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.


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.

  • $\begingroup$ Just to check, here $|\psi\rangle$ is state in a finite dimensional Hilbert space (e.g. $N$ qubits)? How concerned are you about running out of basis states $|n\rangle$ in the particle number basis? $\endgroup$
    – forky40
    Feb 1, 2022 at 22:46
  • $\begingroup$ $|\psi\rangle$ is generally some state, which we know how to prepare (from the $|0^n\rangle$ state via a circuit with $\operatorname{poly}(n)$ gates). However, we may not know the state's amplitudes in advance, and, so, the operator has to act correctly, independently of the form of the state (otherwise, we could just apply $X$ to go from $|0\rangle$ to $|1\rangle$. Sorry, I don't understand your question about "running out of basis states". $\endgroup$
    – mavzolej
    Feb 2, 2022 at 1:21
  • $\begingroup$ I just mean what do you expect (or want) the result of $a^\dagger |2^n-1\rangle$ to be? $\endgroup$
    – forky40
    Feb 2, 2022 at 1:31
  • $\begingroup$ I have updated the question and added some examples. $\endgroup$
    – mavzolej
    Feb 2, 2022 at 3:57

1 Answer 1


I guess the following could work...

Let's just start with the simple case of a single qubit. This would equally well work for picking out a single qubit in an array and acting on that single qubit (and, for fermions, applying the unitary $ZZ\ldots Z$ on some other subset of qubits).

Measure your qubit in the $Z$ basis. If you get the $|1\rangle$ answer, you failed. start again. If you get the $|0\rangle$ answer, you succeeded. Apply an $X$ rotation to that qubit.

To see why this works, notice that if you project on the $|0\rangle$ state, the projection operator is $I+Z$. So, if you pre-multiply by $X$, you have $X(I+Z)=X-iY=a^\dagger$. The possibility of failure is a necessary consequence of the not-normal property you observed.

  • $\begingroup$ Excellent!! I am wondering if the probability of success could be increased by some QSP/Oblivious/Grover-like rotations 🙃 $\endgroup$
    – mavzolej
    Feb 2, 2022 at 8:14
  • $\begingroup$ Provided you have a unitary that can prepare your initial state, then yes, you could do amplitude amplification on the qubit. (As an aside, I would query if you really need to implement this operation. That's obviously a question for you to answer to yourself, but I find it unlikely.) $\endgroup$
    – DaftWullie
    Feb 2, 2022 at 8:25
  • $\begingroup$ do you think using eqs (30)-(33) from here would work? I mean, for such simple forms of the $a^\dagger$ operators as those in the question, wouldn't we be able to implement (33) exactly? $\endgroup$
    – mavzolej
    Feb 7, 2022 at 19:24
  • $\begingroup$ Yes, it will work. But it's something different to what you originally asked (you wanted something non-unitary, this is unitary) so only you can answer if it's doing what you need it to do. $\endgroup$
    – DaftWullie
    Feb 8, 2022 at 7:35
  • $\begingroup$ Is my understanding correct that this method would not work if the operator is bosonic, and multiple qubits are used to encode the occupation? In this case, I would suggest using a CNOT on all the qubit registers and measuring whether all the qubits are in the state, which should be annihilated (and disregarding the state in this case). $\endgroup$
    – mavzolej
    Feb 10, 2022 at 17:17

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