Understanding projective measurements as a special case of POVM measurements ("third postulate" in Nielsen and Chuang) I am working through Nielsen and Chuang's book and am confused about a detail from sections 2.2.3 and 2.2.5. 
On page 88 of my copy (section 2.2.5), they write

Projective measurements can be understood as a special case of Postulate 3. Suppose the measurement operators in Postulate 3, in addition to satisfying the completeness relation $\sum_m M_m^\dagger M_m = I$ also satisfy the conditions that $M_m$ are orthogonal projectors, that is, the $M_m$ are Hermitian, and $M_mM_{m^\prime} = \delta_{m,m^\prime}M_m$.

It seems to me that they're implying that orthogonal projectors are (1) Hermitian and also (2) satisfy $M_mM_{m^\prime} = \delta_{m,m^\prime}M_m$.
My question: My understanding is that a projector is simply an operator which satisfies $P^2 = P$, and for projectors to be orthogonal means that the composition of two distinct ones always yields zero, i.e. $(P_1 \circ P_2)(x) = 0$ for all $x$. But this is all covered by part (2) of the statement alone. So why is (1) necessary?
Edit: Here is a screenshot of their statement of postulate 3  
 A: You're on the right track.  However, if you look closely at what it means for a projector to be orthogonal, then you'll see that an orthogonal projection is necessarily self-adjoint, and vice-versa.
A projector $P$ on a vector space $\mathcal H$ is an operator which satisfies the following conditions:


*

*$P$ is idempotent, so $P\circ P = P$

*The entire Hilbert space can be decomposed as a direct sum $\mathcal H = U\oplus V$ where $P(U) = U$ and $P(V)=\{0\}$

*For all $u\in U$, $P(u) = u$
For a given $x\in\mathcal H$, the decomposition in (2) is given by $x = u + v$ where $u = P(x)$ and $v = x-P(x)$.  Note that $U=Ran(P)$ and $V=Ker(P)$.

When the vector space in question is a Hilbert space, then it's possible to introduce the notion of orthogonality into the picture.  We say that a projector $P$ is orthogonal if, for any $u\in Ran(P)$ and $v\in Ker(P)$, we have that $\langle u,v\rangle = 0$.  
Another way to phrase this is that for any $x,y\in \mathcal H$, we have that
$$\langle \underbrace{Px}_{\in U},\underbrace{(I-P)y}_{\in V}\rangle = \langle\underbrace{(I-P)x}_{\in V},\underbrace{Py}_{\in U}\rangle = 0$$
where $I$ is the identity operator on $\mathcal H$.  This immediately implies that
$$\langle Px, y\rangle - \langle Px,Py\rangle = 0$$
and
$$\langle x,Py\rangle - \langle Px,Py \rangle= 0$$
which together imply that
$$\langle Px, y \rangle = \langle x, Py\rangle$$
We could also have gone the other direction.  If we ignore the question of orthogonality and require that $P$ is self-adjoint, then
$$\langle Px,(I-P)y\rangle = \langle P\circ P x,(I-P)y \rangle = \langle Px, P\circ(I-P)y\rangle $$
$$ = \langle Px, (P-P^2)y\rangle =\langle Px, (P-P)y\rangle = 0$$
The same line of reasoning shows that $\langle(I-P)x,Py\rangle=0$, so we see that a projector is orthogonal if and only if it is self-adjoint.  Therefore, rather than demanding that our projectors be orthogonal, we can equivalently demand them to be self-adjoint, and the orthogonality follows.

To see where your intuition wasn't quite right, look at what you said. 

[...] projectors to be orthogonal means that the composition of two distinct ones always yields zero, i.e. $(P_1\circ P_2)(x)=0$ for all $x$.

By distinct ones, what I assume you mean is that $Ran(P_1)\cap Ran(P_2) = \{0\}$.  If $P_1$ and $P_2$ have that property then certainly $(P_1 \circ P_2)(x)=0$, but that's true by definition of what we mean by a projector, and in fact makes no use at all of the notion of orthogonality.  It also requires a reference to multiple projectors, which is not good if we're trying to talk about a property of a single projector rather than a family of them.
A: https://en.wikipedia.org/wiki/Projection_(linear_algebra)#Orthogonal_projections states:

An orthogonal projection is a projection for which the range U and the null space V are orthogonal subspaces.

Thus, orthogonality is a property of a single projection, not of  a set of projections, as you state it (some kind of mutual orthogonality) -- so the immediate answer to your question is: "You are using the wrong definition of orthogonal projection".
Immediately afterwards, it is shown that:

A projection is orthogonal if and only if it is self-adjoint. 

