5:17 AM
Hello Everyone...

1 hour later…
6:38 AM
@imbAF you should not tag people when you have a question anybody else can answer
@imbAF essentially, yes
@imbAF This has nothing to do with the question you asked above it, and it is already stated correctly, so what are you hoping for?

2 hours later…
8:32 AM
@ACuriousMind What is stopping there from being $\sigma_f \neq 0$ inside the dielectric? In this case, I would think one lets $\sigma_f \mapsto \sigma_f \delta$. Then, one can lump it in with $\rho_f$. But, I think the cucumber is considering such a $\sigma_f$ as surface density.
@SillyGoose The energy needed to assemble $\sigma_f\delta$ would be quite prohibitive. It is slightly more acceptable if it happened at the boundary between two dielectrics.
hm i see
I am not understanding why positive-semidefiniteness is defined in terms of this external Hilbert space $\cal{H}$. Is there a way to just speak of a Banach space $\cal{B}$ and say that $x \in \cal{B}$ is positive-semidefinite? Perhaps in terms of its spectrum?
or, is it just that because we are talking about operators, there is an implied domain of these operators. hence, the $\cal{H}$.
9:10 AM
@SillyGoose The text defined $\sigma_f$ to be the surface density of charges on the surface and $\rho_f$ the volume density of charges in the bulk. I don't make the rules.
@SillyGoose Yes, positive-semidefiniteness is an operator property, not something a random vector in a Banach space has
note that your text isn't talking about a generic Banach space "$B$", it's talking about the space of trace-class operators on a Hilbert space, and that happens to be a Banach space (via induced norm of operators)

2 hours later…
11:20 AM
I see
How can a markovian process describe entanglement between a system and its environment?
In particular, markovianness seems to be a staple assumption in any derivation of a garden variety quantum master equation. but markovianness seems to be precisely equivalent to assuming that the system and environment do not develop correlations. but something like system-environment entanglement requires system-environment correlations, necessarily.
all i can think of is happening is that one can model non-markovian dynamics of one system using markovian dynamics of a distinct system
11:43 AM
@SillyGoose You seem to assume a lot of context here :P Where does "markovianness seems to be precisely equivalent to assuming that the system and environment do not develop correlations" come from? Markovianness of what?
also, there certainly are non-Markovian master equations, e.g. Nakajima-Zwanzig
this text starts a section with the exact expression (due to standard Quantum Mechanics) for open system dynamics (partial trace of unitarily evolved composite initial state) over $\cal{H}_A \otimes \cal{H}_B$. the text then sets out to find a map describing such dynamics: $\cal{E}_{(t,t_0)}: \cal{H}_A \to \cal{H}_A$. This is like searching for an analogue of the "time evolution operator" for an open system.
Okay, then it goes through some theorems and etc. In particular, it states that actually such maps $\cal{E}$ in general depend on the initial system state. E.g., suppose the initial composite state is $\rho(t_0) = \rho_A(t_0) \otimes \rho_B(t_0)$. Then, $\cal{E}$ in general depends on $\rho_A(t_0)$. This is "problematic" in the sense that we cannot get a nice form of an evolution operator that is in general valid.
Okay, so let us specialize to the case in which $\cal{E}$ does not depend on choice of initial system state. The text calls such dynamic maps "Universal Dynamical Maps". They are equivalent to trace-preserving completely positive maps (the "standard" definition of open dynamic maps).
in that case I linked the right thing with Nakajima-Zwanzig: This becomes nicer if you assume the evolution doesn't depend on $\rho_A$ (= Markovian evolution of $\rho_B$) , sure, but it's not necessary
Then, the text further wants to enforce a composition law $\cal{E}(t_2, t_0) = \cal{E}(t_2, t_1)\cal{E}(t_1,t_0)$ where all maps in the aforementioned equation are UDMs. But this of course is not satisfied in general. Hence, markovianness (assuming the composition stated holds) is assumed.
I don't think Markovianness is equivalent to no entanglement, it can also just mean that the environment/bath is only very slowly changing: Consider the usual canonical ensemble heat bath - the density matrix $\mathrm{e}^{\beta H}$ is a mixed state precisely because the system is entangled with the heat bath, but the infinite heat bath by assumption doesn't change its state ever, so near-equilibrium thermodynamics in this sense is Markovian
i don't get how so much can be done with (quantum) master equations that are not like Nakajima-Zwanzig. I.e. assuming markovinness
11:54 AM
see my previous message - you just need an environment that's so big you don't really have to worry about its state
@ACuriousMind hm i can see this. but something still seems quite strange. sure the environment's state doesn't change, but then we seem to contradictorily have that (1) the system and environment are entangling, but (2) the environment's state is not changing. If two systems are entangling it doesn't even seem to make sense to talk about one of the constituent systems anymore
@SillyGoose of course it makes sense, that's what the partial trace does - it gives you the (mixed) state of a constituent system
hm well maybe I am misunderstanding what the markovian assumption actually means
perhaps more directly, i thought that it is assumed that the system and environment remain in a product state for all time (in all Lindblad derivations i have seen). this seems to go by the name "Born approximation". But the text I am reading seems to say it is already a consequence of markovian approximation.
the fiction of the quantum canonical ensemble is that there's a gigantic pure state of system + environment, and the system is small against the environment. When you trace out the environment, you get the mixed $\mathrm{e}^{\beta T}$. Were the system not entangled with the bath, this would be a pure state.
right i get that if we have a mixed system state, then it implies that the composite state actually is entangled
12:01 PM
@SillyGoose Wiki calls the entire step the Born-Markov approximation :P
I am confused because the following facts seem contradictory (1) when modeling open dynamics, assume that $\rho(t) = \rho_S(t) \otimes \rho_E(t)$ for all $t$, (2) $\text{tr}_E(\rho(t)) = \text{mixed state}$ for some time $t$.
right, but this is what the texts I've seen assume to get to the Lindblad master equation
this is from the wiki
@SillyGoose Okay, they're not contradictory, we're just talking about completely different things
you are correct that the Lindblad equation makes these assumptions
but no one claimed that the Lindblad equation describes every quantum system
so where's the problem?
but people use Lindbladian dynamics to describe entanglement. in particular decoherence, and etc.
so? it might still be correct for the systems they're considering
But even aside from that, I do not understand the role Lindblad dynamics would play. Unless the role of entanglement and sys-env correlations is largely overhyped, then it seems like not a useful device
12:10 PM
if the bath is large and fast compared to the scale of the system you're looking at, the entanglement with it is negligible
but if one wants to see decoherence, one needs entanglement to occur between the bath and system
also the $\rho_S(t)$ output of a Lindblad equation in general is mixed...so how does that even happen given the assumptions
i mean the results of the eqn are in contradiction with what went in to create the equation
@SillyGoose there is a difference between "the state remains factorizable" and "there is no interaction"
if we were talking about pure states, this would be true, but note that the assumption here is just "the bath is so big it remains in $R_0$ at all times", so you start with some $\rho(0)\otimes R_0$ and end up with some $\rho(t)\otimes R_0$ to a good approximation
there is not the assumption that $R_0$ has no influence on $\rho(t)$
just that it itself doesn't change (approximately)
but if $\rho(t)$ is mixed, then the composite state (assumed to be) $\rho(t) \otimes R_0$ must be entangled, right?
yes
but this is a blatant contradiction
12:14 PM
but the approximation is precisely that you can ignore that for the purposes of the evolution equation
@SillyGoose it's an approximation, yes :P
also to be precise I mean $S$ and $E$ must be entangled
._.
but this is really not worse than any other contradiction. In essence, the claim here is that time evolution turns $\rho(0)\otimes R_0$ into $\rho(t)\otimes R_0 + \epsilon R$ and that $\epsilon$ is small
and then people argue that this is true for weak coupling to the bath and fast bath-bath interactions
but it seems much stronger than that...in particular, that $\epsilon$ is precisely zero
@SillyGoose No, the claim that it is precisely zero would be that Lindblad dynamics are a 100% accurate representation of reality :P
but i think I still don't understand how if we throw away the necessary condition for $\rho(t)$ to be mixed, that $\rho(t)$ still ends up mixed
@ACuriousMind i do not understand. i thought the lindbladian assumes $\rho(0) \otimes R_0 \mapsto \rho(t) \otimes R_0$ for all t (in particular, $\epsilon = 0$ of your formula)
12:20 PM
@SillyGoose Sure, and correct would be a $+\epsilon R$ on the r.h.s. The assumption here is that $\epsilon \approx 0$ does not incur too much of an error. It's just an approximation that throws away the first order instead of the second order as you might be used to from perturbation theory - but the point is that in perturbation theory you start from already knowing the solution at $\epsilon=0$.
Here we're starting from not even knowing any solution yet, and so we start looking at $\epsilon = 0$
once we have that, we might start to think about what happens for $\epsilon$ small but not zero (but it already turns out the $\epsilon = 0$ version works fine for some situations
but shouldnt this be quite a bad approximation...? it seems remarkable that it works at all or is even trusted to be used
@SillyGoose And no - the necessary condition for that is just the bath and system interaction, and you still have the interaction term $H_\text{BS}$ in your Hamiltonian
@SillyGoose most statistical physics is like that, no :P
no real-world system is in ideal equilibrium, and we always assume infinite baths etc. - and yet equilbrium thermodynamics with a few hacks is all most non-statistical physicists ever need
but i mean shouldn't this break down for a large class of systems. say "mesoscopic" systems. sure if i want to model two qubits I can solve it analytically. if i want to model a million qubits, then maybe it can be argued that lindblad dynamics is good. but if i want to model 2000 qubits i don't expect such physics to accurately model. and i dont have the impression that mesoscopic systems are uncommon
We can set macroscopic systems into superpositions, and so it should still work as long as you can control the interaction strength (esp with regards to environmental noise)
@SillyGoose I don't follow your reasoning - the relevant thing is not the size of the system, but the size of the bath and the strength of the coupling to it
12:32 PM
i mean to specify the system as a few of the qubits and the bath as the remaining
@naturallyInconsistent but presumably one does not have control over these things
In work done $W = F(x) . dx$ the force always be the function of position?
@SillyGoose ...so do you have an example of someone applying this approximation to such a system in that way?
Can we use force $F(t)$ or $F(v)$ in work done?
i am sure i can find one :P i'll have a look sometime later
@123 1. Either you write $\mathrm{d}W = F(x)\mathrm{d}x$ for the infinitesimal work or $W = \int F(x)\mathrm{d}x$, but $W = F(x)\mathrm{d}x$ is inconsistent. 2. $W = \int F(x)\mathrm{d}x$ is specifically for forces independent of time or velocity. The general expression for work is $W = \int F(x(t),v(t),t)\mathrm{d}x(t)$ along the trajectory $x(t)$ of the particle.
12:36 PM
@ACuriousMind thats how basic stat therm is derived
@naturallyInconsistent by applying Lindbladian dynamics to a few 1000 qubits???
@ACuriousMind dx at the last bit
@ACuriousMind definitely not that...
I didn't read in depth, but this 2019 preprint seems to use master equation for three qubits to describe decoherence effects: arxiv.org/pdf/1904.04323
@ACuriousMind I see thanks. I didn't see this general expression of work in any book.
@123 it will usually be written as $\int \vec F\cdot \mathrm{d}\vec s$ or something like that
12:38 PM
although, it doesn't seem to be officially published; though, it is cited still
@ACuriousMind Yes that's why it is not clear. But in K&K they used $F(x)$ to solve $F = m a$.
I thought this is the only way where work use for force as a function of position $F(x)$
@SillyGoose but that's again the size of the system, not of the environment, no?
i think it is the size of the composite system, but i couldn't find the description of the environment. but i will check again
@123 I'm not sure what the equation is. The most general version of Newton's law is indeed $F(x(t),v(t),t) = m a(t)$, i.e an arbitrarily complicated 2nd order differential equation. In many situations you have conservative time-independent forces where $F$ is only a function of $x$, sure. But what's the problem?
hm i guess they just don't specify the environment explicitly. so it could be assumed that the environment is large or what not relative to the system
12:45 PM
@ACuriousMind I was thinking work can only be use the situations where forces are function of position, not others.
this is from knk also in tylor mechanics they also solved newton's law for $F(x)$ to derive work energy theorem.
It means work done formula can only be possible to derive when force only a function of position.
1:02 PM
Hello @RyderRude
@RyderRude I am trying to understand how they compute constants A,B,C,D. I know the calculations. But why they put these 4 conditions to calculate constants.
Also problem in 4th sitation
these r just the physical conditions
condition 1 means that, after the lorentz transform, the observer moving at speed $v$ is now moving at speed 0 (becuz we switched to his frame)
condition 2 means that, after the transform, the observer of the original frame is now moving at -v
condition 3 means that light travels at c in both frames
condition 4 also utilises the invariance of speed of light
1:15 PM
But they sent light to y-axis
yeah
they utilised it in a different way
It means we can use any condition other than these 4 also
maybe
i think ive seen different derivations
but the first 3 conditions r always there
but y = y' why they sent light to y-axis
theyre assuming the two observers have relative motion only in the x direction
so only x and t change after transform. y and z r the same after transform
1:21 PM
@RyderRude Okay it means there is no problem if light pulse sent to y-axis
yes
Thanks
this derivation seems weird tho as it utilises the y axis.
@RyderRude yes
but its correct anyway
lorentz transforms work perfectly fine in 1+1 spacetime.. so there need not be a y axis
maybe wiki has derivations without the y axis
1:23 PM
Thanks
i think the use of y axis is inspired by the light clock derivation of the time dilation formula
there is a famous derivation of time dilation with the light clock where the light moves along y axis while the clock travels along x axjs
this gives u the gamma factor, just like this derivation does
Yes you are right
2:25 PM
mathcal and mathscr work for all letters except r??
@123 no. The force due to viscosity and due to air resistance are functions of velocity and so it cannot be limited to only those simple kinds.
$\mathcal r$
$\mathscr r$
r
$r$
@Relativisticcucumber apparently; unfortunately.
just why
griffiths has a copyright on all curly r's
2:53 PM
𝓇
$𝓇$
This is probably the best that we can do
wait how do u get that
@naturallyInconsistent Thanks. It means work done general equation is $F(x(t), v(t), t) dx(t)$.
@Relativisticcucumber that is a symbol. Just copy pasta
Just be aware that the work done concept is just silly. The thing to care about is energy, and work done is just one way to transfer energy.
3:09 PM
I was confused because work done derived from force as a function of position $F(x)$
Because of that i thought the only use of work done, KE and PE when force is a function of position $F(x)$
3:26 PM
The standard teaching of physics is pretty bad here
yay I have a new Raspberry Pi, which means I have Mathematica again and can use the TV for stuff. And obviously texlive and more
3:52 PM
M I A O ~
I
A
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😁
MＩA O ~
4:41 PM
Hi
for those interested I have done the full mathematical calculation of the paradox
It appears that the resolution is that the force is not equal and opposite as per Newtons law, however the dt over which the force applies is different for each paerticle, allowing for conservation of momentjm
this Is what many of us suspected but the issue was then what about the parallel case. That’s what I still need to check then

2 hours later…
6:31 PM
2
managed to find this after almost two decades
pretty nostalgic

4 hours later…
10:03 PM
Classic.