Let's say we have the following circuit (picture and code shown below), and now the $q_0$ is an ancilla qubit. If the system of interest has only two qubits, Is there a way to use only $q_{1,2}$ as my ansatz and plug it into the built-in VQE in qiskit?
A little more context is the following. I would like to realize a quantum circuit consists of a linear combination of two unitary operators, such as $I+g$ where $I$ is the identity operator and $g$ is the translation operator that moves the qubits as $q_0\rightarrow q_1,q_1\rightarrow q_2,q_2\rightarrow q_0$. According to this paper, https://arxiv.org/abs/1202.5822, it is possible to do so with an ancilla qubit, see Fig. 1 and Eq. 3. But suppose I would like use the resulting circuit as a variational ansatz, how can I do that? If there is an approach to realize $I+g$ without ancilla qubit, I will be very happy to learn that!
from qiskit.circuit import ParameterVector
from qiskit import QuantumCircuit
theta = ParameterVector( 'theta' , 2 )
qc = QuantumCircuit( 3 )
q = qc.qubits
qc.h(q)
qc.crx( theta[0] , q[0], q[1] )
qc.cry( theta[1] , q[1], q[2] )
qc.draw('mpl')
VQE()is extremely inflexible. However, nothing stops you from defining a function taking as input parameterstheta[0]andtheta[1](simply asfloats, with noParameterVectorinvolved) and returning the corresponding circuit, and then usingscipy.minimizeto find the minimum of this function. The only problem with such an approach is that you will generate a new circuit each time you change the parameters. In a real experiment, however, running a circuit is so slow, that this will be a negligible time overhead. $\endgroup$