Let us define a pure vector state of a quantum system as a vector $\psi$ in a Hilbert space $\mathscr{H}$ with norm $\|\psi\| = 1$. Let $\mathscr{B}(\mathscr{H})$ be the Banach space of bounded linear operators on $\mathscr{H}$.

In the $C^{*}$-algebra formulation of quantum mechanics, one defines observables as the $C^{*}$-algebra $\mathscr{B}(\mathscr{H})$ of bounded linear operators on $\mathscr{H}$, and a state as a norm one positive functional $\omega$ on this $C^{*}$-algebra. Strocchi gives a nice interpretation in his book on why these states should be interpreted as average values of observables. In addition, if $A \in \mathscr{B}(\mathscr{H})$ is fixed, with spectrum $\sigma(A)$ and $C^{*}(A)$ is the unit $C^{*}$-algebra generated by $A$, then by Riesz-Markov Representation Theorem, to each state $\omega$ there exists a measure $\mu_{\omega,A}$ such that: $$\omega(A) = \int_{\sigma(A)}\tilde{A}(\lambda)d\mu_{\omega,A}(\lambda)$$ where $\tilde{A}$ is the Gelfand transform of $A$. If the state is $\omega(A) = \langle \psi, A\psi\rangle$, there is a natural interpretation of $\langle \psi, A\psi\rangle$ as the average value of $A$.

In non-algebraic approaches, one usually postulates that given an observable $A$ and a pure vector state $\psi$, the probability that $A$ will take any value inside a Borel set $E \subset \sigma(A)$ is the spectral measure: $$p_{\psi,A}(E) = \langle \psi, \chi_{E}(A)\psi\rangle$$ where $\chi_{E}$ is the characteristic function of the set $E$.

My question is: is there a nice interpretation or justification of this definition? I know that, a posteriori, this leads to the various formulas that physicists use, but I wanted something a priori. Why is this natural? Putting it another way, I understand the spectral measure is a probability measure on $\sigma(A)$, but why should I use it as the probability of obtaining $E$ in the state $\psi$? Notice that, in this case, $A$ can be unbounded too.

  • $\begingroup$ Because it is confirmed by experiments. $\endgroup$
    – jjcale
    Feb 4, 2023 at 14:50
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    $\begingroup$ Nothing in quantum mechanics is "natural". The theory is highly unintuitive and philosophically challenging. $\endgroup$ Feb 4, 2023 at 14:56
  • $\begingroup$ The way I see it we can't actually measure any observable just-like-that. Instead we have a measuring device that has finite precision. So instead we measure a finite spectrum version of the observable, with outcomes given by labels attached to intervals e.g. measuring position $x$ the outcome $x=2$ is captured by 1.5 to 2.5 aka $\chi_{[1.5,2)}(x)$. $\endgroup$ Feb 4, 2023 at 16:13

2 Answers 2


(Small correction: We can take the observables to be the self-adjoint elements of $B(H)$, or any C${}^*$-algebra, and in your whole discussion $A$ should be assumed self-adjoint.)

This can be reduced to the following more fundamental principle. Let $v$ be a unit vector in some Hilbert space $H$ and let $E$ be a closed subspace of $H$. Write $H = E \oplus E^\perp$ and decompose $v = v_1 + v_2$ accordingly; then the probability the state represented by $v$ will be measured to belong to $E$ is $\|v_1\|^2$, the probability it will be measured to belong to $E^\perp$ is $\|v_2\|^2$.

Now if you have an observable (self-adjoint operator) $A$ with finite spectrum, in particular if $H$ is finite dimensional, then we can decompose $H$ as the orthogonal sum of eigenspaces of $A$, $H = E_1 \oplus \cdots \oplus E_n$. If the corresponding eigenvalues are $\lambda_1, \ldots, \lambda_n$, then the probability a state represented by a unit vector $v$ will be measured to be in $E_i$ is $\|v_i\|^2$ where $v_i$ is the component of $v$ in $E_i$, and so the expected value of $A$ for $v$ is $\lambda_1\|v_1\|^2 + \cdots + \lambda_n\|v_n\|^2 = \langle Av,v\rangle$.

That is, the expected value of $A$ for $v$ is the sum of (value of $A$ on $E_i$)(probability $v$ belongs to $E_i$). Make sense?

In the general (infinite dimensional) case you can norm approximate any self-adjoint operator with a self-adjoint operator with finite range, so we can argue by approximation that $\langle Av,v\rangle$ is still the expected value.

  • $\begingroup$ Nik, this is exactly the direction I wanted to go when I posted the question. However, I want to adapt your answer to something a little different. The spectral measure I defined i $\langle \psi, \chi_{E}(A)\psi\rangle$. I know $\chi_{E}(A)$ is bounded and self-adjoint. More importantly, it is a projection operator. So, this is somehow connected to you answer of decomposing into subspaces. However, I still have trouble in interpreting what observable does $\chi_{E}(A)$ represents. $\endgroup$
    – MathMath
    Feb 4, 2023 at 17:51
  • $\begingroup$ @MathMath it is an observable equal to one if $A$ is in $E$ and zero otherwise. No? $\endgroup$ Feb 4, 2023 at 18:40
  • $\begingroup$ @JPMcCarthy in a sense Yes, but what does it mean "if $A$ is in $E$" exactly? $A$ is an operator and $E$ is a Borel subset of its spectrum. I sort of have an almost intuitive understanding of it, but still caneta put in precise terms myself. $\endgroup$
    – MathMath
    Feb 4, 2023 at 18:57
  • $\begingroup$ @MathMath what does it mean that an observable $A$ is measured to be equal to two? $\endgroup$ Feb 4, 2023 at 19:13
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    $\begingroup$ $\chi_E(A)$ and $\chi_{E^c}(A)$ are orthogonal subspaces whose sum is $H$. So $\langle \chi_E(A)v,v\rangle$ is the probability that $v$ will be measured to lie in the range of $\chi_E(A)$. All unit vectors in the range of $\chi_E(A)$, when measured using $A$, will yield a result that lies in $E$, and all unit vectors in $\chi_{E^c}(A)$ , when measured using $A$, will yield a result that lies in $E^c$. $\endgroup$
    – Nik Weaver
    Feb 4, 2023 at 19:40

The algebraic quantum theory literature usually use convex state space approach also known as operational quantum theory as the starting point. I Araki's book has a better explanation of what is going on than Strocchi. Although, to get a good idea, I would recommend G Ludwig's Foundations of Quantum Mechanics I or his former student K Krauss' States, Effects and Operations for a quicker review. I will try to summarise their argument here. I will not write all the details, as that can make this into a far bigger article.

Convex State Space Approach:

The goal is to develop a mathematical model that allows us to study physical systems. The systems of interest to us are those whose properties can be described by sentences of the type:

"An observable R has a value r"

So, the starting point is to get a mathematical model of the notion of an 'observable'. The starting point is instruments we can use to measure the properties of the system and the instruments we prepare our system with. Denote the collection of preparation instruments by $S$, and measuring instruments by $E$. We are interested in mathematically modeling these.

Measuring instruments are capable of undergoing changes upon interaction with the prepared system. The changes occurring are called ‘effects’. Any measuring equipment can be composed of simple yes-no equipment. Relative frequencies of occurrences give us ‘operational statistics’ which are maps, $$\mu: S\times E\to \mathbb{R}, (\rho,X)\mapsto\mu(\rho\vert X)$$ which assigns to each preparation procedure and measurement procedure, its frequency of undergoing the corresponding change. Accounting for the fact that we can combine preparation equipment, we should have, for a collection $\rho_i\in S$, and $\lambda_i \in [0,1]$ with $\sum_i\lambda_i =1$, $$\textstyle\sum_i \lambda_i \rho_i\in S$$ or, it's a convex set. We can embed $S$ inside the formal vector space consisting of formal sums. So, we can think of preparation procedures as linear maps of the space of effects of the form, $$X\mapsto \textstyle\sum_i \alpha_i \mu(\rho_i\vert X).$$ (finite sum) This means each preparation procedure can be modeled as a linear functional on the space of effects, and similarly, each effect can be viewed as a linear functional on the space of preparation procedures. With the weak topology, they are continuous linear functionals. So, each are vector spaces denoted by $\mathcal{S}$ and $\mathcal{E}$ respectively.

So, the measuring equipment and preparation equipment can be mathematically modeled by the above vector spaces. This is however useless by itself. It offers no predictability. We need the relation between different measuring equipments. Which will come in the form of an algebra structure on $\mathcal{E}$.

In classical physics, the space of effects is modeled by commutative algebra. In quantum theory, it's modeled by a C* algebra.

In classical physics, observables, are continuous maps of the form, $$A: \mathcal{M}\to \mathcal{I}.$$ $\mathcal{I}$ is an indexing set, to which the values of the observables belong and $\mathcal{M}$ is a smooth manifold called a phase space. The state space is the collection of all Borel measures on $\mathcal{M}$.

By the 20th century experiments showed the existence of continuous and discrete observables. So $\mathcal{I}$ can be $\mathbb{Z}$ or $\mathbb{R}$. The continuity implies that the topology on $\mathcal{S}$, and $\mathcal{I}$ must be related. But the existence of discrete and continuous variables makes this coexistence impossible. They are topologically incompatible. (think in terms of # of connected components, etc.) So, the collection of measures on $\mathcal{M}$ is not the correct vector space we can use for describing the states.

von Neumann's Idea:

Heisenberg's solution was to use linear operators as a mathematical model for observables. Von Neumann was able to make it mathematically precise as to why this would be a good model. He introduced the notion of a Hilbert space.

Note that a discreet observable has discrete collection of values. Or, it's a sequence. Similarly, a continuous observable is a continuous function to $\mathbb{R}$. We were trying to compare the function space. Von Neumann knew that the collection of square integrable functions and square summable series are isomorphic as Hilbert spaces. So, this unique infinite dimensional Hilbert $\mathcal{H}$ can act as the common state space that can be used to model both discrete and continuous observables. In particular we have an orthonormal basis, and the corresponding projections correspond to the possible events.

Take away: The correct algebraic structure on $\mathcal{E}$ needed for describing observables that can accommodate both discrete, and continuous case is found in $\mathcal{P}(\mathcal{H})$ consisting of projection operators on a separable Hilbert space.

Why Self-Adjoint? What is the physical meaning of Spectral Measures?

Note that we are trying to model measuring instruments. To simplify the procedure, we can describe a collection of effects that can be measured by a single instrument. This can be some thing like a meter, which measures a value for the quantity.

An observable is a collection of effects that can be measured with a single instrument.

Such measuring instruments themselves have a Boolean algebra structure. The values of the quantity being the elements of the Boolean algebra, and the operations $\vee$ and $\wedge$ correspond to 'and' and 'or' operation for the values. Let me denote the Boolean algebra by $\Sigma_A$.

So to model such instruments, we have to respect the expected algebraic structure for this subcollection of effects. What we want is a Boolean algebra homomorphism $$E_A:\Sigma_A\to\mathcal{P}(\mathcal{H}).$$ Here $\mathcal{P}(\mathcal{H})$ is the collection of projection operators on $\mathcal{H}$, with product as the algebra operation.

In physical situations, the measuring scales are real valued. So, the Boolean algebra $\Sigma_A$ is to be taken as $\mathcal{B}(\mathbb{R})$ which is the Borel $\sigma$-algebra on $\mathbb{R}$. The Boolean algebra homomorphism requirements imply that this will be a spectral measure.

So, observables are collections of effects that can be measured by a single instrument and is mathematical modelled by spectral measures. By Spectral theorem, to each self-adjoint operator there corresponds a spectral measure and vice-versa.

From Quantum Logic side, Gleason's theorem [tedious proof] then implies that the linear functionals on the space of effects with "necessary properties" will have to correspond to density operators. So, to each states $\rho$ there is a corresponding density matrix which we will denote by $\rho$ (abusing notation). The probability that the observable corresponding to the spectral measure $E_A$ has a value which lies in the interval $\epsilon\subseteq \mathbb{R}$ is given by $$\mu(\rho,\epsilon)=\mu_\rho(\epsilon)= Tr(\rho E_A(\epsilon)).$$ This will be a measure on $\mathbb{R}$, because of the properties expected of states. The expectation value is given by, $$\langle E_A\rangle=\int_{\mathbb{R}} x d \mu_\rho(x)=\int_{\mathbb{R}} x Tr(\rho E_A(x))=Tr (\rho \underbrace{\int_{\mathbb{R}} x dE_A(x)}_{A}).$$ The underbraced thing is the self-adjoint operator of interest.

One can start with C$^*$-algebras, but if we put the topology on operators such that observables should be close to each other if their expectation values are close to each other, and we expect the algebra to be closed in this topology, then we should expect the C$^*$-algebra to be a von Neumann algebra. (which is the thing we use in algebraic quantum theory)

Although one can view quantum mechanics by starting from functional analysis, I feel that kills all the deeper behind the scene things involved in the formulation of quantum mechanics.

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    $\begingroup$ Thank you for the detailed answer. Many more informations for me, really nice! $\endgroup$
    – MathMath
    Feb 4, 2023 at 17:52

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