What happens to a quantum mechanical system if it is measured using a Projector that is orthogonal to the state of the system? I was wondering what actually happens to a physical system once it is subjected to a projective measurement.
Let's suppose the system is a qubit in the state $|1\rangle$. If I apply $|1\rangle\langle 1|$, the measurement result is $1$ and the qubit remains in the state $|1\rangle$.
But applying $|0\rangle \langle 0|$ gives the result $0$, meaning no state at all. Does that mean that the state is always destroyed when subjected to an orthogonal projection?
Or should I interpret the result as $|0\rangle\langle 0|1\rangle = 0 \times |0\rangle$, meaning that the final state is $|0\rangle$ with a measurement result of $0$?
 A: Quantum measurement is a random process. You cannot "choose" which outcome you get. This is reflected in the fact that a projective measurement always corresponds to a complete set of projectors $\lvert \phi_i\rangle \langle\phi_i\rvert$, such that
$$ \sum_i \lvert \phi_i\rangle \langle\phi_i\rvert = 1. $$
There is no such thing as "measuring a projector". 
Given a two-level system in state $\lvert \psi\rangle$, a measurement in the basis $\{\lvert 0\rangle\langle 0\rvert,\lvert 1\rangle\langle 1\rvert\}$ gives one of the two outcomes $\{\lvert 0\rangle,\lvert 1\rangle\}$ with probability 
$$P_j = \lvert \langle j\rvert\psi\rangle\rvert^2  $$
for $j= 0,1$. In your example, $\lvert \psi\rangle = \lvert 1\rangle$ and therefore the outcome $\lvert 1\rangle$ occurs with probability $P_1 = 1$. The outcome $\lvert 0\rangle$ has probability $P_0= 0$, i.e. it simply does not occur.
A: The answer to both questions is no.
You should interpret the result as telling you that the outcome $|0\rangle$ has probability $0$, hence the outcome is $|1\rangle$ with probability $1$.
A: You have mistaken a “projective measurement” for a “measurement after projection”. If you have an orthogonal projector, your wavefunction indeed ‘disappears’, and you will measure nothing afterwards. A projective measurement will just give you an outcome as described by Mark Mitchison.
