Apa artinya dua qubit dilibatkan?

Saya telah melakukan semacam riset online tentang qubit dan faktor-faktor yang membuat mereka terkenal yaitu membiarkan qubit untuk memegang 1 dan 0 pada saat yang sama dan yang lain adalah bahwa qubit dapat terjerat entah bagaimana sehingga mereka dapat memiliki data terkait di dalamnya tidak peduli seberapa jauh mereka (bahkan di sisi yang berlawanan dari galaksi).

Saat membaca tentang ini di Wikipedia saya telah melihat beberapa persamaan yang masih sulit untuk saya pahami. Inilah tautan ke Wikipedia .

Pertanyaan:

1. Bagaimana mereka terjerat pada awalnya?

2. Bagaimana mereka menghubungkan data mereka?

Sebagai contoh sederhana misalkan Anda memiliki dua qubit di negara tertentu $|0\rangle$ dan $|0\rangle$ . Keadaan gabungan sistem adalah $|0\rangle\otimes |0\rangle$ atau $|00\rangle$ di steno.

Kemudian jika kita menerapkan operator berikut ke qubit (gambar dipotong dari halaman wiki pengkodean superdense ), keadaan yang dihasilkan adalah keadaan terjerat, salah satu status bel .

Pertama pada gambar kita memiliki gerbang hadamard yang bekerja pada qubit pertama, yang dalam bentuk yang lebih panjang adalah $H\otimes I$ sehingga itu adalah operator identitas pada qubit kedua.

Matriks hadamard terlihat seperti mana basis dipesan{| 0⟩,| 1⟩}.

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

Jadi setelah operator hadamard bertindak negara sekarang

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

Bagian selanjutnya dari rangkaian adalah gerbang bukan yang dikontrol, yang hanya bekerja pada qubit kedua jika qubit pertama adalah .$1$

Anda dapat mewakili sebagai | 0 ⟩ ⟨ 0 | ⊗ I + | 1 ⟩ ⟨ 1 | ⊗ X , di mana | 0 ⟩ ⟨ 0 | adalah operator proyeksi ke bit 0 , atau dalam bentuk matriks ( 1 0 0 0 ) . Demikian pula | 1 ⟩ ⟨ 1 | adalah ( 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}$

The operator adalah operator bit sandal direpresentasikan sebagai ( 0 1 1 0 ) .$X$$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

Secara keseluruhan matriks ( 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}$

Ketika kita menerapkan kita bisa menggunakan perkalian matriks dengan menuliskan status kita sebagai vektor ( $CNOT$, atau kita bisa menggunakan bentuk produk tensor.$\begin{pmatrix}\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \\0 \end{pmatrix}$

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

Kami melihat bahwa untuk bagian pertama dari negara bit pertama adalah 0 , sehingga bit kedua dibiarkan saja; bagian kedua dari negara | 10 ⟩ bit pertama adalah 1 , sehingga bit kedua membalik dari 0 ke 1 .$|00\rangle$$0$$|10\rangle$$1$$0$$1$

Keadaan akhir kita adalah yang merupakan salah satu dari empat negara Bell yang maksimal negara terjerat.

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

Untuk melihat apa artinya bagi mereka untuk terjerat, perhatikan bahwa jika Anda mengukur status qubit pertama, jika Anda mengetahui bahwa itu adalah itu akan segera memberi tahu Anda bahwa qubit kedua juga harus menjadi 0 , karena itulah satu-satunya kemungkinan kita.$0$$0$

Bandingkan dengan keadaan ini misalnya:

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

Jika Anda mengukur bahwa qubit pertama adalah nol, maka statusnya runtuh menjadi , di mana masih ada peluang 50-50 qubit kedua adalah0atau1.$\frac{1}{\sqrt{2}}(|00\rangle+|01\rangle)$$0$$1$

Semoga ini memberi gambaran bagaimana negara bisa terjerat. Jika Anda ingin mengetahui contoh tertentu, seperti menjerat foton atau elektron dll, maka Anda harus melihat bagaimana gerbang tertentu dapat diimplementasikan, tetapi Anda masih dapat menulis matematika dengan cara yang sama, dan 1 mungkin mewakili hal-hal yang berbeda dalam situasi fisik yang berbeda.$0$$1$

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Pembaruan 1: Panduan Mini untuk notasi QM / QC / Dirac

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.

In this ordering of the basis we can identify $|0\rangle$ with $\begin{pmatrix} 1\\ 0 \end{pmatrix}$ and $|1\rangle$ with $\begin{pmatrix} 0\\ 1 \end{pmatrix}$. Any single qubit operator then can be written in matrix form using this basis. E.g. a bit flip operator $X$ (after pauli-$\sigma_x$) which should take $|0\rangle\mapsto |1\rangle$ and $|1\rangle \mapsto |0\rangle$, can be written as $\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$

$n$$\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.