Как я могу построить схему для генерации равной суперпозиции 3 результатов для 2 кубитов?

18

При наличии кубитной системы и, следовательно, возможных результатов измерений в базисе , , , , как я могу подготовить состояние, где: $2$ $4$ $\{|00\rangle$ $|01\rangle$ $|10\rangle$ $|11\rangle\}$

возможны только из этих результатов измерений (скажем, , , )? $3$ $4$ $|00\rangle$ $|01\rangle$ $|10\rangle$
эти измерения одинаково вероятны? (например, состояние Белла, но для результатов) $3$

— weekens
источник

1

Вы имеете в виду выписать фактическое состояние или создать схему для подготовки такого состояния с учетом входных данных?

— Хосу Этксезаррета Мартинес

@JosuEtxezarretaMartinez, я имею в виду схему.

— Weekens

@Blue, как ДИТ вам удается преобразовать их 00и 11в обозначениях Дирака? Я пытался $\ket{00}$ и потерпел неудачу.

— Weekens

1

@weekens Если вы нажмете «изменить», вы увидите код MathJax. Также посмотрите это .

— Санчайан Датта

1

Решение от Ниль де Бодрап в Quirk ...

— stestet

10

Разбить проблему по частям.

Скажем , мы уже отправили на $\mid 00 \rangle$ . Мы можем отправить это $\frac{1}{\sqrt{3}} \mid 00 \rangle + \frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle$ по $\frac{1}{\sqrt{3}} \mid 00 \rangle + (\frac{1}{2} (1+i))\frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle + (\frac{1}{2} (1-i))\frac{\sqrt{2}}{\sqrt{3}}\mid 10 \rangle$ . Это удовлетворяет ваши требования со всеми вероятностями $\sqrt{SWAP}$ но с разными фазами. Если вы хотите использовать вентили сдвига фазы на каждой, чтобы получить фазы, которые вам нравятся, если вы хотите, чтобы они были равны. $\frac{1}{3}$

Теперь , как мы получаем от к $\mid 00 \rangle$ ? Если бы это было $\frac{1}{\sqrt{3}} \mid 00 \rangle + \frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle$ , мы могли бы сделать Адамара на второй кубит. Это нелегко, но мы все еще можем использовать унитарный только на втором кубите. Это делается оператором вращения исключительно на втором кубите с помощью $\frac{1}{\sqrt{2}} \mid 00 \rangle + \frac{1}{\sqrt{2}}\mid 01 \rangle$

I d \otimes U : ∣ 0 ⟩ \otimes (∣ 0 ⟩) \to∣ 0 ⟩ \otimes (\frac{1}{\sqrt{3}} ∣ 0 ⟩ + \frac{\sqrt{2}}{\sqrt{3}} ∣ 1 ⟩)

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

работает. Разложите это на более простые ворота, если вам нужно.

U = (\begin{matrix} \frac{1}{\sqrt{3}} & \frac{\sqrt{2}}{\sqrt{3}} \\ \frac{\sqrt{2}}{\sqrt{3}} & - \frac{1}{\sqrt{3}} \end{matrix})

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

In total we have:

∣ 00 ⟩ \to \frac{1}{\sqrt{3}} ∣ 00 ⟩ + \frac{\sqrt{2}}{\sqrt{3}} ∣ 01 ⟩ \to \frac{1}{\sqrt{3}} ∣ 00 ⟩ + (\frac{1}{2} (1 + i)) \frac{\sqrt{2}}{\sqrt{3}} ∣ 01 ⟩ + (\frac{1}{2} (1 - i)) \frac{\sqrt{2}}{\sqrt{3}} ∣ 10 ⟩ \to \frac{1}{\sqrt{3}} ∣ 00 ⟩ + \frac{e^{i θ_{1}}}{\sqrt{3}} ∣ 01 ⟩ + \frac{e^{i θ_{2}}}{\sqrt{3}} ∣ 10 ⟩

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

— AHusain
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How do I construct U from basic gates? Let's say, from those available on IBM Q Experience.

— weekens

1

θ, λ

$\theta, \lambda$

ϕ

$\phi$

U 3 (θ, λ, ϕ) = (\begin{matrix} \cos \frac{θ}{2} & - e^{i λ} \sin \frac{θ}{2} \\ e^{i ϕ} \sin \frac{θ}{2} & e^{i (λ + ϕ)} \cos \frac{θ}{2} \end{matrix}),

$U3 \left( \theta, \lambda, \phi\right) = \begin{pmatrix}\cos\frac{\theta}{2} && -e^{i\lambda}\sin\frac{\theta}{2} \\ e^{i\phi}\sin\frac{\theta}{2} && e^{i\left(\lambda+\phi\right)}\cos\frac{\theta}{2}\end{pmatrix},$ which can be approximated using

θ \approx 1.91, λ = π

$\theta \approx 1.91, \lambda=\pi$ and

ϕ = 0

$\phi = 0$

— Mithrandir24601

To do this in basic gates, it looks like you would need to rotate into the right basis, then do a phase rotation, then rotate back which may require a fair few gates. However, in a sense, the above U3 is basic in that it's a physically implemented gate (i.e. is directly achieved by performing a couple of physical operations on the qubit instead of the many the would be required by stringing lots of 'not-advanced' gates together)

— Mithrandir24601

@ Mithrandir24601, спасибо за объяснение! Я еще не использовал U3, поэкспериментирую с ним в ближайшее время.

— Weekens

@AHusain, implemented your approach in Quirks simulator: here

— weekens

8

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

α | 0 ⟩ \otimes | 0 ⟩ + β | 1 ⟩ \otimes | 1 ⟩ .

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

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 ⟩ = | a_{0} ⟩, U | 1 ⟩ = | a_{1} ⟩

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

Our entangled state is then

α | a_{0} ⟩ \otimes | 0 ⟩ + β | a_{1} ⟩ \otimes | 1 ⟩ .

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

We can similarly apply a unitary to the second qubit.

V | 0 ⟩ = | b_{0} ⟩, V | 1 ⟩ = | b_{1} ⟩

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

which gives us the state

α | a_{0} ⟩ \otimes | b_{0} ⟩ + β | a_{1} ⟩ \otimes | b_{1} ⟩ .

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

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 $|b_1\rangle$ .

— James Wootton
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8

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

| ψ ⟩ = \frac{\sqrt{2}}{\sqrt{3}} | 0 ⟩ | + ⟩ + \frac{1}{\sqrt{3}} | 1 ⟩ | 0 ⟩ .

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

| ψ ⟩

$\ket{\psi}$ as follows:

Prepare two qubits in the state $\ket{00}$ .
Rotate the first qubit so that it is in the state $\tfrac{\sqrt 2}{\sqrt 3}\ket{0} + \tfrac{1}{\sqrt 3}\ket{1}$ .
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.)

— Niel de Beaudrap
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0

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 $\frac{1}{3}$ .

— Martin Vesely
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