As a fundamental component for a quantum computation, the measurement needs to be implemented in a fault tolerant way.

As indicated in Chuang and Nielsen Quantum Computation and Quantum Information, a quantum circuit that allows the implementation of a FT measurement is the one represented in Fig.1:

Fault tolerant Quantum Measurement Quantum Circuit representation of a fault tolerant measurement.

However, three measurements are needed in order to make this implementation actually fault tolerant, due to the possible presence of errors in various part of the circuit (as it is said in the book, between two CNOT gates in a verification step for the cat state).

But how is it actually possible to execute three subsequent set of measurements? Isn't the data (the states on which the measured is performed) modified by the application of the measurement operation? If that is the case, are the measurement operations applied on different data?


2 Answers 2


The three measurements you refer to are composed of two data qubits and an ancilla. They essentially ask the question "is the state of the data qubits in the subspace spanned by $|00\rangle$ and $|11\rangle$ (in which case the measurement result on the ancilla will be $0$) or is the state in the subspace $|01\rangle$ and $|10\rangle$ (in which case the ancilla would output $1$)?"

This means that if you only ever prepare you state in some superposition of $|00\rangle$ and $|11\rangle$, then the measurement will have no effect. It will leave the superposition completely intact, because every part of the state corresponds to the same measurement result.

In the actual circuit here, things are a little more complicated. There are three data qubits in all, prepared in a superposition of $|000\rangle$ and $|111\rangle$. But still, for each pair of data qubits the state should be within the subspace spanned by $|00\rangle$ and $|11\rangle$. So the three measurements (each for a different pair of data qubits) should always return the output $0$. They would have no effect on the superposition in this case.

If they return a different result, it is a sign of error. The majority voting helps us determine what to do to mitigate for that error.


The three measurements you refer to are applied on parts of the circuit where a qubit is first initialized, then acted upon by a controlled gate, and then measured. The fact that they are initialized at the beginning of each of these three processes means that any dependence on their previous history is removed. So we could use the same qubit three times (and use the result of the previous measurement to inform us how to initialize) or we could use three different qubits. We can just choose whatever is practical.

In general, there is not reason why we shouldn't measure a qubit multiple times, despite the fact that measurement modifies the state. For example, in stabilizer codes, we constantly move information about errors that have occurred to certain qubits, and then measure them to get that information out. In this case, the effect of the measurement in collapsing superpositions has the positive benefit of collapsing complex types of error into simpler ones. So if you ever see a multiply measured qubit in a reputable source, you can be sure that the disturbance caused by the measurement has been taken into account, and is probably being specifically made use of.

  • $\begingroup$ I'm not sure I was able to explain me: it's OK to measure multiple times on the verification step (I assume you are referencing them in "where a qubit is first initialized, then acted upon by a controlled gate, and then measured"). The problem (if it's a problem) is the fact that the serie of controlled M' needs to be implemented three times, in order to allow a majority vote on the result. Does this means that you need three results of measurement (set of C-M' gates) on the same set of data qubits? However, thank you for your reply! $\endgroup$
    – D-Brc
    Oct 4, 2018 at 6:58
  • $\begingroup$ I think I misunderstood your question, so I posted another answer. $\endgroup$ Oct 4, 2018 at 8:22

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