# Performing a measurement in the standard computational basis of a three qubit system on two qubits

I often see written "and then we perform measurement in the standard computational basis" but I'm a little fuzzy on what this means as it's never stated what type of measurement we're supposed to take.

Firstly I know what the standard computational basis is and secondly I know that (usually) measurement is take using projective operators.

What I mean is , say we're given some circuit with three qubits for instance and in the circuit at the end is written the measurement symbol on the first and second wire but not the last then how do we know what projective operators to use ( I'm assuming that we have to use a complete set of measurement operators to get a full measurement ). There is no one complete set of measurement operators it all depends on what type of measurement you want to perform so in this case should we perform projective measurement on the end state with all of :

$$P_0 \otimes P_0\otimes I\\ \vdots\\ P_1 \otimes P_1\otimes I$$

Or

$$P_0 \otimes P_0\otimes P_0\\ \vdots\\ P_1 \otimes P_1\otimes P_1$$

Because I would assume that we use the first given that the circuit requires measurement on just the first two but then what becomes of the third qubit if we don't project as we would in the second set I listed. Its wavefunction won't collapse if we don't measure it so do we just discard it if the measurement is not stated for it? Otherwise, what do we do?

P.S. A little bit of a side note but if we want to perform a measurement in say, for example, the bell basis, then do we just take the density operator of the 4 bell states and treat them as projection operators?

How you perform the projection is exactly what you would normally do. If $$|\psi\rangle$$ is a 3-qubit state, you calculate, for example, the probability of getting the 00 result: $$p_{00}=\langle\psi|P_0\otimes P_0\otimes I|\psi\rangle,$$ and the state after measurement is $$\frac{P_0\otimes P_0\otimes I|\psi\rangle}{\sqrt{p_{00}}}.$$ You do not discard the third qubit. It does collapse somewhat - for example, had your initial state been $$(|000\rangle+|111\rangle)/\sqrt2$$, you get the 00 answer with probability 1/2, and the final state is $$|000\rangle$$.
• @bhapi Yes, unless you've already found that $\sum_ip_i=1$ for all the probabilities you've already calculated, because then you know all others have 0 probability. – DaftWullie Aug 14 '19 at 8:59