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Consider an $n$ qubit state $|\psi\rangle$. Let's say I want to implement an $m$ outcome orthogonal measurement on $|\psi\rangle$, where $m \neq n$. Denote the set of $m$ orthogonal measurement operators as

\begin{equation} \mathcal{M} = \{M_1, M_2, \ldots, M_m \}. \end{equation}

From the properties of orthogonal measurements, we know that:

  1. Each $M_i$ can be written as \begin{equation} M_i = \sum_{j = 1}^{k_i} |\lambda_j\rangle \langle \lambda_j|, \end{equation} for a certain choice of $k_i$, with each $|\lambda_j\rangle$ being a normalized pure state, and $\{|\lambda_j\rangle : j = \{1, 2, \ldots, k_i\}\}$ being an orthogonal set. It is also true that \begin{equation} \sum_{i = 1}^{m} k_i = 2^{n}. \end{equation}
  2. Additionally, $m \leq 2^{n}$.

Let us say we can only measure in the standard basis and we want to implement the measurement $\mathcal{M}$. Is it true that there always exists an $n$ unitary $U$ such that we can implement the measurement, without using any ancillary qubits, using the following recipe:

  1. Apply $U$ to the state $|\psi\rangle$.
  2. Measure $\log m$ qubits in the standard basis.

I do not have much intuition on whether or not ancillary qubits are necessary to implement general orthogonal measurements of the type I mentioned. If we consider an example of a famous orthogonal measurement --- the SWAP test (which projects onto either the $2$ dimensional symmetric or the $2$ dimensional antisymmetric subspace, which are orthogonal and span the whole space for $2$ dimensional systems) --- it seems to require ancillary qubits, in its most standard implementation.

Can we get rid of these ancillas?

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    $\begingroup$ Are you sure you don't mean $\sum_{i=1}^mk_i=2^n$? In which case, you just define a mapping of pairs $(i,j)$ where $i$ runs from 1 to $m$ and $j$ runs from $1$ to $k_i$ such that $f(i,j)$ produces a distinct number in the range $0$ to $2^n-1$, which must be possible. Then the operator $U=\sum_{i,j}|f(i,j)\rangle\langle\lambda_j^{(i)}|$ is unitary and transforms into the standard basis. No ancillas are involved. $\endgroup$
    – DaftWullie
    Oct 18 at 7:42
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    $\begingroup$ I think the issue with the swap test is what do you want as an outcome? Do you just want the measurement outcome, or do you want the correct state after measurement? The way I'm proposing potentially projects onto every possible $|\lambda_j\rangle$, which collapses the system more than, for example, the projector onto the symmetric subspace would. $\endgroup$
    – DaftWullie
    Oct 18 at 7:47
  • $\begingroup$ Thanks! Makes sense now. $\endgroup$
    – BlackHat18
    Oct 19 at 0:41
  • $\begingroup$ OK, so if that's what you wanted, I'll turn it into an answer... $\endgroup$
    – DaftWullie
    Oct 19 at 6:54
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If all you care about is the measurement outcome, there's a straightforward way to do it. Essentially, you write out the full list of $\lambda^{(i)}_j$ and make a mapping between them and some number in the range 1 to $2^n$. Let me call this mapping the function $f(i,j)$. In that case, you can write down the unitary $$ U=\sum_{i=1}^m\sum_{j=1}^{k_i}|f(i,j)\rangle\langle \lambda^{(i)}_j|. $$ Apply this unitary, and measure all $n$ qubits in the standard basis, and you have applied the measurement. In fact, you have applied a possibly slightly more detailed measurement because you've actually changed it into a measurement where all the projectors are rank 1, corresponding to the $|\lambda^{(i)}_j\rangle$. There's no problem with "forgetting" some of this information to determine which of the outcomes of a higher rank measurement operator it was.

There is, however, a distinction if you're interested in the state after measurement. Take the SWAP test on two qubits as an example. In the SWAP test, you effectively have two projectors, onto the symmetric and anti-symmetric subspaces. For two qubits, the symmetric subspace has dimension 3 ($|00\rangle, |11\rangle, (|01\rangle+|10\rangle)/\sqrt{2}$) and my proposed measurement would separate out these three, destroying any superposition between them. Instead, if you care about the state after measurement, it is clear (in this case) that an ancilla is necessary. You want a two-outcome measurement, so that means measuring a qubit. But you want to leave behind a 3-dimensional space, which means having two qubits that have not been measured.

This argument suggests a crude bound which would guarantee the requirement of an ancilla: $$ m\times\max_ik_i>2^n. $$

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As far as I can tell, this is just a special case of Naimark's dilation: given an arbitrary POVM $(\mu(a))_{a\in\Sigma}$, there is an isometry $V$ and a projective measurement $(e_a e_a^\dagger)_{a\in\Sigma}$ such that $$\mu(a) = V^\dagger (I\otimes e_a e_a^\dagger) V.$$ In your case, the initial POVM is itself a projective measurement, and thus the above can simplified: no need to enlarge the space, $V$ can be a simple unitary rather than an isometry, and the question is simply to find $V$ that changes basis from the one corresponding to your projective measurement $(M_i)$ to some fixed computational basis $(e_a)_a$.

In other words, you are asking, given a projective measurement $(M_i)_{i=1}^m$, what is/how to find a unitary $V$ such that $M_i = V^\dagger \Pi_i V$, so that $\langle M_i,\rho\rangle = \langle \Pi_i, V\rho V^\dagger\rangle$, and $(\Pi_i)_{a=1}^m$ is a projective measurement in the "standard basis". For this, any unitary $V$ sending the eigenbasis of $M_i$ to that of $\Pi_i$ will do. For example, $$V = \sum_{j=1}^{k_1} |j\rangle\!\langle \lambda_j^{(1)}| + \sum_{j=1}^{k_2} |k_1 + j\rangle\!\langle \lambda_j^{(2)}| + ...$$ Note that there is no need to refer to a multipartite structure here. If you are dealing with a system of qubits, then a projective measurement on $\log m$ qubits will have $\sim m$ outcomes, thus the second requirement is satisfied naturally.

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