20
$\begingroup$

To me, as a stupid mathematician, a random variable is a measurable function from some probability space $(\Omega, \sigma, \mu)$ to $(\Bbb{R}, B(\Bbb{R}))$. This makes sense. You have outcomes, events, and probabilities of these events. A random variable is just assigning these numbers.

I took QM as an undergrad and I remember computing eigenvalues, expectations, etc. of various operators and I never quite got what an operator is in QM. A (quantum) random variable is a Hermitian operator on some Hilbert space. You can compute probabilities and expectations by some formulas involving eigenvectors and orthogonal projections.

I must admit, even after my QM class, I don't get this. Is this random variable in any way related to my ignorant definition? Could we model say a coin flip or dice roll using this? Or is this type of random variable only for quantum things? Why not model all of quantum mechanics using my stupid definition? I know in quantum mechanics we can allow negative probabilities, so why not just have some signed measure space and forget all the business about Hilbert spaces?

$\endgroup$
  • 6
    $\begingroup$ As a comment: it does not make sense to think about quantum mechanics with probabilities because the free dynamics is not stochastic. It's perfectly reversible. The only place where "probabilities" sort of come into play is the measurement process and the reason why that produces a stochastic result is because it starts with an unknown state of the measurement device and it has to end with an infinite fixed state of the measurement device, i.e. it has to be irreversible. Strictly speaking it's not quantum theory that is probabilistic, but quantum measurement theory. $\endgroup$ – CuriousOne Jun 14 '16 at 2:53
  • $\begingroup$ @CuriousOne As a continuation/response to your comment: I completely agree that the quantum dynamics is usually deterministic and reversible; however it is the deterministic evolution of either random variables (Heisenberg picture) or probabilities (Schrödinger picture). I understand why many people (including myself) do not like the term "randomness" related to QM, since it gives the impression that there is some fundamental unpredictability in (the evolution of) quantum systems, and that is not true. $\endgroup$ – yuggib Jun 14 '16 at 4:05
  • 2
    $\begingroup$ On the other hand however, the mathematical framework used for quantum mechanics - that allows for predictions in almost perfect agreement with the experimental observations - is exactly that of non-commutative probability theory. Therefore in some sense it is the correct mathematical terminology to use, when dealing with quantum mechanics in a mathematical fashion (a thing that I know you don't like so much to do, but it is very useful ;-) ). $\endgroup$ – yuggib Jun 14 '16 at 4:07
  • 2
    $\begingroup$ I don't think any of the answers so far is that good. You probably need at least several pages of explanation to figure out what's going on. I'd recommend looking at the first section of lecture notes for a quantum computing courses ... maybe chapters 2 and 3 of John Preskill's These courses try to explain quantum mechanics to non-physicists. You seem to have confused density operators (the quantum analog of probability distributions) and measurement operators (operators that describe how you measure a quantum state). $\endgroup$ – Peter Shor Jun 14 '16 at 11:59
  • 2
    $\begingroup$ I recommend ValterMoretti's answer to a similar question. $\endgroup$ – ACuriousMind Jun 14 '16 at 12:33
9
$\begingroup$

I'm going to try to explain why and how density operators in quantum mechanics correspond to random variables in classical probability theory, something none of the other answers have even tried to do.

Let's work in a two-dimensional quantum space. We'll use standard physics bra-ket notation. A quantum state is a column vector in this space, and we'll represent a column vector as $\alpha|0\rangle + \beta |1 \rangle.$ A row vector is $\gamma \langle 0 | + \delta \langle 1 |\,$.

Now, you might think that a probability distribution is a measure on quantum states. You can think of it that way, but it turns out that this is too much information. For example, consider two probability distributions on quantum states. First, let's take the probability distribution

$$ \begin{array}{cc} |0\rangle & \mathrm{with\ probability\ }2/3,\\ |1\rangle & \mathrm{with\ probability\ }1/3. \end{array} $$

Next, let's take the probability distribution $$ \begin{array}{cc} \sqrt{{2}/{3}}\,\left|0\right\rangle +\sqrt{1/3}\, \left|1\right\rangle & \mathrm{with\ probability\ }1/2,\\ \sqrt{{2}/{3}}\,\left|0\right\rangle -\sqrt{1/3}\, \left|1\right\rangle & \mathrm{with\ probability\ }1/2. \end{array} $$

It turns out that these two probability distributions are indistinguishable. That is, any measurement you make on one will give exactly the same probability distribution of results that you make on the other. The reason for that is that $$ \frac{2}{3} |0\rangle\langle0| +\frac{1}{3}|1\rangle\langle 1| $$ and $$ \frac{1}{2}\left(\sqrt{2/3}\left|0\right\rangle +\sqrt{1/3}\, \left|1\right\rangle\right) \left(\sqrt{2/3}\left\langle 0\right| +\sqrt{1/3}\, \left\langle 1\right|\right) +\frac{1}{2}\left(\sqrt{{2}/{3}}\left|0\right\rangle -\sqrt{1/3}\, \left|1\right\rangle\right) \left(\sqrt{2/3}\left\langle 0\right| -\sqrt{1/3}\, \left\langle 1\right|\right) $$ are the same matrix.

That is, a probability distribution on quantum states is an overly specified distribution, and it is quite cumbersome to work with. We can predict any experimental outcome for a probability distribution on quantum states if we know the corresponding density operator, and many probability distributions yield the same density operator. If we have a probability density $\mu_v$ on quantum states $v$, we can predict any experimental outcome from the density operator $$ \int v v^* d \mu_v \,. $$

So for quantum probability theory, instead of working with probability distributions on quantum states, we work with density operators instead.

Classical states correspond to orthonormal vectors in Hilbert space, and classical probability distributions correspond to diagonal density operators.

$\endgroup$
  • 3
    $\begingroup$ A most wonderfully clear embedding of CM into QM. Many thanks, awesome! $\endgroup$ – WetSavannaAnimal Jun 14 '16 at 23:41
  • $\begingroup$ Thank you this is good, along with your suggestion of lecture notes given in comments. You have given me a lot of material to work on, thank you! $\endgroup$ – user79317 Jun 15 '16 at 5:02
  • 2
    $\begingroup$ I haven't even mentioned measurements here, which is another place that both probability and operators come up in quantum mechanics, and which some of the other answers are talking about. A von Neumann measurement, which you probably learned about in your quantum course, is simply a positive operator. The most general measurement is a POVM, or positive operator valued distribution, which is defined by a measure on positive operators. $\endgroup$ – Peter Shor Jun 15 '16 at 10:55
  • $\begingroup$ This answer is the subject of a new question. $\endgroup$ – Nat Oct 28 '18 at 2:19
12
$\begingroup$

Quantum mechanics is indeed a probability theory, but it is a non-commutative probability theory.

So it is not just a matter of having signed/complex measures, but really of having a non-commutative probabilistic framework. Quantum mechanics was developed, historically, before non-commutative probability theories and I think that people in probability modelled non-commutative probability theory on quantum mechanics and not vice-versa. One mathematical example of non-commutative probability is the free probability introduced by Voicolescu (it is similar to quantum mechanics, but in quantum mechanics some of Voicolescu's axioms about freeness are not necessary).

The idea of non-commutative probability is to extend usual probability theory exploiting the fact that random variables usually form an abelian algebra. So you start directly from a $C^*$ or $W^*$ algebra $\mathfrak{A}$ of random variables, possibly non commutative, and introduce (non-commutative, complex) measures as the topological dual $\mathfrak{A}'$. The interpretation in quantum mechanical terms is that states are the non-commutative probabilities, i.e. the positive and norm one elements of $\mathfrak{A}'$, while the observables are usually taken to be the self-adjoint elements affiliated to $\mathfrak{A}$ (i.e. possibly unbounded operators $a$ whose spectral family $\bigl(P_{t}(a)\bigr)_{t\in\mathbb{R}}\subset\mathfrak{A}$ ). The usual concepts of probability extend, mutatis mutandis, to this framework; e.g. the evaluation $\mathbb{E}_\omega(a\in [0,1])$, giving the probability of finding a value in the interval $[0,1]$ for the observable $a$, in the state $\omega$, is given by $\omega\bigl(P_1(a)-P_0(a)\bigr)$.

$\endgroup$
8
$\begingroup$

You could certainly model any one quantum observable as a random variable.

The problem comes in when you have multiple observables, which you might attempt to model as classical random variables with some joint distribution. From this joint distribution, you can compute various probabilities (like $\textrm{Prob}(Y\neq X)$, for example), according to the standard rules you learned in your undergraduate probability classes.

The problem is that in general, no joint distribution can yield the probabilities that are predicted by quantum mechanics (and observed in the laboratory).

For example, for classical random variables, it's easy to prove that no matter what the joint distribution of $X,Y$ and $Z$ might be, you have $$\textrm{Prob}(X\neq Z)\leq \textrm{Prob}(X\neq Y)+\textrm{Prob}(Y\neq Z)$$

For quantum observables, such inequalities can be violated. Therefore you need a different formalism.

$\endgroup$
5
$\begingroup$

I believe it is misguided to think that classical probability makes sense any more than quantum mechanics, with its "peculiar" probability calculations, makes sense.

I'm going to be slightly mischievous here and make a friendly attack your first paragraph: does really make sense?

Of course it makes perfect sense as a measure-theoretic definition, but how do you know it represents probabilities for real world random events? What does "probability" even mean? Do you take a frequentist, or subjectivist standpoint in bringing meaning to the word? I do think that we only "understand" classical probability insofar that we are simply accustomed to it.

The way wherein the first paragraph makes sense is due, I believe, mainly to Kolmogorov. His grand contribution was to understand that measure theory gives us a way to rigorously, set theoretically define events and show that the calculation of "probabilities" through their measure gives us a mathematical system that reproduces Pascal's, Laplace's and so forth intuitions about probability.

You can look at Kolmogorov as a physicist here: he's making postulates that events won't be represented by things such as Vitali sets, and that Pascal's intuitions, in keeping with the law of excluded middle that, are reasonable.

But along come the experimental physicists and show, experimentally, that this framework does not model all situations in the experimental physics world. In particular, the Bell Inequality, which is a Fréchet Inequality, can be violated. There are propositions that cannot be classically joined by the "and" operator: $X$ has momentum $p$ AND $X$ has position $x$ has no meaning in Nature. $\sigma$- and Boolean algebras simply cannot describe real world systems, and this is an experimental fact. See Valter Moretti's answer to the Physics SE question "Classical logic in concern with QM Mathematics".

Quantum observables make sense because they foretell experimentally measured results, whereas classical probability does not. The latter is experimentally falsified.

$\endgroup$
5
$\begingroup$

Whilst it is certainly true that Quantum Probability Theory (QPT) is an entirely different framework from Classical (Kolmogorovian) Probability Theory (CPT) (specifically because the event structure is non-Boolean and the random-variable structure is non-commutative), we can still identify enough formal similarity to borrow the classical terminology. In particular, we can still give a satisfying answer to the OP's main question, which in my reading is:

Is this random variable [i.e. Hermitian operator] in any way related to my ignorant definition [i.e. measurable function]

The answer we'll see is a resounding yes. The reason this isn't obvious is because physicists do not tend to express the QM formalism in a probabilistic language. So let's do that now...

First note that whereas the underlying measurable space in CPT has the form $\langle\Omega,\Sigma(\Omega)\rangle$, the underlying measurable space in QPT has the form $\langle\mathscr{H},\Pi(\mathscr{H})\rangle $, for some complex Hilbert Space $\mathscr{H}$ and corresponding projection lattice $\Pi(\mathscr{H})$.

Second, note that whereas we use a Kolmogorov measure $\mu$ to make the classical probability space $\langle\Omega,\Sigma(\Omega),\mu\rangle$, we use a Gleason measure $\gamma$ to make the quantum probability space $\langle\mathscr{H},\Pi(\mathscr{H}),\gamma\rangle $. (A result called Gleason's Theorem establishes the relationship between these measures and the conventional density operators.)

But what about random variables?

Here we need to be a little bit sneaky and note that when it comes to calculating probabilities in CPT, the guy doing all the work is not really the measurable function $X : \Omega \to \mathbb{R}$, but rather its inverse, considered as a set function (let's call it $\sigma$):

$$ \sigma: \mathscr{B}(\mathbb{R}) \ni \bigtriangleup \mapsto X^{-1}(\bigtriangleup) \in \Sigma(\Omega) $$

Specifically, if you want to calculate the probability of $X$ having a value in some subset $\bigtriangleup\in\mathscr{B}(\mathbb{R})$, you first pull that subset back into $\Sigma(\Omega)$ and then apply $\mu$.

In other words, instead of working with the random variable $X$, we can work with its sister $\sigma : \mathscr{B}(\mathbb{R}) \to \Sigma(\Omega)$, which satisfies the axioms of what's called a Set Valued Measure (SVM).

What's the big deal about this alternative formulation of a classical random variable?

Well, this formulation has a perfect analogy in quantum mechanics; namely, that of a Projection Valued Measure (PVM), which is a map $ \pi : \mathscr{B}(\mathbb{R}) \to \Pi(\mathscr{H}) $, satisfying some simple axioms analogous to the properties of a SVM (e.g. disjoint Borel sets map to orthogonal projectors).

But now we can employ the Spectral Theorem of functional analysis to construct an equivalent self-adjoint operator $A : \mathscr{H} \to \mathscr{H}$ for this PVM. It is the self-adjoint operator $A$ that turns out to be more computationally convenient for calculating statistics than the underlying PVM, which is more easily interpreted as a random variable.

I've left out a few details that you can easily find in any good book on functional analysis, but the main takeaway is this. You can express the version of QM traditionally taught in physics courses in more measure-theoretic clothes, and when you do, it is the most natural thing in the world to think of a density operator as a probability measure and a self-adjoint operator as a random variable.

(By the way, your assumption that QM deals in negative probabilities is incorrect. The Gleason measures this algorithm admits all generate ordinary probabilities between 0 and 1, inclusive, which are used to predict relative frequencies of experimental outcomes as per the standard Born Rule.)

$\endgroup$
0
$\begingroup$

A quantum system can be described by a set of evolving quantum mechanical observables. This is not the same as describing a system in terms of a stochastic quantity described by a single number chosen at random. A quantum system really does have multiple values of any unsharp observable, see

https://arxiv.org/abs/quant-ph/0104033.

Those different versions of the system can influence the outcomes of experiments such as interference experiments and EPR experiments.

In an interference experiment, the quantities used to predict the probabilities of outcomes of measurements do not obey the rules of probability as noted in some of the answers above. This is a result of multiple versions of the system evolving and then being recombined to produce an outcome.

In EPR experiments, each system is measured and there are multiple versions of each measurement result. As a result, correlations between the outcomes of measurements can be established when they are compared rather than when each individual measurement takes place, see

https://arxiv.org/abs/quant-ph/9906007

https://arxiv.org/abs/1109.6223.

$\endgroup$
0
$\begingroup$

Maybe you can be interested in another interpretation of Hermitian matrices. In a recent paper we have proposed to see them as gambles on a quantum experiment. We have then enforced rational behaviour in the way a subject accepts/rejects these gambles by introducing few simple rules.

These rules yield, in the classical case, the Bayesian theory of probability via duality theorems.

In the quantum setting, they yield the Bayesian theory generalised to the space of Hermitian matrices. This theory is quantum mechanics: in fact, we have derived all its four postulates from the generalised Bayesian theory. It also leads us to reinterpret the main operations in quantum mechanics as probability rules: Bayes' rule (measurement), marginalisation (partial tracing), independence (tensor product).

To say it in a nutshell, we have obtained that quantum mechanics is the Bayesian theory in the complex numbers.

http://arxiv.org/abs/1605.08177

$\endgroup$