1
$\begingroup$

First of all I would like to point out that when it comes to quantum physics, I have very poor knowledge so please excuse me if I misuse some words to describe what I mean.

My question is based on Isaac Arthur's example in this video. To save you time, he is giving an example to show that if something is possible, it will happen no matter how improbable it is. His example is about a mad scientist who vaporizes himself in a box and I quote him: "yet at the same time, so to speak, at some time and place in the multiverse, random quantum chances assemble that box and out steps a clone of the doctor or one close enough to make no difference"

In a previous example in the same video he says: "there is a universe right now where you flicked on your light switch and a random chunk of furniture in your house turned into a bunch of gold ingots"

Disregarding the multiverse and the many worlds hypotheses now as the question is not about them, and disregarding the exact details of the examples above to prevent the violation of mass and energy and momentum conservation, could quantum physics have any probability (even if extremely low) to change say trillions of atoms from one element to another?

In this article, the author answers it clearly with a NO saying: "The uncertainty principle applies only to individual quantum objects and small collections of quantum objects. When trillions of quantum objects are involved, the uncertainty goes completely away"

So which of those arguments is the correct one?

$\endgroup$

1 Answer 1

0
$\begingroup$

In mainstream physics at this day and age, the underlying level of all nature is quantum mechanical. From this level the classical level emerges , and this can be proven with the appropriate mathematics. The quantum mechanical level is probabilistic,i.e. the (x,y,z,t) coordinates of a quantum mechanical particle will follow a probability distribution if a measurement is made . A probability distribution is the same for quantum and classical physics. For classical, think of the probability distribution for the face of a dice to give a number from 1, to 6. It is flat. In quantum mechanics there exist differential equations that determine the shape of the probability distribution. It needs many throws of the dice to get its distribution and it takes many repetitions of any experiment with exactly the same boundary conditions to give the quantum distribution.

The many worlds interpretation you describe your question, is assuming that the mathematics that describes how the probability functions are found, instead of just being mathematical tools, they are real, and each "throw of the quantum mechanical dice" branches off into another world. The predictive results are the same. Using gold ingots takes it out of the mathematically plausible interpretation into the science fiction realm, as you also realize, because momentum energy angular momentum conservation are absolute, and also a lot of quantum number conservation laws too.

The statement :

if something is possible, it will happen no matter how improbable it is

is a matter of belief, not of physics, which is the discipline that searches for mathematical models that will describe existing data and observations, and will predict future ones.

could quantum physics have any probability (even if extremely low) to change say trillions of atoms from one element to another?

Lets make it much simpler: Is there a probability for the simple helium atom, to be two deuterium atoms? In quantum mechanical terms, does the wavefunction of a helium atom have any overlap to a sum of two deuterium atoms' wavefunction? The answer is no, not spontaneously, not in mainstream physics. Helium atoms are stable, it needs extra interactions , i.e. new wave functions, to be able to fit two deuterium atoms, extra energy and momentum have to be supplied.

$\endgroup$
1
  • 1
    $\begingroup$ So to understand you correctly, is it the conservation of energy and momentum that prevents entire objects from changing? or is it the macro vs. micro argument (article in the original post) that prevents this? $\endgroup$ Commented Dec 3, 2019 at 21:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.