2
$\begingroup$

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.

Ok

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?

$\endgroup$
1

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.