# Comment puis-je construire un circuit pour générer une superposition égale de 3 résultats pour 2 qubits?

18

Étant donné un système à $2$$2$ qubits et donc $4$$4$ mesures possibles, on obtient la base $\left\{|00⟩$$\{|00\rangle$ , $|01⟩$$|01\rangle$ , $|10⟩$$|10\rangle$ , $|11⟩\right\}$$|11\rangle\}$ , comment puis - je préparer l'état, où:

1. seulement $3$$3$ de ces $4$$4$ résultats de mesure sont possibles ( par exemple, $|00⟩$$|00\rangle$ , $|01⟩$$|01\rangle$ , $|10⟩$$|10\rangle$ )?

2. ces mesures sont tout aussi probables? (comme l'état de Bell mais pour $3$$3$ résultats)

1
Vous voulez écrire l'état réel ou créer un circuit pour préparer un tel état avec une entrée?
Josu Etxezarreta Martinez

@JosuEtxezarretaMartinez, je veux dire le circuit.
2018

@Blue, comment dit - vous réussi à les convertir 00et 11à la notation de Dirac? J'ai essayé $\ket{00}$et échoué.
2018 à 10h02

1
@weekens Si vous cliquez sur "modifier", vous pouvez voir le code MathJax. Voir aussi ceci .
Sanchayan Dutta

1
La solution de Niel de Beaudrap dans Quirk ...
stestet

Réponses:

10

Brisez le problème en plusieurs parties.

Disons que nous avons déjà envoyé à $\mid 00⟩$$\mid 00 \rangle$. Nous pouvons l'envoyer à$\frac{1}{\sqrt{3}}\mid 00⟩+\frac{\sqrt{2}}{\sqrt{3}}\mid 01⟩$$\frac{1}{\sqrt{3}} \mid 00 \rangle + \frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle$par un$\frac{1}{\sqrt{3}}\mid 00⟩+\left(\frac{1}{2}\left(1+i\right)\right)\frac{\sqrt{2}}{\sqrt{3}}\mid 01⟩+\left(\frac{1}{2}\left(1-i\right)\right)\frac{\sqrt{2}}{\sqrt{3}}\mid 10⟩$$\frac{1}{\sqrt{3}} \mid 00 \rangle + (\frac{1}{2} (1+i))\frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle + (\frac{1}{2} (1-i))\frac{\sqrt{2}}{\sqrt{3}}\mid 10 \rangle$ . Cela satisfait vos exigences avec toutes les probabilités$\sqrt{SWAP}$$\sqrt{SWAP}$ mais avec différentes phases. Si vous voulez utiliser des portes de déphasage sur chacune pour obtenir les phases que vous souhaitez, si vous voulez les rendre toutes égales.$\frac{1}{3}$$\frac{1}{3}$

Maintenant , comment pouvons-nous obtenir de à $\mid 00⟩$$\mid 00 \rangle$? Si c'était$\frac{1}{\sqrt{3}}\mid 00⟩+\frac{\sqrt{2}}{\sqrt{3}}\mid 01⟩$$\frac{1}{\sqrt{3}} \mid 00 \rangle + \frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle$, nous pourrions faire un Hadamard sur la deuxième qubit. Ce n'est pas facile avec cela, mais nous pouvons toujours utiliser un unitaire uniquement sur le deuxième qubit. Cela se fait par un opérateur de rotation uniquement sur le deuxième qubit en factorisant comme$\frac{1}{\sqrt{2}}\mid 00⟩+\frac{1}{\sqrt{2}}\mid 01⟩$$\frac{1}{\sqrt{2}} \mid 00 \rangle + \frac{1}{\sqrt{2}}\mid 01 \rangle$

$Id\otimes U:\phantom{\rule{thickmathspace}{0ex}}\mid 0⟩\otimes \left(\mid 0⟩\right)\to \mid 0⟩\otimes \left(\frac{1}{\sqrt{3}}\mid 0⟩+\frac{\sqrt{2}}{\sqrt{3}}\mid 1⟩\right)$

$U=\left(\begin{array}{ccc}\frac{1}{\sqrt{3}}& \frac{\sqrt{2}}{\sqrt{3}}& \\ \frac{\sqrt{2}}{\sqrt{3}}& -\frac{1}{\sqrt{3}}& \end{array}\right)$
works. Decompose this into more basic gates if you need to.

In total we have:

$\mid 00⟩\to \frac{1}{\sqrt{3}}\mid 00⟩+\frac{\sqrt{2}}{\sqrt{3}}\mid 01⟩\phantom{\rule{0ex}{0ex}}\to \frac{1}{\sqrt{3}}\mid 00⟩+\left(\frac{1}{2}\left(1+i\right)\right)\frac{\sqrt{2}}{\sqrt{3}}\mid 01⟩+\left(\frac{1}{2}\left(1-i\right)\right)\frac{\sqrt{2}}{\sqrt{3}}\mid 10⟩\phantom{\rule{0ex}{0ex}}\to \frac{1}{\sqrt{3}}\mid 00⟩+\frac{{e}^{i{\theta }_{1}}}{\sqrt{3}}\mid 01⟩+\frac{{e}^{i{\theta }_{2}}}{\sqrt{3}}\mid 10⟩$

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

1
@weekens There's an 'advanced' gate called U3 that allows you to implement any single qubit unitary - you input the values for $\theta ,\lambda$$\theta, \lambda$ and $\varphi$$\phi$ to implement
$U3\left(\theta ,\lambda ,\varphi \right)=\left(\begin{array}{ccc}\mathrm{cos}\frac{\theta }{2}& & -{e}^{i\lambda }\mathrm{sin}\frac{\theta }{2}\\ {e}^{i\varphi }\mathrm{sin}\frac{\theta }{2}& & {e}^{i\left(\lambda +\varphi \right)}\mathrm{cos}\frac{\theta }{2}\end{array}\right),$
which can be approximated using $\theta \approx 1.91,\lambda =\pi$$\theta \approx 1.91, \lambda=\pi$ and $\varphi =0$$\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, thanks for your explanation! I haven't used U3 yet, will experiment with it in nearest time.
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

$\alpha \phantom{\rule{thinmathspace}{0ex}}|0⟩\otimes |0⟩+\beta \phantom{\rule{thinmathspace}{0ex}}|1⟩\otimes |1⟩.$

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

$U|0⟩=|{a}_{0}⟩,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}U|1⟩=|{a}_{1}⟩$

Our entangled state is then

$\alpha \phantom{\rule{thinmathspace}{0ex}}|{a}_{0}⟩\otimes |0⟩+\beta \phantom{\rule{thinmathspace}{0ex}}|{a}_{1}⟩\otimes |1⟩.$

We can similarly apply a unitary to the second qubit.

$V|0⟩=|{b}_{0}⟩,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}V|1⟩=|{b}_{1}⟩$

which gives us the state

$\alpha \phantom{\rule{thinmathspace}{0ex}}|{a}_{0}⟩\otimes |{b}_{0}⟩+\beta \phantom{\rule{thinmathspace}{0ex}}|{a}_{1}⟩\otimes |{b}_{1}⟩.$

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$$U$ and $V$$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}⟩$$|a_0\rangle$ and $|{a}_{1}⟩$$|a_1\rangle$. The eigenstates for the second qubit will be $|{b}_{0}⟩$$|b_0\rangle$ and $|{b}_{1}⟩$$|b_1\rangle$. You'll also find that $|{a}_{0}⟩$$|a_0\rangle$ and $|{b}_{0}⟩$$|b_0\rangle$ will have the same eigenvalue, which is ${\alpha }^{2}$$\alpha^2$. The coefficient $\beta$$\beta$ can be similarly derived from the eigenvalues of $|{a}_{1}⟩$$|a_1\rangle$ and $|{b}_{1}⟩$$|b_1\rangle$.

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 $|\psi ⟩=\frac{1}{\sqrt{3}}\left(|00⟩+|01⟩+|10⟩\right)$$\ket{\psi} = \tfrac{1}{\sqrt 3} \bigl( \ket{00} + \ket{01} + \ket{10} \bigr)$. Note the normalisation of $1/\sqrt{3}$${\small 1}/\small \sqrt 3$, which is necessary for $|\psi ⟩$$\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 $|+⟩=\frac{1}{\sqrt{2}}\left(|0⟩+|1⟩\right)$$\ket{+} = \tfrac{1}{\sqrt 2}\bigl(\ket{0}+\ket{1}\bigr)$, or in the state $|0⟩$$\ket{0}$, by using some conditional operations. This motivates considering the decomposition

$|\psi ⟩\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}\frac{\sqrt{2}}{\sqrt{3}}|0⟩|+⟩\phantom{\rule{thickmathspace}{0ex}}+\phantom{\rule{thickmathspace}{0ex}}\frac{1}{\sqrt{3}}|1⟩|0⟩.$
Taking this view it makes sense to consider preparing $|\psi ⟩$$\ket{\psi}$ as follows:
1. Prepare two qubits in the state $|00⟩$$\ket{00}$.
2. Rotate the first qubit so that it is in the state $\frac{\sqrt{2}}{\sqrt{3}}|0⟩+\frac{1}{\sqrt{3}}|1⟩$$\tfrac{\sqrt 2}{\sqrt 3}\ket{0} + \tfrac{1}{\sqrt 3}\ket{1}$.
3. Apply a coherently controlled operation on the two qubits which, when the first qubit is in the state $|0⟩$$\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$$X$ and the Hadamard are self-inverse to find as simple a decomposition as possible in step 3.)

0

Here is an implementation of a circuit producing state $$|ψ⟩=13√(|00⟩+|01⟩+|10⟩)|ψ⟩=13(|00⟩+|01⟩+|10⟩)|\psi\rangle = \frac{1}{\sqrt{3}}(|00\rangle + |01\rangle + |10\rangle)$$ on IBM Q: Note that $$θ=1.2310θ=1.2310\theta = 1.2310$$ for $$RyRy\mathrm{Ry}$$ on $$q0q0q_0$$. $$θ=π4θ=π4\theta = \frac{\pi}{4}$$ and $$θ=−π4θ=−π4\theta = -\frac{\pi}{4}$$ for first and second $$RyRy\mathrm{Ry}$$ on $$q1q1q_1$$.

The $$RyRy\mathrm{Ry}$$ on $$q0q0q_0$$ prepares qubit in superposition $$|q0⟩=23−−√|0⟩+13√|1⟩|q0⟩=23|0⟩+13|1⟩|q_0\rangle = \sqrt{\frac{2}{3}}|0\rangle + \frac{1}{\sqrt{3}}|1\rangle$$. $$RyRy\mathrm{Ry}$$ gates on $$q1q1q_1$$ and $$CNOTCNOT\mathrm{CNOT}$$ implements controlled Hadamard gate. When $$q0q0q_0$$ is in state $$|0⟩|0⟩|0\rangle$$ the Hadamard acts on $$q1q1q_1$$ thanks to negation $$XX\mathrm{X}$$. This happens with probability $$2323\frac{2}{3}$$. Since Hadamard turns $$|0⟩|0⟩|0\rangle$$ to $$|+⟩|+⟩|+\rangle$$, i.e. equally distributed superposition, final states $$|00⟩|00⟩|00\rangle$$ and $$|01⟩|01⟩|01\rangle$$ can be measured with probability $$1313\frac{1}{3}$$. When $$q0q0q_0$$ is in state $$|1⟩|1⟩|1\rangle$$, controled Hadamard does not act and state $$|10⟩|10⟩|10\rangle$$ is measured. Since $$q0q0q_0$$ is in state $$|1⟩|1⟩|1\rangle$$ with probability $$1313\frac{1}{3}$$, $$|10⟩|10⟩|10\rangle$$ is measured also with probability $$1313\frac{1}{3}$$.

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