4
$\begingroup$

I don't quite understand how the probability language of sample spaces, $\sigma-$algebra, random variables, etc, fit into the quantum mechanics' formalism.

To wit, we usually say that an observable is a linear operator in a Hilbert space, and afterwards we define the "expected value" of this operator. But what sense does this make in the probability framework? This is not a random variable in any sense. When we compute the expected value of position or moment, what is our sample space, our algebra, our probability measure?

Also, it is always said that $\vert \psi \vert ^{2}$ gives the density probability function (with $\psi$ a solution of the Schrödinger equation)... the density of what random variable? And what would be the density or distribution of the random variables "position" and "moment"? Certanly, there must be such random variables. Or they only exist formally as operators in a Hilbert space? But, if so, What sense does it have, formally, to compute its expected value?

Thanks

$\endgroup$
  • 1
    $\begingroup$ The thing is, there are generalisations of probability theory in which we need no such thing as sample spaces, or $\sigma$-algebras. These do encompass the usual probabilistic descriptions, but they use "random variables" and "expectations" as the fundamental entities and not sample spaces, algebras and probability measures. $\endgroup$ – Raskolnikov Nov 21 '14 at 12:15
  • 1
    $\begingroup$ This is where the work of Dan Voiculescu, for instance, fits in. He studied what is called free probability in which the notion of independence of random variables is replaced by the notion of freeness, which leads to a new type of statistics. Maybe this intro of the subject can give you a feel for what it is like. $\endgroup$ – Raskolnikov Nov 21 '14 at 12:16
  • $\begingroup$ Now, it has to be said that QM is less radical than free probability. While it involves non-commuting variables, we usually still work with the ordinary notion of indepence. Therefore, you still get the usual statistical distributions like the normal distribution. Only when non-commuting variables are involved does one get stranger things. $\endgroup$ – Raskolnikov Nov 21 '14 at 12:19
  • 1
    $\begingroup$ Best as I can tell, the random variable involved takes values in the spectrum of the operator corresponding to the observable quantity and the probability measure is the spectral projection of the wavefunction. I'm not an expert, though, so I'll leave this as a comment. (Relevant Wikipedia: en.wikipedia.org/wiki/…) $\endgroup$ – Neal Nov 21 '14 at 12:23
  • 1
    $\begingroup$ @Qwertuy For simplicity, let the spectrum of $T$ be discrete with unit eigenfunctions $\phi_i$ and eigenvalues $\lambda_i$.So $\langle \psi,T\psi\rangle = \langle \sum_j \langle\psi,\phi_j\rangle\phi_j, T(\sum_i\langle\psi,\phi_i\rangle\phi_i)\rangle = \sum_{i,j}\lambda_i \langle \psi,\phi_j\rangle\langle \psi,\phi_i\rangle\langle \phi_i,\phi_j\rangle = \sum_i \lambda_i\langle \psi,\phi_i\rangle^2$. So I guess the probability density should be the square of the norm of the projection onto the eigenspace, because that would make this the expected value. $\endgroup$ – Neal Nov 22 '14 at 16:27
4
$\begingroup$

The probability model of quantum mechanics is different from the Kolmogorov model, is due to von Neumann, and actually predates the Kolmogorov model. The basic model goes something like this: events are represented by the lattice of projectors on a Hilbert space. The elementary outcomes are the one-dimenstional projectors. Probabilities are assigned to events $P$ as

$$\text{Prob}(P) = \omega(P)$$

with $\omega$ satisfying the following requirements

  • $\omega(0)=0$
  • $\omega(\mathbb{I})=1$ with $\mathbb{I}$ the identity.
  • $\omega(\sum_j P_j) =\sum_j \omega(P_j)$ whenever $\{P_1,P_2,P_3,\ldots\}$ is an at most countable collection of mutually orthogonal projectors.

You can see that this is analogous to the Kolmogorov axioms. The functionals $\omega$ have a convex structure and the extreme states, called pure states, play the role of Dirac measures on the quantum probability space.

Gleason's theorem then states that for a separable Hilbert space of dimension greater than or equal to three, any state on the lattice of projectors on that Hilbert space is a convex combination of pure states and any pure state is of the form

$$\text{Prob}(P)=\langle \phi , P \phi\rangle$$

with $\phi$ in the Hilbert space and $\|\phi\|=1$.

Now, the Spectral Theorem tells you how to compute the probabilities of observing an observable within a certain range of values $S$. If

$$A=\int_{\mathbb{R}}\lambda dE^A(\lambda)$$

in which $E^A$ is the spectral measure of $A$, then

$$\text{Prob(Observed value of } A \text{ belongs to } S) = \langle \phi , E^A(S) \phi\rangle$$

Similarly, the expected value of $A$ can be written as

$$\langle A \rangle_{\phi} = \int_{\mathbb{R}}\lambda \langle \phi, dE^A(\lambda)\phi\rangle = \langle \phi, A \phi \rangle$$

which is the usual expectation formula we encounter in basic quantum mechanics textbooks.

Now, to see how the classical model fits in, take the triplet $(\Omega,\Sigma,\mu)$ to represent a usual Kolmogorov probability model. Then, $L^2(\Omega,\mu)$ is the space of complex-valued square integrable functions w.r.t. the measure $\mu$ forms a Hilbert space. The observables on this space are given by functions out of the same Hilbert space that act as multiplication operators on the Hilbert space:

$$\phi \mapsto (M_f \phi)(x):= f(x)\phi(x), \;\; \phi,f \in L^2(\Omega,\mu) \; .$$

The projectors are given by indicator functions on elements $S$ of the sigma algebra $\Sigma$:

$$I_S: x \mapsto \begin{cases}1 & \text{if } x \in S \\ 0 & \text{else}\end{cases}$$

or rather, a projector is an observable $M_{I_S}$.

The functional $\omega$ is defined as

$$\omega(M_f) := \int_{\Omega} f(x) \mu(dx) = \langle {\bf 1}, f {\bf 1} \rangle \; .$$

In which ${\bf 1}$ is the constant function, a normalized vector in $L^2(\Omega,\mu)$. Thus for a projector we have

$$\omega(M_{I_S}) := \int_{\Omega} I_S(x) \mu(dx) = \mu(S) = \langle {\bf 1}, I_S {\bf 1} \rangle \; .$$

$\endgroup$
  • $\begingroup$ Nice answer. I'm sorry I didn't see it sooner. Any good references on this subject? $\endgroup$ – Qwertuy Dec 1 '14 at 22:19
  • $\begingroup$ I used the book of my thesis advisor to write down the answer, it's almost verbatim (Quantum Dynamical Systems by Alicki and Fannes). But the description of this construction in the book is somewhat too succinct. Good for a reply on SE, but not good enough to study the matter. However, here are some of the references contained in the book: Peres A. (1993) Quantum Theory: Concepts and Methods. Kluwer Academic Publishers Dordrecht. I strongly recommend this book. $\endgroup$ – Raskolnikov Dec 2 '14 at 10:01
  • $\begingroup$ There are also the book by von Neumann (1932) Mathematische Grundlagen der Quantenmechanik, Springer, Berlin. The paper by Gleason (1957) Measures on closed subspaces of a Hilbert space, J. Math.Mech. 6, 91-10. My book also mentions some monographs, but those are already pretty specialized works. I'll mention the book by Voiculescu about free probability if the concept interests you: Voiculescu et al. *Free Random Variables", volume 1 of CRM Monograph series, AMS, Providence. And also look for monographs on Quantum probability if you want to dig deeper. $\endgroup$ – Raskolnikov Dec 2 '14 at 10:10
  • $\begingroup$ But Asher Peres' book is definitely a good start, especially if you are more mathematically minded and want to see how the axioms of Quantum mechanics are built up from experiment. $\endgroup$ – Raskolnikov Dec 2 '14 at 10:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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