Confusion about the relation between POVMs and projective measurements

3

I'm a little confused about the terminology of measurement.

So say that we have the single qubit state $|\phi \rangle=c_0|0\rangle+c_1|1\rangle$.

If we perform the projective measurement $P_0=|0\rangle\langle 0|$. Then we say that the probability of obtaining the measurement result $|0\rangle$ is $\langle \phi|P_0|\phi \rangle$. So in this case we're talking about a possible state that can occur in the collapse of the wavefunction of $|\phi\rangle$.

In the context of P.O.V.M. measurement where we assume it's not a projective measurement, and we label the measurement operators $E_i$ then we say that if the result of the measurement is $E_i$ then we assume that the state sent by Alice was state-x. The probability of the measurement result $\langle\phi_j| E_i |\phi_j\rangle$.

So, in this case, it seems as though we're talking about the measurement of an operator instead , but if $E$ can be projective operator and in the case of a projective operator we're measuring the probability of the state collapsing into some state, then why does it seem like that's not what's happening in the more general case ?

bhapi

Posted 2019-08-13T22:43:25.860

Reputation: 769

Answers

2

POVMs are more general than projective measurements. Thus, every projective measurement is also a POVM, by choosing $E_i=P_i$.

Norbert Schuch

Posted 2019-08-13T22:43:25.860

Reputation: 3 740

Perhaps I didn't phrase my question very well. I know that POVM's can be projective operators and just a more general case of measurement operator my confusion is about what is actually being measured. For in the case of projective measurements we're measuring a state but in the context of projective measurements it seems we're measuring an operator . Yet if POVM's can be projectors then why aren't we measuring an operator in that case too ? – bhapi – 2019-08-14T00:51:50.313

1

In the case of projective measurements, we have a set of projectors $\{P_i\}$ satisfying the completeness relation $$ \sum_iP_i=I. $$ Note that this also means they satisfy $\sum_iP_i^\dagger P_i$, which I would argue is more relevant. If you have a state $|\phi\rangle$, then with probability $p_i=\langle\phi|P_i^\dagger P_i|\phi\rangle=\langle\phi|P_i|\phi\rangle$, you get outcome $i$. If you get outcome $i$, the state after measurement is $P_i|\phi\rangle/\sqrt{p_i}$.

For a general measurement, you have operators $\{M_i\}$ satisfying a completeness relation $$ \sum_iM_i^\dagger M_i=I $$ If you have a state $|\phi\rangle$, then with probability $p_i=\langle\phi|M_i^\dagger M_i|\phi\rangle$, you get outcome $i$. If you get outcome $i$, the state after measurement is $M_i|\phi\rangle/\sqrt{p_i}$. So you can see how projective measurements are a special case of general measurements.

POVMs use a slightly different formalism. They make the assumption that you are only interested in the probability of measurement outcomes, and not the state after measurement. So, they choose to set $E_i=M_i^\dagger M_i$, so they satisfy $\sum_iE_i=I$, and the probability of outcome $i$ is $p_i=\langle\phi|E_i|\phi\rangle$, but the formalism is deliberately chosen so you cannot answer the question of what the post-measurement state is.

DaftWullie

Posted 2019-08-13T22:43:25.860

Reputation: 35 722