From Qiskit document, the Statevector can be used to specify the oracle of Grover Search. After seeing the source code of the operators, it seems a statevector is converted to a rank-1 projector operator. In this case, the operator may not be unitary.

The code below can be run on a real IBMQ machine.

from qiskit import *
from qiskit.quantum_info import Statevector
from qiskit.aqua.algorithms import Grover
from qiskit import QuantumCircuit

oracle = Statevector([0,0,0,0,0,1,1,0])
good_state = ['110','101']
grover = Grover(oracle=oracle, good_state=good_state)
# backend = provider.get_backend('ibmq_lima')
backend = BasicAer.get_backend('qasm_simulator')
result = grover.run(quantum_instance=backend)
print('Result type:', type(result))
print('Success!' if result.oracle_evaluation else 'Failure!')
print('Top measurement:', result)

However, it does not make sense to me because the operator here is not unitary. enter image description here

I am wondering if Statevector of Qiskit has actual speed up or it is just for illustration of Grover Search. Besides, how does it realize when the operator is not unitary?

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    $\begingroup$ Hi, thanks for your question! In future, please copy and paste code rather than posting screenshots, as it makes it easier for other people to run it if they want/need to $\endgroup$ – met927 May 27 at 7:51
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    $\begingroup$ Yes, I have edited the post $\endgroup$ – Jiawei Ren May 27 at 12:58

The reason for this is that the actual oracle used in the algorithm is not obtained from oracle.to_operator(). If you look up the code for Grover's algorithm, you can see the following:

grover_operator = GroverOperator(oracle=oracle,

Then, you can take a look at the GroverOperator source code to find the following:

if isinstance(oracle, Statevector):
    from qiskit.circuit.library import Diagonal  # pylint: disable=cyclic-import
    oracle = Diagonal((-1) ** oracle.data)

What this does is explained in the documentation: it builds a diagonal matrix whose entries on the diagonal are all equal to $(-1)^{x_i}$, where the $x_i$ are the entries of your Statevector. In your case, those entries on the diagonal will all be equal to $1$ except for the ones corresponding to the states $|101\rangle$ and $|110\rangle$ which will be equal to $-1$. You can easily see that:

  1. This operator is unitary
  2. This operator flips the good states as it is supposed to

Since it is used for initialiazing the Grover's algorithm, I don't think it has any kind of speedup when compared to more "traditional" initialisation methods.

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    $\begingroup$ Adding to this very good answer: Using the full statevector to construct the oracle would invalidate any quantum speedup since you need an exponential amount of data to represent the oracle. $\endgroup$ – Cryoris May 27 at 11:22
  • $\begingroup$ It makes sense to me. Thanks $\endgroup$ – Jiawei Ren May 27 at 12:53

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