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

18

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

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

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

minggu
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1
Anda bermaksud menulis keadaan aktual atau membuat sirkuit untuk mempersiapkan keadaan seperti itu diberi input?
Josu Etxezarreta Martinez
@ YosuEtxezarretaMartinez, maksud saya sirkuit.
weekens
@Blue, bagaimana Anda bisa mengonversinya 00dan 11notasi Dirac? Saya mencoba $\ket{00}$dan gagal.
weekens
1
@weekens Jika Anda mengklik "edit", Anda dapat melihat kode MathJax. Juga, lihat ini .
Sanchayan Dutta
1
Solusi dari Niel de Beaudrap di Quirk ...
stestet

Jawaban:

10

Pecahkan masalah di beberapa bagian.

Katakanlah kita telah mengirimkan ke 100. Kami dapat mengirimkannya ke11300+23011300+(12(1+i))2301+(12(1i))2310 by a SWAP. That satisfies you're requirements with all probabilities 13 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 00 to 1300+2301? If it was 1200+1201, 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

IdU:0(0)→∣0(130+231)

U=(13232313)
works. Decompose this into more basic gates if you need to.

Secara total kami memiliki:

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
1
@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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8

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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