Is it Possible that Heisenberg's Uncertainty Principle has Observable Macroscopic Consequences?

Question

HUP is generally discussed (specially in traditional books) as a consequence of quantum mechanics that's generally better (or even only) illustrated with microscopic physics. Even though I'm not the most careful of people, I've been careful enough to consider this question under the light of a particularly strong formulation of the principle. Namely, the so-called Schrödinger-Robertson version of it (for two arbitrary observables $\hat{A}$ and $\hat{B}$: $$ \sigma_{A}^{2}\sigma_{B}^{2}\geq\left|\frac{1}{2}\langle\{\hat{A},\hat{B}\}\rangle-\langle\hat{A}\rangle\langle\hat{B}\rangle\right|^{2}+\left|\frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle\right|^{2} $$ Generally, quantum commutators are $\hbar$-order expressions of their classical Poisson brackets. The first term can be safely ignored for systems that are deep into the classical regime --I think. The reason being that the anti-commutator term has correcting terms proportional to expected values, so that the expected value of the anti-commutator can be taken to be largely suppressed by the product of the corresponding expected values, say a second-order quantity when compared to the commutator term, (which is not suppressed) at least for quasi-classical observables.

So let's centre on the typically quantum term, the one given by the commutator. Taking as examples the standard operators for orbital angular momentum, and ignoring the anti-commutator, we get, $$ \triangle J_{x}\triangle J_{y}\geq\frac{\hbar}{2}\left|\left\langle J_{z}\right\rangle \right| $$ Now, the first thing that strikes someone like me is that these are generally very small quantities in comparison to $\left\langle J_{x}\right\rangle $ and $\left\langle J_{y}\right\rangle $. Except in case the expected value $\left\langle J_{z}\right\rangle$ is humongous! I'm thinking pulsars, magnetars, black holes, and the like.

Here's the question: Do you know of any examples in which these uncertainties of purely quantum origin could have macroscopically-observable consequences?

I do remember reported inconsistencies in the spectrum from quickly-rotating stellar objects (in directions other than their spin) back in the early 2000s that were attributed to dust. I'll try to dig that out to make the question more complete. But the question should make sense in and of itself.

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Reformulation of the question

Apparently, I haven't been understood; and I have only myself to blame. I will re-edit it to everyone's satisfaction, I hope. Some people want to talk about dead cats; other people don't want to talk about dead stars. Fair enough.

Here's the thing, then. Let's say I'm talking about a "quantum quasar". What is it? For the purposes of discussion, let's say it's a quantum system that's in an eigenstate of spin, of such an expected high value that we can take the only relevant observables to be, $$ J_{x}=10^{120}\sigma_{x} $$ etc. All sigmas re-scaled to $10^{120}$, which is my lucky number. (I've removed the offending $\hbar$ factors, which play no part in the argument.) I've also removed any reference to Poisson brackets, which play no really essential part either.

Because of previously stated hypothesis, $$ \left|\psi\right\rangle =\left(\begin{array}{c} 1\\ 0 \end{array}\right) $$ Now, allow me to be laconic --because I've taken enough of your time already--.

$$ \left\langle J_{y}\right\rangle _{\left|\psi\right\rangle }=\left\langle \psi\left|10^{120}\sigma_{y}\right|\psi\right\rangle =\left(\begin{array}{cc} 1 & 0\end{array}\right)\left(\begin{array}{cc} 0 & -i10^{120}\\ i10^{120} & 0 \end{array}\right)\left(\begin{array}{c} 1\\ 0 \end{array}\right)=0 $$ $$ \left\langle J_{x}\right\rangle _{\left|\psi\right\rangle }=\left\langle \psi\left|10^{120}\sigma_{x}\right|\psi\right\rangle =\left(\begin{array}{cc} 1 & 0\end{array}\right)\left(\begin{array}{cc} 0 & 10^{120}\\ 10^{120} & 0 \end{array}\right)\left(\begin{array}{c} 1\\ 0 \end{array}\right)=0 $$ But, $$ \left\langle J_{z}\right\rangle _{\left|\psi\right\rangle }=\left\langle \psi\left|10^{120}\sigma_{z}\right|\psi\right\rangle =\left(\begin{array}{cc} 1 & 0\end{array}\right)\left(\begin{array}{cc} 10^{120} & 0\\ 0 & -10^{120} \end{array}\right)\left(\begin{array}{c} 1\\ 0 \end{array}\right)=10^{120}$$ Wow!

And what's worse:

$$ \left\langle J_{y}^{2}\right\rangle _{\left|\psi\right\rangle }=\left(\begin{array}{cc} 1 & 0\end{array}\right)\left(\begin{array}{cc} 0 & -i10^{120}\\ i10^{120} & 0 \end{array}\right)\left(\begin{array}{cc} 0 & -i10^{120}\\ i10^{120} & 0 \end{array}\right)\left(\begin{array}{c} 1\\ 0 \end{array}\right)=10^{240} $$ $$ \left\langle J_{x}^{2}\right\rangle _{\left|\psi\right\rangle }=\left(\begin{array}{cc} 1 & 0\end{array}\right)\left(\begin{array}{cc} 0 & 10^{120}\\ 10^{120} & 0 \end{array}\right)\left(\begin{array}{cc} 0 & 10^{120}\\ 10^{120} & 0 \end{array}\right)\left(\begin{array}{c} 1\\ 0 \end{array}\right)=10^{240} $$ Dispersion analysis: $$ \triangle_{\left|\psi\right\rangle }^{2}J_{x}\triangle_{\left|\psi\right\rangle }^{2}J_{y}=10^{480}\gg\frac{1}{4}10^{120}=2.5\times10^{118} $$ which is only ridiculously true. Measurable in any context? That is my question. So, please, understand me; I don't mean a state in which $J\left(J+1\right)$ is fixed, but you can play around with $J_x$, $J_y$, and $J_z$. I mean a $J_z$ eigenstate with very high expected value.

Note: I've removed my tag "astrophysics". Sorry, @CosmasZachos, and thank you for the illuminating aspects.

Answer 1

In electrical engineering, we have a device known as a Zener diode. This lets current flow one way, but not the other. However, if the voltage is high enough across a diode in the wrong direction, the diode will suddenly start conducting, something known as "avalanche breakdown."

The physics of this breakdown is that a diode has a "depletion region" where electrons cannot travel through. At higher voltages, this depletion region gets narrower but more depleted, ensuring electrons do not get through. However, at some voltage, they do.

The reason for this is that the depletion region gets smaller and smaller until it is on the scale of the position uncertainty of the electrons. At this point, there is a statistical probability that an electron on one side will simply start existing on the other side -- quantum tunneling.

The trick, of course, is that this effect is repeated for billions of electrons, creating a predictable macroscopic effect. These macroscopic effects are visible in basically every electronic device you have in your house. Zener diodes, in particular, are crafted to control this avalanche breakdown so that it occurs at a chosen voltage. You find them in basically every voltage regulator we make, including every wall wart in your house!

Answer 2

Motivated by your comment, I will ignore modeling stellar objects (what a relief!) and such, and interpret the relation for a simple quantum state you might be visualizing, which has enormous j. That is, I will take $j(j+1) \approx j^2$. In that case, if you are talking about enormous 2j×2j matrices satisfying the algebra $[J_x,J_y]= i J_z$, etc, where I've absorbed the irrelevant dimensionful $\hbar$ into the normalization of the matrices, since it enters homogeneously there, then:

• Indeed, the maximum eigenvalues of all three matrices are $j$, and hence likewise their maximum expectation values.

• Indeed, $\langle J_x^2+ J_y^2+ J_z^2\rangle \approx j^2$.

• Indeed, $$ \left ( \langle J_x^2 \rangle - \langle J_x \rangle^2 \right ) \left ( \langle J_y^2 \rangle - \langle J_y \rangle^2 \right ) \geq {1\over 4} \langle J_z \rangle ^2 . $$ So the dispersion of these matrices is bounded below, as they still obey that commutation relation.

As a consequence, no matter how large these matrices get, they still suffer from the "quirks" Robertson identified in his paper, namely the customary $j\geq |m|$ and that only one of these matrices can have arbitrarily precise expectation values: If you take an eigenstate of $J_x$ or $J_y$ for your state, then you know the right-hand side must vanish, let alone have a huge eigenvalue. (You see that for humdrum small js, but you can easily prove it for arbitrary j as well.)

Soft stuff: I think the psychological slippage occurs when you assume at some level the above commutator "behaves the same", or does the same thing as the Poisson Bracket, Dirac's PhD thesis, which obeys an identical Lie-algebraic relation. But the one thing the PB fails miserably in is the uncertainty principle, whence all that follows here. This is easier to see in the phase-space formulation of QM, and outranges the question, but is most often what underlies such aspirational inferences...

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Comment on the amended/explained question:

There is a misunderstanding here. Your redefined $$ J_{x}=10^{120}\sigma_{x}, $$ etc... are just two-by-two matrices, merely normalized in a freaky way, so their commutators entering the Robertson relation are now just $$ [J_x,J_y]= i(2\cdot 10^{120} ) J_z. $$

Eliminating the $ 10^{240}$s in both sides of the resulting Robertson relation nets merely $$ \left ( \langle \sigma_x^2 \rangle - \langle \sigma_x \rangle^2 \right ) \left ( \langle \sigma_y^2 \rangle - \langle \sigma_y \rangle^2 \right ) \geq \langle \sigma_z \rangle ^2 , $$ so basically spin 1/2 in a freaky inconsequential normalization. I don't know what this has to do with macroscopic systems (it doesn't) but it is comfortably satisfied and hardly surprising. For your m=1/2 state, you have $$ \langle \sigma_x^2 \rangle \langle \sigma_y^2 \rangle = \langle \sigma_z \rangle ^2 \\ \leadsto ~~~~~1=1. $$

Answer 3

Every quantum mechanical experiment is carefully designed to yield macroscopically observable consequences of quantum mechanical relations. At least I have never heard of wave-like micro-physicists doing experiments directly on the Angstrom scale, and publishing their observations in journals that are subject to energy-time uncertainty or something equally weird.