Martin Bauer Profile picture
Nov 14 7 tweets 1 min read Read on X
The collapse of the wavefunction doesn't break causality

Consider a lottery where many people buy tickets. Before drawing the winning numbers, the probability distribution of winners is a complicated function (depending on the number of participants, tickets per person,..)

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At the moment of drawing the winning numbers this probability distribution instantly collapses to a delta function (assuming just one winner)

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Nobody is concerned about causality here, because everyone can be informed about the result only with the speed of light

In quantum mechanics, a measurement ‚collapses the wavefunction‘ just like that. At no point is there a violation of causality

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For the lottery probability distribution, everyone agrees it’s not a physical object and the ‚collapse‘ only updates our knowledge. In case of the wavefunction views differ.

But the collapse has no operational effect outside the lightcone*, it is completely in agreement with special relativity

*more carefully: spacelike separated fields commute

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So what really is the difference? The difference is there is no fundamental, classical model underlying the wavefunction. In the case of the lottery we don't worry, because we know the probability distribution is just a way to parameterise our ignorance, there is an underlying classical microstate

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In the case of quantum mechanics not only is the wavefunction not parameterising our ignorance, there cannot possibly be an underlying classical microstate

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If you're coming from classical physics this makes Quantum mechanics look like a floating table cloth without a table underneath, but it really shows that a classical theory isn't the only way to construct a consistent, relativistic theory

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More from @martinmbauer

Oct 8
Short explanation of the physics Nobel Prize 2025

In classical mechanics you can know where a particle is and its momentum at the same time. In Quantum mechanics you can't. All information is in the wavefunction. Even if a particle is trapped, part of the wavefunction..

🧵1/7
..is on the other side of the barrier (unless its an infinitely strong force field)

If you measure the position of the particle there is a finite probability it's outside the trap even if it never had enough kinetic energy to overcome the barrier. It 'tunnels' through the barrier

2/7Image
This effect isn't uncommon in nature

It explains radioactive alpha-decay where a whole Helium nucleus is emitted from a heavy decaying element even if the binding forces wouldn't allow this process classically

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Read 7 tweets
Sep 7
Despite the 2012 Higgs boson discovery, we still don't know whether it is responsible for the masses of first generation fermions (up&down quarks and the electron)

Higgs couplings to gauge bosons, 3rd and 2nd generation fermions agree well with the Standard Model, but..

🧵1/5 Image
..we have never measured the Higgs boson decaying into first generation fermions.

The reason is the way mass generation via the Higgs mechanism works. The heavier a particle, the stronger it's interactions with the Higgs field, the more likely it is for a Higgs boson to interact with it

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But the first generation fermions are the lightest fermions. They interact weakly with the Higgs field and the Higgs boson will very rarely interact with them

3/5
Read 5 tweets
Aug 1
In 3 (or more) dimensions, all fundamental particles are either fermions and bosons. But why?

This is a direct consequence of the properties of the configuration space for identical particles

🧵 1/14
If you have 2 indistinguishable particles, the configuration in which particle 1 is at r and particle 2 is at -r is completely equivalent to the configuration where they swap positions

So in 2D, the configuration space is R^2 with opposite points identified (and excluding the origin which corresponds to both particles being at the same place)

This space is a cone w/o a tip (e.g. upper half-plane glued along the x-axes)

2/14Image
Any exchange of these two particles corresponds to a loop on this cone. You can think of this as rotations of the line connecting the two particles through their centre of mass.

Two loops that can be continually transformed into each other are equivalent. But because the tip of the cone is missing, any number of rotations around the cone can't be transformed into one with fewer or more windings

3/14Image
Read 14 tweets
Jun 26
Gravity is weaker than all other forces, but is there a reason why it should be? Maybe.

This is a short thread on the weak gravity conjecture (minimal math)

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There are two parts to the argument why gravity should be the weakest force: First, black holes can carry charges and can evaporate

Consider now a black hole with mass M and total charge Q. It can only fully evaporate if there exist particles with mass m and charge q so that M/m > Q/q

2/5
Because a charged black hole with charge in q-units larger than its mass in m-units can't distribute its total charge into particles of mass m and charge q

A charged remnant would remain, which can result in a 'naked singularity' floating through the Universe (not hidden behind an event horizon)

3/5
Read 5 tweets
May 22
Are neutrinos their own anti-particles?

They are the only fundamental fermions that could have this unique property. But they're so elusive, how could we measure that?

A brilliant experiment could answer this question without detecting a single neutrino

🧵1/11 https://room.eu.com/news/the-mystery-of-neutrino-mass-to-be-solved-by-a-calcium-isotope
You only need two facts to understand this experiment.

1. Neutrinos are produced when neutrons decay into protons (beta decay). That's how they were discovered.

Because of charge conservation a neutron decays into an electron and a proton

2/11 Image
It's incredibly difficult to detect neutrinos, but if they weren't present in this decay, momentum conservation would only allow back-to-back electrons and protons, but in reality there is always another momentum component: the (anti-)neutrino

3/11 Image
Read 11 tweets
May 4
Does quantum mechanics really need to be complex?

Why can't we just use real wavefunctions?

🧵1/15 *some math, non-relativistic QM Image
First of all, not all wavefunctions are complex. The Schrödinger equation tells us that time-independent (stationary) solutions can be real. These are energy eigenstates, e.g. a bound state in an hydrogen atom

But even if you start with a real wavefunction, time-evolution introduces an imaginary part

2/15Image
Time-evolution for a real-valued wavefunction correspond to a global phase. But global phases don't have observable consequences

3/15 Image
Read 15 tweets

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