One way or another, I would like to implement the action of the second-quantized creation operator on a quantum state: $|\psi\rangle\mapsto a^\dagger|\psi\rangle$. The motivation, of course, comes from physics, but for simplicity let's forget about fermions and bosons, and just assume a single-qubit Hamiltonian and $a^\dagger=X-iY$.
For normal operators, one way to accomplish such a task amounts to using QSP, as described in appendix A of the qubitization paper. Clearly, $a^\dagger$ is not normal, and something else is required.
Note that if one needs to measure something like $\langle \psi_1 |a^\dagger |\psi_2\rangle$, where $|\psi_1\rangle$ and $|\psi_2\rangle$ are the states, which one knows how to prepare, then one could simply use the definition of $a^\dagger$ and measure $\langle \psi_1 |X |\psi_2\rangle - i \langle \psi_1 |Y |\psi_2\rangle$. However, this approach becomes increasingly complex if measurements of the form $\langle\psi|\ldots a^\dagger U_2 a^\dagger U_1|\psi\rangle$ are considered, and fails in the case if the compact mapping is used, and no efficient qubit representation of $a^\dagger$ exists.
CLARIFICATION
I am not talking here about an operator turning an $|n\rangle$ qubit state into a $|n+1\rangle$ qubits state, for any value of $n$.
Below are some examples of how the mapping of a second-quantized $a^\dagger$ operator onto qubits may look like for multi-particle systems, if the direct mapping is used:
- Hard-core bosons: $1\otimes\ldots\otimes1\otimes(X-iY)\otimes1\otimes\ldots\otimes1$.
- Bosons, binary encoding of a bosonic mode (max occupancy 3): $1\otimes\ldots\otimes1\otimes(|01\rangle\langle00|+\sqrt{2}|10\rangle\langle01|+\sqrt{3}|11\rangle\langle10|)\otimes1\otimes\ldots\otimes1$.
- Fermions, Jordan-Wigner: $1\otimes\ldots\otimes1\otimes(X-iY)\otimes Z\otimes\ldots\otimes Z$.
Similarly one can consider other cases of bosonic or fermionic mappings of creation and annihilation operators on qubits.
