# How to implement an unitary operator expressed as a linear combination of unitaries without qubits ancilla

Let's say that I know the decomposition of a unitary operator $$\hat{A}$$ in terms of other unitary operators $$U_{k=0, \dots, M}$$, i.e:

$$\hat{A} = \sum_k \alpha_k U_k$$

I know how to implement in circuit form $$U_k$$ but not $$\hat{A}$$. Is there a way to obtain a decomposition:

$$\hat{A} = \sum_k \alpha_k U_k = B_1 B_2 \dots B_M$$

where $$B_{i=1,\dots, M}$$ are unitary operators and are obtained from $$\alpha_k$$ and $$U_k$$ ? I'd like the $$B_i$$ to use the same number of qubits as $$U_k$$, so no ancillary register (no LCU method).

• Maybe this question is too broad. I mean the clearly the optimal answer is to write $\sum \alpha_kU_k = AI\cdots I$ or $\sqrt[M]{A}$ (multiplied by itself $M$ times), maybe you should restrict the $B_k$s a little bit more. Sep 9, 2022 at 17:27
• Mmm ok. Being more precise. Let's say I know to implement in a circuit the unitary gates $U_k$. I do not know how to implement the operator $\hat{A}$ in the circuit. There is a way that, from knowing $\alpha_k$ and $U_k$ I can implement the operator $\hat{A}$ on the circuit? PS: Thanks for your answer. I'd change the word 'optimal' to 'trivial'. I was not precise enough, but I didn't mention these solutions since they were trivial. Sep 12, 2022 at 12:33
• Maybe you should indicate that in your post (you can edit it). If I understand correctly you want $B_{i}$ just in terms of $U_k$ and $\alpha_k$ (unless you allow for a series of other gates then you should indicate all of them). Sep 12, 2022 at 12:45
• I think something is still missing. Imagine that you want to write $H=(X+Z)/\sqrt{2}$. Clearly it is not possible to write $H$ as just products of $X$ and $Z$. So if there is a solution to your answer it is very restricted to some particular cases. Sep 12, 2022 at 13:45
• Here is a proof of no-go for single qubits: every unitary matrix $U$ can be written as a linear combination of Pauli matrices. Pauli matrices form a group. Products of Pauli matrices stay in the Pauli group. So you are only able to decompose $U$ if $U$ is in the Pauli group (which is incredibly restrictive). Sep 12, 2022 at 14:35