Suppose we wanted to measure the observable $Z_{1} + Z_{2} + \cdots + Z_{N}$ in a stabilizer state. Is it possible to do this using only Clifford operations, and possibly adding some auxiliary qubits?

For $N=1$, this is trivially true. For $N=2$, this is possible using one ancilla, as follows:

  1. Initializing the ancilla in state $|0\rangle$, perform two CNOTs, one from each system qubit to the ancilla. Measuring the ancilla in the $Z$ basis will then reveal the parity $Z_{1} Z_{2}$.
  2. a) If the result of the $Z_{1} Z_{2}$ measurement is $-1$, then we know we are in the $Z_{1} + Z_{2} = 0$ sector. b) If the result of the $Z_{1} Z_{2}$ measurement is $+1$, then we are either in the $Z_{1} + Z_{2} = -2$ sector or the $Z_{1} + Z_{2} = +2$ sector. We can then find out which sector by measuring either $Z_{1}$ or $Z_{2}$ (it does not matter which).

However, my suspicion is that these are special cases, and that for $N>2$ it is impossible to measure $Z_{1} + Z_{2} + \cdots + Z_{N}$ using only Clifford operations and auxiliary qubits. I would appreciate it if anyone could either provide a proof of this, or a counterexample.

  • $\begingroup$ I assumeyou don't want to just measure $Z_1$, and$Z_2$, and$Z_3$ etc.? If not, can you formalise a condition that clarifies what is not OK about that solution, but is OK about the solution you've given for $N=2$? (Is it something to do with the fact that you're not disturbing the superposition within a given measurement subspace?) $\endgroup$
    – DaftWullie
    Nov 11 '21 at 16:26
  • $\begingroup$ Yes, that's right: individually measuring all the $Z_{i}$ reveals too much information. Denote by $P_{Q}$ the projector on to the eigenspace of $\sum_{i} Z_{i}$ with eigenvalue $Q$. I want to apply the projector $P_{Q}$ to the stabilizer state $|\psi\rangle$ with Born rule probability $p_{Q} = \parallel P_{Q} |\psi\rangle \parallel^{2}$. As you say, in principle this can preserve superpositions of states within the same eigenspace, whereas individually measuring the $Z_{i}$ will collapse the state to a product state. $\endgroup$
    – anon1802
    Nov 11 '21 at 17:30

No, it's not possible. For example, being able to directly measure $X+Y$ would allow you to prepare T states and thereby perform T gates, which are not stabilizer operations.

If the fact that $X$ and $Y$ don't commute bothers you, then note that preparing $|+\rangle^{\otimes n}$ and then measuring $\sum_{k=0}^{n-1} Z_k$ and getting a result of $n-2$ would indicate you prepared a $W_n$ state. But $W_n$ states aren't stabilizer states, so you shouldn't be able to prepare them with a stabilizer circuit even probabilistically. For example, you can turn $W_4$ into a $|CCZ\rangle$ state, which you can then teleport through to perform Toffoli gates:

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  • $\begingroup$ Thanks! I suppose this gives another way of seeing why $N=2$ is special: $W_{2}$ is a stabilizer state, but this is not true for higher $N$. $\endgroup$
    – anon1802
    Nov 11 '21 at 18:30
  • $\begingroup$ Nice counterexamples. Note that the $(n-2)$-eigenspace of $\sum_{k=0}^{n-1}Z_k$ is $n$-fold degenerate so the collapsed state depends on the input. That said, you can use a stabilizer state like $|+\dots +\rangle$ to get $W_n$ indeed. $\endgroup$ Nov 11 '21 at 18:52
  • $\begingroup$ @AdamZalcman Ah good point, I forgot to say the starting state. $\endgroup$ Nov 11 '21 at 18:54

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