I'm reading Nielsen and Chuang's famous book on Quantum Computation and Information. In section 2.2 on the postulates of quantum mechanics, they talk about quantum measurements starting with

Postulate 3

Quantum measurements are described by a collection $\left \{ M_{m} \right \}$ of measurement operators. These are operators acting on the state space of the system being measured. The index $m$ refers to the measurement outcomes that may occur in the experiment. If the state of the quantum system is $|\psi \rangle$ immediately before the measurement then the probability that result $m$ occurs is given by $$ p\left ( m \right ) = \left\langle \psi | M_{m}^{\dagger}M_{m} |\psi \right\rangle $$ and the state of the system after the measurement is $$\frac{M_{m}|\psi\rangle}{\sqrt{ \left \langle \psi | M_{m}^{\dagger}M_{m} |\psi \right\rangle}}$$ The measurement operators satisfy the completeness equation $$\sum_{m} M_{m}^{\dagger}M_{m} = I$$

After that, the authors talk about the well-known Projective measurements and POVMs. They say that: "Section 2.2.5 explains a special case of Postulate 3, the projective or von Neumann measurements. Section 2.2.6 explains another special case of Postulate 3, known as POVM measurements."

I uderstand the Projective measurement as being a special case of these generalized measurements from Postulate 3. But it is hard to see how their description of POVMs may be a special case of Postulate 3:

POVM Measurements

Suppose a measurement described by measurement operators $M_m$ is performed upon a quantum system in the state $|\psi \rangle$. Then the probability of outcome $m$ is given by $ p\left ( m \right ) = \left\langle \psi | M_{m}^{\dagger}M_{m} |\psi \right\rangle $. Suppose we define $$ E_m = M_{m}^{\dagger}M_{m} $$. Then from Postulate 3 and elementary linear algebra, $E_m$ is a positive operator such that $\sum_{m} E_{m} = I$ and $ p\left ( m \right ) = \left\langle \psi | E_{m} |\psi \right\rangle $. Thus the set of operators $E_m$ are sufficient to determine the probabilities of the different measurement outcomes. The operators $E_m$ are known as the POVM elements associated with the measurement. The complete set ${E_m}$ is known as a POVM.


And, finally, in Box 2.5 in page 91 they end with

Where do POVMs fit in this picture? POVMs are best viewed as a special case of the general measurement formalism, providing the simplest means by which one can study general measurement statistics, without the necessity for knowing the post-measurement state. They are a mathematical convenience that sometimes gives extra insight into quantum measurements.

My questions are:

  1. Defining the POVM as they did (with $ E_m = M_{m}^{\dagger}M_{m} $) there is nothing as From Postulate 3 and some linear algebra, they just changed names and Postulate 3 is intact. I can't see how this is a "special case". They just renamed it. So, what am I missing? How these POVMs are different from what they wrote in Postulate 3 as a general measurement?
  2. From the question above I continue with: since they just renamed things, whet do they meant with "without the necessity of knowing the post-measurement state"?

I mean, the difference between the the general measurement and the POVM is that the first is the set ${M_m}$ and the second is the positive operator $ M_{m}^{\dagger}M_{m} $ and I can see that I can't rewrite the post-measurement state as something like ${E_{m}|\psi\rangle}$. Are the POVM "just" a mathematical tool for computing statistics? Are there anything important that I am missing or that's it?


1 Answer 1


The POVM Measurements are a strict subset of the full postulate 3, because $M_m$ uniquely determines $E_m$, but $E_m$ does not uniquely determine $M_m$ (for example transforming $M_m \rightarrow U_mM_m$ leaves $E_m$ unchanged if $U_m$ is unitary). $E_m$ is sufficient to determine the measurement probabilities (via $P(m) = \langle \psi|E_m|\psi\rangle$), but do not determine the state of the system after measurement. For the latter, you need the full postulate 3, in particular that the state after measurement is

$$\frac{M_m|\psi\rangle}{\sqrt{\langle\psi|M^\dagger_m M_m|\psi\rangle}} \neq \frac{U_mM_m |\psi\rangle}{\sqrt{\langle\psi|M_m^\dagger U_m^\dagger U_m M_m|\psi\rangle}}$$

To illustrate this distinction, think of measuring $\sigma_z$ for a two-level atom. A projective measurement has $M_0 = E_0 = |0\rangle\langle 0|$ and $M_1 = E_1 = |1\rangle\langle 1|$. But if instead use on spontaneous emission of the $|0\rangle$ state to decay to the $|1\rangle$ state and detect the emitted photon, then we have $M_0 = |1\rangle\langle 0|$, $E_0 = |0\rangle\langle 0|$, and $M_1 = E_1 = |1\rangle\langle 1|$ as before. As the $E$s are the same, these measure the same physical quantity, but the physical process and final states are different.


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