# Measurement in QM - positivity criterion

Let me describe my understanding of measurements in quantum mechanics before getting to the part that I'm unclear about.

If we have a quantum state $$\rho$$ and we do a projective measurement in some basis, then our measurement can be described in the following way. We have a set of measurement elements $$\{F_i\}$$ where each $$F_i = M^\dagger_i M_i = \vert i\rangle\langle i\vert$$. The measurement elements are clearly positive and Hermitian and we also have that $$\sum_i F_i = I$$.

Moving onto POVMs, the difference now is that we don't require that the $$F_i$$ satisfy orthogonality. However, do they still need to be positive operators? In my textbook, this condition is not asked for. As long as $$\sum_i F_i = I$$, it is a valid measurement.

In contrast, on Wikipedia, it is stated that a POVM is

... a measure whose values are non-negative self-adjoint operators on a Hilbert space..

yet later on in the same article, when talking about POVMs seen as the system coupled to an ancilla with projective measurements, it says

The operators of the resulting POVM are given by $$F_i = M^\dagger_iM_i$$. Since the $$M_i$$ are not required to be positive...

Does it mean that $$F_i$$ is still positive even if the $$M_i$$ are not? And if yes, how does one show this? I can see that $$M^\dagger_i M_i$$ is Hermitian so it has real eigenvalues but I cannot see why it should have positive eigenvalues.

I would appreciate it very much if someone could clarify what is going on.

To answer the last part first: If $$F_i=M_i^\dagger M_i$$, it is automatically positive, as it is hermitian and $$\langle \phi|F_i|\phi\rangle = \langle \phi|M_i^\dagger M_i|\phi\rangle=(M_i|\phi\rangle)^\dagger (M_i|\phi\rangle) \ge 0$$.

As to the first part: The basic idea for general measurements is that probabilities should be given by linear operators on $$\rho$$ (otherwise the whole ensemble interpretation breaks down), $$p_i = \mathrm{tr}[F_i\rho]\ .$$

Now first, it is easy to see that you can choose $$F_i$$ hermitian: Since $$\rho$$ is hermitian and $$\mathrm{tr}[F_i\rho]\in\mathbb R$$, $$\mathrm{tr}[F_i\rho]=\mathrm{tr}[F_i^\dagger\rho]$$, and we can replace $$F_i$$ by $$\tfrac12(F_i+F_i^\dagger)$$.

Second, since $$\sum p_i=1$$, it follows that $$\sum F_i=\mathrm{Id}$$.

Finally, if $$F_i$$ were not positive, there would be a $$|\phi\rangle$$ such that $$\langle \phi|F_i|\phi\rangle<0$$. Then, with $$\rho=|\phi\rangle\langle\phi|$$, it follows that $$p_i = \mathrm{tr}[F_i\rho]= \mathrm{tr}[F_i |\phi\rangle\langle\phi|]<0\ ,$$ which is impossible and thus, $$F_i$$ must be positive semidefinite.

• "Since $\rho$ is Hermitian, only the Hermitian part of $F_i$ contributes". Do you say this because you know that $p_i$ is a probability and hence real or is there some other reason? – user1936752 Dec 26 '18 at 20:32
• @user1936752 Hm, you are right, it is indeed because it is real, which implies $tr(\rho F)=tr(\rho F^\dagger)$. Thanks. – Norbert Schuch Dec 26 '18 at 20:54