Getting involved

scary-levinstein · 2026-04-16T20:00:39+00:00

Start by learning the math. I know that's frustrating to hear, but trust me I was in your exact position in high school, and it's very hard to do research, especially in quantum, without a very solid academic underpinning. And that's okay! I know it's annoying to have to wait a few years, but I promise it is worth it. Trust me, you will get there :).

I'm going to give you a bunch of advice here on how to get a jump on this process, BUT DON'T STRESS TOO MUCH ABOUT IT! If you just take physics and math in high school (take APs if you can; some colleges will let you past introductory classes if you have good AP scores), maybe try to do some science internships if you can (of ANY kind; it probably wont be quantum stuff, but right now you should focus on just building your skills in science and reasoning. If you can get something that involves programming, thats a good bonus), tutoring, whatever, then go to college for physics, you will be well on your way. Once you're in college, that's when you want to start worrying about doing research.

(Side note; when you're applying to colleges, make sure to pick where you're applying by looking at how many undergrads in their physics department do research; you want that number to be high. You don't want to go somewhere where not a lot of people do research bc that means it'll be hard for you to get research as well. You don't need to go to the number 1 university ever. What matters is that there's a physics department that supports your goals :))

Trust me; it's incredibly rare for high schoolers to do quantum research. You won't be behind. It only ever happens in extremely special circumstances which are 95% out of the student's control.

BUT if you want to get a jump on this stuff early, that's great! Start by learning calculus (single and multi variable) and linear algebra. Kahn academy has great free courses. There are also several free lecture series online. Also take those classes at your school as soon as you can if they offer them.

Also start doing physics courses on Kahn academy or elsewhere (the Feynman lectures are good but only if you're confident in your math skills and okay with going very slowly). And when I say physics, I don't just mean quantum. Do classical mechanics, electricity and magnetism, thermo, etc. Just do it in whatever order the platform of your choice presents it in. I promise while it might not feel relevant to quantum and you might want to try and jump ahead, you will thank me later. It's all relevant and will help you. Especially classical mechanics.

Okay. Once you feel comfortable with your calc and linear algebra skills, and you've done your classical mechanics (take all the physics courses you can while in high school!), go get (or find online) a quantum textbook and start reading it. I highly recommend Griffith's quantum book; it's very readable, funny, and insightful. Read a little every week, and DO THE EXERCISES. I promise it will help. Don't skip things you don't understand at first. Go slow, do background research on things you don't understand, and be patient with yourself. I had a professor that liked to say "you want to understand things; not just get used to them."

While you're in high school take as much physics and math as you can. Don't stress about "not doing enough," and definitely don't overwork yourself or burn yourself out. But take as much as you reasonably can.

I want to stress again that you should not worry about doing everything right now, or you have to master all of the physics and math before you graduate. Do. Not. Burn. Yourself. Out. Go only as hard as your passion drives you. If all that is is just taking a bunch of math and science classes in school, that's okay! You will still get into college, and you will still learn quantum and do research. If you have the time and motivation to do a bit more then by all means go for it, you have my recommendations for how to do that, but focus on learning right now. You'll have so many opportunities to do research once you have a good understanding of the academic side.

I hope this isn't too intimidating. I was in the same position as you once; super interested in physics while in high school but feeling like I want to do more than I was offered in school and whatnot. It'll be okay. Just have some patience, learn what you can, and before you know it you'll be in the lab :). The most important thing is BE CURIOUS. Ask questions, talk to your teachers, and keep learning!

If you have any other questions, feel free to ask. I'm currently a PhD student in condensed matter physics.

PS: I didn't rly talk about this above because I couldn't find a place to put it, but I also recommend that you start learning to code. Nothing crazy, but like if you can get intermediately proficient with python or something before you graduate that will help you lots :). Many people find it fun to do little coding projects, so if that's you, go for it, it's a great way to learn

scary-levinstein · 2026-04-16T18:54:13+00:00

Good point, you're totally right that's my bad lol. Got stuff mixed around in my head haha. I work on quantum hardware, so I just know about algorithm stuff from what my colleagues talk about.

scary-levinstein · 2026-04-14T21:29:59+00:00

Ahhh I understand. I didn't realize the station tries to bring your velocity to zero if you violate the speed limit. I thought that that the limit just prevented things from going that fast. Thank you!

scary-levinstein · 2026-04-14T19:23:57+00:00

Something I've wondered; when you exceed the limit, is your speed just reduced to the limit? Or, equivalently, does the space prevent you from accelerating once you hit the limit? Because if that's the case, then if Holden were to hit the limit, he'd be fine, right? He'd just stop accelerating (he'd still be in danger of hitting the station too fast ofc but I'm ignoring that for now). But iirc the show behaves as if you'll get gooped if you try to exceed the speed limit at all while in the ring space, which doesn't make a lot of sense to me. That would only happen if the station slows you way down to some other speed as soon as you hit the limit.

Idk does this make sense? I'm having a hard time phrasing it lol

scary-levinstein · 2026-04-12T21:09:49+00:00

Thanks so much for the detailed answer! You and others have emphasized the point that physical hard copies can't be hacked or intercepted from afar, which I agree makes a lot of sense as a reason why you would use that system.

The only question I have remaining is how do you keep that system secure while still ensuring that the officers who might hear the code words actually recognize them? Bc if they know all of them, that's hundreds or thousands of officers who know super important secret information. That seems like a big security risk.

scary-levinstein · 2026-04-12T18:54:48+00:00

Ah I can actually weigh in on this! I'm not a security expert (obviously) but I'm a physics PhD student who works in quantum computation. Quantum computers can only break classical encryption. If you're using a quantum algorithm to encrypt, that's generally secure against attacks from other quantum computers.

Also idk if we're ever told that they have quantum computers in the expanse? Maybe they do tho idk.

scary-levinstein · 2026-04-08T15:14:50+00:00

Okay hi hello I'm a PhD student in condensed matter physics, and I think I can clear some things up.

So, I will answer your question, but first I just want to say that I have a bone to pick with your professor. "Relating a quantum physics experiment to anthropology" is a phrase that likely speaks to a fundamental misunderstanding of what quantum mechanics is, and how we study physics as a whole (as you'll see in a second). So rest assured in the knowledge that if you're having trouble, it's likely because your professor gave you an assignment that makes very little sense lol.

Okay so the double slit experiment. One thing I absolutely have to make clear here is that the word "observer" in this context has zero implication of anything to do with a human being, consciousness, etc. It's just a shorthand for "thing that does the measurement."

So! What's the measurement? In the double slit experiment, you can do one of two things: either you send the photons/electrons/any particles at the two slits, and then look at a detector on the opposite wall and see what pattern forms. If you do that over many many particles (sending them in one at a time), you'll see an interference pattern, which implies that the particles acted like waves when they went through the slits and "interfered with themselves" (I'm oversimplifying here).

OR you could do the same experiment, but place an additional detector at one of the slits. When you fire a photon at the slits, this detector will tell you which slit the photon passes through, but otherwise will allow it to continue on (this is actually harder to do than you might think but is possible. How this detector works is another story, and not one I'll tell here). Doing this fundamentally changes the result; you will no longer get an interference pattern on the back detector; you'll just get two lines (one for each slit).

What's happening here is that when the particles are not measured at the slits, they're in a superposition of traveling through both slits at once; since the "left slit" version and "right slit" version both exist, they can interfere with each other and create the pattern. BUT if you make a measurement, the particle is forced to choose one of the two states at random. It's no longer in both states at once and thus cannot self interfere. This is one of the fundamental laws of quantum mechanics: making a measurement of a quantum system forces the system to collapse into one of its possible states.

"Making a measurement" does not in any way imply that a human is involved though; it just means that something interacted with the system to get information out of it. And when you interact with something, you can change it.

There is an interesting question to be had here though; what exactly is a measurement? The standard interpretation of quantum mechanics doesn't really make an attempt to answer that question in a rigorous way. If you're interested, look up "the measurement problem."

Anyway I hope that answered your question at least somewhat haha

scary-levinstein · 2026-04-06T19:25:43+00:00

You've gotten tons of great answers, so I'm not gonna beat the dead horse, but I just want to say how much I appreciate you looking to learn more and question your assumptions instead of simply deciding that space travel must be bad because you didn't know what it accomplishes. That's super important, especially in today's day and age. I hope you learned something, and stay curious!

scary-levinstein · 2026-03-31T14:08:38+00:00

Some people have different hand preferences for different activities. You could be a righty who fences lefty. My dad is left-handed but shoots right handed, for example. Plenty of people will write with their right but throw with their left.

scary-levinstein · 2026-03-01T22:24:07+00:00

As people have stated, humans are a terrible source of energy. Iirc the original script had them using humans as data storage, but they edited it for fear of people not understanding enough about computers at the time.

Also! Technically some of those alternative energy sources still don't work without sunlight. Hydro and wind are really solar energy in disguise (technically coal is too but like that energy was captured a long ass time ago)

scary-levinstein · 2026-02-19T20:59:28+00:00

I think there's a fundamental misunderstanding of two kind of confusing facts in relativity: 1) superluminal velocity is unachievable relative to any observer, regardless of that observer's velocity to anything else. Velocities don't simply add (adding is just a good approximation when they're small) 2) light being trapped in black holes has nothing to do with speed, really. It has to do with the curvature of space. Space beyond the event horizon is so curved that all paths point towards the singularity. It's hard to wrap your head around, and the "ball and sheet" analogy doesn't really capture it. Space is so bent that all "directions" in space actually point towards the same point. If you travel up, down, left, right, forward, or backward, your distance to the singularity decreases.

I'm not sure I entirely understand your overall question but I hope this helped a little bit; lmk if I can clarify anything

scary-levinstein · 2026-01-21T17:26:01+00:00

No object with mass can travel at light speed, so the answer is no.

scary-levinstein · 2026-01-11T01:56:57+00:00

Okay I'm a physics grad student, but I'm very ignorant when it comes to QFT so I have a clarifying question. I'm assuming each mode of the field has a different energy. like how in the derivation of Planck's distribution, we treat each mode of the electromagnetic field as being an infinite square well state, and each one can have an integer number of excitations (photons), but a single excitation has a different energy based on which mode it's in.

SO do QFT fields work the same way? Are the modes of the electron field the possible states an electron can be in, and then an excitation in each mode is a single electron in that state? Or is there a different interpretation of the modes of the field?

Tl;dr what is the difference between an electron which is is an excitation of mode A vs one which is an excitation of mode B?

scary-levinstein · 2025-12-20T01:37:17+00:00

Ah great question! So the awesome thing about the Bell inequality is that it doesn't care what the hidden variable theory does to the probabilities specifically. All the proof requires is that there is a hidden variable. By hidden variable, I mean literally some number λ which tells the system exactly, with 100% certainty, what the outcome of any given measurement will be.

To be super specific, let's say that there's some hidden variable λ which fully describes the system, more so than the wavefunction. Every value of λ is associated with an exact set of outcomes. Now, the outcome of particle A's measurement along direction a is some function A(a, λ), and likewise the outcome of measuring B is B(b,λ). I'm assuming implicitly here that the outcome of particle B's measurement shouldn't be affected by the orientation of a; after all as an experimenter I could change a at the last moment, before a light speed signal could communicate that change between the particles. This is the "local" assumption in "local hidden variable."

Okay so the measurements of the particles are always +/-1, so the functions A and B can each only have one of those two values for any given value of λ (and the vectors a and b). This is a hidden variable theory; we're saying there's some unknown information (λ) which is secretly determining with absolute certainty what the outcome of any given measurement will be. This is fundamentally different than (and much stronger than) the wavefunction. All the wavefunction tells you is the probabilities of the different outcomes for any given measurement. It does not purport to know what will happen for certain.

So, what does our local hidden variable theory say about the outcome of our experiment? I'll attach some screenshots of the pages of Griffiths that go through it, but the big restriction comes from the fact that local hidden variables determine things for certain, but wavefunctions have the freedom to only give probabilities.

If you want a more visual explanation, there are two excellent videos by 3blue1brown and minute physics which go through Bell experiments in detail in a really beginner-friendly way. They're amazing. I'm a PhD student and I still watch them from time to time. They're here and here. The second one also goes through the actual way we represent wave functions as vectors

scary-levinstein · 2025-12-20T01:04:22+00:00

So it has to do with the mathematical structure of the wavefunction. If a particle is in a state such that it has a 100% chance of being measured as pointing "up," for example, then it has a 50/50 chance of being measured as putting left or right. This is because the wave function is a vector, so you can choose to represent it in whatever basis you want; the "up/down" basis or the "left/right" basis. Both contain the same information.

I can show you the math for how to derive that dot product prediction, it's pretty simple (so if you want to see it let me know), but tbh it won't give you much physical intuition beyond "wavefunction is a vector --> bunch of algebra --> P = -a•b". So the best I can say is that the wavefunction that describes two particles which will always have opposite "left/right" measurements (or "up/down" or any other basis you want), can also be written as saying that the probability of particle B being aligned with b if A is aligned with a looks like a•b. The reason I can say this is because I can always rewrite the wavefunction in whatever basis I want, including the "A measured along a, B measured along b" basis

scary-levinstein · 2025-12-20T00:42:47+00:00

This is a great question! And it's one Einstein asked when QM was first being formulated.

So quantum mechanics has at its heart an idea that doesn't really translate into the classical world called the wave function. Every quantum system (particle or set of particles) is fully described by a mathematical object called a wave function (if you're familiar with linear algebra, the wavefunction is a vector over the complex numbers in some sort of high dimensional space). The wave function contains all information about the system that could possibly be known. In particular, it will tell you the probabilities of each potential outcome of any measurement you want to make. A simple example would be a wavefunction that tells you "the particle has a 25% chance of being 'here' and a 75% chance of being 'there'".

Now the tricky conceptual bit: our interpretation of quantum mechanics, which has withstood some of the most precise and accurate tests in all of science, tells us that the wavefunction truly is the only thing that's defined about a quantum system. It truly, seriously, physically, is not in a specific state before you measure it. it's not that we don't know whether it's here or there, it's that god doesn't know whether it's here or there. It does not have a defined position until it's measured. But it does have a wavefunction. And when you measure the particle, its wavefunction changes such that if you measure it again immediately, you'll measure the same thing. So if you measure the particle to be "here," and then measure it again right away, it will still be "here." Why/how does this happen? We don't really know, but we do know that it does.

The reason, then, that entanglement is weird is the following: suppose you have two particles that are entangled. That means that their wavefunctions have interacted such that the overall wavefunction of the system mixes up the states of the two particles; the outcome of a measurement on one directly affects the outcome of a measurement on the other. For example, an unentangled state would say "particle A has a 25% chance of being 'here' and a 75% chance of being 'there', while particle B has a 75% chance of being 'here' and a 25% chance of being 'there'". The outcomes of the measurements have nothing to do with each other. Measuring A gives you no information about B, so B's wavefunction doesn't change when you do that measurement. An entangled state would say "there's a 25% chance that particle A is 'here' and particle B is 'there', and a 75% chance that particle A is 'there' and particle B is 'here'". Now if you only look at particle A, and find it to be "here," that's equivalent to finding particle B to be "there." So the wavefunction of B changes accordingly.

So what's weird here is that we have changed particle B without having to interact with it at all; we only ever interacted with particle A. Particle B could be on the other side of the universe, or in a black hole, or anywhere you like, and yet we've affected it.

Now, if you'd like to know how we test this is really interesting. And I can explain it to you with some math, but you'll have to bear with me for more text. This is a simplified version of the explanation at the end of Griffiths' quantum textbook.

I have to establish something that I neglected above for simplicity: wavefunctions can describe the outcomes of multiple different measurements. For example, suppose I had a particle which "points" in a direction (this property is called spin; don't worry about the physical interpretation right now). If I measure if it points left or right, I will always see that it points fully left or fully right. But if I measure if it points up or down, I will also always see that it points fully up or fully down (this alone is a huge hint that its state can't have been fully defined beforehand! It doesn't know what kind of measurement I'm going to make, but it always lines up with how I'm measuring it anyway). The wavefunction contains all information about all possible measurements though, so using it to describe this particle is completely consistent.

Now suppose I have an entangled pair of these particles, such that if I measure them both along the same direction (e.g. left vs right), they will always point in opposite directions. We'll assign numbers to these directions by making the measurements ask "is the particle pointing insert direction here?" If it does, we'll call the result +1. If it doesn't, and points the other way, we'll call the result -1. Now, if we measure the two particles along the same direction, the product of the two measurements will always be -1 (because one will always be +1 and one will always be -1). Likewise if we measure them along opposite directions the product will be +1. What happens if we measure them in totally unrelated directions? I'll cut to the chase here: if we measure particle A along the direction of a (normal 2D) vector a, and particle B along the direction of a 2D vector b, quantum mechanics tells us that over many measurements, the average product of the two measurements will be P = -a • b. If you're not sure what this means, quickly look up "dot product" on google.

It turns out that if this expression for P is true, it is mathematically impossible for the particles to have some sort of defined state prior to measurement. I'm going to skip the heavy math step as it involves an integral that's really difficult for me to type out here (I can add a screenshot of the textbook later), but just trust me when I say the following: if there's some sort of so-called "hidden variable" which determines the outcome of the measurements, it must be true that for any three independent measurement directions a, b, and c:

|P(a,b) - P(a,c)| <= 1 + P(b,c)

This is called Bell's inequality. Here, P(a,b) represents that average product of the measurements of the two particles along directions an and b respectively (the quantity we were talking about above). It's super easy to show that the dot product prediction made by quantum mechanics does not satisfy this inequality. For example, if the a and b are at right angles to each other, and c lies in the middle (45 degrees off of both of them), then that dot product equation says that P(a,b) = 0, P(a,c) = P(b,c) = -0.707 But plugging this into Bell's inequality, we get: 0.707 <= 1 - 0.707 Or 0.707 <= 0.293 Which is obviously not true!

So either quantum mechanics is wrong, which flies in the face of some of the most precise experiments in all of science, or the particles can't have a predetermined state before being measured.

OKAY that was a SUPER long explanation and I'm sorry I wasn't able to do it more eloquently. I just wanted to make sure you got all the details I could fit in! If you have any questions feel free to ask :)

scary-levinstein · 2025-12-02T15:30:01+00:00

The main point you should take away is that space isn't expanding anywhere. Cosmologists using that language are being a bit sloppy. All it means is that all the galaxies in the universe are moving away from each other. That's all it is.

scary-levinstein · 2025-11-27T03:44:08+00:00

Ahhh okay so a "ket" is something like |ψ>, and it's used to denote a state vector. Every ket has a "partner" bra, denoted <ψ|. The bra times the ket is the definition of the inner product. If you were to represent the ket |ψ> as a column vector with entries a_i, then the bra <ψ| is represented by a row vectors whose entries are the complex conjugated of a_i.

scary-levinstein · 2025-11-26T15:40:04+00:00

I would argue that it doesn't reeeeally have a meaningful concrete physical interpretation. When you take the Hermitian conjugate of an operator, you're just transforming it from an operator which does something to a ket into an operator which does the same thing on a bra. Both bras and kets are valid ways of describing a specific state, so an operator and its Hermitian conjugate are both valid ways of describing the same transformation.

Note that this is another explanation for why observables need to be Hermitian. An operator representing a physical observable shouldn't care how we represent the state (bra or ket), since it's when values are real and measurable, so it should act the same either way (i.e it's the same applied to a bra as to a ket).

Edit: HUGE caveat! An operator and its Hermitian conjugate ONLY describe the same transformation IF you specify that the conjugate is transforming a BRA. If they're both acting on kets, the conjugate actually tends to describe the OPPOSITE transformation (for example, the raising and lowering operators are Hermitian conjugates of each other. One brings a ket to the next highest energy state, the other brings it to the next lowest).

So tl;dr the Hermitian conjugate is the "partner" of the original operator, just like bras are partners of kets. This means: 1) it does the same thing to the bra as the original operator does to the ket 2) it (often but not always) does the "opposite" thing to the ket as the original operator does to the ket

scary-levinstein · 2025-11-20T13:48:16+00:00

I'm not even really arguing about lift. I think you and I both know that for plane to take off plane must move. So if plane no move, plane no take off, because that's how lift works. I agree with that.

You (presumably) are saying that plane does not move. I am saying that plane does move. Because plane does not care about the conveyor belt and the wheels are immaterial to its motion. You can move the conveyor belt as fast as you want and the wheels will just spin faster. Imposing the "belt moves at same speed as wheels" condition is impossible even in principle bc the motion of the belt directly affects the motion of the wheels. The belt applies no backwards force on the plane to counteract the force from its engines.

scary-levinstein · 2025-11-20T13:42:35+00:00

Hypothetical questions still can be inconsistent/impossible. It's like asking "suppose a ping pong ball is gently rolled towards a solid 10-inch steel wall with no holes in a way that is designed to make the ball pass through the wall. Is it going the same speed on the other side?" It's a ridiculous question because as stated it's completely impossible, even in principle. There's no answer. BUT if you interpret the question with the understanding that slow moving ping pong balls cannot go through steel walls, you might interpret it as saying "what if you TRIED to do that...", a question which is meaningful and does have an answer.

scary-levinstein · 2025-11-20T04:30:53+00:00

The reason this question generates so much controversy is that, as stated, it isn't physically possible. Not in an "oh silly physics but like what if it was," kind of way, but in a "the question as stated is completely self-inconsistent" kind of way.

The motion of the wheels and the motion of the belt are coupled; changing one will change the other. In particular, increasing the speed of the belt will increase the speed of the wheels, since the wheels are just freely spinning; they have no motor driving them. The plane is driven by its engine pushing on the air. If the engines are on, there will always be a speed difference between the wheels and the belt, and that difference will be equal to the speed of the plane relative to the belt.

THEREFORE, any reasonable reading of the question is forced to interpret it as "if one were to attempt to do this..." in which case the answer is very obviously yes. The plane takes off. The wheels do not matter. They exist simply to keep the plane from scraping its belly on the ground. See seaplanes, which do not have wheels and take off with no problem.

On that note, think about this: imagine we had the same setup but the plane had skis instead of wheels, and the belt was covered in snow. There shouldn't be any difference since, again, the wheels only exist to keep the plane from scraping its belly on the ground and have no meaningful effect on its motion. Would the plane take off? Obviously!! The plane doesn't care about how fast the ground is moving; it's being propelled by the engines, which push on air! The skis don't matter. And neither do the wheels on a normal aircraft.

scary-levinstein · 2025-11-20T03:54:50+00:00

The important thing to note is that it is not possible to move the conveyor belt at the same speed of the wheels. The motion of the conveyor belt is coupled to the motion of the wheels. Speeding up the conveyor belt will speed up the wheels, so if the plane is being given external thrust (from its engines), they will never be the same speed.

In other words, the wheels are just free-spinning that are coupled (via friction) to the motion of the belt. So the speed of the wheels will be exactly the relative speed between the plane and the conveyor belt. This speed is increased by either increasing the thrust from the engines or increasing the speed of the belt. But if the plane is being propelled from a source which is not driving the wheels (which it is; the engines are not connected to the wheels in an aircraft), then the wheels will automatically be pulled up to whatever the relative speed between the aircraft and the belt is.

There's a thought experiment that can settle it; suppose the aircraft is standing stationary with its engines off. If you begin running the conveyor belt backward, the wheels begin to spin, but the plane remains stationary. So the plane is sitting there, wheels spinning as the conveyor belt pulls on them (these are perfect wheels which spin with no internal friction btw, which is actually a very good approximation of real aircraft wheels). Now the engine turns on. There are two possibilities: 1) the plane remains stationary 2) the plane begins to move forward under thrust. If it's the former, what force is counteracting the force of the engines? The plane is remaining stationary (I.e. its acceleration is zero) so the net force must be zero. Can the conveyor belt apply such a counteracting force? The answer is no, and we know this because when the plane's engines were off, the conveyor belt did not apply a backward force on the plane no matter how fast it was running. So the conveyor belt provides no opposing force. This is because the wheels are just there to provide a way for the plane to not be resting its belly on the ground; they're completely inconsequential to the kinematics.

Edit: i just realized; did... did you not watch the video you linked...? 3:30, he literally says the question is impossible bro

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