I've had this question for a while:

Is a state space $\mathcal{H}$ for a quantum system just a sample space in a probability space?

The question arises because i can't really tell a difference between a die and a quantum superposition. A die is an object that can be perfectly modeled through conventional probability via Kolmogorov axioms, however, this seems to be the same phenomena that we see in quantum-mechanical systems:

Let $\mathcal{H}$ be the state space of a quantum system, and $(\mathcal{H},\sigma)$ a measurable space of it. Let $S\in \sigma$ be a subset of $\mathcal{H}$, and define the function $m:\sigma\rightarrow\mathbb{R}$ as:

$$m(S)=\sum_{|\psi\rangle\in S}\langle\psi|\psi\rangle$$

This should satisfy all the Kolmogorov axioms, and make $(\mathcal{H},\sigma,m)$ into a probability space. Of course, i'm assuming here $m(\mathcal{H})=1$

But if these models resemble so much, why do we have quantum interpretations for quantum systems, and not for dice? Why is the measurement problem important? I've never heard of space-time branches for the throwing of a die, nor the die being simultaneously in many faces at the same time. While i get why you could argue this about the die, i don't understand why these arguments are presented as exclusively quantum phenomena...


6 Answers 6


"Is a state space H for a quantum system just a sample space in a probability space?"

No. Random variables defined on a sample space have a joint probability distribution. Quantum observables, in general, do not. This is the essential content of Bell's Theorem.

Example: Let $X,Y,Z,W$ be random variables defined on a sample space. Then the existence of a joint probability distribution implies that
$$Prob(X\neq W)\le Prob(X\neq Y)+Prob(Y\neq Z)+Prob(Z\neq W)$$ but some collections of quantum observables violate this inequality.

The most obvious paradoxes are avoided by the fact that only two of the four observables can be observed simultaneously. Without that restriction, the above would yield paradoxical conclusions (e.g. there would be occasions when $X=Y,Y=Z,Z=W$ and $W\neq X$.) If $X,Y,Z$ and $W$ were random variables, there would be nothing in principle to prevent you from observing them all simultaneously and the paradox would be unavoidable. Hence they can't be random variables.

In particular, you write "This should satisfy all the Kolmogorov axioms." I don't know what "should" means here, but your proposal most definitely violates the Kolmogorov axioms.


The function you defined does not satisfy Kolmogorov's axioms for many reasons. First of all it is not even additive!

A quantum state is, in fact, a generalized probability measure on the orthocomplemented lattice of propositions of the considered quantum system.

In the absence of constraints, like superselection rules or a gauge groups, these propositions or events are represented by the orthogonal projectors in the Hilbert space $H$ of the system. $$\rho : {\cal L}(H)\to [0,1]$$ The lattice is above indicated by ${\cal L}(H)$.

By definition

(i) $\rho(I)=1$ and

(ii) a suitably $\sigma$-additivity condition holds: (the series of projectors is computed in the strong operator topology) $$\rho\left( \sum_{j \in \mathbb{N}} P_j \right) = \sum_{j\in \mathbb{N}} \rho(P_j)\quad if \:P_k \in {\cal L}(H)\quad and \quad P_iP_j=0\quad for \quad i\neq j\:.$$

If $H$ is separable with dimension $\neq 2$, then the above probability measures are one-to-one with the statistical operators, i.e., positive, trace-class operators, with unit trace $T: H\to H$.

The afore-mentioned correspondence is that

$\rho$ corresponds to $T$ if and only if $\rho(P)= tr(TP)$.

That is the basic statement of the celebrated Gleason theorem.

Pure states, i.e., unit vectors $\psi$ up to phases, are nothing but the extremal elements $\rho_\psi$ of the convex space of the above probability measures and are of the form $\rho_\psi(P) = \langle \psi| P\psi\rangle$ for every $P \in {\cal L}(H)$.

What is the interplay of these generalized probability measures and the ones in the standard theory by Kolmogorov?

${\cal L}(H)$ includes a plethora of maximal sets ${\cal L}_0$ of pairwise commuting projectors. Each of these maximal sets turns out to have the (abstract) structure of a $\sigma$-algebra when restricting the lattice operations to ${\cal L}_0$. Furthermore, a state $\rho \equiv T_\rho$ becomes a true Kolmogorov measure when restricted to every ${\cal L}_0$.

In that sense every quantum state, i.e., a generalized probability measure, boils down to a standard probability measure when the space of events is restricted to a "classical" (commuting) space of events.


(1) There are many different maximal sets ${\cal L}_0\subset {\cal L}(H)$ with non-trivial intersection (that is because the commutativity relation is not transitive). They give rise to a very complex structure. In this sense, there are many incompatible "classical" words embedded a quantum system.

(2) In view of an elementary topological obstruction (if $dim H >2$), commonly known as the Bell-Kochen-Speker theorem, there are no sharp probability measures on ${\cal L}(H)$. I mean measures that assume only deterministic values $\rho(P) \in \{0,1\}$ if $P\in {\cal L}(P)$. That is quite different from what happens in classical theories. Quantum theory, from this perspective, encompasses a family of intrinsically undeterministic "classical" theories.

In summary, a quantum state can be equivalently defined as an assignment of a Kolmogorov probability measure $\rho_{k} : {\cal L}_k \to [0,1]$ to every maximal subset of pairwise commuting projectors ${\cal L}_k \subset {\cal L}(H)$, such that the coherence condition is satisfied $$\rho_k(P) =\rho_{j}(P) \quad if \quad P \in {\cal L}_k\cap {\cal L}_j\:.$$

(See also an old answer of mine Why is the application of probability in Quantum Mechanics fundamentally different from application of probability in other areas?)

  • $\begingroup$ Wow, thanks. A lot to unpack over there. I just have two questions: in your answer, you speak about a lattice $\mathcal{L}(P)$, but in your linked answer you speak about the set of orthogonal projectors $\mathcal{P}(H)$ how are these two related? Also, how are these projectors mathematically defined? $\endgroup$ Commented May 27 at 15:04
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    $\begingroup$ There are the same structure, just different names/symbols. A lattice is a partially ordered set such that every couple of elements $a,b$ admits $a\vee b := \sup\{a,b\}$ and $a\wedge b := \inf\{a,b\}$ which belong to the set... $\endgroup$ Commented May 27 at 16:00

The quantum state space is a probability space, but it is a much larger space than for classical probabilities. I will explain only the case for a two-level system, and for the big picture, refer you to one of my favourite textbooks: Geometry of Quantum States, an introduction to quantum entanglement.

A classical two-level probability space is the unit interval, and every event either occurs with probability $p \leq 1$ or doesn't occur with probability $1-p$. That is all there is to it.

The quantum two-level system is much richer than that. First, a two-level quantum system is fully described in terms of a $2\times2$ hermitian matrix $\rho$ for which we have:

  • $\text{Tr}(\rho) = 1$

  • $\text{Tr}(\rho^2) \leq 1$

  • all eigenvalues of $\rho$ are positive

A matrix for which these conditions hold is called a density matrix. In its most general form, it takes the form $$ \rho = \frac{1}{2}\begin{pmatrix}1+z&x-iy\\x+iy&1-z\end{pmatrix} $$ where $x,y,z\in\Bbb R$ fully describe the state. The above conditions imposed on $\rho$ imply that $x^2 + y^2 + z^2 \leq 1$, so that, if they are interpreted ad components of a vector in $\Bbb R^3$, they are all part of the three-dimensional unit ball, the Bloch sphere. Now this ball consists of infinitely many classical two-level probability spaces, between any two of which you will in general not find any correlation, as opposed to the states on a classical two-level probability space.

  • $\begingroup$ Your answer suggests, that you talk a bout a general 2-state-quantum system. However your density matrix does consists of a weighted superposition of the Pauli-matrices, which can be identified with a spacial direction each. You then talk about a 3 dimensional Bloch sphere. Could you elaborate on where exactly the 3-dimensionality is imposed on/emerges from your considerations? I know it doesn’t make sense to talk about spin in less then 3 dimensions, but what about a general 2 state quantum system? $\endgroup$
    – Zaph
    Commented May 21 at 13:45
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    $\begingroup$ the answer to this is way longer than I can fit into a comment, but the gist of it is that if you take a general pure state $\vert\psi\rangle$ in a $2$-level system, you can associate to it a density matrix by $\vert\psi\rangle\mapsto\vert\psi\rangle\langle\psi\vert$. You can then form linear combinations of these pure state density matrices to represent general probabilistic mixtures of states. The three-dimensionality arises naturally from the algebraic structure of $\text{SU}(2)$ and its associated Lie algebra. Try to get your hands on the book I linked, it explains everything. $\endgroup$
    – paulina
    Commented May 21 at 13:54
  • $\begingroup$ @Zaph I'm sure you can also find the answer here on PSE, and if you can't, post a seperate question. I (or someone else) will be happy to answer. $\endgroup$
    – paulina
    Commented May 21 at 13:57
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    $\begingroup$ Sorry the quantum space is not a probability space in Kolmogorov sense, just because there are quantum incompatible events. physics.stackexchange.com/questions/116595/… $\endgroup$ Commented May 21 at 14:45
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    $\begingroup$ @ValterMoretti well, yeah, that's the point. you need to radically alter your idea of a probability space to deal with quantum mechanics. the kolmogorov space is explicitly classical. thank you for linking this answer though, it really completes the picture. $\endgroup$
    – paulina
    Commented May 21 at 15:54

Probabilities in quantum theory are calculated using the Born rule. In a state $|\psi\rangle$ if you measure an observable $\hat{A}$ with eigenvectors $|a\rangle$ the probability of getting the $a$th eigenvalue is the square amplitude $|\langle\psi|a\rangle|^2$. However, those amplitudes don't always evolve in a way that satisfies the rules of probability because the probability of a result depends on the phase the quantum system picks up while it is evolving in all of the possible states:


So then there has to be an explanation of the circumstances under which those rules and why they apply. The rules can only apply when interference is negligible: that is, in decoherent systems


Given that there will be some particular set of outcomes as a result of decoherence there is a further issue of why the probability rule has the form given above. A couple of different explanations of the form of the probability rule have been given in one in terms of decision theory and another in terms of symmetries of the quantum state called envariance:





One of the most basic experiments that shows the difference between quantum mechanics and classical physics is the double slit experiment. The core non-classical phenomenon in that experiment is that there are regions where probabilities subtract rather than add.

Let $A$ be the event that the photon goes through the first slit, and $B$ be the event that it goes through the second. Now pick a point and let $C$ be the event that the photon lands there. We can write the probability that it goes through the first slit and lands at our selected point as $P(A \& C)$, and the probability that it goes through the second slit and lands there as $P(A \& C)$. We then expect the total probability that it lands at that point to be $P(A \& C)+P(A \& C)$. This is the additive property of probability: if two events are mutually exclusive, then the probability of one or the other happening is the sum of the individual probabilities. Even if they weren't mutually exclusive, we would still expect the probability to be monotonic; that is, the probability of them together should at least be no less than the separate probabilities.

But in fact in the double slit experiment, the probabilities can not only fail to be additive, but they can fail to be monotonic; the probability of a photon at a particular spot can go down when another path is made available to it.


When we talk about a quantum system, we also mean that the result of an interaction cannot be measured precisely enough.
With your cube, you always see exactly one result - provided that the cube always tilts onto one of the six faces. However - even with a round die that has 6 results printed on it, you will be able to determine a clear result in the vast majority of throws. Your measuring equipment is simply good enough for this.

The situation is different with interactions for which we do not have an adequate measurement tool. The state of a photon polarized in a random direction is inevitably influenced during its measurement; a new measurement does not work. And in this first and only measurement, the polarization foil - if it is designed for this wavelength - only allows the photon to pass through in two 90° segments. If the photons are always polarized in the same direction, simply rotate the polarizer (and thus carry out a multiple measurement). In this case, your measurement setup is sufficiently accurate and your knowledge of the state is very close to 100%.

What you calculate with your formulas is the conversion of uncertainty about the final state into probabilities of what will come out at the end of the experiment (which must always be carried out statistically, means many times).


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