Is what statisticians call a "random variable" what physicists call an "observable" in QM? [duplicate]

Question

I read at http://www.statlect.com/fundamentals-of-probability/random-variables that

A random variable is a variable whose value depends on the outcome of a probabilistic experiment. Its value is a priori unknown, but it becomes known once the outcome of the experiment is realized.

That sounds to me like the definition of an observable in quantum mechanics modeled by hermitian operators. In addition it seems to me what statisticians call realization of a random value is what physicists call eigenvalue of a hermitian operator. The set of realizations of a random variable would then be the spectrum (set of eigenvalues) of an operator. So could I tell a statistician that a "random variable" is in fact an operator?

Answer 1

Random variables satisfy the Kolmogorov axioms for probability; quantum observables do not. In particular, any four-tuple of binary random variables (with any joint distribution) satisfies Bell's Inequality, while there are four-tuples of quantum observables that don't.

Answer 2

An observable in quantum mechanics is an operator (say $\widehat{\mathcal{O}}$) on the Hilbert space (Say $\mathcal{H}$) of physical states, such that eigenkets in (say $\widehat{\mathcal{O}}$) in $\mathcal{H}$ span $\mathcal{H}$. The eigenvalues of $\widehat{\mathcal{O}}$ are then the observable values of some classical variable $\mathcal{O}$, even though classical mechanics might predict a more inclusive set of allowed values. A finite-dimensional matrix only has finitely many eigenvalues. Finding the eigenvalues (of which there may be infinitely many) of $\widehat{\mathcal{O}}$ is in general a more difficult problem, often solved with Sturm-Liouville theory.

But there's a lot more to quantum observables than just being a random variable with a distribution over whatever support such an analysis predicts. (Note that the physical state determines the distribution.) A key point often overlooked is that "classical probability" obeys somewhat different axioms from those of quantum probability; the latter allows "interference". For an introduction to the difference between these two kinds of probability, see here. Statisticians almost always concern themselves with classical probability. Classical probability emerges in the many-particle limit, but I think that's beyond the scope of that paper.

Answer 3

What you say is quite reasonable. At the risk of being slightly more pedantic, I would say that physical observables are only those random variables that are Hermitean. Any operator (Hermitean or not) is a random variable -- in quantum mechanics these might be various properties of a particular state like spin, energy, etc. In quantum field theory, the random variables are the field values at each point in spacetime (there exist formalisms where those are treated as operators, though there are other formalisms too).

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To some extent, this is a matter of semantics: what does one mean by a random variable? To me, the term does not carry the connotation of classical probability -- a random variable is just something that can spit out different numbers under different observations. Whether those observations form an ensemble of realizations, or successive measurements in time, is a matter of details: whether your system is ergodic, etc. And sure, classical random variables might behave differently from quantum random variables -- that's like saying that mixed states are different from pure states.

Answer 4

An observable in quantum mechanics is a measurable quantity in an experiment or observation. A postulate for building the mathematical theory of quantum mechanics is that

2.With every physical observable q there is associated an operator Q, which when operating upon the wavefunction associated with a definite value of that observable will yield that value times the wavefunction.

The postulates can be found in the second page in this link.

These postulates are in the foundations of building up mathematical theories of quantum mechanics , all of them, from Hilbert spaces to quantum field theory.

It is simpler when talking of probabilities of getting a value of an observable to discuss it in the original wave function framework of probabilities. A single measurements picks up an instant from the predicted by the theory probability distribution. This is the same as with the statistical mechanics probability distributions.

Where the two diverge are on the mathematics behind them. For random variables one has Gaussian, or Poisson or ...distributions that are an average over dynamical functions derivable from classical mechanics, for example, with a few assumptions on the way the sample under study behaves.

For quantum mechanics the probabilities are the dynamical functions determined from first principles. There exist no further internal dynamics that generate the probability distributions.

Thus statistical probabilities with assumptions of randomness cannot be equivalent with probabilities determined dynamically.

The set of realizations of a random variable would then be the spectrum (set of eigenvalues) of an operator.

One would have to define the mathematical space on which this operator would work, giving the continuum you call "spectrum".

So could I tell a statistician that a "random variable" is in fact an operator?

Statistical mechanics is a meta level on classical mechanics. Quantum mechanics is the first level , no underlying structure deterministic or not, has been discovered.

In addition, it is not the variable that is an operator, the value of the variable instance of a variable, measurment, is an entry in the probability distribution .

You could say an "operator will give me the measurement of temperature in this sample", but there is no mathematical theory of operators that will act on the collective sample , for example a "temperature operator" to give the probable temperature on a specific thermometer measurement.

Probabilities in quantum mechanics are predictive because they are not uniform, giving discrete spectra.

So the analogy does not hold.