# Can you measure sums of Paulis in the stabilizer formalism?

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.

• 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?) Nov 11 '21 at 16:26
• 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. 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:
• 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 $N$. Nov 11 '21 at 18:30
• 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. Nov 11 '21 at 18:52