Wie kann ich eine Schaltung aufbauen, um eine gleiche Überlagerung von 3 Ergebnissen für 2 Qubits zu erzeugen?

17

Bei einem 2 Qubit-System und damit 4 möglichen Messungen ergibt sich die Basis {|00 , |01 , |10 , |11} , wie kann ich den Zustand herzustellen, wobei gilt:

  1. nur 3 davon 4 Messergebnisse sind möglich ( zum Beispiel |00 , |01 , |10 )?

  2. Diese Messungen sind gleich wahrscheinlich? (Wie Bell State, aber für 3 Ergebnisse)

Wochen
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1
Sie wollen den aktuellen Zustand ausschreiben oder eine Schaltung erstellen, um einen solchen Zustand bei einem Eingang herzustellen?
Josu Etxezarreta Martinez
@ JosuEtxezarretaMartinez, ich meine die Schaltung.
Weekens
@Blue, wie schaffen Sie es, diese 00und 11die Dirac-Notation zu konvertieren ? Ich habe es versucht $\ket{00}$und bin gescheitert.
Weekens
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@weekens Wenn Sie auf "Bearbeiten" klicken, wird der MathJax-Code angezeigt. Siehe auch dies .
Sanchayan Dutta
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Die Lösung von Niel de Beaudrap in Quirk ...
stestet

Antworten:

10

Brechen Sie das Problem in Teile.

Sagen wir bereits gesendet bis 100. Wir können das an1senden1300+2301durch ein1300+(12(1+i))2301+(12(1i))2310 . Das erfüllt Ihre Anforderungen mit allen Wahrscheinlichkeiten 1SWAP13 aber mit unterschiedlichen Phasen. Wenn Sie Phasenverschiebungs-Gatter für jede Phase verwenden möchten, erhalten Sie die gewünschten Phasen, wenn Sie alle gleich machen möchten.

Nun , wie kommen wir aus bis 100? Wenn es1 wäre1300+2301, könnten wir eine Hadamard auf dem zweiten Qubit tun. Dies ist nicht einfach, aber wir können eine Einheit nur für das zweite Qubit verwenden. Dies geschieht durch einen Rotationsoperator rein auf dem zweiten Qubit durch Faktorisierung als1200+1201

IdU:0(0)→∣0(130+231)

U=(13232313)

Insgesamt haben wir:

001300+23011300+(12(1+i))2301+(12(1i))23101300+eiθ1301+eiθ2310
AHusain
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How do I construct U from basic gates? Let's say, from those available on IBM Q Experience.
weekens
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@weekens There's an 'advanced' gate called U3 that allows you to implement any single qubit unitary - you input the values for θ,λ and ϕ to implement
U3(θ,λ,ϕ)=(cosθ2eiλsinθ2eiϕsinθ2ei(λ+ϕ)cosθ2),
which can be approximated using θ1.91,λ=π and ϕ=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

α|0|0+β|1|1.

Then you can apply an arbitrary unitary, U, to the first qubit. This rotates the |0 and |1 states to new states that we'll call |a0 and |a1,

U|0=|a0,U|1=|a1

Our entangled state is then

α|a0|0+β|a1|1.

We can similarly apply a unitary to the second qubit.

V|0=|b0,V|1=|b1

which gives us the state

α|a0|b0+β|a1|b1.

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 |a0 and |a1. The eigenstates for the second qubit will be |b0 and |b1. You'll also find that |a0 and |b0 will have the same eigenvalue, which is α2. The coefficient β can be similarly derived from the eigenvalues of |a1 and |b1.

James Wootton
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7

Here is how you might go about designing such a circuit. Suppose that you would like to produce the state |ψ=13(|00+|01+|10). Note the normalisation of 1/3, which is necessary for |ψ 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 |+=12(|0+|1), or in the state |0, by using some conditional operations. This motivates considering the decomposition

|ψ=23|0|++13|1|0.
Taking this view it makes sense to consider preparing |ψ as follows:
  1. Prepare two qubits in the state |00.
  2. Rotate the first qubit so that it is in the state 23|0+13|1.
  3. Apply a coherently controlled operation on the two qubits which, when the first qubit is in the state |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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