# Marginal and joint probability in quantum mechanics

Why is it not possible to find a joint probability distribution that can provide information on all relevant marginals in quantum mechanics? What is exactly the meaning of this statement? Could someone suggest some example to clarify this statement?

• What about the Wigner function: en.wikipedia.org/wiki/… ? Jul 21, 2017 at 12:46
• @ZeroTheHero If there were a joint probability distribution for position and momentum, it would be the Wigner function. The Wigner function can be negative though, and is therefore not a probability distribution. Jul 21, 2017 at 14:41
• @Noiralef ok you're reading this in a narrower way that I did. Try this aip.scitation.org/doi/pdf/10.1063/1.524501 instead: "We demonstrate the existence of positive phase space density functions which yield the quantum mechanical marginal distributions of position and momentum. " There is a string of excellent papers by Leon Cohen, dating back to the mid 60's all the way to the 90's, on this topic. Jul 21, 2017 at 14:45
• @ZeroTheHero Cool, I didn't know about that :) and I guess I take back what I said in the previous comment. Still, the functions defined in your reference only yield the correct marginal distribution for q and p, and not for all possible marginals. I'll write an answer explaining. Jul 21, 2017 at 14:56
• @Noiralef no worries... the work is technical and as you point out doesn't work beyond $p$ and $q$. Cohen's work is the standard on this. Jul 21, 2017 at 15:05

## 2 Answers

This is very well explained in "Breuer, Petruccione: Theory of Open Quantum Systems" (section 2.1.4) and I don't really know how it could be explained any better. I will just give a quick overview of their line of reasoning. The basic idea is the following: Given two classical random variables $R_1$ and $R_2$, we can calculate expectation values such as $\langle R_1 R_2 \rangle = \langle R_2 R_1 \rangle$ from their joint distribution function. Quantum mechanically, if $[\hat R_1, \hat R_2] \neq 0$, that is obviously more problematic.

Let me first introduce some notation. An observable $R$ in QM corresponds to a self-adjoint operator $\hat R$ with spectral representation $$\hat R = \int_{-\infty}^\infty r\, dE(r) ,$$ where $dE(r)$ is the spectral-valued measure (e.g. in finite dimensions $dE(r) = \sum_{k=1}^N \delta(r - \lambda_k) | k \rangle\langle k |\, dr$).

As long as we are considering only one observable, we can describe the outcome of a measurement of $R$ if the system is in the state $|\psi\rangle$ as a classical random variable. The cumulative distribution function of that random variable is $$F_R(r) \equiv P(R \leq r) = \langle \psi | E(r) | \psi \rangle = \int_{-\infty}^r \langle \psi | dE(r') | \psi \rangle .$$

Let us now turn to the case where we have two classical random variables $R_1$ and $R_2$. Their joint distribution function is $$F(r_1, r_2) = P(R_1 \leq r_1 \text{ and } R_2 \leq r_2) .$$ Knowledge of $F(r_1, r_2)$ is equivalent of knowledge of $$F_{\vec k \cdot \vec R}(r) = P(k_1 R_1 + k_2 R_2 \leq r) ,$$ because the characteristic function $G(k_1, k_2) = \int_{\mathbb R^2} \mathrm e^{\mathrm i \vec k \cdot \vec r} dF(\vec r)$ can be calculated from $$G(k_1, k_2) = \int_{-\infty}^\infty \mathrm e^{\mathrm i r} dF_{\vec k \cdot \vec R}(r) .$$

We say that two quantum observables have a joint probability distribution if there are two classical random variables such that $$F_{\vec k \cdot \vec R}(r) = \langle \psi| E_{\vec k \cdot \hat{\vec R}}(r) |\psi\rangle .$$ Obviously, the characteristic function must then be $$G(k_1, k_2) = \langle\psi| \mathrm e^{\mathrm i \vec k \cdot \hat{\vec R}} |\psi\rangle. \tag{*}$$

Nelson's Theorem states that two observables have a joint probability distribution if and only if they commute. The "if" part is clear, we can just set $F(r_1, r_2) = \langle\psi| E_{\hat R_1}(r_1) E_{\hat R_2}(r_2) |\psi\rangle$.

The "only if" part is more difficult, but consider the example of $\hat Q$ and $\hat P$. Calculating the Fourier transform of (*), we get that the joint probability density is - if there is one - equal to the Wigner function. However, the Wigner function can be negative and is therefore not a probability density.

Note that - as stated in the comments to your question - there are positive functions $f(q,p)$ similar to the Wigner function such that $\int f\, dp = |\psi(q)|^2$ and $\int f\, dq = |\tilde \psi(p)|^2$. This is not a contradiction, because those functions only produce the correct marginal distributions for $\hat Q$ and $\hat P$, not for all $k_1 \hat Q + k_2 \hat P$.

My favourite explanation is via by way of Bell's Theorem. Alice and Bob each have a box in front of them into which they can insert either a zero of a one. If Alice inserts a zero she gets a random variable $A_0$, and if she inserts a one she gets $A_1$. If Bob inserts a zero he gets a random variable $B_0$, and if he inserts a one she gets $B_1$. All these are $\pm 1$-valued Bernoulli random variables. The probability that Alice and Bob's outputs equal $(-1)^{ab}$ given $a$ and $b$ is the Bell-CHSH correlation $V$, where $a$ and $b$ are Alice and Bob's inputs correspondingly.

Bell's Theorem states that $V\leq 0.5$ for a local hidden variables theory. But in quantum mechanics, $V$ can be as big as about $0.85$, and this is supported by experiments. How come?

Ignoring the metaphysics of it all, $V\leq 0.5$ follows from the existence of a joint distribution between all pairs and triples of outcomes, $A_0$, $A_1$, $B_0$, and $B_1$, as shown in:

A. Fine, Hidden Variables, Joint Probability, and the Bell Inequalities, Phys. Rev. Lett. 48, 291 – Published 1 February 1982.

Violation of Bell's inequality thus demonstrates that such joint probabilities don't exist, although the marginal certainly do.

... And why should they? Either Alice measures $A_0$ or she measures $A_1$. One is physical reality, the other is the proverbial path not taken. What physical meaning would there even be to a joint probability between two mutually exclusive events?