mwacksen

Karma: 82

mwacksen 5 Nov 2020 9:29 UTC
3 points
in reply to: Linch’s comment on: Covid-19 IFR from a Bayesian point of view
Thanks for the comments!

I agree with all of these points. In fact, with respect to (4) it is even plausible that some “once infected” people never go on to develop the kind of antibodies that are being tested for. Point (3) is why I control for age/sex, but of course there are a number of further complexities.

These further complexities, along with (1) and (2) are currently “un-modelable complexities” for me. There are just so many selection effects in play that it isn’t clear if you gain anything from trying to take them into account. Given that there are a number of papers that try to calculate some kind of IFR making a number of basic mistakes, I wanted to set out to see what the data + simple models gets if you do the maths correctly, as opposed to doing ridiculous things like caring about the median study like the Ioannadis meta-analysis does. After all, it seems like the IFR is what everyone cares about, so it would be nice if we were doing things “less wrong” here.

mwacksen 5 Nov 2020 11:56 UTC
14 points
on: Multiple Worlds, One Universal Wave Function
I’m very confused by the mathematical setup. Probably it’s because I’m a mathematician and not a physicist, so I don’t see things that would be clear for a physicists. My knowledge of quantum mechanics is very very basic, but nonzero. Here’s how I rewrote the setup part of your paper as I was going along, I hope I got everything right.

You have a system $S$ which is some (seperable, complex, etc..) Hilbert space. You also have an observer system O (which is also a Hilbert space). Elements of various Hilbert spaces are called “states”. Then you have the joint system $S \otimes O$ of which $Ψ$ is an element of, which comes with a (unitary) time-evolution $E : S \otimes O \to S \otimes O$ . Now if $S$ were not being observed, it would evolve by some (unitary) time-evolution $E_{S} : S \to S$ . We assume (though I think functional analysis gives this to use for free) that $(v_{i})_{i}$ is an orthonormal basis of eigenfunctions of $E_{S}$ , with eigenvalues $(λ_{i})_{i}$ .

Ok, now comes the trick: we assume that observation doesn’t change the system, i.e. that the $S$ -component of $E$ is $E_{S}$ . Wait, that doesn’t make sense! $E$ doesn’t have an “ $S$ -component”, something like an $S$ -component makes sense only for pure states, if you have mixed states then the idea breaks down. Ok, so we assume that $E$ , when acting on pure states, is equal to $E_{S}$ . So this would give $E : φ \otimes ψ \mapsto (E_{S} φ) \otimes ψ_{φ}$ , where $ψ_{φ}$ is defined so that this holds. Presumably something goes wrong if we do this, so we instead require the weaker $E : φ_{i} \otimes ψ \mapsto (E_{S} φ_{i}) \otimes ψ_{i}$ . And bingo! Since the $φ_{i}$ are eigenfunctions, we get that $E (φ_{i} \otimes ψ) = φ_{i} \otimes (λ_{i} ψ_{i})$ , and let’s redefine $ψ_{i}$ to include the $λ_{i}$ term because why not. Now, if we extend by linearity we get that $E : ϕ \otimes ψ \mapsto \sum_{i} a_{i} φ_{i} \otimes ψ_{i}$ . Applying $E$ again gives $\sum_{i} a_{i} φ_{i} \otimes (ψ_{i})_{i}$ , and the same for further powers.

Ok, let’s interpret that last part in terms of “observations”. If we take states of the combined system $S \otimes E$ , then time-evolution maps pure states with only a $v_{i}$ component to pure states with only a $v_{i}$ component. Wait, that’s exactly what we assumed, why should we be surprised? Well yeah, but if you started out with some linear combination of eigenfunctions, these will be mapped to a linear combination of pure states, and each pure state in this linear combination evolves as assumed, which may or may not be abig deal to you. In a mixed state that is a linear combination of pure states, we call each pure state a “separate observer” or something like this. Of course, mixed states in a tensor product state cannot be uniquely be written as a sum of pure states. However, if we take our preferred basis $(v_{i})_{i}$ and express our mixed states as pure states with respect to that basis in the $S$ -component, this again makes sense.

So it’s super important that we have already distinguished the eigenfunctions of $E_{S}$ at the start, we unfortunately don’t get them out “naturally”. But I guess we learn something about consistency, in the sense that “if eigenfunctions are important, then eigenfunctions are important”.

Ok, now assume our system $S$ is itself a tensor-product of $N$ subsystems $S_{1} \otimes \dots \otimes S_{N}$ , which we think of as “repeating a measurement”. Now what we get if we start with some pure-state is (in general) a mixed state which can be written as a linear combination of pure states of eigenfunctions. As the eigenfunctions of the different systems are different (they are elements of different spaces), if you start out with some non-eigenfunction in each subsystem, you’ll end up with some mixed state that contains different eigenfunctions for the different systems. The “derivation of the Born rule” doesn’t need this step with multiple systems. Basically, we can see this already with just one system. If we start with a non-eigenfunction $\sum_{i} a_{i} φ_{i}$ , then this gets mapped to some linear combination of pure states via the time-evolution. As the time-evolution is unitary, and the |a_i|^2 sum to 1, we can see that each pure state has “length” |a_i|^2.

Thanks for the great paper! I think I’ve finally understood the Everett interpretation.I think the basic point is that if you start by distinguishing your eigenfunctions, then you naturally get out distinguished eigenfunctions. Which is kind of disappointing, because the fact that eigenfunctions are so important is what I find weirdest about QM. I mean I could accept that the Schrödinger equation gives the evolution of the wave-function, but why care about its eigenfunctions so much?

mwacksen 5 Nov 2020 16:10 UTC
1 point
in reply to: paragonal’s comment on: Multiple Worlds, One Universal Wave Function
Right, but (before reading your post) I had assumed that the eigenvectors somehow “popped out” of the Everett interpretation. But it seems like they are built in from the start. Which is fine, it’s just deeply weird. So it’s kind of hard to say whether the Everett interpretation is more elegant. I mean in the Copenhagen interpretation, you say “measuring can only yield eigenvectors” and the Everett interpretation, you say “measuring can only yield eigenvectors and all measurements are done so the whole thing is still unitary”. But in the end even the Everett interpretation distinguishes “observers” somehow, I mean in the setup you describe there isn’t any reason why we can’t call the “state space” the observer space and the observer “the system being studied” and then write down the same system from the other point of view...

The “symmetric matrices<-> real eigenvectors” is of course important, this is essentially just the spectral theorem which tells us that real linear combinations of orthogonal projections are symmetric matrices (and vice versa).

Nowadays matrices are seen as “simple non-commutative objects”. I’m not sure if this was true when QM was being developed. But then again, I’m not really sure how linear QM “really” is. I mean all of this takes place on vectors with norm 1 (and the results are invariant under change of phase), and once we quotient out the norm, most of the linear structure is gone. I’m not sure what the correct way to think about the phase is. On one hand, it seems like a kind of “fake” unobservable variable and it should be permissible to quotient it out somehow. On the other hand, the complex-ness of the Schrödinger equation seems really important. But is this complexness a red herring? What goes wrong if we just take our “base states” as discrete objects and try to model QM as the evolution of probability distributions over ordered pairs of these states?

mwacksen 5 Nov 2020 23:51 UTC
1 point
in reply to: paragonal’s comment on: Multiple Worlds, One Universal Wave Function
Isn’t the whole point of the Everett interpretation that there is no decoherence? We have a Hilbert space for the system, and a Hilbert space for the observer, and a unitary evolution on the tensor product space of the system. With these postulates (and a few more), we can start with a pure state and end up with some mixed tensor in the product space, which we then interpret as being “multiple observers”, right? I mean this is how I read your paper.

We are surely not on the same page regarding decoherence, as I know almost nothing about it :)

The arxiv-link looks interesting, I should have a look at it.

mwacksen 6 Nov 2020 0:01 UTC
1 point
in reply to: paragonal’s comment on: Multiple Worlds, One Universal Wave Function
Ok, but OP of the post above starts with “Suppose we have a system S with eigenfunctions {φi}”, so I don’t see why (or how) they should depend on the observer. I’m not claiming these are just arbitrary functions. The point is that requiring the the time-evolution on pure states of the form $ψ \otimes φ_{i}$ to map to pure states of the same kind is arbitrary choice that distinguishes the eigenfunctions. Why can’t we chose any other orthonormal basis at this point, say some ONB $(w_{i})_{i}$ , and require that $w_{i} \otimes ψ \mapsto E_{S} w_{i} \otimes ψ_{i}$ , where $ψ_{i}$ is defined so that this makes sense and is unitary? (I guess this is what you mean with “diagonalization”, but I dislike the term because if we chose a non-eigenfunction orthonormal basis the construction still “works”, the representation just won’t be diagonal in the first component).

mwacksen 6 Nov 2020 0:15 UTC
1 point
in reply to: SymplecticMan’s comment on: Multiple Worlds, One Universal Wave Function
Yes, I know all of this, I’m a mathematician, just not one researching QM. The arxiv link looks interesting, but I have no time to read it right now. The question isn’t “why are eigenvectors of Hermitian operators interesting”, it is “why would we expect a system doing something as reasonable as evolving via the Schrödinger equation to do something as unreasonable as to suddenly collapse to one of its eigenfunctions”.

mwacksen 6 Nov 2020 10:07 UTC
3 points
in reply to: SymplecticMan’s comment on: Multiple Worlds, One Universal Wave Function
I think my question isn’t really well-defined. I guess it’s more along the lines of “is there some ‘natural seeming’ reasoning procedure that gets me QM ”.

And it’s even less well-defined as I have no clear understanding of what QM is, as all my attempts to learn it eventually run into problems where something just doesn’t make sense—not because I can’t follow the math, but because I can’t follow the interpretation.

If we accept that mutually exclusive states are represented by orthogonal vectors, and we want to distinguish mutually exclusive states of some interesting subsystem, then what’s unreasonable with defining a “measurement” as something that correlates our apparatus with the orthogonal states of the interesting subsystem, or at least as an ideal form of a measurement?

Yes, this makes sense, though “mutually exclusive state are represented by orthogonal vectors” is still really weird. I kind of get why Hermitian operators here makes sense, but then we apply the measurement and the system collapses to one of its eigenfunctions. Why?

mwacksen 6 Nov 2020 10:09 UTC
1 point
in reply to: SymplecticMan’s comment on: Multiple Worlds, One Universal Wave Function
We don’t lose unitarity just by choosing a different basis to represent the mixed states in the tensor-product space.

mwacksen 6 Nov 2020 20:03 UTC
1 point
in reply to: paragonal’s comment on: Multiple Worlds, One Universal Wave Function
Well yeah sure. But continuity is a much easier pill to swallow than “continuity only when you aren’t looking”.

mwacksen 6 Nov 2020 20:05 UTC
1 point
in reply to: SymplecticMan’s comment on: Multiple Worlds, One Universal Wave Function
Then I have misunderstood Everett’s proof of the Born rule. Because the tensor product structure seems absolutely crucial for this, as you just can’t get mixed states without a tensor product structure.

mwacksen 9 Feb 2021 17:09 UTC
2 points
on: Quadratic, not logarithmic
Say you meet in a group of $n$ people that all care about each other. Then, by your reasoning, each of the $n$ people is responsible for $n^{2}$ risk, so in fact (by double counting once more as in the original post), the total risk is $n^{3}$ . If however, we share the responsibility equally each person is responsible for $n$ risk which is intuitive. So this quadratic growth assumption is a bit questionable, I’d like to see it done more formally because my intuition says it is not complete nonsense, but it’s obviously not the whole truth.

I feel like this is almost too obvious to state, but question is really not about the marginal risk, but about the marginal benefit. Meeting 0 people probably is really bad your mental health in comparison to meeting 2. Meeting 98 people is probably not much worse for your mental health compared to meeting 100. Meeting 2 people might be more than twice as good as meeting one. But since we don’t just care about ourselves, we should also think about the other person’s benefit. So even if you are already meeting 98 people but the 99th meets nobody else the benefit you provide to that person may (depending on your age, preconditions, etc...) be worth more than the extra bit of risk.

mwacksen 9 Feb 2021 17:18 UTC
1 point
in reply to: J Mann’s comment on: Quadratic, not logarithmic

On the other hand, the other problem is that even if the person is accepting the risk for themselves, I’m not sure they’re processing the risk that somebody else gets seriously ill or dies.

Well maybe, but are you thinking of the fact that (trivially) P(you infect someone else) < P(you are infected)?

mwacksen 20 Aug 2021 18:13 UTC
1 point
in reply to: ADifferentAnonymous’s comment on: Quadratic, not logarithmic
Apologies for the really late reply, but I don’t think “marginal risk” in this context is well-defined. The marginal risk to yourself grows linearly in the number of people to first order. You could feel responsible for the marginal risk to all the other party goers, but they are people with their own agency, you aren’t (in my opinion) responsible for managing their risk.

mwacksen 20 Aug 2021 18:14 UTC
1 point
in reply to: cistran’s comment on: Quadratic, not logarithmic
Sorry for the (very) late reply, but I do not understand this comment and suspect maybe my point didn’t come accross clearly, cf. also my other reply to the comment below this one.

mwacksen 20 Aug 2021 18:18 UTC
1 point
in reply to: cistran’s comment on: Quadratic, not logarithmic
Sorry for the late reply. I’m assuming you need to be “infected” in order to infect someone else (define “infected” so that this is true). Since being infected is a neccessary precondition to infecting someone else,

P(you infect someone else) ⇐ P(you are infected),

and it’s clear you can replace “<=” by “<”.

This is basic probaility theory, I can’t follow your notation but suspect that you are using some different definition of “infected” and/or confusing probabilities with expected values..

mwacksen 24 Aug 2021 9:54 UTC
1 point
on: COVID/Delta advice I’m currently giving to friends

though Delta is a bigger risk than Alpha, ignoring effects of vaccination

I keep seeing claims like this get thrown around, but I feel like the evidence isn’t really there. Could you comment on why you think this?

mwacksen 25 Sep 2021 11:24 UTC
2 points
in reply to: MondSemmel’s comment on: Covid 9/23: There Is a War

I’m from Germany but haven’t bought any of the rapid tests myself, so I don’t have any first-hand experience with the situation. From what I understand, you can use these tests to test yourself, but they aren’t considered accurate enough to fulfill a condition of having to get a validated test for the bureaucracies.

I spent some time in Germany recently, and this is (or at least “was”, when I was there) wrong. The tests themselves are considered accurate enough, but the bureacracies usually don’t trust you to do it (or do it correctly). In some German states you can do a test at the restaurant to be let in. You can buy a test, take it to a pharmacy, and let the pharmacy do it and then validate it. In some German states it is ok for some jobs to test yourself regularly. Back then I think employers also had to provide take-home tests for free (or if not required, strongly encouraged).

mwacksen 4 Dec 2021 11:33 UTC
−4 points
in reply to: wunan’s comment on: Omicron Post #3
It’s the terminology you use to signal that you believe the vaccine is safe and effective™ and therefore cannot be fully “evaded”.

mwacksen 10 Dec 2021 20:13 UTC
5 points
on: COVID and the holidays

An average vaccinated 30yo now loses about 6 weeks of expected life from contracting COVID instead of 2 weeks, because of waning vaccine efficacy.

The counterfactual here is “you never get covid”, so I’d take this number with a large grain of salt. If, on the other hand, the counterfactual is “you get covid a few years later”, then the loss of expected life does not occur. Additionally, if you do get covid, you’re (probably) super immune for a while, which presumably increases your quality of life.

mwacksen 30 Dec 2021 14:39 UTC
7 points
in reply to: CuriousApe’s comment on: COVID Skepticism Isn’t About Science
Funnily enough, I have yet to read a single not-completely-ridiculous cost-benefit analysis that goes this way. We must live in different bubbles.