There are a lot of different ways of looking at qubits, and the state vector formalism is just one of them. In a general linear-algebraic sense a measurement is projection onto a basis. Here I will provide insight with an example from the Pauli observable point of view, that is the usual circuit model of QC.

Firstly, it's of interest which basis the state vector is being provided in-- every measurement operator comes with a set of eigenstates, and whatever measurements you look at (eg. $X,Y,Z, XX, XZ$, etc.) determine the basis that might be best for you to write the state vector in. The easiest way to answer your question is if you know which basis is of interest to you, and more importantly, *whether it commutes with the measurement you just made*.

So for simplicity's sake, let's say you start with two coupled qubits in an arbitrary state written in the $Z$-basis for both qubits:

$$| \psi \rangle = a | 0_{Z} \rangle \otimes | 0_{Z} \rangle +b | 0_{Z} \rangle \otimes | 1_{Z} \rangle + c | 1_{Z} \rangle \otimes | 0_{Z} \rangle + d | 1_{Z} \rangle \otimes | 1_{Z} \rangle $$

The simplest possible measurements you could make would be $Z_{1}$, that is the $Z$ operator on the first qubit, followed by $Z_{2}$, the $Z$ operator on the second qubit. What does measurement do? It projects the state into one of the eigenstates. You can think of this as eliminating all possible answers that are inconsistent with the one we just measured. For instance, say we measure $Z_{1}$ and obtain the outcome $1$, then the resulting state we would have would be:

$$| \psi \rangle = \frac{1}{\sqrt{|c|^{2} +|d|^{2}}} \left(c | 1_{Z} \rangle \otimes | 0_{Z} \rangle + d | 1_{Z} \rangle \otimes | 1_{Z} \rangle \right) $$

Note that the coefficient out front is just for renormalization. So our probability of measuring $Z_{2}=0$ is $\frac{1}{|c|^{2} +|d|^{2}} |c^{2}|$. Note this is different from the probability we had in the initial state, which was $|a|^{2}+|c|^{2}$.

Suppose the next measurement you make does not commute with the previous one, however. This is trickier because you have to implement a change of basis on the state vector in order to understand the probabilities. With Pauli measurements, though, it tends to be easy since the eigenbases relate in a nice way, that is:

$$| 0_{Z} \rangle = \frac{1}{\sqrt{2}} (|0_{X}\rangle + |1_{X} \rangle )$$

$$| 1_{Z} \rangle = \frac{1}{\sqrt{2}} (|0_{X}\rangle - |1_{X} \rangle )$$

A good way to check your understanding: What is the probability of measuring $X= +1$ after the $Z_{1}=1$ measurement above? What is the probability if we have not made the $Z_{1}$ measurement? Then a more complicated question is to look at product operators that act on both qubits at once, for instance, how does a measurement of $Z_{1}Z_{2}=+1$ affect the initial state? Here $Z_{1}Z_{2}$ measures the product of the two operators.

2Nice and simple answer. I think it is important to note, that what you describe is only true if you a) perform projective measurements and b) you

knowthe outcome of the measurement. Just keep in mind that in general you will need mixed states to describe the post-measurement state. – M. Stern – 2018-03-20T00:35:58.720