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

18

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

  1. możliwe są tylko z wyników pomiarów (powiedzmy, , , )?4 | 00 | 01 | 10 34|00|01|10

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

weekendy
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1
Masz na myśli zapisanie stanu faktycznego lub stworzenie obwodu, aby przygotować taki stan, biorąc pod uwagę dane wejściowe?
Josu Etxezarreta Martinez
@JosuEtxezarretaMartinez, mam na myśli obwód.
weekendy
@Blue, jak dit uda się przekształcić te 00i 11notacji Diraca? Próbowałem $\ket{00}$i nie udało mi się.
weekendy
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@weekens Jeśli klikniesz „edytuj”, zobaczysz kod MathJax. Także, zobaczyć to .
Sanchayan Dutta
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Rozwiązanie od Niel de Beaudrap w Quirk ...
stestet

Odpowiedzi:

10

Złam problem na części.

Że mamy już wysłane do 100. Możemy wysłać to do11300+2301przez1300+(12(1+i))2301+(12(1i))2310 . To spełnia twoje wymagania ze wszystkimi prawdopodobieństwami 1SWAP 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.13

Teraz w jaki sposób dostać się z do 100? Jeśli to był11300+2301moglibyśmy 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 as1200+1201

IdU:0(0)→∣0(130+231)

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

U=(13232313)

W sumie mamy:

001300+23011300+(12(1+i))2301+(12(1i))23101300+eiθ1301+eiθ2310
AHusain
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Jak zbudować U z podstawowych bram? Powiedzmy, że z tych dostępnych w IBM Q Experience.
weekendy
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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.

UV

|a0|a1|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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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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0

Here is an implementation of a circuit producing state |ψ=13(|00+|01+|10) on IBM Q:

Circuit

Note that θ=1.2310 for Ry on q0. θ=π4 and θ=π4 for first and second Ry on q1.

The Ry on q0 prepares qubit in superposition |q0=23|0+13|1. Ry gates on q1 and CNOT implements controlled Hadamard gate. When q0 is in state |0 the Hadamard acts on q1 thanks to negation X. This happens with probability 23. Since Hadamard turns |0 to |+, i.e. equally distributed superposition, final states |00 and |01 can be measured with probability 13. When q0 is in state |1, controled Hadamard does not act and state |10 is measured. Since q0 is in state |1 with probability 13, |10 is measured also with probability 13.

Martin Vesely
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