Bagaimana saya bisa membangun sirkuit untuk menghasilkan superposisi 3 hasil yang sama untuk 2 qubit?

Diberikan sistem $2$ qubit dan dengan demikian $4$ kemungkinan pengukuran menghasilkan basis $\{|00\rangle$ , $|01\rangle$ , $|10\rangle$ , $|11\rangle\}$ , bagaimana saya dapat mempersiapkan negara, di mana:

1. hanya $3$ ini $4$ hasil pengukuran yang mungkin (katakanlah, $|00\rangle$ , $|01\rangle$ , $|10\rangle$ )?

2. pengukuran ini sama-sama mungkin? (seperti Bell state tetapi untuk $3$ hasil)

Pecahkan masalah di beberapa bagian.

Katakanlah kita telah mengirimkan ke $\mid 00 \rangle$. Kami dapat mengirimkannya ke$\frac{1}{\sqrt{3}} \mid 00 \rangle + \frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle$$\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$ by a $\sqrt{SWAP}$. That satisfies you're requirements with all probabilities $\frac{1}{3}$ but with different phases. If you want use phase shift gates on each to get the phases you want like if you want to make them all equal.

Now how do we get from $\mid 00 \rangle$ to $\frac{1}{\sqrt{3}} \mid 00 \rangle + \frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle$? If it was $\frac{1}{\sqrt{2}} \mid 00 \rangle + \frac{1}{\sqrt{2}}\mid 01 \rangle$, we could do a Hadamard on the second qubit. It is not a easy with this but we can still use a unitary only on the second qubit. That is done by a rotation operator purely on the second qubit by factoring as

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

$$U = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{\sqrt{2}}{\sqrt{3}} & \\ \frac{\sqrt{2}}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & \\ \end{pmatrix}$$ works. Decompose this into more basic gates if you need to.

Secara total kami memiliki:

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

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

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

Due to the Schmidt decomposition, it is possible to express any pure state of two qubits in the form above. This means that any pure state of two qubits, including the one you want, can be created by this procedure. You just need to find the right rotation around the x axis, and the right unitaries $U$ and $V$.

To find these, you first need to get the reduced density matrix for each of your two qubits. The eigenstates for the density matrix of your first qubit will be your $|a_0\rangle$ and $|a_1\rangle$. The eigenstates for the second qubit will be $|b_0\rangle$ 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 $\frac{1}{3}$.