Is there an efficient way to create entangled states using Qiskit?

Question

I am trying to implement Grover's search using Qiskit. And I would like to create entangled states so I could search for the index of a specific value in a list. For example, if the third element in the list has value 7, then the corresponding state should be $|3\rangle|7\rangle$. One way to do that is to construct a particular circuit for each element after the hadamard transform but it seems to be hard. Is there an efficient way in Qiskit to do that so I will not need to construct the circuit for every element in the list? Thank you for your help!

Answer 1

Entanglement

As to how to create entangled states, the most easy way to do this is with the following circuit:

qc = QuantumCircuit(4)
qc.h(0)
qc.cx(0, 1)
qc.cx(1, 2)
qc.cx(2, 3)

This is an example for $n=4$, and you get the state $\frac{1}{\sqrt{2}}(|0\rangle^{\otimes 4} + |1\rangle^{\otimes 4})$; as confirmed by the statevector_simulator.

svsim = Aer.get_backend('statevector_simulator')
result = svsim.run(qc).result().get_counts()

Output: {'0000': 0.5, '1111': 0.5}

Of course, you can extend this circuit to any number of qubits and get $\frac{1}{\sqrt{2}}(|0\rangle^{\otimes n} + |1\rangle^{\otimes n})$. Now, this is not a efficient way of doing this since the number of CNOT gates is $n-1$, thus error on these gates plus decoherence time of the qubits on real devices will affect your state. However, this is the way to do it. For more information on this, check this answer and the GHZ state page on Wikipedia.

Oracle

Now for the oracle, you want to implement a unitary $U|x\rangle |0\rangle = |x\rangle |f(x)\rangle$ such that $f(x)$ is the element of a list/array at index $x$. However, for Grover's algorithm, we want to implement a unitary of one of the following two forms:

$$ U|x\rangle = \left\{ \begin{array}{ll} |x\rangle & \mbox{if } x\not=\omega \\ -|x\rangle & \mbox{if } x = \omega \end{array} \right. $$

where $\omega$ is the "winner" state. The other form is

$$ U|x\rangle = (-1)^{f(x)}|x\rangle. $$

In the case you are describing, it is pretty easy to construct the first type of oracle. Suppose that you want $|\omega\rangle = |7\rangle = |111\rangle$, where the last one is in binary. We just need to make

$$ U = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ \end{bmatrix}. $$

This works because the column where the element is $-1$ corresponds to the $|111\rangle$ state. You can do the same unitary for any number, just set the element in the corresponding element to $-1$. To translate this unitary matrix into a circuit, check out this answer.

I believe that the unitary as you describe it is not very useful in Grover's algorithm, but the method described above should work out. Just access the index $x$ of the list, classicaly build the unitary matrix by setting the element of the appropiate column to $-1$, and finally translate the matrix into a Qiskit circuit.