# confusion on the LCU method regarding the normalization

Let $$A = \sum_{k} a_k U_k$$ where $$a_k$$ are real, positive coefficients $$U_k$$ are unitary matrices. I have realized that $$\sigma = A \rho A$$ can be implemented on a quantum computer by using the LCU method where $$\rho$$ is a density matrix that I can prepare on a quantum computer. However, it seems like the normalization of this state may be wrong: $$\text{tr}(\sigma) = \text{tr}(A \rho A)$$ is not always equal to $$1$$ unless we divide $$\sigma$$ to $$\text{tr}(\sigma)$$ (think about $$A$$ being a projection operator onto some subspace $$H$$ and $$\rho \notin H$$).

To be more specific, this reference says that we can implement $$\frac{A}{\lambda}\lvert \psi \rangle$$ using LCU where $$\lambda = \sum_k {a_k}$$. However, consider $$A = \frac{1}{2}(I + Z)$$ and $$\rho \notin \text{span}(\lvert {0} \rangle)$$. Then, $$A \rho A$$ doesn't have trace $$1$$ (since $$\lambda = 1$$), so according to the reference, you're preparing an invalid density matrix. I should be missing something here. If anyone could help me to understand what I'm missing, and describe how to actually prepare $$\frac{1}{\text{tr}(A\rho)}A \rho A$$ where $$A = \frac{1}{2}(I + Z)$$ on a quantum computer using LCU, I would be much appreciated.

You're correct, LCU will prepare a state proportional to $$A|\psi\rangle$$, with whatever renormalization is required for that to be true. Its possible that the normalization factor was omitted because it is sort of assumed to be present for LCU to describe valid quantum operation. I'm not familiar with the reference you're using, but Lemma 6 of (CKS, 2015) is very straightforward: LCU allows you to prepare a state (containing $$m$$ additional ancilla qubits) of the form, $$\frac{1}{\lambda} |0^m\rangle A|\psi\rangle + |\Phi^\perp\rangle \tag{1}$$ where $$|\Phi^\perp\rangle$$ does not contain any amplitude for its ancillas to be found in the $$|0^m\rangle$$ state. In general, if you have a normalized state $$|\phi\rangle = |0\rangle |u\rangle + |0^\perp\rangle |v\rangle \tag{2}$$ for complex vectors $$|u\rangle, |v\rangle \in \mathbb{C}^d$$ (not necessarily normalized!) and $$\langle 0 | 0^\perp \rangle = 0$$, the probability of measuring $$0$$ in the first register is given by $$\lVert | u \rangle \rVert^2$$. So in Eq. (1), if you perform a projective measurement $$\{ |0^m \rangle \langle 0^m| \otimes \mathbb{I}, \left(\mathbb{I} - |0^m \rangle \langle 0^m|\right) \otimes \mathbb{I} \} \tag{3}$$ on the system as a whole, the working register will be projected onto a state proportional to $$A|\psi\rangle$$ with probability $$\text{Pr}(0^m) = \bigl\lVert \frac{1}{\lambda}A|\psi\rangle\bigr\rVert^2$$. In this case the final state of the original system becomes $$\frac{1}{\lambda} |0^m\rangle A|\psi\rangle + |\Phi^\perp\rangle \rightarrow \frac{1}{\sqrt{\text{Pr}(0^m)}} \left(\frac{1}{\lambda}A|\psi\rangle\right) = \frac{1}{ \bigl\lVert \frac{1}{\lambda}A|\psi\rangle\bigr\rVert} \left(\frac{1}{\lambda}A|\psi\rangle\right).$$ You can check that this is normalized, or that this state is never prepared at all if $$A|\psi\rangle = 0$$ (e.g. if you apply $$\frac{1}{2}(I + Z)$$ to $$|1\rangle$$ in your example).