# What do normalization term and partial measurement represent when tracing out ancillary qubits?

I am reading a paper and I am having trouble following some equations.

The system in this paper has $$N$$ qubits, with $$N_A$$ ancillary and the rest ($$N - N_A$$) as data qubits. For the purpose of this question, any subscript $$t$$ can be ignored. $$|\Psi(z)\rangle$$ represents the quantum state of the entire system.

We then take the partial measurement $$\Pi_{\mathcal A}$$ on the ancillary subsytem $$\mathcal A$$ of $$|\Psi(\mathcal z)\rangle$$, i.e., the post-measurement quantum state $$\rho_t(\mathcal z)$$ is $$\rho_t(\mathcal z) = \frac{\text{Tr}_{\mathcal A}(\Pi_{\mathcal A}|\Psi_t(\mathcal z)\rangle\langle\Psi_t(\mathcal z)|)}{\text{Tr}(\Pi_{\mathcal A}\otimes\mathbb{I}_{2^{N-N_{\mathcal A}}}|\Psi_t(\mathcal z)\rangle\langle\Psi_t(\mathcal z)|)}$$ An immediate observation is that state $$\rho_t(\mathcal z)$$ is a nonlinear map for $$|\mathcal z\rangle$$, since both the nominator and denominator of Eq. (13) [the one above] are the function of the variable $$|\mathcal z\rangle$$.

What do the top and bottom lines mean?

I understand the top line is trying to trace out the ancillary qubits. However, I don't understand the purpose of the partial measurement $$\Pi_\mathcal{A}$$. The point is to disregard the ancillary qubits, so is $$Tr_\mathcal{A}(|\Psi(z)\rangle \langle \Psi(z)|)$$ not sufficient? (How to get subspace of quantum circuit?)

Regarding the bottom line, does this act as some form of normalisation?

Many thanks for the help!

Tracing out without measuring is not equivalent to simply tracing out. Consider the state: $$\frac{|00\rangle+|11\rangle}{\sqrt{2}}.$$ Tracing out the last qubit, you would get: $$\rho=\frac12I,$$ that is the fully mixed state. Measuring the last qubit, noting down the result $$y$$ and tracing it out yields the state: $$\rho=|y\rangle.$$ Thus, tracing out may not be sufficient depending on your needs and depending on whether your registers are entangled.