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From this SE question and this Qiskit tutorial, I understand how to compute the expectation such in the form of $\langle \psi|H|\psi\rangle$ or $\langle 0^{\otimes n} |e^{iA} H e^{-iA}|0^{\otimes n} \rangle$ . But what about efficiently computing $\langle 0^{\otimes n} | e^{iA} H e^{-iB} |0^{\otimes n} \rangle$ when $A$ and $B$ are different and have more than single Pauli term in a quantum computer? $A$, $B$, and $H$ can be assumed to be Hermitian since that is the usual condition in such problem. Also, let us say we know $H$ as the linear combination of Pauli basis (i.e., Pauli decomposition of $H$).

I can certainly use Hadamard test to achieve my goal, but the circuit depth will be very large when $A$ and $B$ are composite of multiple Pauli terms, even use order-1 Suzuki trotterization to trotter $e^{iA}$ and $e^{-iB} $.

Also, if I want to use Qiskit functions like PauliExpectation() and CircuitSampler to do the computation, it seems like $e^{iA} H e^{-iB}$ is never involved with any quantum techniques but computed classically. Here is why I am saying this:

from qiskit.opflow import X, Y, Z, I
from qiskit import QuantumCircuit
from qiskit.opflow import CircuitStateFn,PauliExpectation,CircuitSampler
from qiskit.opflow import StateFn,PauliTrotterEvolution,Suzuki
from qiskit.providers.aer import AerSimulator
from qiskit.utils import QuantumInstance 

## Backend
backend = AerSimulator(method='statevector')
q_instance = QuantumInstance(backend, shots=1024)
sampler = CircuitSampler(q_instance)

## Operators
A = (0.6*Z^X) + (0.1*Z^Y)
H =  (0.3*I^Z) + (0.6*Y^Z)
B = (0.1*I^Y) + (0.6*Y^X)
op = A.exp_i().adjoint() @ H @ B.exp_i() # e^{iA} H e^{-iB}
obs = StateFn(op).adjoint() # let Qiskit consider this as the observable

## Initial State |00>
init_state = QuantumCircuit(2)
init_state = CircuitStateFn(init_state)

## Compute expectation
measurable_expression = obs @ init_state
trotterized_op = PauliTrotterEvolution(trotter_mode=Suzuki(order=1, reps=1)).convert(measurable_expression)
expectation = PauliExpectation().convert(trotterized_op)
sampled_exp_op = sampler.convert(expectation) 
print("Expectation:", sampled_exp_op.eval())

The printout will be

Expectation: (0.008744335460630945-0.015886672611238318j)

It looks good so far, but if I check how many and what circuits executed in the simulator

print(len(list(sampler._circuit_ops_cache.values())))

It will give

1

which is just the circuit for state $|00\rangle$

enter image description here

So in this case, I suppose Qiskit only use the quantum circuit to obtain init_state, and no quantum techniques, like qubit-wise commutativity, is applied. Of course, if one changes A and B to a single Pauli term, then sampler._circuit_ops_cache.values() will store multiple circuits and qubit-wise commutativity is indeed applied.

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  • $\begingroup$ @forky40 Thank you for your nice suggestion! I think it should be fine to assume A, B and H are Hermitian, which should be a common case for a such problem in practice. I think we can also assume the knowledge of the $H$ as the linear combination of Pauli basis. I will add this condition to the question. Thus, as long as the expectation value is real, I can compute the value for each Pauli string and do the linear combination to sum them up. $\endgroup$
    – Firepanda
    Jul 16 at 5:23
  • $\begingroup$ @forky40 how can we compute $|\langle 0 | X | 0 \rangle|^2$ if $X$ is not a projector, nor it is Hermitian? $\endgroup$
    – MonteNero
    Jul 16 at 6:03
  • $\begingroup$ yes maybe it doesn't work actually. $\endgroup$
    – forky40
    Jul 16 at 22:29

1 Answer 1

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You can use swap test to calculate $\langle 0^{\otimes n} | e^{iA} P_j e^{-iB} |0^{\otimes n} \rangle$ for each term $P_j$ in Pauli decomposition of $H$, then do the weighted sum classically:

enter image description here

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    $\begingroup$ I upvote. But note that this will not calculate the inner product the OP wants (the product may be a complex number). The swap test gives the squared magnitude of the inner product which is a real number. $\endgroup$
    – MonteNero
    Jul 17 at 20:17
  • $\begingroup$ Thanks for your answer. I think it should work if the expected value is real, in the cost of $2n+1$ qubits. Let me code it in Qiskit and check the circuit properties like connectivity of qubits, circuit depth and so on. I used to think finding the expectation in this form is not a rare demand, but seems like people don't really deal with this type of problems in practice. $\endgroup$
    – Firepanda
    Jul 18 at 4:40
  • $\begingroup$ I notice a question while I do the implementation: I suppose we cannot assume $\langle 0^{\otimes n} | e^{iA} P_j e^{-iB} |0^{\otimes n} \rangle \geq 0$, right? Or is there any way to know the sign of expected value? $\endgroup$
    – Firepanda
    Jul 18 at 18:01

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