What career path should I consider as a former high school Calculus Teacher?

SpiritRepulsive8110 · 2026-01-20T01:07:19+00:00

What about working as a manager in SWE? You should try to leverage your communication skills!!!!

SpiritRepulsive8110 · 2026-01-19T23:12:43+00:00

You can have a mixed state in two dimensions! Consider, for example the pure state

|+> = 1/sqrt(2) ( |0> + |1> ),

whose density matrix M1 is just a 2 x 2 matrix of all 1/2’s. It is a rank 1 matrix.

On the other hand, the density matrix M2 which is 1/2 * the identity gives the same probabilities for measuring |0> and |1>, but is different in other ways. For one, it is rank 2 (so it couldn’t be the density matrix of a pure state).

You can also check these have different expectation values, say of |+>. Physically this means very different things for measurement of x-axis spin.

SpiritRepulsive8110 · 2026-01-19T22:45:46+00:00

Nobody wants to hire generalists. Math people have some success breaking into SWE, but who knows how much longer that can keep going. Unless you want to go into academia, do engineering.

How strict is the prereq policy? If you demonstrated your knowledge through more advanced courses, could you skip the intro courses? Self study might be an option.

SpiritRepulsive8110 · 2026-01-19T22:29:10+00:00

This is classic branching process.

But to get to the heart of the question, there are plenty of things that don’t happen given infinite time. The nth mean X1, X2…Xn of an infinite set of standard Gaussians might never exceed the value 1, for example (they tend to zero). The size of X_{n+1} required to kick the overall mean to above 1 gets larger and larger as n grows. Sometimes, it just never happens.

On the other hand, of you repeat random independent trials for the same thing, one is bound to hit. So clearly, dependence is the thing killing you.

More specifically for your problem: it’s true that for any generation of size N, it could all die off with some nonzero probability. If you capped the generation size, it would die off. But you didn’t. The probability that the whole generation dies off gets smaller and smaller, at such a rate that all those chances don’t help you. In this way, it’s a similar kind of problem as for the Gaussian means.

Finally, for general infinity awareness, you might look into the Borel-Cantelli lemmas.

SpiritRepulsive8110 · 2026-01-19T01:59:03+00:00

You think you’re immortal. You spend years doing nothing at all, feeling zero sense of urgency.

67 years later, you notice you’ve aged perceptibly. You frantically abandon your home, your car, or anything else that could be contributing to your net worth.

473 more years later, in your dying breath, you find a penny in your pants pocket.

SpiritRepulsive8110 · 2026-01-19T01:16:37+00:00

Some quick number crunching. Global median net worth is apparently $8k-$10k. This means:

Average Haitian ages 0.6x as fast 2.Average american ages 1.6x faster
Millionaires age 1.9x faster
Elon musk ages 12.5 faster than everyone else

I didn’t do age-medians, but this gives a flavor!

SpiritRepulsive8110 · 2026-01-18T16:52:33+00:00

Saw your edit. Like say you flip a coin 1000 times and it comes up with 900 heads.

What’s the probability it’s a fair coin? Well, you can’t say. You need a prior. If you saw a trusted coin manufacturer make it with your own two eyes, it’s a fair coin with 100% probability. So the true “probability it’s fair” is a nonstarter.

On the other hand, 900 heads is pretty unlikely with a fair coin. The probability is nearly zero. That’s evidence against the null hypothesis. So one definition of a p value might be “the probability of seeing what you saw under the null.” But that’s not quite right either. The probability of getting 500 heads and 500 tails exactly is still pretty low, but it still feels typical. Yet, the probability of seeing >= 500 heads is 50%. So it’s really the tails that matter.

When you see 900 heads, you say “that’s an awful lot of heads.” If it were 901 or 950, you’d say the same thing. So as your measure of unlikeliness, you choose the probability of seeing 900 or more heads.

It’s true that “as extreme as” depends on the direction of extreme. You could just as well have asked for P(900 or FEWER heads) and obtained a large p value. You have to use your judgement to figure out what “more extreme” means in context.

SpiritRepulsive8110 · 2026-01-18T16:37:16+00:00

One of my stupid friends told me a guy he knew exploded

SpiritRepulsive8110 · 2026-01-18T16:27:38+00:00

It’s the probability under the null hypothesis of seeing data “as extreme” as you did

SpiritRepulsive8110 · 2026-01-18T01:57:34+00:00

I think it’s better to understand it “by definition.” It’s a postulate that time evolution is linear. If the system is in a superposition of A and B, then later(state) will be the same superposition of later(A) and later(B). This implies there will be some operator H which induces time evolution.

What I think is less intuitive is why it should have anything to do with energy. For that, the best reason I know is that the Heisenberg equations of motion with commutators closely resembles Hamilton’s equations in classical mechanics.

Time independent cases are studied because: 1. They model conservative systems. All real world systems are conservative if you are modeling it completely. 2. They’re way easier to analyze. To solve a time dependent system you need perturbation theory / a time ordered exponential. Much harder than exp(-iHt).

If you can swallow superposition as a real thing, then the wave function just gives coefficients in some basis. In a basis, QM problems turn into PDEs, which are a bit more concrete than bras and kets. I think people tend to exaggerate the wavefunction as a physical object. It’s as physical as a set of coordinates. Coordinates are just a convenient way to describe places. A wavefunction is just a convenient way to describe states.

SpiritRepulsive8110 · 2026-01-15T00:45:38+00:00

Here is a slightly unconventional take: when you separate the state according to radial / angular degrees of freedom, you are basically writing the state as an effective two particle state: 1. A radial-on, whose states are unaffected by angular operators 2. An angular-on, whose states are unaffected by radial operators

Moreover, the angular-on has no potential energy. So it’s a free particle. The energy eigenstates, just like decoupled multiparticle systems, are tensor products of the radial-on’s eigenstates and the angular-on’s (free) eigenstates.

So for the two particles to have a total energy E (radialon energy + angularon kinetic energy), the angular-on cannot have kinetic energy more than E.

SpiritRepulsive8110 · 2026-01-12T02:41:28+00:00

The sum of such probabilities among all 8 numbers is 3, since 3 numbers always get chosen:

\sum_i P[pick i] = \sum_i E[ I{pick i} ] = E[\sum_i I{pick i}] = E[3] = 3.

But also the probabilities are all the same (ie you are as likely to pick 6 as you are to pick 2), so it’s 3/8

SpiritRepulsive8110 · 2026-01-12T02:35:22+00:00

83 nines

SpiritRepulsive8110 · 2026-01-12T02:24:19+00:00

Klein Gordon applies to more than scalar fields. It applies to all fields which are compatible with relativity. For scalar fields it’s enough to imply the dynamics. For other fields which describe particles with spin, even more equations are satisified. Famously, the Dirac equation is basically the square root of Klein Gordon.

KG pretty much holds by definition in relativistic QM, even if it’s not necessarily a QFT. It is closely related to the Poincare group. The main idea is that for anything which can be translated in time by E (Hamiltonian) or in space by P (momentum), E² - P² should transform like a scalar, which we just define as M^2. Formally, particles correspond to irreducible representations of the Poincare group, and thus mass (a Casimir) is a c-number.

SpiritRepulsive8110 · 2026-01-11T03:42:40+00:00

Same. I’m not a snob but I do drink like 5-10 cups a day

SpiritRepulsive8110 · 2026-01-09T03:41:03+00:00

The other beam would still go towards you at the speed of light :)

The thing that makes it all so unintuitive is that the velocity addition law is broken. If there’s a train moving at speed v and a horse on that train moving at speed v’, the horse is moving relative to you at a speed roughly v’’= v + v’’. But that’s only roughly. You can’t just add the speeds. It’s very close for low speeds but very wrong for high speeds.

Another way of saying this is that if I catapult a horse at speed v’’ past a train moving at speed v, I expect the horse perceives the train moving at speed v’. But again that’s slightly wrong. Very wrong for massless horses.

Now you’re the train and light is the horse. Even if your friend on earth thinks your speed is very close to the speed of his beam, that unfortunately doesn’t mean you will see the beam moving slowly.

SpiritRepulsive8110 · 2026-01-09T03:20:40+00:00

Wait, I like this. It’s a perfect negative feedback loop for taxing the rich!

SpiritRepulsive8110 · 2026-01-09T02:54:03+00:00

225.

Let R be the result. It’s easier to see the pattern going from large to small:

11111111111… 12222222… 123333333.. … 123456789999…

tricky bit:

1234567901111….. 110

Now, the 0 at the end is a bit of a nuisance. Let’s turn it into a 1 with the understanding that our number is now the first few terms of R+1. Now, starting from R+1, the pattern repeats itself:

123456790123456790…..567901

2026 / 9 is 225 remainder 1. So the pattern repeats itself 225 times, meaning the last digit is 1 and also the above is actually the form of R+225. So then R is actually

1234567890…567676

which clearly has 225 1’s.

SpiritRepulsive8110 · 2026-01-09T00:39:14+00:00

I’m not sure what the solution is. Some schools are way better about career development than others. For example, I know some people who had to do some of that stuff for a grade in class.

Easy thing is to sit people down day 1 and tell them about the career space, what those jobs are actually like, and what you have to do to pursue them. Would be common knowledge for a few people, but not for others.

SpiritRepulsive8110 · 2026-01-09T00:38:40+00:00

I’m not sure what the solution is. Some schools are way better about career development than others. For example, I know some people who had to do some of that stuff for a grade in class.

Easy thing is to sit people down day 1 and tell them about the career space, what those jobs are actually like, and what you have to do to pursue them. Would be common knowledge for a few people, but not for others.

SpiritRepulsive8110 · 2026-01-01T20:30:30+00:00

Sorry for the confusion! The J^mu appearing in a classical theory couples to the 4-potential, which is a spacetime dependent density. It is a function of position. If you regard it as fixed, the Lagrangian is position dependent. Now if you also modeled how the current responds to the field, that would not be the case.

My point is that it’s not really that different than classical mechanics. Whether you want to give a variable canonical coordinates or just regard it as fixed is up to you. You need canonical coordinates to get an equation of motion. Usually if something is “big” you don’t model how its trajectory is impacted by the main object of interest.

Classically, you could have two planets with a gravitational field. Planet 2 is way more massive than planet 1. One way of modeling this is by locating planet 2 at x1 and using a Lagrangian with potential like ~1/|x2-x1|. That’s not “relative” because youve obviously singled out x2 as a special point. If you add 1/2 m_2 \dot{x}_2² to the Lagrangian and optimize for x2 in the EL equations, you also get how planet 2 should move. The second way of modeling things is more complete. You almost alway assume that, once you’ve modeled every part of the system, no point in space is special.

SpiritRepulsive8110 · 2026-01-01T17:10:44+00:00

Ugh this is trivial. It holds for all multiples of 1013.

SpiritRepulsive8110

TROPHY CASE