Propensity, Probability, and Quantum Theory

Abstract

Quantum mechanics and probability theory share one peculiarity. Both have well established mathematical formalisms, yet both are subject to controversy about the meaning and interpretation of their basic concepts. Since probability plays a fundamental role in QM, the conceptual problems of one theory can affect the other. We first classify the interpretations of probability into three major classes: (a) inferential probability, (b) ensemble probability, and (c) propensity. Class (a) is the basis of inductive logic; (b) deals with the frequencies of events in repeatable experiments; (c) describes a form of causality that is weaker than determinism. An important, but neglected, paper by P. Humphreys demonstrated that propensity must differ mathematically, as well as conceptually, from probability, but he did not develop a theory of propensity. Such a theory is developed in this paper. Propensity theory shares many, but not all, of the axioms of probability theory. As a consequence, propensity supports the Law of Large Numbers from probability theory, but does not support Bayes theorem. Although there are particular problems within QM to which any of the classes of probability may be applied, it is argued that the intrinsic quantum probabilities (calculated from a state vector or density matrix) are most naturally interpreted as quantum propensities. This does not alter the familiar statistical interpretation of QM. But the interpretation of quantum states as representing knowledge is untenable. Examples show that a density matrix fails to represent knowledge.

Appendix: The Law of Large Numbers

Because propensity obeys some, but not all, of the axioms of probability theory, it is important to prove this theorem using only a minimum of assumptions.

Lemma (Chebyshev’s Inequality): Let p(x) be a non-negative normalized density,

$$\begin{aligned} p(x) \ge 0\ ,\ \ \ \int _{-\infty }^{\infty } p(x) dx = 1\ . \end{aligned}$$

(72)

Denote the mean, or centroid, of the density as $\mu = \int _{-\infty }^{\infty } x p(x) dx$, and define

$$\begin{aligned} M_n = \int _{-\infty }^{\infty } |x-\mu |^n p(x) dx \end{aligned}$$

(73)

to be the n’th absolute moment of the density. The second moment is $M_2 = \sigma ^2$, where $\sigma$ is the standard deviation. Next we introduce the function

$$\begin{aligned} J_{\delta }(x-\mu )= & {} 0\ \ \ \text{ if }\ \ |x-\mu | < \delta \nonumber \\= & {} 1\ \ \ \text{ if }\ \ |x-\mu | > \delta \end{aligned}$$

(74)

It is apparent that $J_{\delta }(x-\mu ) \le |x-\mu |^n / \delta ^n$ for $n>0$ and all x. Hence

$$\begin{aligned} \int _{-\infty }^{\infty } J_{\delta }(x-\mu ) p(x) dx \le \frac{M_n}{\delta ^n} \end{aligned}$$

(75)

and, in particular, we have

$$\begin{aligned} \int _{-\infty }^{\infty } J_{\delta }(x-\mu ) p(x) dx \le \frac{\sigma ^2}{\delta ^2}\ \ \ \text{ for } \text{ any }\ \ \delta >0 \end{aligned}$$

(76)

This, Chebyshev’s inequality, says that the total “weight” of the density in the outer region, $|x-\mu |>\delta$, is bounded above by $\sigma ^2/\delta ^2$.

Notice that the proof of Chebyshev’s inequality does not depend on any concepts that are specific to probability. It requires only a non-negative normalized density, and so also applies, for example, to a mass density.

For application to probability theory, we take p(x) to be a probability density for a continuous variable x. Specifically,

$$\begin{aligned} p(x) dx = P\big (X\in [x, x+dx]\bigm |S\bigm ) \end{aligned}$$

(77)

is the probability that X lies in the range x to $x+dx$, under some condition S. (To treat a discrete variable we take p(x) to be a sum of Dirac $\delta$-functions.) From the definition (74) it is apparent that

$$\begin{aligned} \int _{-\infty }^{\infty } J_{\delta }(x-\mu ) p(x) dx = P\big (|X-\mu |>\delta \bigm |S\bigm ) \end{aligned}$$

(78)

is the probability that X differs from $\mu$ by more that $\delta$. Hence Chebyshev’s inequality (76) yields

$$\begin{aligned} P\big (|X-\mu |>\delta \bigm |S\bigm ) \le \frac{\sigma ^2}{\delta ^2} \end{aligned}$$

(79)

Notice that by writing (78) we have implicitly invoked the additivity rule (14) for mutually exclusive values of X.

To derive the Law of Large Numbers (following the method of [30]), we consider a sequence of n repetitions of a certain experiment, all carried out under the same condition (or state) S. Let $R_1, R_2, \cdots R_n$ denote the sequence of outcomes, and introduce the corresponding indicators: $J_{A,i} = 1$ if $R_i = A$; $J_{A,i} = 0$ if $R_i \ne A$. Recall that, from (11), we have $\;\langle J_{A,i}\rangle _S\; =\; P\big ( R_i=A \bigm | S \bigm )$.

The variable $f_n = \frac{1}{n} \sum _{i=1}^n J_{A,i}$ is the relative frequency of the particular outcome A in the sequence of n experimental runs. Its probabilistic average is

$$\begin{aligned} \langle f_n \rangle _S= & {} \frac{1}{n} \sum _{i=1}^n \langle J_{A,i}\rangle _S \nonumber \\= & {} \frac{1}{n} \sum _{i=1}^n P\big ( R_i=A \bigm | S \bigm ) \ \ =\ \ P(A|S) \end{aligned}$$

(80)

(The last equality is permitted because the probability of outcome A is the same in each experimental run.) The variance of $f_n$ is

$$\begin{aligned} \sigma _{f_n}^{\ 2}= & {} \langle f_n^{\ 2} \rangle _S - \langle f_n \rangle _S^2 \nonumber \\= & {} n^{-2}\sum _{i=1}^n\sum _{j=1}^n \left( \langle J_{A,i}\,J_{A,j}\rangle _S - \langle J_{A,i}\rangle _S\,\langle J_{A,j}\rangle _S \right) \end{aligned}$$

(81)

Now $\langle J_{A,i}\, J_{A,j}\rangle _S = P\big ((R_i=A)\cdot (R_j=A)\bigm |S\bigm )$ is the joint probability of two variables that describe the outcomes of two independent experimental runs. Therefore we have $\langle J_{A,i}\,J_{A,j}\rangle _S = \langle J_{A,i}\rangle _S\,\langle J_{A,j}\rangle _S$ for $i\ne j$. Notice that here we have used the factorization rule (17) for the joint probability of uncorrelated variables. Since $J_{A,i}^{\ \ 2} = J_{A,i}\;$, Eq. (81) now becomes

$$\begin{aligned} \sigma _{f_n}^{\ 2}= & {} n^{-2}\sum _{i=1}^n \big (\langle J_{A,i}\rangle _S - \langle J_{A,i}\rangle _S^{\ 2} \big ) \nonumber \\= & {} n^{-1} \left\{ P(A|S) - [P(A|S)]^2 \right\} \le \frac{1}{4n} \end{aligned}$$

(82)

Finally, we substitute $X = f_n$ into (79) and obtain the Law of Large Numbers:

$$\begin{aligned} P\big (|f_n - p| > \delta \bigm | S \bigm ) \le \frac{1}{4n\delta ^2} \end{aligned}$$

(83)

where $p = P(A|S)$.