è $|0⟩\otimes |0⟩$$|0\rangle\otimes |0\rangle$ o $|00⟩$$|00\rangle$ in stenografia.

Quindi se applichiamo i seguenti operatori ai qubit (l'immagine è tagliata dalla pagina wiki di codifica superdensa ), lo stato risultante è uno stato impigliato, uno degli stati di campana .

Innanzitutto nell'immagine abbiamo la porta hadamard che agisce sul primo qubit, che in una forma più lunga è $H\otimes I$$H\otimes I$ modo che sia l'operatore di identità sul secondo qubit.

$H=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$
dove viene ordinata la base .$\left\{|0⟩,|1⟩\right\}$$\{|0\rangle,|1\rangle\}$

Quindi, dopo che l'operatore Hadamard agisce, lo stato è ora

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

La parte successiva del circuito è un gate non controllato, che agisce sul secondo qubit solo se il primo qubit è .$1$$1$

È possibile rappresentare come , dove è un operatore di proiezione sul bit o in forma di matrice . Allo stesso modo è .$CNOT$$CNOT$$|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes X$$|0\rangle\langle0|\otimes I+|1\rangle\langle1|\otimes X$$|0⟩⟨0|$$|0\rangle\langle0|$$0$$0$$\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$|1⟩⟨1|$$|1\rangle\langle1|$$\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$$\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$

L' operatore è l'operatore di inversione bit rappresentato come .$X$$X$$\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

Nel complesso la matrice è $CNOT$$CNOT$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\end{array}\right)$$\begin{pmatrix} 1 & 0 &0 & 0 \\ 0 & 1 &0 & 0 \\ 0 & 0 &0 & 1 \\0 & 0 &1 & 0 \\\end{pmatrix}$

Quando applichiamo il possiamo usare la moltiplicazione della matrice scrivendo il nostro stato come vettore $CNOT$$CNOT$, oppure possiamo semplicemente utilizzare il modulo del prodotto tensore.$\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\ 0\\ \frac{1}{\sqrt{2}}\\ 0\end{array}\right)$$\begin{pmatrix}\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \\0 \end{pmatrix}$

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

Lo vediamo per la prima parte dello stato il primo bit è , quindi il secondo bit è lasciato solo; la seconda parte dello stato il primo bit è , quindi il secondo bit viene capovolte provenienti da a .$|00⟩$$|00\rangle$$0$$0$$|10⟩$$|10\rangle$$1$$1$$0$$0$$1$$1$

Il nostro stato finale è

$\frac{1}{\sqrt{2}}\left(|00⟩+|11⟩\right)$
che è uno dei quattro stati di Bell, che sono massimamente stati entangled.

Per vedere cosa significa per loro essere intrecciati, nota che se dovessi misurare lo stato del primo qubit dire, se hai scoperto che era uno ti dice immediatamente che anche il secondo qubit deve essere uno , perché questa è la nostra unica possibilità.$0$$0$$0$$0$

Confronta con questo stato per esempio:

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

Se si misura che il primo qubit è zero, lo stato si riduce a , dove c'è ancora una possibilità 50-50 il secondo qubit è uno un.$\frac{1}{\sqrt{2}}\left(|00⟩+|01⟩\right)$$\frac{1}{\sqrt{2}}(|00\rangle+|01\rangle)$$0$$0$$1$$1$

Spero che questo dia un'idea di come gli stati possano essere intrappolati. Se vuoi conoscere un esempio particolare, come intrappolare fotoni o elettroni ecc., Allora dovresti esaminare come possono essere implementate determinate porte, ma potresti comunque scrivere la matematica allo stesso modo, lo e potrebbero rappresentare cose diverse in diverse situazioni fisiche.$0$$0$$1$$1$

Aggiornamento 1: Mini guida alla notazione QM / QC / Dirac

Di solito esiste una base computazionale standard (orto-normale) per un singolo qubit che è , dire $\left\{|0⟩,|1⟩\right\}$$\{|0\rangle,|1\rangle\}$ è lo spazio vettoriale.$\mathcal{H}=\mathrm{span}\left\{|0⟩,|1⟩\right\}$$\mathcal{H}=\operatorname{span}\{|0\rangle,|1\rangle\}$

In questo ordinamento della base possiamo identificare con e con . Qualsiasi singolo operatore qubit può quindi essere scritto in forma di matrice usando questa base. Ad esempio un bit capovolgere l'operatore (dopo pauli- ) che dovrebbe prendere e , può essere scritta come $|0⟩$$|0\rangle$$\left(\begin{array}{c}1\\ 0\end{array}\right)$$\begin{pmatrix} 1\\ 0 \end{pmatrix}$$|1⟩$$|1\rangle$$\left(\begin{array}{c}0\\ 1\end{array}\right)$$\begin{pmatrix} 0\\ 1 \end{pmatrix}$$X$$X$${\sigma }_{x}$$\sigma_x$$|0⟩↦|1⟩$$|0\rangle\mapsto |1\rangle$$|1⟩↦|0⟩$$|1\rangle \mapsto |0\rangle$ , la prima colonna della matrice è l'immagine del vettore della prima base e così via.$\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$$\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$

$n$$n$${\mathcal{H}}^{\otimes n}:=\stackrel{n-times}{\stackrel{⏞}{\mathcal{H}\otimes \mathcal{H}\otimes \cdots \otimes \mathcal{H}}}$$\mathcal{H}^{\otimes n}:=\overbrace{\mathcal{H}\otimes\mathcal{H}\otimes\cdots\otimes \mathcal{H}}^{n-times}$$|0⟩\otimes |1⟩\otimes |1⟩\otimes \dots \otimes |0⟩$$|0\rangle\otimes|1\rangle\otimes |1\rangle\otimes\ldots \otimes|0\rangle$$|011\dots 0⟩$$|011\ldots0\rangle$

${\mathcal{H}}^{\otimes 2}=\mathcal{H}\otimes \mathcal{H}$$\mathcal{H}^{\otimes 2}=\mathcal{H}\otimes \mathcal{H}$, is $\left\{|0⟩\otimes |0⟩,|0⟩\otimes |1⟩,|1⟩\otimes |0⟩,|1⟩\otimes |1⟩\right\}$$\{|0\rangle\otimes|0\rangle,|0\rangle\otimes|1\rangle,|1\rangle\otimes|0\rangle,|1\rangle\otimes|1\rangle\}$ or in the shorthand $\left\{|00⟩,|01⟩,|10⟩,|11⟩\right\}$$\{|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$$3$ qubits you could order the basis as

$\left\{|000⟩,|001⟩,|010⟩,|011⟩,|100⟩,|101⟩,|110⟩,|111⟩\right\}.$

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:

$|0⟩\otimes |0⟩=\left(\begin{array}{c}1\\ 0\end{array}\right)\otimes \left(\begin{array}{c}1\\ 0\end{array}\right):=\left(\begin{array}{c}1\cdot \left(\begin{array}{c}1\\ 0\end{array}\right)\\ 0\cdot \left(\begin{array}{c}1\\ 0\end{array}\right)\end{array}\right)=\left(\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right)$

and

$|0⟩\otimes |1⟩=\left(\begin{array}{c}1\\ 0\end{array}\right)\otimes \left(\begin{array}{c}0\\ 1\end{array}\right):=\left(\begin{array}{c}1\cdot \left(\begin{array}{c}0\\ 1\end{array}\right)\\ 0\cdot \left(\begin{array}{c}1\\ 0\end{array}\right)\end{array}\right)=\left(\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right)$

and similarly

$|1⟩\otimes |0⟩=\left(\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right),\phantom{\rule{1em}{0ex}}|1⟩\otimes |1⟩=\left(\begin{array}{c}0\\ 0\\ 0\\ 1\end{array}\right)$

If you have an operator e.g. ${X}_{1}{X}_{2}:=X\otimes X$$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:

${X}_{1}{X}_{2}=X\otimes X=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\otimes \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}0\cdot \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)& 1\cdot \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\\ 1\cdot \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)& 0\cdot \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\end{array}\right)=\left(\begin{array}{cccc}0& 0& 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\end{array}\right)$

If we look at the example of $CNOT$$CNOT$ above given as $|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes X$$|0\rangle\langle0|\otimes I+|1\rangle\langle1|\otimes X$.${}^{\ast }$$^*$ This can be computed in matrix form as $\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\otimes \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)+\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)\otimes \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$$\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$$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}$$2^n$ for $n$$n$-qubits, which means for three cubits you have $8×8$$8\times 8$ matrices, $4$$4$-qubits you have $16×16$$16\times 16$ matrices and it quickly becomes less than practical to convert to matrix form.

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

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

I've failed to see through the complete calculations part, as I don't have the fundamentals to simplify. But it helped me to get an idea!
Arshdeep Singh

@ArshdeepSingh I can try add in anything that helps in understanding. I could probably add a bit more about entangled states. Glad it was a bit helpful anyway :)
snulty

@snulty maybe if you use the vector notation for the qubits the calculations become more transparent? Just a suggestion.
Kiro

1
@Kiro I've added a small bit about vector/matrix notation, only you might want to move away form that notation where possible in order to avoid multiplying large matrices by hand.
snulty

5

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

$|\psi ⟩=\frac{1}{\sqrt{2}}\left(|01⟩-|10⟩\right)$
where Alice holds one qubit of $|\psi ⟩$$\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’.

2

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.

I have two blank pieces of paper. I flip a coin and write the outcome on both of them and fold them. I hand you one of the two pieces and I keep the other. This process creates two random variables. You may not know the value of either, but if you measure one, you immediately know the other. Does this process entangle the pieces of paper?
Andrea

Great question! The analogy may seem valid at first, but there's one problem, once qubits become entangled you can perform additional operations on them, modifying their internal state simultaneously. This behavior can be used for instance to implement quantum teleportation. In your case we end up with a classical deterministic system in which states are pre-determined, and further operations that take advantage of physical entanglement phenomenon are not possible.
Thomas C. G. de Vilhena