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!


1 Answer 1


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.

Concerning the actual equation, you were right: the first line traces out the ancillary qubits after having measured them, while the second one is a normalization coming from the measurement of the ancillary qubits.

  • 1
    $\begingroup$ Notably, the normalization is equivalent to the probability of measuring this result $\endgroup$ Commented Dec 3, 2021 at 14:33

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