True. The em field is significantly more energetic, meaning that it contributes more to the stress energy tensor than the force of gravity
But recall that the force of gravity is the momentum change “due to” the curvature of spacetime. The fact that the force of gravity can bend spacetime at all is a really weird second-order perturbative effect.
These perturbative effects are typically described with a quantum field theory, but gravity has been thus far notably difficult to quantize.
Thank you!!
People keep looking at pitch, but there are cis women with deep, unquestionably feminine voices.
When I thought about it for a while, I think it’s all in the vowels: to make your voice more feminine, make sure your close vowels like your /y/‘s and /u/‘s have the same fundamental frequency as your open vowels like /ɒ/ and /ε/. I think this is what the article means by ‘resonance.’
Keep your voice box in the same ‘place’ in your throat as you switch vowels, and try to use only your tongue to shape them.
I can’t attest to this method too much since I’m not really out (and thus not using a ‘girl voice’), but when I tried to subtly make this one change to my voice (even at the same pitch), the first sentence out of my mouth made a cis coworker flinch. Take that how you will. I never really tried again after that, though.
There’s probably a lot more motor control going on than you would expect when you need to think (writing, fidgeting, etc.).
Your brain wants sugars, so when you run out of immediately-available glycogen to break down, you will want to eat more in order to keep thinking. Breaking down fats wont supply energy fast enough (in the short term) to keep complex thought running continuously.
This seems reasonable to me, though perhaps not perfect. If you don’t want your money going towards ai development at all… IDK. I don’t think they train their own models for what it’s worth, so I think you can avoid paying ai companies by simply not using the ai parts of kagi.
"I should really get back around to responding to that comment about time-evolution in quantum systems"
3 months ago
"hmm..."
Well, better late than never.
I understand quantum computing, though, and that evolution of a circuit is a unitary (linear) operator/matrix. So, wouldn’t continuous evolution be a one-parameter Lie subgroup of the unitary operators over your Hilbert space?
I am in quite the opposite situation; my experience is in the raw physics without much of the application to quantum computing. From what I understand, though, I think this is largely correct.
In general, observables (such as the Hamiltonian) are Hermitian (self-adjoint), which is neither a superset nor a subset of unitary operators. You are not, of course, restricted to only applying observables to your quantum state (In fact, you could apply any operator you want to your quantum state; the physical meaning behind most operators involves fundamentally changing the system, but it's not strictly forbidden to do this). We require observables to be represented by Hermitian operators because the eigenvalues of Hermitian matrices are always real (since the value you observe is an eigenvalue of the observables' matrix representation, you don't want any of the eigenvalues to be non-physical complex numbers).
I had to look up why specifically quantum circuits require unitary operators, and I found this Stack Exchange response, which describes how unitary operators are used to find the time-varying component of the wavefunction that solves Schrödinger's equation. I think we were kinda describing the same thing, ultimately: continuous evolution of a quantum system is dictated by the Hamiltonian (as shown by it's presence in Schrödinger's equation), and the time-varying component of the solution is unitary (the non-time varying component is a linear combination of eigenstates of the Hamiltonian whose associated eigenvalues need not be a root of unity).
Basically, when you say:
So, wouldn’t continuous evolution be a one-parameter Lie subgroup of the unitary operators over your Hilbert space?
... the answer is 'yes,' but the Hilbert space itself is defined by the eigenstates of the Hamiltonian, which itself could be changing in time,^†^ meaning that a complete description of the time evolution requires slightly more careful consideration.
In your example of a standing wave on a string, unitary operators would take you from one mode to another, but if the length of string is changing, those unitary operators are changing too.
^†^A time-varying Hamiltonian implies a time-varying energy in the system. This sounds like breaking energy conservation (and it kinda is), but it is used to describe any system that is being pumped from the outside.
Is this calculated by assuming the wavefunction is static?
Typically sorta? The way the Schrödinger equation is typically solved is by taking linear combinations of eigenfunctions (of the Hamiltonian) and making them time-dependent with a time-dependent phase out front.
The eigenfunctions are otherwise time-independent since you can usually make the Hamiltonian be time independent.
If the problem is easier to think about with a time-dependent Hamiltonian, you can use the Heisenberg formulation of quantum mechanics, which makes the wavefunctions static and lets the operators evolve in time. This can be helpful in a number of situations—typically involving light.
Like, maybe a steady-state eigenfunction of the system's evolution with an eigenvalue that's 1, or another root of unity.
I assume you mean eigenfunction of the Hamiltonian here, but the eigenvalue associated with that eigenfunction would be the energy of the state, so you can’t really make it be a root of unity (it must, in fact, be fully real since energy is an observable)
I'm basically a tourist in quantum physics with no more than approximate understanding of several concepts, and I don't think I'll ever fully understand a field that took dozens of Nobel prize winners and multiple lifetimes to formulate
That’s everyone, honestly. Physics is big enough these days that I don’t think anyone could get all of it.
I always thought of "position" as simply a point in Euclidean space described by a vector, but I'm guessing that doesn't translate directly to quantum mechanics because the uncertainty principle gets introduced with having to account for momentum.
That very much still is the case (though it’s technically Minkowski space once you introduce special relativity); when you measure the position of an electron, you will get a single point as far as we can tell. It’s just that there is a range of locations you might see it in when you observe it.
Does that mean that two electrons can, at the instant they are observed, have no meaningful distance between them, only different momentum?
Hmm… yes?
I believe that two electrons ‘occupy the same space’ (down to some uncertainty) when they scatter off of each other. As stated above, they are point-like, though, so you would need infinite precision to make them properly overlap.
But there is a less finicky way to do it:
If you observe position (down to some accuracy), you can’t observe momentum (down to a related accuracy)—that is the core of the uncertainty principle. That being said, if you have perfect knowledge of their momentum, you will have no knowledge of their position, which will allow them to be ‘in the same place’ insofar as they both are everywhere all at once.
This can actually be done practically by cooling them down: if you constrain their temperature/energy/momentum, you can get them to ‘overlap’ through uncertainty. When this happens, they actually pair up, adding their one-half spins up to either 0 or 1. This integer spin makes the pair a boson and allows them to occupy the same states as other pairs (note that the electrons themselves cannot occupy each others’ states, but the pairs can, and these ‘Cooper pairs’ become the principle particles of interest). This lets them (the pairs) flow through each other without scattering, which is how superconductors work.
The electric field curves it insofar as any energy curves it. The force exerted by the em field is independent of this curvature, though.
“Upends” is a strong word for “a bit more interconnected than we thought.”
It looks like the previous model is still broadly correct—there are just more signal crossings than initially accounted for.
Which, like… yeah. No shit.
This is good progress on understanding the brain. It isn’t disproving phlogiston or anything, though.
True. The em field is significantly more energetic, meaning that it contributes more to the stress energy tensor than the force of gravity
But recall that the force of gravity is the momentum change “due to” the curvature of spacetime. The fact that the force of gravity can bend spacetime at all is a really weird second-order perturbative effect.
These perturbative effects are typically described with a quantum field theory, but gravity has been thus far notably difficult to quantize.
It does. The energy density of the em field will contribute to the stress-energy tensor that changes the metric of spacetime.
Though the force itself is irrelevant
Leave… the stove… on….
Got it
Daily Akari 😊 429 2026-03-10 (Tue) ✅ Solved! 🕓 5:06 https://dailyakari.com/
K is potassium
Or is there a K that refers to phosphorus?
Thank you!! People keep looking at pitch, but there are cis women with deep, unquestionably feminine voices.
When I thought about it for a while, I think it’s all in the vowels: to make your voice more feminine, make sure your close vowels like your /y/‘s and /u/‘s have the same fundamental frequency as your open vowels like /ɒ/ and /ε/. I think this is what the article means by ‘resonance.’
Keep your voice box in the same ‘place’ in your throat as you switch vowels, and try to use only your tongue to shape them.
I can’t attest to this method too much since I’m not really out (and thus not using a ‘girl voice’), but when I tried to subtly make this one change to my voice (even at the same pitch), the first sentence out of my mouth made a cis coworker flinch. Take that how you will. I never really tried again after that, though.
There are probably 2 reasons for this:
There’s probably a lot more motor control going on than you would expect when you need to think (writing, fidgeting, etc.).
Your brain wants sugars, so when you run out of immediately-available glycogen to break down, you will want to eat more in order to keep thinking. Breaking down fats wont supply energy fast enough (in the short term) to keep complex thought running continuously.
Strong led or (very) weak incandescent. It’s about 20 Watts (at least, that’s what the popsci meme reports)
I’ve used kagi’s ai stuff like… twice, maybe. They are fully opt-in and unobtrusive. For reference, this is their ai philosophy: https://help.kagi.com/kagi/why-kagi/ai-philosophy.html
This seems reasonable to me, though perhaps not perfect. If you don’t want your money going towards ai development at all… IDK. I don’t think they train their own models for what it’s worth, so I think you can avoid paying ai companies by simply not using the ai parts of kagi.
If you want breasts, why are you on serms?
"I should really get back around to responding to that comment about time-evolution in quantum systems"
"hmm..."
Well, better late than never.
I am in quite the opposite situation; my experience is in the raw physics without much of the application to quantum computing. From what I understand, though, I think this is largely correct.
In general, observables (such as the Hamiltonian) are Hermitian (self-adjoint), which is neither a superset nor a subset of unitary operators. You are not, of course, restricted to only applying observables to your quantum state (In fact, you could apply any operator you want to your quantum state; the physical meaning behind most operators involves fundamentally changing the system, but it's not strictly forbidden to do this). We require observables to be represented by Hermitian operators because the eigenvalues of Hermitian matrices are always real (since the value you observe is an eigenvalue of the observables' matrix representation, you don't want any of the eigenvalues to be non-physical complex numbers).
I had to look up why specifically quantum circuits require unitary operators, and I found this Stack Exchange response, which describes how unitary operators are used to find the time-varying component of the wavefunction that solves Schrödinger's equation. I think we were kinda describing the same thing, ultimately: continuous evolution of a quantum system is dictated by the Hamiltonian (as shown by it's presence in Schrödinger's equation), and the time-varying component of the solution is unitary (the non-time varying component is a linear combination of eigenstates of the Hamiltonian whose associated eigenvalues need not be a root of unity).
Basically, when you say:
... the answer is 'yes,' but the Hilbert space itself is defined by the eigenstates of the Hamiltonian, which itself could be changing in time,^†^ meaning that a complete description of the time evolution requires slightly more careful consideration.
In your example of a standing wave on a string, unitary operators would take you from one mode to another, but if the length of string is changing, those unitary operators are changing too.
^†^A time-varying Hamiltonian implies a time-varying energy in the system. This sounds like breaking energy conservation (and it kinda is), but it is used to describe any system that is being pumped from the outside.
Daily Akari 😊 Sun Aug 31, 2025 ✅Solved in 1:17✅ https://dailyakari.com/
Exquisite luck
Typically sorta? The way the Schrödinger equation is typically solved is by taking linear combinations of eigenfunctions (of the Hamiltonian) and making them time-dependent with a time-dependent phase out front.
The eigenfunctions are otherwise time-independent since you can usually make the Hamiltonian be time independent.
If the problem is easier to think about with a time-dependent Hamiltonian, you can use the Heisenberg formulation of quantum mechanics, which makes the wavefunctions static and lets the operators evolve in time. This can be helpful in a number of situations—typically involving light.
I assume you mean eigenfunction of the Hamiltonian here, but the eigenvalue associated with that eigenfunction would be the energy of the state, so you can’t really make it be a root of unity (it must, in fact, be fully real since energy is an observable)
That’s everyone, honestly. Physics is big enough these days that I don’t think anyone could get all of it.
That very much still is the case (though it’s technically Minkowski space once you introduce special relativity); when you measure the position of an electron, you will get a single point as far as we can tell. It’s just that there is a range of locations you might see it in when you observe it.
Hmm… yes?
I believe that two electrons ‘occupy the same space’ (down to some uncertainty) when they scatter off of each other. As stated above, they are point-like, though, so you would need infinite precision to make them properly overlap.
But there is a less finicky way to do it:
If you observe position (down to some accuracy), you can’t observe momentum (down to a related accuracy)—that is the core of the uncertainty principle. That being said, if you have perfect knowledge of their momentum, you will have no knowledge of their position, which will allow them to be ‘in the same place’ insofar as they both are everywhere all at once.
This can actually be done practically by cooling them down: if you constrain their temperature/energy/momentum, you can get them to ‘overlap’ through uncertainty. When this happens, they actually pair up, adding their one-half spins up to either 0 or 1. This integer spin makes the pair a boson and allows them to occupy the same states as other pairs (note that the electrons themselves cannot occupy each others’ states, but the pairs can, and these ‘Cooper pairs’ become the principle particles of interest). This lets them (the pairs) flow through each other without scattering, which is how superconductors work.