is a boost an infinitesimal acceleration?

@Allie yes, you could say that

yay!!!

@SillyGoose Why would it be?

well let's consider a boost that infinitesimally changes the velocity of an object. if you keep applying such boosts (in a time-dependent manner) then you effectively get acceleration

I have no idea what kind of mathematical operation you are trying to describe here by "applying such boosts (in a time-dependent manner)"

Acceleration is the second time derivative of a worldline/curve $x(t)$. What do boosts have to do with this at all?

i just remembered the weirdest thing

well i can do something like $K^{\text{floor function}(t)}$ where $t$ is time. then every $0.5$ units of time, the system gets boosted by $K$ again. in this sense, the velocity is being made time-dependent by just applying boosts

what a creepy thing to see a group of 7 year olds being forced to chanting that

@SillyGoose I don't know how you expect anyone to understand what $K^{\text{floor function}(t)}$ is supposed to be.

(also, the floor function $\lfloor x\rfloor$ is \lfloor x \rfloor)

let's work in classical mechanics. let $K$ be a boost operator. Then, I am considering the action of the operator $K^{\lfloor t \rfloor}$ where $t$ is time.

why would you do that

what does this have to do with a trajectory $x(t)$ or its acceleration $\ddot{x}(t)$

Because it does what I was describing conceptually: it first acts like $\text{id}$, then like $K$, then like $K^2$, ... in a time dependent way. So it gradually boosts the thing it acts on as time passes.

What do you mean by "boosts the thing"?

what thing?

what is this $K$ acting on?

well i mean i can only speak on qm i don't know the classical analogue. if i have a quantum state with momentum $p$ then i write this eigenstate as $\lvert p \rangle$. then, the action of a boost is asserted to be $K \lvert p \rangle = \lvert K p \rangle$ where $K$ has a fundamental action on the vector $p$.

so an example of a situation i am describing would be $\lvert \psi (t) \rangle = K^{\lfloor t \rfloor} \lvert p \rangle$

which at time $t = 0.4$ looks like $\lvert p \rangle$, at time $t = 1.1$ looks like $\lvert K p \rangle$, etc.

mew

@SillyGoose You're not doing a good job of motivating this construction at all. Why are we discretizing time? Why would you consider a $\psi(t)$ of that form? What is this supposed to model?

This is definitely not the time evolution under any Hamiltonian I've ever seen :P

lololol

i was gonna say

isnt the state like not evolving between 0-1s

but this is way out of my

well i couldn't think of a simple way to mathematically implement the time dependent action $\text{id}, K, K^2, K^3,...$

the simplest way i could think of was with the floor function. i am ignoring coing up with a hamiltonian that does this

@SillyGoose But why are you trying to "implement that action"???

because i am interested in if this can model a notion of acceleration

I don't knoe what that means

what I'm asking you is precisely to explain what you think any of this has to do with acceleration

classically, acceleration is $dv/dt$ or rate of change of velocity. i.e., the presence of a changing velocity means the presence of an acceleration.

again, acceleration is $\ddot{x}(t)$ of a classical trajectory or of the Heisenberg picture operator

it does not have anything to do with boosts

i mean if i have a screen in front of me and it tells me at $t = 0$, $v = 0$ at $t = 1$, $v= 1$ at $t = 2$, $v = 2$,... where $v$ is velocity, then a notion of acceleration (independent of all other details) can be $a = 1 = \Delta v / \Delta t$

mrow

i am just attempting to make the observation that i can then consider a sequence of states $\lvert p \rangle, \lvert K p \rangle, \lvert K^2 p \rangle, ...$ which naively looks like the momentum of the particle is changing from $p$ to $Kp$ to $K^2 p$ to ...

so if this boost is infinitesimal, then concretely we can naively talk of the state's momentum going from $p$ to $p + \alpha$ to $p + \alpha + \alpha$ to ...

@SillyGoose You're not making an observation here, the sequence of boosted vectors has nothing to do with something accelerating. Note that if $p=0$, then your sequence is identically $\lvert p\rangle$ by elementary linear algebra. But certainly things that are at rest can accelerate. The application of a boost has nothing to do with accelerating a state.

hm well that is a fatal flaw

a separate question is: can one have a Lie group a subset of which forms a Lie algebra?

I don't understand that question either, I'm afraid :P

so you can obtain a string of structures $\mathfrak{g} \xrightarrow{\exp} G_0 \subseteq H \xrightarrow{\exp} K$

well say I have a Lie group $G$. then can i naturally construct a lie algebra out of elements of $G$?

that is, is there some subset $H \subset G$ such that $H$ can be equipped (in some natural way, or at this point even artificially) with a Lie algebra structure (that is not trivial)?

@Relativisticcucumber this sounds like a bit about US patriotism I'd have seen in something like The Simpsons decades ago :P but sorry that happened to you

@SillyGoose No. Where would the addition operation of the Lie algebra as a vector space come from?

i hated pledging allegiance every morning in school

The natural way to construct a Lie algebra out of a Lie group is to...take its Lie algebra, i.e. the space of invariant vector fields or the tangent space at the identity or whatever else you want to take as the definition

is it necessarily the case that the density operator of a pure state only has one eigenvector? namely, the state vector

yes

@Allie no - a self-adjoint operator in a space of dimension N has N eigenvectors, always

oh

but the density operator of a pure state is the projector onto it, and so all other eigenvectors are eigenvectors of eigenvalue zero

okay i guess yeah the other eigenvectors are eigenvalue $0$, but right there are multiple eigenvectors