Let's say we have a system coupled with an ancilla $\vert \psi_{SA}\rangle = \vert\psi_S\rangle\otimes\vert\phi_A\rangle$. The unitary evolution of this state is given by $U_{SA}$. If we perform a projective measurement on the ancilla, the $i^{th}$ outcome occurs with probability

$p_i = \langle \psi_{SA} \vert U_{SA}^{\dagger}(\mathbb{1}\otimes\vert m_i\rangle\langle m_i\vert) U_{SA}\vert\psi_{SA}\rangle$.

The state after this outcome is


where we have $M_{i} = \langle m_i\vert U_{SA}\vert\phi_{A}\rangle$, an operator that only acts on the system $S$. We can verify that $\sum_i M^\dagger_i M_i = \mathbb{1}_S$. The proof uses the fact that $U_{SA}$ is unitary.

How does one reverse this argument (essentially the proof of Neumark's theorem)? I would like to start with $M_i$ that satisfy $\sum_i M^\dagger_i M_i = \mathbb{1}_S$ and prove that there exists an ancilla system and corresponding unitaries $U_{SA}$ and projective measurements that follow the argument above. I'm not sure how to do this and the proof on Wikipedia isn't very illuminating.


1 Answer 1


Without loss of generality, I will assume $|\phi_A\rangle=|0\rangle$ -- otherwise, just rotate your ancilla space into that basis first as part of $U$. Similarly, I will assume $|m_i\rangle=|i\rangle$.

The condition $\sum M_i^\dagger M_i=1\!\!1$ says that the matrix $$ V=\begin{pmatrix}M_1\\M_2\\\vdots\\M_k\end{pmatrix} $$ has orthogonal columns. You can thus complete it to a unitary matrix $U$ by adding further columns to it (e.g., by taking states from your favorite basis and orthonormalizing them, unless they are in the span of the existing columns).

This way, you arrive at a unitary $U$ with $$ V=\begin{pmatrix} M_1 & * & * & \cdots \\ M_2 & * & * & \cdots \\ \vdots\\ M_k & * & * & \cdots \ , \end{pmatrix} $$ this is, $$ \langle m_i|U|0\rangle = M_i\ . $$

  • $\begingroup$ Thank you - a quick question. How many such columns must one add to V to obtain U? For instance, if $M_i$ are already projectors, I should have to add no columns. Is this correct? $\endgroup$ Commented Dec 17, 2018 at 11:34
  • 2
    $\begingroup$ Unitary matrix are always square. $\endgroup$ Commented Dec 17, 2018 at 15:07

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