It's possible to prove the Born Rule of quantum mechanics

[@mathemaniac]

To address a lot of the questions here, there are a few assumptions here in the video: (1) If the coefficients of the two states are equal, then they have the same probability. This is a reasonable assumption - you want the definition of probability to be consistent. Then this video provides a method on how to make sense of the probability with unequal coefficients. Essentially, you break those states down into "finer" states so that all coefficients of these "finer" states are now the same, which you know how to interpret it as probabilities, and then add those probabilities up to get the probability of the original "coarser" states. (2) Treating those states as "vectors", we require each state to have "length" 1. (3) (A hidden assumption) Those different states are "orthogonal" in some sense, which is why you can literally use Pythagorean theorem to add the squares up to 1. For (2), Mithuna has mentioned the use of Schrodinger equation. However, the notion of "length" of a state / vector requires a notion of inner product (at least that's what I am aware of). This in turn means that it is somewhat necessary to bring in the "bra" notation. Similarly, an "orthogonal" state requires the notion of inner product. Then you have to treat the "bra" notation purely as a mathematical tool for defining inner products (i.e. "bra" being a dual "vector" to a "ket"). But then once you try to interpret inner products, you are back to Born rule all over again. Of course, the question is how fundamental do you want all these to be? The main point of this video is to try to make it somewhat more fundamental - if you define "probability" reasonably, i.e. same coefficients = same probability, then you can derive Born rule. While previously you might just say "those coefficients are essentially defined to be the probability", now you can say "if we assume that when the states with equal coefficients have equal probability, then we are forced to interpret that in the general case, those coefficients squared *are* probabilities."

[@Impatient_Ape]

As for (2), just a reminder to those with some experience... remember that the Hilbert space used in quantum mechanics is actual a *projective* Hilbert space, meaning that all co-linear vectors are lumped into the same class. Normalization can be thought of as the process of identifying which vector from the original Hilbert space is selected as a representative for the projective Hilbert space. To do this for the purposes of computing lengths or probability amplitudes, *we* make a choice about which co-vectors form the dual space. It's kind of a two-for-one deal. When you normalize a representative in the projective Hilbert space, you automatically define one of the co-vectors from the dual space.

[@Achrononmaster]

Note the point made by @'Impatient Ape' => projective Hilbert space is the setting for QM. ALSO, on (3) the "hidden assumption" is not really "orthogonality" it is rather that we _choose_ a basis. There is nothing more to it. You can choose a non-orthogonal basis too, it just makes the linear algebra more awkward, since then eigenvalues/eigenkets of the operators will not correspond to your _chosen_ basis states, but you do not lose anything physical.

[@renedekker9806]

Don't you need the Born rule for assumption (1) ? Without the Born rule, the coefficients could many anything at all, and may mean something different for different states. The assumption that if coefficients are equal then the probabilities are equal, implies that the coefficients can be interpreted as probabilities. Doesn't the Born rule basically _define_ that meaning of the coefficients?

[@Packbat]

It feels like the Born rule is a kind of side effect of the Pythagorean theorem - if you add orthogonal vectors, the length is the square root of the sum of the squares of those vectors' lengths (and the square root of 1 is 1). So a ratio of 3 to 4 in the coefficients is actually 9 to 16 in probability, because that's how much they contribute to the final amplitude of the vector. I'm not super satisfied yet either but it does feel like a valuable step. Thank you!

[@enterprisesoftwarearchitect]

The squaring is to get rid of the Complex numbers - Mithuna simplified it by making the amplitudes only real numbers (1/3, 2/3). The complex numbers must be eliminated because all measurements are produce Real (non-complex) numbers.

[@NuclearCraftMod]

Gleason's theorem is closely related to the idea that if you want to relate amplitudes to probabilities which consistently add up to one, then in general you are forced to interpret the squares of the absolute values of the amplitudes as the associated probabilites.

[@Achrononmaster]

@@enterprisesoftwarearchitect That's not it. They need to be real because they represent probability density, a la Kolmogorov. Measurement results can be geometric, hence do not have to be real scalars. The geometric amplitudes are the weakest generalization of Kolmogorov probability, which is obtained by dropping the distributive rule. So why the distributive rule fails is the "real" proof. (Alternatively, failure of contextuality, a la Gleason.) Interestingly, the distributive rule (and contextuality) fails if there is a single closed timelike curve, which is possible in GR, and likely unavoidable at the Planck scale, so in GR you already have quantum logic. See the work of Mark Hadley, and more recently Lenny Susskind. To diabuse yourself of the notion "complex number" is somehow intrinsic to QM read the papers by Doran, Lasenby, Gulland Hestenes "Complex Numbers are Not Real" geometry.mrao.cam.ac.uk/wp-content/uploads/2015/02/ImagNumbersArentReal.pdf

[@enterprisesoftwarearchitect]

@@Achrononmaster That’s not what Max Born said - read his 1954 Nobel Prize speech. N-dimensional polytope volumes also can represent probability amplitudes, so there are geometric representations without squaring! The complex numbers are there due to a wave description, but to remove them you have two choices … and equivalence class saying they don’t matter or account for their contribution by taking the square - because the empirical rule for finding a photon was the square of the EM wave, Born said he generalized that rule. That’s where the Born rule comes from.

[@Quarky_]

Maybe this observation makes more sense once you realise the different quantum states denoted by the kets (|bla>) are orthogonal basis vectors in a Hilbert space by definition.

[@enterprisesoftwarearchitect]

Sad that Born’s beautiful granddaughter has passed this week - RIP Olivia Newton-John

[@LookingGlassUniverse]

I didn’t realise they were related!

[@Jabber_Wock]

Thanks for that info! I didn’t know ONJ was related to Born.

[@ftumschk]

Olivia came from a rather gifted family all round. Her uncle was a renowned professor of pharmacology and her father, a linguist and MI5 agent, was one of the "code-breakers" at Bletchley Park.

[@enterprisesoftwarearchitect]

Great genes in that family. I am one of tens of millions of ONJ fanboys

[@岡安一壽-g2y]

I pray for her soul.

[@Jim6593]

Let me provide a counter: The first assumption is that if the coefficient of mutually independent (after orthonormalisation) states is equal, then they are equally likely? It’s not clear what’s behind this assumption. Where is this probability interpretation coming from? If it’s an assumption not rooted in any rigorous structure, then it’s far far worse than Born rule. The second assumption is that the ratio of probabilities need to be integers. That’s catastrophic. Actual probabilities are mostly irrational numbers. This method is incapable of establishing a generalised Born rule even if I accept your first assumption. I hope you see my point.

[@lepidoptera9337]

None of those assumptions are valid. The probability "interpretation" is Copenhagen, but Copenhagen can't actually give you the correct branching ratios for e.g. atomic scattering processes. For that we need, at least, quantum electrodynamics because the line widths (and shapes!) are non-trivial. Most people who are into "fundamentals of quantum mechanics" don't even understand that much about actual physics. They seem to have forgotten everything about their undergrad atomic physics classes where all of this was discussed ad nauseam.

[@gabitheancient7664]

1- I mean, you can just turn this assumption into "the probability is a function of the coefficients" I'm pretty sure that's what being assumed 2- this wasn't assumed at all, idk where did you get that idea, but the idea is that when you have irrational probabilities you can prove it assuming continuity

[@MrJaffjunior]

This was the only satisfactory proof I seen so far. The way you explain things is really clear. Your explanation is so good it makes things look obvious. This video was totally worth watching. Thank you so much.

[@brothermine2292]

A key step seems to have been skipped. The state Z is itemized as a superposition of apparently artificial states B & C with no regard to whether B & C are physically possible, and B & C are arbitrarily assigned equal probabilities (1/2) even though their sum can be 1 without them being equal. Why is it okay to treat a state that's physically possible as if it's a sum of fictional states that could be physically impossible (note that physical impossibility implies probability = zero) or that might physically need to have nonequal probabilities (such as pB = 2/3 and pC = 1/3 instead of pB = pC = 1/2)?

[@TheLastScoot]

This comment asked my question better than I could have.

[@FermionPhysics]

Because the states here are eigenstates of the Schrodinger equation (or any other operator). There wouldn’t be any reason to add together different (unphysical) states from different operators since they don’t share the same basis.

[@ericvilas]

It's less of an assumption and more of a "this situation is possible, so we can use it to investigate what the coefficients mean" Once you know what the coefficients mean for that specific situation you know how to handle, you can go "ah, so this is how to interpret wavefunction coefficients!" and apply it to the more general case

[@TheLastScoot]

@@ericvilas Just because it's possible for the coefficients to be the same (or for the sub-coefficients to be the same) doesn't mean that logic can be applied to situations with different coefficients, especially if it can only be resolved because the coefficients (or the sub-coefficients) are the same. Because of that, your answer doesn't feel very useful.

[@NuclearCraftMod]

There is no such thing as a "fictional" state. All states are perfectly legitimate: superposing any two states gives another state, and they're all legitimate, possible states of the quantum system. There is one possible detail that complicates this a little though, which are "superselection" rules (the Wikipedia article is quite good).

[@LookingGlassUniverse]

Sorry it's a week late...

[@harshnandan9553]

Np. Awesome content as always.

[@danielwcrompton]

No worries, thank you!

[@michaelsommers2356]

Don't be sorry. Your videos are always a pleasure, no matter how long between them.

[@brothermine2292]

Did we have a contract that you broke?

[@Zishen3333]

I think we automatically assume Born rule is right when doing the normalization. From experimental perspective, quantum state is a function that maps the measurements to probabilities. We can’t tell what a quantum state looks like. If and only if we make an assumption that quantum state in finite dimensional system can be considered as normalized vector in complex Euclidean space, we can define that the square of the component gives us the probability and this is Born rule.

[@周品宏-o7w]

11:04 The broken down states needs to be orthogonal to eachother. So the proof only works for infinite dimensional Hilbert spaces, right?

[@rv706]

It would be very interesting if you made a video on how to interpret *probabilities in the Many Worlds* interpretation of QM. That is one of the (few) nontrivial criticisms that MWI has to grapple with.

[@Achrononmaster]

There is no such theory. No one can define a measure for Many Worlds. Every attempt to do so requires additional ad hoc assumptions, such as what counts as a branch. This is never going to be empirical science imho, it is pure philosophy... unless you can say how to detect branching and thus experimentally explore the branching process. But who is to even say that is even a discrete phenomenon? No one seems to even consider continuum branching, although no one can rule it out, not without some new ad hoc postulates.

[@CyberMongoose]

I don't think it's fair to say that the Born rule is tacked on. While it's true that there are proofs showing that it is the only consistent way to assign probabilities in QM. It's only the Born rule that tells you that there should be probabilities in QM. Historically it was thought that the wavefunction of an electron represented a charge density and it wasn't until Born's work that the statistical interpretation was developed.

[@FermionPhysics]

That’s what “tacked on” means

[@CyberMongoose]

Not at all. Something that is tacked on is the 4th postulate of QM that follows necessarily from the others. The Born rule is literally what gives QM meaning.

[@FermionPhysics]

@@CyberMongoose again, that’s what “tacked on” means. Quantum mechanics derived from whatever quantization technique used is just a mathematical wave equation. The born rule needs to be artificially tacked on in order for it to make predictions.

[@NuclearCraftMod]

Agreed. As someone else mentioned, Gleason’s theorem is essentially the best known way to cut down the necessary details of the postulated relationship between amplitudes and probabilities.

[@Achrononmaster]

@@FermionPhysics but CyberMongoose makes a good point. Non-contextuality provides the Born Rule via Gleason's theorem. So yeah, it is not "tacked on." What is tacked on is the postulate of non-contextual observables. MWI does not solve this conundrum --- why is QM non-contextual?

[@岡安一壽-g2y]

I support Born's probability interpretation. I also support the idea of Schrödinger's cat paradox. The Copenhagen interpretation was created to reconcile Einstein's idea of wave packets with Born's probability interpretation, but I am not satisfied with it. If we abandon the idea of wave packets, there is nothing strange.

[@chiepah2]

It seems I really need to learn the maths behind quantum mechanics, since I sort of intuited the Born rule from your previous video, although I suspect that was more an aspect of your teaching style than my own intuition. I'm just glad I was able to follow where you were leading with my rudimentary understanding. 95% of what I know about quantum mechanics is from youtube videos, and probably around 50% from this channel.

[@misterlau5246]

If you know linear algebra you have more than 50% of the maths you need for quantum. And it's not too difficult. Something really horrendous is Navier-Stokes for turbulent fluid regime like for airplanes. Non linear, only aprox. with numerical methods. 🥺🤬 . Quantum is linear.. 🥰🤓

[@theempireofhand]

a quick question, do you believe in Copenhagen interpretation? Also can you make a vid on the collapse models, I understand those but not enough

[@jezer8325]

Yess. I’ve been waiting for this :)

[@GeoffryGifari]

hmmm correct me if i'm wrong here, but here it is explained where we can get the form of the born rule (assuming QM works like linear algebra), but we need quantum mechanics to already be probabilistic?

[@LookingGlassUniverse]

Yup! I should have said it but here I’m assuming measurement collapse etc is true, but you don’t necessarily know the value of the probabilities

[@GeoffryGifari]

@@LookingGlassUniverse i think the the problem with born rule being "add on" has something to do with measurement and quantum system evolution (schrodinger eq) apparently being separate (which leads to collapse being proposed etc), i believe open quantum systems guys are working on this

[@docopoper]

The step that's still missing for me is going from the Schrodinger equation vectors being possible to treat as length 1 to calling that probability. It seems more reasonable to call it proportion instead of probability. Like, this electron is X% spin up and Y% spin down.

[@enterprisesoftwarearchitect]

Probabilities are just proportions that add to 1, so there’s nothing wrong with that.

[@pukpukkrolik]

Whatever computes like a proportion that adds to 1, mathematicians can call a probability. Frequencies and beliefs can (arguably) both be described as probabilities, despite feeling dissimilar enough that it can clash with out intuitions and feel like an abuse of language. Still, the math is the same and correct.

[@ericvilas]

If you have an electron that's 50% up and 50% down, what that means is that if you measure its spin you will get up 50% of the time and down 50% of the time. You have no way of knowing beforehand what spin you're gonna get, so that 50% represents the probability of you getting that result if you do the measurement.

[@Alexandra-l2v4z]

Hello!! I’m from Belarus, have a very important, as I consider it), message . I have much more clear interpretation of quantum mechanics, made by Igor Bayak. I’m Sasha Bayak( and actually the daughter of this writer)) You are doing a great content! I have finished the 10 grade and just love to watch your videos, and repeat the experiments))). I will really appreciate you reading this article ( I can send you the link, because I can’t put it here)

[@grumpytroll6918]

So this assumes that the sum of squares have to add to 1. Ie normalized. So if you assume this normalization this is the only consistent way to assign probabilities that depend on the amplitudes… it feels almost circular.

[@mc4444]

Oh congrats, so cool of Brilliant to sponsor :)

[@mrcc9589]

Overall good explanation. Thanks. BUT: In your last video you explained why branch counting isn't the solution and though in this video you use branch counting to prove the born rule. If |A> has an amplitude sqrt(1/3) and |Z> has an amplitude sqrt(2/3), where |Z> can be rewritten (as you have shown) as |Z> = sqrt(1/3) |B> + sqrt(1/3) |C>, then this means that there are two branches in which |B> and |C> occurs, whereas there is only one branch where |A> occurs, which is exactly what branch counting means. Two make this always possible (even with irrational numbers as you pointed out), we have to assume that branches are uncountable in an absolute sense and infinitly divisible. This becomes clear by including decoherence. If we have ( sqrt(1/3) |A> + sqrt(1/3) |B> + sqrt(1/3) |C> ) |E> |M> where |E> is the state of the environment and |M> is the state of the measurement apparatus, then decoherence leads to sqrt(1/3) |A> |E_A> |M_A> + sqrt(1/3) |B> |E_B> |M_B> + sqrt(1/3) |C> |E_C> |M_C> so that the measurement apparatus will have a probability of 1/3+1/3=2/3 to be in branch B or C (and measuring |M_B> or |M_C>), which is the same as to say, that there are "more" branches for Z (consisting of B and C) as there are for A, which in some sense is "branch counting". Maybe it would be more precise to talk about "branch thickness".

[@vacuumdiagrams652]

It's possible, yes, provided you assume a measurement rule that's isomorphic to it. If it is at all possible without it, nobody has been able to prove it, and there are excellent conceptual reasons why it shouldn't be possible.

[@Achrononmaster]

Good comment. It is something so many youtube physics geeks miss. If QM postulates from any textbook did not have the Born Rule they'd still need one other postulate in some new context --- and it would be equivalent to the BR in the given context.

[@TheDavishers]

Hi, I'm really confused about which playlist to watch first - I've finished the LA refresher course - superb - so which series should I watch next? Ideally a single playlist with all your videos in order would be awesome though :D

[@saifuddinkayal98]

Please make a video on mechanics

[@damo87araimo]

Awesome, thanks.

[@duggydo]

You worked for Brilliant? That’s brilliant! 😂 It’s fitting though. You really are brilliant. 😉 👍🏻

[@GeoffryGifari]

oh and btw mithuna, about measurement collapse and interaction vs measurement: how much does a quantum system need to interact with a probe for a measurement to be done and the system "collapses"? can a probe "kind of" interact with a quantum system and not causing an entire collapse? maybe becoming partially unitary still? i've heard about a branch of quantum physics called weak measurement that might fit into this channel's theme of quantum interpretation

[@PhilMoskowitz]

Before the Born Rule (most) physicist viewed the Schrodinger equation as describing the properties of matter as waves, not particles.

[@Achrononmaster]

They were confused though. They probably had not absorbed Einstein's photoelectric effect paper of 1905 pre-dating Schrödinger, which first showed the phenomenology is all particle, no waves.

[@yodaqwq]

What if the probility amplitude are complex?

[@FermionPhysics]

When probability amplitude is squared (to get probability) the complex conjugate is used making the result real. For a number z=a+bi complex conjugate makes it z=a-bi

[@blueckaym]

I have trouble following the logical sequence ... first we assign coefficients dressed in square roots, then undress them by squaring them, and magically we get the numbers we put there as probabilities ... wasn't the goal to measure these probabilities rather than assume them? Also I've heard many times that squaring the coefficients gives you the probability, but I don't think that anyone explained why. Perhaps a more real-world case would clarify? In any case the quantum Randomness is a concept I struggle in the last years. Most quantum physicists insist that the randomness isn't result from our ignorance (in the purely technical meaning of not-knowing enough about something), and is actually what Reality is down there. I completely agree with the statistical approach of quantum mechanics, and given the lack of data, and the difficulties to observe quantum particles precisely it's completely logical (and smart!) to start by trying to establish what are the probabilities for this & that. But just because we don't know something it doesn't mean that it's inherently random, or that it turns in a cloud of randomness when we don't look. Many physicists claim that randomness of Reality is a fact, but so far I haven't found single experiment that shows it. If you know of it, please let me know!

[@HT-rq5pi]

I think this "symmetry" argument works when the coefficients are real. What if one of the coefficients are complex e.g. 1/sqrt(2) * (a + i*b). in this case the states 'a" and "b" are not symmetric, yet there doesn't seem to be a way to argue that the two states must have equal probability by appealing to symmetry.

[@whatitis4872]

I think it works so long as |A|=|B| for a state v such that v=A + B. If Bob gives you v = cA + dB where c,d are unit complex nimbers, rename A=cA and B=dB and youre good if the size of c,d are not the same then youre in the second part of her argument with unequal coefficients and you pull the same trick.

[@RazorBaze]

Hi) That's very interesting.. gotta think about that more

[@stoncjoesperanto8632]

It seems circular thinking when you say Psi=(1/√3)|A>+(1/√3)|B>+(1/√3)|C> because (1/√3) seems cherry picked. Say, if the Rule was P=|a| instead of P=|a|^2, then we'd have Psi=(1/3)|A>+(1/3)|B>+(1/3)|C>

[@BooBaddyBig]

I *think* that something similar happens in electromagnetism where the intensity of the wave is the square of the E or M component. IMO that's probably the Born rule disguised because the probability of the photon(s) of any given frequency and the intensity of the E-M wave is the same thing- the more photons you have the more intense the E-M wave is as it is consisting of billions of photons on average. At least I *think* so.

[@Achrononmaster]

Nope. You are talking about the Poynting vector and EM energy density, which are dimensioned quantities, not a probability. It's silly to take mathematical analogies literally. The mathematics is not the physics. Also, the EM theory is a field theory, not a theory of photons, so you are confusing the principle of complementarity. You cannot treat the phenomena as both wave and particle _at the same time,_ for any situation you can only use one or the other. Intensity of the EM wave depends also upon wavelength, so is not the same as probability density for photon quanta.

[@anthonynault]

Are the more corse grained measurements really physically equivalent to the other style of measurement? The two different ways of doing the measurements still seem to yield radically different branching structures (number of branches) and I can’t help but think that has got to affect our credences. If you just go back to simple spin up vs spin down case, on many worlds why should I ever be surprised to see spin down if its coefficient is 10^-27? There will be two copies of me and one will with certainty observe the (vanishingly improbable) spin down outcome. Should I be surprised because on a counterfactual scenario if I had done the measurement in a more corse grained way that way would have yielded relatively many more copies of “me” that saw spin up? I guess (if I understand right) that explanation is not satisfying. I’m not interested in why I would be surprised in the counterfactual scenario. I want to know why I should be surprised in the actual case that I just measure spin up vs spin down and MW says there are just two branches in that case.

[@adelelopez1246]

The coarse-graining into 2 parts can be done explicitly by tensoring the original Hilbert space with a Hilbert space containing a single qubit (which can even be in another galaxy, if you want to be sure this doesn't affect anything about the original set-up). Then the joint amplitudes will have normalized amplitudes of sqrt(1/2) times the original amplitude, because the normalization happens according to the inner-product of the Hilbert space, which is just the Euclidean one. You can extend this to any amount of course-graining by using qudits, which can have an arbitrary number of quantum states. It's important to remember branching structure isn't "real" at this level, it only becomes meaningful when we have observers living in a quantum universe and want to describe what parts of the wavefunction they can still see.

[@Achrononmaster]

You can never "do the experiment" in "two different ways". Each experiment only measures one set of complementary observables at a time. MWI is likely baloney btw. You can get a bias towards spin-up by altering the magnetic field in the typical Stern-Gerlach apparatus. The measurement apparatus is setting the boundary on all possible spacetime cobordisms, so it is a global notion (Lagrangian or SOH view), it cannot be branching sequentially in time. MWI is pure speculative metaphysics mumbo jumbo (imho). It tells a story, nothing more. There are far more parsimonious ways of accounting for the types of cobordism we observe in nature, quite naturally, e.g., from general relativity with closed timeliek curves on the Planck scale. (If I get a vote for beauty, QM=(GR+CTCs) is far more beautiful than the grotesqueness of MWI.)

[@erinm9445]

So is the implication here that any measurement you would take could always be split into multiple finer-grained measurements with equal probabilities? That doesn't seem right. To take your electron orbital example, given that electron excitement levels are quantum, with discreet steps, what finer measure could there be of electron orbital level could there be? Or does it mean that you can add on measuring the probability of X orbital level, plus spin up vs spin down? That seems like it would break the rules. But even if it doesn't, it doesn't help with the case where you want to break the measurement into 3 finer equal-probability measurements, or 7, or 1,000,001.

[@lepidoptera9337]

It means that she doesn't understand quantum mechanics. A measurement in physics is an irreversible energy transfer. In spectroscopy that energy transfer goes from the atom to the free electromagnetic field and then later we are detecting the energy in the electromagnetic field with a spectrometer. The actual experiment is a scattering process from a light source (at infinity) to a bunch of atoms to an angle (momentum), polarization (angular momentum) and energy sensitive detector that we call "spectrometer". It's actually worse than that because in case of atoms we even have to throw a local magnetic and electric field in to distinguish the symmetries of atomic states. And if you remember that there is also a nucleus in an atom, that also has a magnetic moment, then you also need an RF field to distinguish nuclear spin states and the resulting spin-spin coupling. Ouch. This is all well documented in atomic physics textbooks, which no KZbinr who calls him or herself "physicist" ever seems to remember. What you are getting here is half-information and wrong information. You are never getting correct physics.

[@xjuhox]

How it makes sense to invent a wave function _without_ the Born rule? Also, it's meaningless to invent an equation _without_ understanding the function that it represent! No wonder that students are lost in their QM class...

[@Miguel_Noether]

What other topics are you working on now?

[@victorrielly4588]

As you state at the end of the video one of the assumptions used to prove the Born rule is a state can always be broken down into equal probability states. For instance, a superposition of 2 unequal energy states could actually be a superposition of 3 equal energy states in disguise. It is clear in all cases where a state can be broken down into an equal probability state, the Born rule must apply. Suppose however there are some states that cannot physically ever be broken down into equal probability states. Consider for instance the situation where the state is a superposition of 2 unequal states, and neither state can ever be physically subdivided due to for instance their quantum nature. Would it be possible to observe a deviation from Born’s rule in this case?

[@Achrononmaster]

The Born Rule should still hold. It is a simplification for pedagogical purposes only to assume the states are equi-probable, hence is a tad misleading. The Born Rule follows from generalizing Kolmogorov probabilities to the next simplest non-distributive rule case (non-contextuality). The BR can be derived for any combination of states in general from General Relativity with closed timelike curves (assumed on the Planck scale, so as in "spacetime foam" models). Whether that means QM=(GR+CTCs) is however only a conjecture at this point in time. But it does yield the Born Rule, and is more satisfying (imho) than Mithuna's presentation.

[@absolute___zero]

did she say "quantum mechanics is random" ? that is not true. Randomness is a complex function, there is no way the Universe would implement randomness considering the amount of resources this algorithm requires. It is that we see it sort of random because we are macroscopic, but wave function is not random, it is is fully deterministic.

[@anthonynault]

Another point worth stressing... In these kind of derivations of the Born Rule initially there is great emphasis on the uncontroversial case where the coefficients are equal. My feeling is that this really is a bait and switch that plays with our intuitions. If a prize is behind one of two doors and the same number is on each door, sure you’re going to give equal credence to the prize being behind either door. That’s the “uncontroversial” case. But (all else being equal) you would have done that even if the numbers on the doors were different! What makes the probability uncontroversial is the number of doors, not whatever number happens to have been painted on the doors. The reason we believe the probabilities are equal in the “uncontroversial” case is because on MW there are two outcomes and they both happen. The equal coefficients were never doing any work at all in our reasoning! You would have rationally come to that conclusion even if you had no idea what the coefficients were. The bait and switch is pumping our intuitions to think that coefficients have a chance at playing a relevant role in our credences at all. I think MW defenders need to show in the first step that if we assume we don’t know what the coefficients are (they are hidden from us) then we _must be absolutely clueless_ about what credences to assign. But on the assumption of MW that’s deeply implausible since we know a copy of “me” will experience each branch. MW defenders argue "naive" branch counting leads to inconsistencies but I haven’t seen a convincing proof of that yet :) But even if we grant that assumption I think MW still has work to do in justifying why the coefficients should play a role in our reasoning at all and I don't think the "uncontroversial" case does it without begging the question.

[@NuclearCraftMod]

Maybe I'm not following, but derivations of the Born rule are independent of which interpretation of quantum mechanics is subscribed to. The issue underlying many worlds is that we don't experience or see any evidence for the existence of the other "branches". We don't necessarily need to insist on the physical "reality" of other outcomes to get around wavefunction collapse... you could just keep the part of the state which didn't correspond to the measurement outcome, and you'll find it just evolves according to the Schrodinger equation independently of the rest.

[@anthonynault]

​@@NuclearCraftMod The Born rule is not _derived_ on other theories of QM, it's usually stated as its own standalone axiom of the theory. We take it to be an empirical fact that probabilities are given by the Born Rule. But that is the challenge for Many Worlds. It stands apart from other interpretations in that measurements do not have a unique outcome. Every possible measurement result is with certainty going to be observed on some resulting branch. So it has the unique challenge of explaining why we see the measurement frequencies or "chances" that we do in the real world. Because other theories have unique outcomes (collapse or hidden variables) they can use the Born rule. But because every outcome happens on MW the contention is that we can confidently rule it out by experiment. MW makes wildly wrong predictions about what frequencies we should expect to see. Hence the need for defenders of MW to try to find a way to justify assigning credences according to the Born rule. (The issue isn't that we don't observe the other branches or worlds. It's that MW makes really wrong predictions about what we should expect to see within our own world)

[@macronencer]

@@anthonynault "But that is the challenge for Many Worlds. It stands apart from other interpretations in that measurements do not have a unique outcome. Every possible measurement result is with certainty going to be observed on some resulting branch. So it has the unique challenge of explaining why we see the measurement frequencies or "chances" that we do in the real world." People keep on saying this sort of thing and I still can't understand it. You jump from "every possible measurement is with certainty observed..." to "why do we see chances?" as if we were some sort of multiverse-straddling superbeings who could *see* every outcome. But we can't, obviously! The whole point is that our minds are part of the branching. That's where the probabilities come from. This seems completely obvious and easy to me, and I struggle to understand your position. This may be because I don't know enough, of course - I'm an amateur, not a real physicist - but I yearn for a clear "layman's explanation" of why this is so controversial, and so far I haven't seen one.

[@NuclearCraftMod]

@@anthonynault I should have been clearer. When I say "derived", I mean through arguments like the one in this video which show that the postulate can be quite a bit weaker and still imply the exact Born rule. Many worlds doesn't have the issue you say it does. On any given branch, the measurement result was unique, and the probabilities tell you the likelihood of being on the branch that you now are.

[@anthonynault]

​@@macronencer I think the disagreement is fundamentally about what the predictions of MW really are and if those are plausibly consistent with what we observe in the lab. By experiment we know the Born Rule must be satisfied. But MW predicts every outcome happens so at least at face value it seems we have to give equal probability to each branch. The coefficients seem irrelevant (so the argument goes). "Chance" only seems sensible in a world were measurements have just one outcome. Sure we are not super beings that can see directly every outcome in the MW multiverse. But if the world just evolves according to the Schrodinger equation then we can just use the laws of physics to extrapolate that those other branches exist and see what the equations tell us about observers on those branches. From that point of view I think it's plain you can get probabilities from MW. Self locating uncertainty makes sense. But when you look at the resulting branch structure the actual frequencies of measurement outcomes deviate wildly from the Born Rule. i.e. it seems the overwhelming majority of observers in any experiment should expect to see statistics that are radically different from the Born Rule.

[@michaelwerkov3438]

Please don't let the success of your latest popsci video convince you to make less of these harder video

[@Tititototo]

Hello, you started with a linear combination that is already normalized, implying that the Born rule has already been applied. Kind of circular logic.

[@whatitis4872]

I think that Mithuna skipped a bunch of steps here. Her presentation was still suggestive but I think some steps were skipped. One example is that the states that one ends up cutting the other states into-- I believe they have to be orthogonal. If v = #(1/3)A+#(2/3)B and I say that B =#(1/2)(C+D) for everything to work out the states C, D have to be orthogonal and of size 1. By the way I made a video regarding Mithunas "measurement without collapse video" and one of the Susskind lectures on the double slit and tying them together. kzbin.info/www/bejne/pZ-qkqFqq7R8mMU I think that video of hers was very inspirational, and so is this one.

[@Kraflyn]

but all eigenstates are mutually orthogonal, aren't they?

[@LookingGlassUniverse]

Great question! What you're doing in these "fine grade" measurements is moving to a higher dimension space. That's how the eigenstates are still mutually orthogonal

[@Kraflyn]

@@LookingGlassUniverse Bur if Z, B and C are all mutually orthogonal, then they are linearly independent and none can be written as a combination of the other two.

[@NoBorschtForYou]

I emailed you a while back about doing an interview for a YT video; I was wondering if you had received it 🙂

[@edwinagnew6800]

I'm a little confused how you're always allowed to add more fine-graining to end up with equal amplitudes. Isn't a fundmental part of the discreteness of quantum mechanics that can't fine grain forever? For example, if you're working with qubits then you can only have |0> or |1> (or some linear combination). So how you fine-grain a case like sqrt(1/3)|0> + sqrt(2/3)|1> to make the amplitudes equal?

[@whatitis4872]

I have the same feeling, I think youre allowed to fine grain until you run out of subspaces to break up into. In 2d though I think the subspaces dont need to be orthogonal so that saves you there. I think 2d is special here.

[@steelmagnum]

What do you do in a scenario where Z can't be broken up into a number of substates that cancels out the numerator of the coefficient? ie. Sqrt(3/5) Z, but Z can only have 2 possible states?

[@NuclearCraftMod]

Then just use sqrt(3/10).

[@ericvilas]

You're not assuming anything about Z, other than the fact that the full spectrum of possible results of your experiment is fully determined by the coefficients, and not by anything else. if you have a scenario where the coefficients are sqrt(3/5) and sqrt(2/5) because those are the True Values, and you have a different scenario where the coefficients are sqrt(3/5) and sqrt(2/5) because you can't properly distinguish all the 5 Actual True Values, those two scenarios are described by identical result possibilities, which means you will get the same result in one scenario as in the other, because all that matters is the wavefunction. So you can use the results from the scenario that's easier to analyze to get results in the scenario that's harder to figure out

[@jakubtvrdy4934]

But why is the probability equal to absolute value being squared? Why squared......? I can get real values from complex numbers just by taking the absolute value. It seems to me that no one understand this, and it seems to me that the answer is just: because we made it that way. Can you please explain why it is SQUARED absolute value of complex number? My quantum mechanics 1 exam will be on Thursday and I am kinda hopeless about it, because I hate to memorize something without actually knowing why it like that. If you answer and help me, thank you very much for that answer...

[@lepidoptera9337]

Because it isn't like that. It's much, much more general. Why? Quantum mechanics is an ensemble theory, just like probability theory. Both are based on Kolmogorov's axioms. One of those axioms says that the probability of P(x) + P(x̄) = 1, where x is an arbitrary set of events (outcomes) and x̄ is the complement of x (i.e. all the events that are not in x). The union of x and x̄ is the set of all possible events and we require that P(x or x̄)=1, which is given by the above formula. This equation can be satisfied with either p + (1-p) = 1, which is what we have in ordinary probability theory or it can be satisfied with something like sin^2(x)+cos^2(x)=1. The latter form derives from a linear scalar product of two vectors. It represents a circle, which is formed by vectors of length 1. In quantum mechanics this vector is called a "normalized wave function". It can be multiplied with arbitrary complex numbers if we define the scalar product as the multiplication of a complex number with the complex conjugate of another complex number. Scalar products are linear, so we can integrate over the simple complex product using an arbitrary complex function and voila... there is the Hilbert Space! We can even mix and match the scalar product form and the linear p +(1-p) form and then we get an event space that is a mix of quantum mechanics and ordinary probability theory. That's called the "density matrix" in physics. One step above that is second quantization where we replace the functions with linear operators, because even that satisfies Kolmogorov! So now we have quantum field theory. Integrate one more time over a (microscopic) length s and you have... string theory! Integrate twice and you have branes. Sum all possible such terms over a graph and you get loop quantum gravity. Needless to say, you can do all of this pointwise on top of an arbitrary Riemannian manifold and construct the god of all theories that nobody will ever be able to solve. But you know what? At the end of the day a hundred thousand theoretical physicists have simply been playing with a bunch of possible implementations of statistically independent events that an experimental physicist calls a "click" in one of his detectors. Because THAT is all the physics that's behind all of this. These are simply the possible descriptions of Geiger-Mueller-counter clicks. ;-)

[@TheWyrdSmythe]

It doesn’t feel as if you’ve proved the Born rule so much as illustrated several cases of figuring probabilities given the starting odds. To me the Born rule is just another example of Pythagoras. When a “measurement” occurs the current state vector has a projection onto the basis axes comprising the measurement interaction. The state vector has length one, and the projections form sides to its hypotenuse. Pythagoras requires the squares of the sides to give us the (square of the) hypotenuse. QM state vectors use complex numbers, so we need the square modulus, and there’s the Born rule. But trying to make sense of the Born rule in the context of the MWI is a can of worms.

[@lepidoptera9337]

In other words, you don't know what it means and where it comes from. ;-)

[@TheWyrdSmythe]

@@lepidoptera9337 That would depend on what you're referring to as "it".

[@lepidoptera9337]

@@TheWyrdSmythe The Born Rule. ;-)

[@TheWyrdSmythe]

@@lepidoptera9337 Reread my first comment. I explained _exactly_ what the Born rule is and where it comes from.

[@lepidoptera9337]

@@TheWyrdSmythe Yes, you gave us a random string of words that means nothing. ;-)

[@En_theo]

Quick question : if we try the double slit experiment but with two sources of photons (we assume that they're emitted at the same moment), will we see a difference in the interference pattern ? If not, why ?

[@Achrononmaster]

If the sources emit the same wavelength then the interference pattern is unchanged. The double slit experiment is a single photon interference pattern. Each photon interferes with itself. The pattern builds up from the collective, since each photon only registers a discrete pixel, but it is not the collective that are interfering. The wave motion is an illusion (it is epistemological, not ontological).

[@En_theo]

@@Achrononmaster And why does QM consider that a photon can interact only with itself ? It's kinda contradictory in the principle. Before any measurement (in this case, the screen behind the slits), QM consider that the photon is in an undetermined state. And yet, it does have a behavior that implies an interaction (the addition of the "peaks" of the wave). Is there any explanation behind or is it just an observation and we simply don't know why it's like that ?

[@mrcc9589]

@@En_theo This is just slappy wording in much of the educational videos, which just want to point out the strange behaviour of single photons. "It seems" like one photon interacts with itself, because we assume that there is only one photon in the system, as we measure only one. A more precise description is, that the wave function leads to interference and the modulus square of the amplitude just yields the probability to measure the photon in a certain state.

[@En_theo]

@@mrcc9589 Well, then if the interference pattern is not linked to any real interference, then one should wonder why the final result is an interference pattern.

[@mrcc9589]

@@En_theo This is one of the reasons why so many are not happy with the Copenhagen Interpretation and prefer Bohmian Mechanics or MWI.

[@vadimuha]

What's the point at going extra step and taking modulus squared if squaring alone is kinda making the same result?

[@volbla]

Because the coefficient might be a complex number.

[@patrickmeehan6856]

What if when the 2/3 Z is a superposition of 1/7 B and 2/7 C? Wouldn't this make the probability of A, B, and C not all 1/3?

[@rv706]

She didn't say it works for every choice of B and C

[@mrcc9589]

Splitting |Z> into sqrt(1/7) |B> + sqrt(2/7) |C> would be of no use. Moreover it would be even wrong, as 1/7+2/7 is unequal 1. The hole point of dividing |Z> into further states is to come up with states for A, B and C, which all have equal probabilities, to show how branching can explain the "experience" of differing probabilities in the MWI.

[@nothingspecial8261]

My problem with many worlds and quantum mechanics in general isn't philosophical or mathematical. I just don't believe it fits the observation. We can see what an electron is when it hits the screen, or what an alpha particle is when uranium decays in a cloud chamber. A particle. A tiny piece of something. Modern quantum theory is all about trying to make reality fit the math. We should be making theory fit reality, not warping our perspectives to fit math that doesn't make "physical" sense but very accurately describes the behavior of systems. The physics community needs to come to terms with the fact the electron IS a persistent particle with a definite momentum and position. The question should be WHY do these particles behave according to wavefunctions, not whether or not they're really particles.

[@NuclearCraftMod]

The question of whether particles have a definite position and momentum, which quantum mechanics clearly predicts is impossible, is different to the question of what the correct interpretation of quantum mechanics is.

[@nothingspecial8261]

@@NuclearCraftMod That's the problem with quantum mechanics though- it predicts that particles can't have a definite momentum and position when undetected. But they do. Just because the math we use to predict the system appears to indicate that the particles don't have a position and momentum when unobserved, and the math works every time, doesn't mean we should be attempting to reinterpret physical facts which are clearly observable. It simply means we're missing something big. And physics wasn't ready to handle suddenly not knowing anything about the universe when only fifty years earlier they thought they were almost finished with the whole understanding reality business. Thus, the Copenhagen interpretation. Collapse of "the wavefunction" is ridiculous. It's like treating the parabola that describes the arc of a ball flying through space as a physical object simply because it perfectly describes the path of every ball flying through space. Extending that interpretation of the wavefunction as a physical object is what has given rise to all of these different competing interpretations trying to mold a perception of reality to fit the math, and stopped science from finding the theory of quantum gravity. Don't tell me the math is weird. Tell me why the particle, which when we observe it is always in fact a particle, moves according to a wavefunction.

[@NuclearCraftMod]

​@@nothingspecial8261 Again, it has has nothing to do with wavefunction collapse or any other property of any interpretation of QM. It's just a fact about the position and momentum bases that they're related by a Fourier transform, which immediately gives rise to an uncertainty principle. We can experimentally verify the correctness of this with double slit experiments, or even just the stability of atomic orbitals in the case of angular position and momentum. To say that they really do have a definite position and momentum is just in contradiction to QM - double slit experiments would not yield interference patterns and atomic orbitals would not be stable. We know it is correct because of its ability to make accurate predictions. It is simply not a fact that we know particles actually have definite positions and momenta when not observed, as explained above. The uncertainty principle is not a property of measurement, but instead a property of the definitions of position and momentum. As a side note, none of this is what makes quantum gravity hard. What makes quantum gravity hard is that we have no immediately available physical systems that might exhibit its features, and so instead it's a theoretical battle to try to discover possible ideas based on the other physics we know.

[@nothingspecial8261]

And yet if the particles didn't have a definite momentum and position they couldn't resolve themselves as a single particle when detected. What's your explanation for kzbin.info/www/bejne/m2TJmmmrrt6le9U ? Those particles appear to move on a trajectory as though they are real persistent objects with a real position and velocity. Are you saying those particles only come into existence because the cloud chamber lets you see them? That if they were invisible they wouldn't actually exist except as a set of probabilities, but because they leave a trail they "become" particles? Show me an experiment that has actually verified the electron wavefunction as being a physical object. There are already experiments that can demonstrate how a physical particle can move according to a wavefunction. This kzbin.info/www/bejne/pJ6mYaydp5Vrqqc is a great example. The single electron double slit experiment can easily be reproduced with this sort of set up. Edit: Also, It's impossible to verify that the particles have a definite position and momentum when unobserved, because in order to find their position or momentum we have to observe them. The point is that every time we observe them, the resolve as a single particle, as though they really did have a definite position and momentum all along. To say that it's more plausible that these particles simply don't exist as particles until they're observed than that there's some other mechanism governing quantum movement that we don't understand is preposterous.

[@NuclearCraftMod]

​@@nothingspecial8261 1. Being able to resolve the position or momentum of a particle to some precision defined by the properties of a detector isn't evidence that either the position or momentum are definite. If you calculate the experimental imprecision in the position and momenta of cloud chamber observations, you will find that their product is way, way above the minimum given by the uncertainty principle. 2. The wavefunction (or quantum state) is unobservable in QM. What is observable are the observables of the quantum system, represented as Hermitian operators. The Schrodinger equation tells us how the quantum state evolves and Born's rule tells us how the probabilities of measurements depend on what the quantum state is. 3. Pilot-wave theory still has the uncertainty principle. Non-relativistically, it produces exactly the same predictions as any other QM interpretation. You can still not measure either the position or momentum of a particle with certainty, the difference being that the uncertainty principle is built into the fact that the initial conditions of the evolution of any particle is unknown. You also can't avoid the fact that the pilot wave is still unobservable, so you haven't actually solved the issue of removing unobservable quantities from the theory. 4. When did I say they don't exist as particles while not being measured? QM just tells us is that a particle doesn't have the properties that we thought it did classically.

[@MrNopperz]

Where is the proof that the Born rule takes the square of the probability amplitude?

[@Jabber_Wock]

Can states be broken into substates arbitrarily? Does each state not have to be an eigenstate?

[@LookingGlassUniverse]

Mathematically what these proofs do is go to a higher dimensional space. That’s why these states can still be orthogonal even though there’s more of them

[@Jabber_Wock]

@@LookingGlassUniverse thanks!

[@Jabber_Wock]

@@LookingGlassUniverse Thinking about this some more: each time a dimension is added, I can see that it will always be possible to find an orthonormal basis in the new dimension. But are we assured that there will always be at least one eigenvector in this new dimension? The new decomposition still needs to be made up of a linear combination of eigenvectors.

[@narfwhals7843]

Do you even need A1 and A2 to be different? Can't we just say A=1/2A+1/2A?

[@johnrichardson7629]

In these cases, you are taking the square root of x and squaring it to get x. How is this not trivial?

[@whatitis4872]

If you do the proof right you'll find its not trivial. Granted Mithuna did not get to the hardest, hairiest parts of Zureks treatment of this, but her explanation is a light beer version of Zureks that is nicely paced enough that I was able to get the proof. If v = #{2/3}A + #{1/3}B is your state write as #{1/3}( #{2}A + B) now find two vectors C, D orthogonal to B and to each other, and that both have the same norm as B and such that C+D = #{2}A. Then our state is v= #{1/3}( #{2}A + B) = #{1/3}( C+D+ B) these are all orthogonal and same size and therefore have equal probability of 1/3. By this I mean that the projections of v onto C or onto B or onto D all have the same size and so the properties associated with the subspaces spanC, span B, spanD should be equally probable, and since theres 3 space here the prob should be 1/3. Kapish?

[@extraterrestrialcontent]

I love watching smart people talk and pretending I understand what they’re talking about

[@enterprisesoftwarearchitect]

The redundancy/degenerate states you are introducing are used in gauge theory and Lagrangians too, but many people consider it an issue instead of a clarification- introducing more hidden parameters (states in this case) that are irrelevant or can’t be detected and asserting by fiat that they represent the same physical state. Or not?

[@KitagumaIgen]

If you make the comparison with electronically excited states in atoms you have the situation that some states are degenerate when there's no magnetic field - however when the B-field is cranked up the Zeeman-effect shifts the energy-levels apart. So: at least some states that are degenerate in some conditions represent clearly different physical states.

[@enterprisesoftwarearchitect]

@@KitagumaIgen hmm good point

[@NuclearCraftMod]

Even in that context though, it is possible to first fix the gauge and then consider only physically distinct states.

[@a.hardin620]

Gleason’s theorem is the only real derivation. That says nothing though about which interpretation is correct or likely correct. The rest of these that you discuss are either true for some select cases but not universal or circular/question begging or just wrong.

[@NuclearCraftMod]

Definitely think Gleason's theorem is worth a mention :)

[@finalfan321]

maybe first explain what the Born rule is. i was jsut thinking of the movies

[@mlerma54]

The way I have seen probability introduced in MWI is by interpreting it in a frequentist way, i.e., by looking at what happens when we repeat the same experiment many times. The technical quality of the following video is not good but the explanation of why the wave collapse postulate is unnecessary and the Born rule can just be derived is excellent (the interesting parts start at 40 min 25 sec and 59 min 10 sec respectively) - kzbin.info/www/bejne/aZPUp3R_gK1liMk

[@lepidoptera9337]

MWI can't tell you anything about the Born rule. It doesn't have a physics model for the measurement apparatus, which is what the Born rule models. That quantum mechanics is "frequentist" is obvious already in the Copenhagen interpretation. What else would it be?

[@mlerma54]

@@lepidoptera9337 The point is that Born's rule does not need to be introduced as a postulate since it can be derived from the other postulates of QM - Mithuna does it by using Laplace's interpretation of probability (number of favorable cases divided by total number of cases), but that requires to work with cases that must be equiprobable by symmetry considerations. The link I posted above provides a more general derivation in which equiprobability of cases is not required. About the model of a measurement device, we don't need it to be essentially different of any other physical system, it is just a system that react in a particular way to an interaction (it "flashes" in a way if the spin is up, and in a different way if it is down), but other that that it is a physical system as any other - see the video I posted the link to above at 56 min 32 sec. Mithuna makes the point that the requirement of a system to work as measurement device and cause an apparent "collapse" of the wave function does not require it to be macroscopic, even a single particle can do it - kzbin.info/www/bejne/rnPPoYJ9mdBgZrc

[@lepidoptera9337]

​@@mlerma54 Of course we need the Born rule or some equivalent formulation. Every measurement depends on two systems: the system under measurement and the system that does the measurement. In non-relativistic quantum mechanics the Schroedinger Equation only models the system under measurement. It CAN NOT incorporate the actual measurement system dynamics. That's why you absolutely need a second formula. The people who think otherwise can't even count the number of systems involved in these scenarios. In relativistic field theory there is no measurement system to begin with, but we are changing the ontology of what "a measurement is" considerable (and to the better). I am not even sure the critics of the Born rule understand any of the underlying physics.

[@GarryBurgess]

Tell me that this haunted you for a week or more, thinking about it all day long and doing no end of thinking about it, or else I might feel dumb.

[@timpreece945]

Does this argument work irrespective of one's interpretation of quantum mechanics? Does many worlds ( or bohemian mechanics) simple mean that one doesn't need to postulate wave function collapse?

[@ericvilas]

It assumes that when you do an experiment, you obtain a single measurement result. The interpretation of that can be anything from "copenhagen collapse probabilities" to "probability that you end up in a specific branch of the multiverse" to "probability that the pilot wave happens to be in that configuration given your lack of knowledge"

[@Jkauppa]

do you enjoy statistics

[@Jkauppa]

just pure statistics, nothing more, nothing less

[@RalphDratman]

Cool, very cool.

[@ranga386fe]

uncontraversial spelt that way is controversial..

[@ashishpatel350]

a comment for the algo. =P

[@misterlau5246]

I'm watching this video after the new ones... You are writing in Dirac notation so my contribution is useless. You are not a noob in quantum and everything 🧐😐

[@jim3977]

Clearly you have done PhD and still don’t understand Born rule.

[@jim3977]

Without Born rule, how do you connect amplitude of waveform to the probability of finding the particle? Clearly, you are among the ignorant crowd who don’t understand the central principle of physics. How do you intérprete the physical world? Born rule gives us the answer.

[@AndreyLomakin]

I you make such bold clame you need to elaborate one. Otherwise it worth nothing.

[@jim3977]

@@AndreyLomakin Hello, without Born rule, the probability of measurement doesn’t make any sense. Quantum probability is only sensible when Born rule conjectures a correspondence between probability amplitude and actual measurement possibilities.

[@AndreyLomakin]

@@jim3977 Seems like you are not completely right, according to the articles listed in description. But I do agree that provided explanation in reality still uses the Born rule in the sense that it calculates probability based on amount of states with equal amplitude. So it can not be the prove that Born rule is not needed.

[@Jim6593]

@@AndreyLomakin Give me the actual argument

[@marhawk6468]

😍

[@jamescollier3]

Why are e sad? 🙃

[@LookingGlassUniverse]

Electrons are just so negative

[@ernestomamedaliev4253]

@@LookingGlassUniverse i have not liked very much the vid, but this comment worth it 100% 😂😂

[@g_leuenberger]

So the origin of the Born rule is just the “normalization thing”, which she didn’t explain further. Therefore the video is useless.

[@whatitis4872]

I dont think her video is useless. So long as it gets you thinking its done something. Also its not just a normalization thing. If v=A+B+C and |A|=|B|=|C| and A,B,C mutually perp, then I think you get something like the Born rule but that needs a normalizing factor. Also the languge people typically use for the Born rule I think is bad. Not general enough and depends on an operator and arbitrary choice of eigenvalues. The born rule I like is: Given a set of mutually orthogonal subspaces E_i that sum to the whole hilbert space, and a vector v we assign prob(E_i, v) = / Where the P_{E_i} are otrthogonal projection operators onto E_i and < , > is the Hilbert space inner product.

[@bnouadam]

Garbage

[@markxxx21]

You explain things so well. If you are looking for ideas, I have one. With the current trend toward switch to electric, there is a lot of news of lithium batteries catching fire. Yet, I have not come across a good video explaining "why"? And what alternatives to lithium like sodium (salt battery) or magnesium would be good. I know that "basic" physics than your usual videos, but I think you'd do an excellent job, weeding through the misinformation.

[@spiritualevolvedlightbeing8597]

What do you think if I say that : Quantum Physics are very limited if humans try to understand how reality really works and when a say reality im talking about all phenomenon that happens in our reality the physical energy particles waves and the metaphysical paranormal supernatural events!i mean you can understand and explain something so big and amazing like how galaxies are born and evolve, But with the dame Quantum Physics you can not understand or explain paranormal phenomenon that happen in our own planet for example can you understand the mechanics of a ghost or a demon? Why!? Because Quantum physics as the concept says are about the energy, subatomic particles that made the matter in our 3D physical everyday reality but those laws don't apply for the metaphysical 4D reality phenomenon energies, that's why I say that in order to understand both dimensions of Reality the 3d physical and 4D metaphysical humans needs to use (Hyperdimensional Quantum Physics) What do you think!?