I have a very simple question on Generalized measurements.

The definition I was given is the one below:
A set of operator $\left \{ M_{\alpha} \right \}$ is called generalized measurements for an measurement operator $M_{\alpha}$ (not necessary projections operator) if $\sum_{\alpha} M_{\alpha}^H M_{\alpha}$

But does anyone have an other definition to help me understand?

  • What does it means that $M_{\alpha}$ is an operator?
  • It just has to be a linear operators ?
  • Moreover all the different operator of the set $\left \{ M_{\alpha} \right \}$ has to measure the same quantity $\alpha$ but what does it mean? That means that each $M_{\alpha}$ is susceptible to give us a different possible value of the quantity $\alpha$ ? It is strange because when we only want operators to be Hermitian the same operator $H$ apply on a state can give all the possible values of the quantity (that are the eigenvalues of $H$) that the operator $H$ represent? (At least this is what I've learned).

More generally any more precise, even longer definition/explanation on what is Generalized measurements will be great.


2 Answers 2


$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}$ I will try to explain the generalized measurements, and I hope I address all of your questions in my explanation.

Quantum measurements are described by a set $\{M_k\}$ of measurement operators satisfying one constraint $$\begin{align} \sum_k M_k^\dagger M_k = I \,.\tag{1} \end{align}$$

Given a state $\ket{\psi}$, immediately after measurement, it becomes $\ket{\psi_k}$, \begin{equation} \ket{\psi} \xrightarrow[]{\{M_k\}}\ket{\psi_k} \equiv \frac{M_k \ket{\psi}}{\sqrt{p_k}}\,,\tag{2} \end{equation}

with probability

\begin{align} p_k &= \bra{\psi}M^\dagger_kM_k\ket{\psi}\tag{3.1}\,,\\ &= \lVert M_k \ket{\psi} \rVert^2\tag{3.2}\,,\\ &\geq 0 \tag{3.3}\,. \end{align} Measurement outcome is the index $k$ of the state that resulted.

Assume if you perform measurement corresponding to $\{ M_1, M_2, \cdots, M_k \}$ on state $\ket{\psi}$, then there can be $k$ different outcomes which can occur. Any of these outcomes can occur at random with probabilistic odds given by Eq.$(3.1)$.

\begin{equation} \ket{\psi}\xrightarrow[]{\{ M_1,M_2, \cdots, M_k \}}\begin{cases} \ket{\psi_1}, & \text{with probability $p_1$ ----- Getting outcome 1}\\ \ket{\psi_2}, & \text{with probability $p_2$ ----- Getting outcome 2}\\ \vdots\\ \ket{\psi_k}, & \text{with probability $p_k$ ----- Getting outcome $k$} \end{cases} \end{equation}

You can think of these outcomes as some pointer device in your laboratory. If you have $k$ different possible outcomes for the measurement, in the lab, your apparatus will point to one of the $k$ possible points, and then you can say that $k$-outcome has happened, and then you can deduce that state of your system now is $\ket{\psi_k}$.

So, in general, these operators $\{M_k\}$ can be any linear operators, i.e. matrices, which satisfy the condition given in Eq. $(1)$.

However, there are special cases of this general mathematical framework of generalized measurements where, in each special case, apart from Eq.$(1)$, there are some additional conditions on measurement operators. For example, in Projective (Vonn Neumann) measurements, these measurement operators form a complete set of orthogonal projectors.

  • $\begingroup$ Thank a lot for your time and your details answer! $\endgroup$
    – OffHakhol
    Commented Feb 10 at 10:11
  • $\begingroup$ So if I understood you well given a Hermitian operator is a special case. The set of $(M_{\alpha})$ of this operator will be the set of $ (\sqrt{P_{\lambda_i}}) $ with $P_{\lambda_i}$ a projector into the sub space corresponding to the e.v. $\lambda_i$? $\endgroup$
    – OffHakhol
    Commented Feb 10 at 11:28
  • 1
    $\begingroup$ You are correct. We call that measuring an observable, but underlying mathematics is the same. Every hermitian operator is an observable. We can write the spectral decomposition of the $H$ as $$H = \lambda_k |k\rangle \langle k| \,,$$ where now, your possible outcomes are distinct eigenvalues $\{\lambda_k\}$ and measurement operators are the outer product of eigenvectors corresponding to that eigenvalue $\lambda_k$. $$M_k = P_k = |k\rangle \langle k|\,. $$ If there is degenercy in eigenvalues, then the corresponding projection measurement operator would be more than dimension 1. $\endgroup$
    – FDGod
    Commented Feb 10 at 19:22
  • $\begingroup$ OK and so not all measurement are observable? How can it be? Physically speaking how can we have a measure of smthg that it is not observable? $\endgroup$
    – OffHakhol
    Commented Feb 10 at 19:40
  • $\begingroup$ Yes, broadly speaking, there are two ways of measurement. One is measuring a state, and the other is measuring an observable. $\endgroup$
    – FDGod
    Commented Feb 11 at 8:04

Forgetting for a second the way you formalise generalised measurements (that is, POVM), ask yourself how a "measurement" should be formalised in quantum mechanics. A measurement is effectively some rule sending any quantum state into a probability distribution describing the probability with which you should see each of the possible outcomes. In other words, for each outcome, call it $k$, you need a rule (that is, a function) sending each input state $\rho$ into a corresponding probability. Let us denote this function with $p_k$, so that $p_k(\rho)\in[0,1]$ for all states $\rho$.

As it happens, this function is linear. So for each outcome $k$, we need a linear function $p_k$ sending density matrices $\rho$ (which are linear operators) into numbers $p_k(\rho)$. As per usual considerations about duality, any such linear functional can be represented as $p_k(\rho)=\langle \mu_k,\rho\rangle\equiv \operatorname{tr}(\mu_k\rho)$ for some Hermitian $\mu_k$. Due to the bijection between $p_k$ and $\mu_k$, we can directly say that a "measurement" is the operator $\mu_k$. Or better said, a "measurement" is the set of such operators $\mu_k$. The other conditions on $\{ \mu_k\}$ follow from the properties of $p_k$: we want $\sum_k p_k(\rho)=1$, which corresponds to $\sum_k \mu_k=I$, and we want $p_k(\rho)\in[0,1]$, which corresponds to $0\le \mu_k \le I$.

The operators $\{\mu_k\}$ form a so-called POVM, which is the most general way to describe a measurement in quantum mechanics. But note that this is meant in the sense of: the most general way to describe the relation between quantum states and measurement probabilities. In particular, POVMs do not deal with/describe the potential states you might still have after the measurement. If you want to describe post-measurement states, the better approach is imo to think of it in terms of a measurement channel. The "generalised measurements" discussed in N&C are a (IMO) somewhat cryptic way to do this. They relate with the POVM as described above via the simple relation $\mu_k=M_k^\dagger M_k$. However, the operators $M_k$ also describe a way to get post-measurement states via the usual rule also mentioned in the other answer. More formally, you can think of $M_k$ as the Kraus operators of the measurement channel describing the situation.

  • $\begingroup$ Thank a lot for your time and answer $\endgroup$
    – OffHakhol
    Commented Feb 11 at 14:11

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