Here, we provide more details of the construction mentioned in Roy Simpson's answer. The question (v1) really has two parts:
How are Hermitian observables $A:H \to H$ and projective measurements $P_m:H \to H$ related? This issue is discussed in many places, e.g. in this answer.
How are projective measurements $P_m$ and general measurements $M_m$ related? This will be the main topic of this answer.
Let $H$ be the Hilbert space of the system. (We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.)
I) Projective measurement operators $P_m:H \to H$, $m\in I$, satisfy by definition
$$P_n P_m ~=~ \delta_{nm} P_m, \qquad P^{\dagger}_m~= P_m , \qquad \sum_{m\in I} P_m ~=~{\bf 1}_{H}. \tag{1}\label{eq:1} $$
Here $I$ is an index set.
II) General measurement operators $M_m:H \to H$, $m\in I$, satisfy by definition
$$ \sum_{m\in I} M^{\dagger}_m M_m ~=~{\bf 1}_{H}.\tag{2}\label{eq:2} $$
Consider a fixed $m\in I$. The probability $p(m)$ to measure outcome $m$ for the [density operator (https://en.wikipedia.org/wiki/Density_matrix) $\rho:H\to H$ is
$$ p(m)~=~ {\rm tr}_H (M^{\dagger}_m M_m\rho).\qquad\qquad(3) $$
The collapse $\rho\longrightarrow \rho^{\prime}$ of the density operator, due to the general measurement, is
$$ \rho^{\prime}~=~ \frac{M_m\rho M^{\dagger}_m}{p(m)}.\qquad\qquad(4) $$
This is essentially Postulate 3 in Ref.1.
III) On one hand:
On the same Hilbert space $H$, the projective measurement operators $\eqref{eq:1}$ are a very special case of general measurement operators $\eqref{eq:2}$.
It is not hard to construct examples of general measurement operators that are not projective measurement operators (if the Hilbert space $H$ is fixed, cf. Section IV below). Note, in particular, that if we repeat the general measurement $M_m$ with the same $m$, the doubly collapsed density operator
$$ \rho^{\prime\prime}~=~ \frac{M_m\rho^{\prime} M^{\dagger}_m}{{\rm tr}_H (M^{\dagger}_m M_m\rho^{\prime})}\qquad\qquad(5) $$
might in general be different from $\rho^{\prime}$. On the other hand, for projective measurement operators $\rho^{\prime\prime}=\rho^{\prime}$ always, mainly because of idempotency of $P_m$.
IV) On the other hand:
There is a way to realize general measurement operators $M_m:H \to H$, $m\in I$, which live on a Hilbert space $H$, as projective measurement operators $P_m:L \to L$, $m\in I$, on a larger Hilbert space $L:=H\otimes K$ by introducing a so-called ancilla Hilbert space $K~\cong~\mathbb{C}^{I}$ with an orthonormal basis $|m\rangle\in K$, $m\in I$, labeled by the same index set $I$.
This is explained in Section 2.2.8 of Ref. 1. The construction relies, in particular, on Postulate 4 for tensor products in Ref. 1. A sketched proof goes as follows.
Choose a fixed normalized state $|a_0\rangle\in K$. Call the corresponding density operator $\rho_K:=|a_0\rangle\langle a_0|: K \to K$.
Introduce an isomorphic copy $\tilde{H}$ of $H$ inside $L$ as
$$H~\stackrel{\cong}\longrightarrow~ \tilde{H}~:=~H\otimes |a_0\rangle ~\subseteq ~H\otimes K~=:~L.\qquad\qquad(6) $$
Define an isometry $U:\tilde{H} \to L$ as
$$U~:=~ \sum_{m\in I} M_m \otimes |m\rangle \langle a_0 |.\tag{7}\label{eq:7}$$
It is an isometry mainly because of eq. $\eqref{eq:2}$ and because $|m\rangle\in K$, $m\in I$, is an orthonormal basis.
Extend the isometry $U:\tilde{H} \to L$ from eq. $\eqref{eq:7}$ to a unitary operator $\tilde{U}:L \to L$.
Define projective measurement operators $Q_m:L\to L$, $m\in I$, as
$$Q_m~:=~ {\bf 1}_H\otimes|m\rangle \langle m|. \qquad\qquad(8) $$
Define unitarily equivalent projective measurement operators $P_m:L\to L$, $m\in I$, as
$$P_m~:=~\tilde{U}^{\dagger}Q_m\tilde{U} . \tag{9}\label{eq:9} $$
Recall the fixed density operator $\rho_K:=|a_0\rangle\langle a_0|: K\to K$ from above. Consider an arbitrary density operator $\rho_H: H\to H$. Define the product density operator $\rho_L:=\rho_H\otimes\rho_K: L\to L$.
Consider for fixed $m\in I$ the projective measurement operator $P_m:L\to L$ defined in eq. $\eqref{eq:9}$. Now apply Postulate 3 for projective measurement operator $P_m:L\to L$.
The probability is
$$p(m)~=~ {\rm tr}_L (P^{\dagger}_m P_m\rho_L)~=~\ldots
~=~{\rm tr}_H (M^{\dagger}_m M_m\rho_H).\tag{10}\label{eq:10} $$
The collapse $\rho_L\longrightarrow \rho^{\prime}_L$ of the density operator, due to the projective measurement, is
$$ \rho^{\prime}_L~=~\frac{P_m\rho_L P^{\dagger}_m}{p(m)}
~=~\ldots~=~\rho^{\prime}_H~\otimes~\rho^{\prime}_K,\qquad\qquad(11) $$
where
$$\rho^{\prime}_H~:=~\frac{M_m\rho_H M^{\dagger}_m}{p(m)}~=~{\rm tr}_K\rho^{\prime}_L,\tag{12}\label{eq:12} $$
and
$$\rho^{\prime}_K~:=~|m\rangle \langle m|~=~{\rm tr}_H\rho^{\prime}_L.\tag{13}\label{eq:13} $$
Equations $\eqref{eq:10}$ and $\eqref{eq:12}$ reproduce Postulate 3 for the initially given general measurement operator $M_m:H\to H$. (The last equality of eqs. $\eqref{eq:12}$ and $\eqref{eq:13}$ uses partial trace.)
References:
- M.A. Nielsen and I.L. Chuang, Quantum Computation and Quantum Information, 2011.
