5
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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 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 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 at 1:31
  • $\begingroup$ I have updated the question and added some examples. $\endgroup$
    – mavzolej
    Feb 2 at 3:57

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Excellent!! I am wondering if the probability of success could be increased by some QSP/Oblivious/Grover-like rotations 🙃 $\endgroup$
    – mavzolej
    Feb 2 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 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 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 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 at 17:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.