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I've started reading Brian Hall's Quantum Mechanics for Mathematicians. He gives a motivation for the operator formalism for quantum mechanics. If you think of position of a particle as a random variable $X$, and if $\psi(x)$ is the wave function, then $|\psi(x)|^2$ gives the density of $X$, and so $$ \mathbb{E}(X)=\int_{\mathbb{R}}x|\psi(x)|^2dx. $$ At the same time, this is $\langle\psi, \hat{X}\psi\rangle$, where $\hat{X}$ is the position operator. I understand this is just motivation, but why is this necessary? It's just a way to rewrite an expectation. I know there is a non-commutative probability theory for computing probabilities in quantum mechanics, but I'd like to know why this is necessary. Why does the interpretation that position and momentum are classical random variables fail? What experiments demonstrate this?

There are four examples that come to my mind.

In the double-slit experiment, if you consider a sequence of electrons as independent, identically distributed Bernoulli random variables, the interference pattern seems to just be a consequence of the central limit theorem, so I don't see why classical probability is not suited in that example.

The uncertainty principle puts a lower bound on the product of variances of position and momentum, but would that contradict an interpretation of position and momentum as random variables?

I read somewhere that the Stern-Gerlach experiment shows that if you consider $X$ and $Z$ as random variables measuring the $x$ and $z$ spin of an electron, then $\mathbb{E}(XZ)\not=\mathbb{E}(ZX)$. This would be what I'm looking for, showing the random variables do not commute. However, I'm not sure I understand the argument/experiment. If this is indeed correct and someone could explain this in some detail, I would appreciate that.

Bell's inequality seems to say that if you have three random variables $X_1,X_2,$ and $X_3$ defined on a probability space taking values in $\{0,1\}$, then their inner products satisfy $$ |\langle X_1,X_2\rangle+\langle X_2,X_3\rangle|\leq 1-\langle X_1,X_3\rangle $$ Is this not satisfied in the Stern-Gerlach experiment?

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Let's say $H$ is a Hilbert space, $Q, P$ are self-adjoint operators on $H$ and $\psi$ a unit vector in $H$ describing a state of a system. The expectation is defined as $$\Bbb{E}_\psi(X) = \langle \psi, X\psi\rangle$$ for an operator $X$. Now, $$\begin{split} \Bbb{E}_\psi(QP) = \langle \psi,QP\psi\rangle = \langle Q\psi, P\psi\rangle\\ \Bbb{E}_\psi(PQ) = \langle \psi,PQ\psi\rangle = \langle P\psi, Q\psi\rangle=\overline{\langle Q\psi,P\psi\rangle} \end{split}$$ and generally, $\langle Q\psi, P\psi\rangle\ne \overline{\langle Q\psi,P\psi\rangle}$. In fact $$\Bbb{E}_\psi(QP) - \Bbb{E}_\psi(PQ) = \langle \psi, [Q,P]\psi\rangle$$ where $[Q,P]$ is the commutator which for position and momentum is $[Q,P]=i\hbar$ and then $\langle \psi, [Q,P]\psi\rangle = i\hbar\langle\psi,\psi\rangle = i\hbar$.

From the Cauchy-Schwarz inequality you get $$ |\langle Q\psi, P\psi\rangle| \le \|Q\psi\|\|P\psi\| = \sqrt{\langle \psi,Q^2\psi\rangle\langle \psi,P^2\psi\rangle}=\sqrt{\Bbb{E}_\psi(Q^2)\Bbb{E}_\psi(P^2)} $$ but also $$\Im(\langle Q\psi,P\psi\rangle) = \frac{1}{2i}(\Bbb{E}_\psi(QP) - \Bbb{E}_\psi(PQ)) = \frac{1}{2i}\Bbb{E}_\psi([Q,P])$$ and since $|\Im(\langle Q\psi,P\psi\rangle)|\le |\langle Q\psi, P\psi\rangle|$ we get $$\frac{1}{2}|\Bbb{E}_\psi([Q,P])|\le \sqrt{\Bbb{E}_\psi(Q^2)\Bbb{E}_\psi(P^2)}$$ which is the Heisenberg uncertainty principle and specifically for $Q, P$ position and momentum says that $$\frac{\hbar}{2}\le \sqrt{\Bbb{E}_\psi(Q^2)\Bbb{E}_\psi(P^2)}$$

Now, let $\mu^Q, \mu^P$ be the projection-valued measures associated with $Q, P$ from the spectral theorem. If $I_1, I_2\subseteq \Bbb{R}$ are any intervals, then in general, when $Q, P$ don't commute, the projections $\mu^Q(I_1), \mu^P(I_2)$ don't commute either. This means $\Bbb{E}_\psi(\mu^Q(I_1)\mu^P(I_2))\ne\Bbb{E}_\psi(\mu^P(I_2)\mu^Q(I_1))$. But in classical probability theory, $\Bbb{E}_\psi(\mu^Q(I_1))$ is the probability of finding $Q$ in the interval $I_1$, $\Bbb{E}_\psi(\mu^P(I_2))$ is the probability of finding $P$ in the interval $I_2$ and $\Bbb{E}_\psi(\mu^Q(I_1)\mu^P(I_2))$ is supposed to be the joint probability of finding $Q$ in the interval $I_1$ and $P$ in the interval $I_2$. But this probability is not well-defined, because when $Q, P$ don't commute, $\mu^Q(I_1)\mu^P(I_2)$ is generally not a projection so this expectation is not necessarily positive. It can be complex. You get a different value in the opposite order $\mu^P(I_2)\mu^Q(I_1)$. So the joint probability distribution of two non-commuting self-adjoint operators is not defined. In classical probability, when you have two random variables on the same measure space, their joint probability is always defined.

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    $\begingroup$ I think their confusion is earlier. They are going for first principles explanation of operator formalism in the first place. I am not sure what axioms they are willing to start from. But there is literature could point to that show how usual formalism is inevitable from more innocent sounding axioms. $\endgroup$
    – AHusain
    Commented Dec 15, 2023 at 19:47
  • $\begingroup$ For the meantime, see the axioms given in the quantum logic wikipedia and Solèr's theorem. $\endgroup$
    – AHusain
    Commented Dec 15, 2023 at 19:58
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    $\begingroup$ @user124910: In the double-slit experiment, if you cover the left hole, you get a stain on the screen without a diffraction pattern. If you cover the right hole, you get the same stain slightly shifted. When both holes are open, if the electron is like a billiard-ball, you expect to get the sum of the of the stains. Instead you get a diffraction pattern. How does the electron know the other hole is open so there are places it's not allowed to go? This is explained for light by the wave theory, but a wave is not localized. Is an electron spread out? $\endgroup$
    – Chad K
    Commented Dec 15, 2023 at 20:31
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    $\begingroup$ @user124910: In the Stern-Gerlach experiment, spin operators for different directions don't cummute. You can align SG devices serially for detecting spin in different direction. When you do, each detector randomizes the spin in the prior direction - showing that they don't commute and there is no joint probability of spin in different directions. You can't nail spin down simultaneously in different directions. $\endgroup$
    – Chad K
    Commented Dec 15, 2023 at 20:33
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    $\begingroup$ @user124910: The Schrödinger equation also shows quantum tunneling in its solutions and how to compute transmission and reflection probabilities for a potential barrier. Most of solid-state physics is based on quantum tunneling. For example, flash memory uses floating-gate mosfet. There's a charged metal bubble in the middle of an insulator. You wouldn't be able to charge or discharge flash memory cells without quantum tunneling. Classically, it would need a voltage so high that would destroy the insulator. $\endgroup$
    – Chad K
    Commented Dec 15, 2023 at 20:37

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