Jak zbudować obwód, aby wygenerować równą superpozycję 3 wyników dla 2 kubitów?

Biorąc pod uwagę $2$ qubit-system, a zatem możliwe wyniki pomiarów, w podstawie , , , , jak mogę przygotować stan, gdzie:{ | 00 ⟩ | 01 ⟩ | 10 ⟩ | 11 ⟩ $4$$\{|00\rangle$$|01\rangle$$|10\rangle$$|11\rangle\}$

1. możliwe są tylko z wyników pomiarów (powiedzmy, , , )?4 | 00 ⟩ | 01 ⟩ | 10 $3$$4$$|00\rangle$$|01\rangle$$|10\rangle$

2. te pomiary są równie prawdopodobne? (jak stan Bella, ale dla wyników)$3$

Złam problem na części.

Że mamy już wysłane do $\mid 00 \rangle$. Możemy wysłać to do$\frac{1}{\sqrt{3}} \mid 00 \rangle + \frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle$przez$\frac{1}{\sqrt{3}} \mid 00 \rangle + (\frac{1}{2} (1+i))\frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle + (\frac{1}{2} (1-i))\frac{\sqrt{2}}{\sqrt{3}}\mid 10 \rangle$ . To spełnia twoje wymagania ze wszystkimi prawdopodobieństwami$\sqrt{SWAP}$ ale z różnymi fazami. Jeśli chcesz użyć bramek przesunięcia fazowego na każdej z nich, aby uzyskać fazy, które chcesz, jeśli chcesz, aby wszystkie były równe.$\frac{1}{3}$

Teraz w jaki sposób dostać się z do $\mid 00 \rangle$? Jeśli to był$\frac{1}{\sqrt{3}} \mid 00 \rangle + \frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle$moglibyśmy zrobić Hadamard na drugim qubitu. Nie jest to łatwe, ale nadal możemy używać jednostki tylko na drugim kubicie. Dokonuje tego operator obrotu wyłącznie na drugim kubicie przez faktoring as$\frac{1}{\sqrt{2}} \mid 00 \rangle + \frac{1}{\sqrt{2}}\mid 01 \rangle$

$$Id \otimes U : \; \mid 0 \rangle \otimes (\mid 0 \rangle) \to \mid 0 \rangle \otimes (\frac{1}{\sqrt{3}} \mid 0 \rangle + \frac{\sqrt{2}}{\sqrt{3}} \mid 1 \rangle)$$

działa. Rozłóż to na bardziej podstawowe bramy, jeśli potrzebujesz.

$$U = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{\sqrt{2}}{\sqrt{3}} & \\ \frac{\sqrt{2}}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & \\ \end{pmatrix}$$

W sumie mamy:

$$\mid 00 \rangle \to \frac{1}{\sqrt{3}} \mid 00 \rangle + \frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle\\ \to \frac{1}{\sqrt{3}} \mid 00 \rangle + (\frac{1}{2} (1+i))\frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle + (\frac{1}{2} (1-i))\frac{\sqrt{2}}{\sqrt{3}}\mid 10 \rangle\\ \to \frac{1}{\sqrt{3}} \mid 00 \rangle + \frac{e^{i \theta_1}}{\sqrt{3}}\mid 01 \rangle + \frac{e^{i \theta_2}}{\sqrt{3}}\mid 10 \rangle$$

I'll tell you how to create any two qubit pure state you might ever be interested in. Hopefully you can use it to generate the state you want.

Using a single qubit rotation followed by a cnot, it is possible to create states of the form

$$\alpha \, |0\rangle \otimes |0\rangle + \beta \, |1\rangle \otimes |1\rangle .$$

Then you can apply an arbitrary unitary, $U$, to the first qubit. This rotates the $|0\rangle$ and $|1\rangle$ states to new states that we'll call $|a_0\rangle$ and $|a_1\rangle$,

$$U |0\rangle = |a_0\rangle, \,\,\, U |1\rangle = |a_1\rangle$$

Our entangled state is then

$$\alpha \, |a_0\rangle \otimes |0\rangle + \beta \, |a_1\rangle \otimes |1\rangle .$$

We can similarly apply a unitary to the second qubit.

$$V |0\rangle = |b_0\rangle, \,\,\, V |1\rangle = |b_1\rangle$$

which gives us the state

$$\alpha \, |a_0\rangle \otimes |b_0\rangle + \beta \, |a_1\rangle \otimes |b_1\rangle .$$

$U$$V$

$|a_0\rangle$$|a_1\rangle$$|b_0\rangle$ and $|b_1\rangle$. You'll also find that $|a_0\rangle$ and $|b_0\rangle$ will have the same eigenvalue, which is $\alpha^2$. The coefficient $\beta$ can be similarly derived from the eigenvalues of $|a_1\rangle$ and $|b_1\rangle$.

Here is how you might go about designing such a circuit.$\def\ket#1{\lvert#1\rangle}$ Suppose that you would like to produce the state $\ket{\psi} = \tfrac{1}{\sqrt 3} \bigl( \ket{00} + \ket{01} + \ket{10} \bigr)$. Note the normalisation of ${\small 1}/\small \sqrt 3$, which is necessary for $\ket{\psi}$ to be a unit vector.

If we want to consider a straightforward way to realise this state, we might want to think in terms of the first qubit being a control, which determines whether the second qubit should be in the state $\ket{+} = \tfrac{1}{\sqrt 2}\bigl(\ket{0}+\ket{1}\bigr)$, or in the state $\ket{0}$, by using some conditional operations. This motivates considering the decomposition

$$\ket{\psi} \;=\; \tfrac{\sqrt 2}{\sqrt 3}\ket{0}\ket{+} \;+\; \tfrac{1}{\sqrt 3}\ket{1}\ket{0}.$$ Taking this view it makes sense to consider preparing $\ket{\psi}$ as follows:

1. Prepare two qubits in the state $\ket{00}$.
2. Rotate the first qubit so that it is in the state $\tfrac{\sqrt 2}{\sqrt 3}\ket{0} + \tfrac{1}{\sqrt 3}\ket{1}$.
3. Apply a coherently controlled operation on the two qubits which, when the first qubit is in the state $\ket{0}$, performs a Hadamard on the second qubit.

Which specific operations you would apply to realise these transformations — i.e. which single-qubit transformation would be most suitable for step 2, and how you might decompose the two-qubit unitary in step 3 into CNOTs and Pauli rotations — is a simple exercise. (Hint: use the fact that both $X$ and the Hadamard are self-inverse to find as simple a decomposition as possible in step 3.)

Here is an implementation of a circuit producing state $|\psi\rangle = \frac{1}{\sqrt{3}}(|00\rangle + |01\rangle + |10\rangle)$ on IBM Q:

Note that $\theta = 1.2310$ for $\mathrm{Ry}$ on $q_0$. $\theta = \frac{\pi}{4}$ and $\theta = -\frac{\pi}{4}$ for first and second $\mathrm{Ry}$ on $q_1$.

The $\mathrm{Ry}$ on $q_0$ prepares qubit in superposition $|q_0\rangle = \sqrt{\frac{2}{3}}|0\rangle + \frac{1}{\sqrt{3}}|1\rangle$. $\mathrm{Ry}$ gates on $q_1$ and $\mathrm{CNOT}$ implements controlled Hadamard gate. When $q_0$ is in state $|0\rangle$ the Hadamard acts on $q_1$ thanks to negation $\mathrm{X}$. This happens with probability $\frac{2}{3}$. Since Hadamard turns $|0\rangle$ to $|+\rangle$, i.e. equally distributed superposition, final states $|00\rangle$ and $|01\rangle$ can be measured with probability $\frac{1}{3}$. When $q_0$ is in state $|1\rangle$, controled Hadamard does not act and state $|10\rangle$ is measured. Since $q_0$ is in state $|1\rangle$ with probability $\frac{1}{3}$, $|10\rangle$ is measured also with probability $\frac{1}{3}$.