@EnthusiastiC It's similar to ordinary multi variable calculus. The action is a function of a path, i.e. of infinitely many variables. The classical path is a peak or trough or a saddle point of the action.
So the first order term in the Taylor series expanded around the classical path is zero.
How does a physicalist resolve this dilemma (and what is it known as)?
Consider a physicalist who wants to solve the symbol grounding problem. He postulates that a particular brain configuration corresponds to a particular meaning. He gives this postulate the status of a law.
I don't think this s...
@MoreAnonymous Sure. In your case it's the *human*/gas paying the cost, not the demon.
like, the human's physical configuration has already changed from the experience they're describing to the computer when they're done describing it
"I have just experienced X" and "I have just told a computer that I have just experienced X" are obviously different states, so this "measurement" is not a free operation
that's a fundamental rule of lossless compression: If you manage to use fewer than n bits to convey information of n bits, then there are some messages of n bits for which you have to use more than n bits to convey them (an easy application of the pidgeonhole principle)
you can't wave a magic wand and say "I can encode all information in arbitrarily few bits"
@MoreAnonymous I'm not interested in what exactly the demon needs to violate the law
I'm interested in how you think the computer in your idea can use the brain state information at all to violate the 2nd law
in particular, since, as I already said, the act of conveying that information to the computer changes the brain state of the human, so you don't even have information about the current brain state of the human, just about a past one
@MoreAnonymous in order to predict deterministically what the current brain state is, you would need to have complete information about the past brain state, but you conceded above that we really will only communicate partial information about the state with finite messages
Okay I disagree. If I tell u the probability distribution of a physical system. There's no reason I can't tell u how this probability distribution evolve
(also: this "probability distribution" is purely epistemic, it is not a frequentist description of the world. you're just saying "I have partial information which means the brain is in one of these states and I'm randomly going to assume the probability for them is uniform/normal/whatever you decide" Evolving that distribution is just evolving your lack of knowledge into the future. )
@ACuriousMind because I have I have a probability distribution of the physical state without paying the measurement thermodynamic cost? (As I said above system gas/human is this time paying the measurement cost and not the demon)
What is in my mind previously. We know the momentum $\Delta P = m \Delta v$, which is for single object. If we want to write the total momentum of two different objects we use Newton's 3rd law for it. What about KE and PE relation for two objects?
@ACuriousMind (without getting into what thermodynamic probabiliies are I think my point still stands I have a better probability distribution because I have accounted for my system being happy :p )
@MoreAnonymous I have no idea what you mean by that. Why would I need to "pay a cost" to get a probability distribution? Everyone nowadays knows an ideal gas is Maxwell-Boltzmann distributed without ever making a single measurement! Are we all continually violating the 2nd law when we think about thermodynamics?
@ACuriousMind I mean Maxwell isnt the only distribution that describes the gas. I mean imagine dirac delta functions where I know the exact position and momentum of each particle. This may appear like a cheat. But let's say I now "cheat" and smear the dirac delta function with some kind of gaussian?
I'm not sure how that would work. I have a mug right next to me. I know exactly where the mug is just by looking at it, I expend no work to make a complete measurement of its state
Yet you wouldn't claim that my knowledge of the mug violates the 2nd law, would you?
My explanation for why my knowledge of the mug doesn't violate the 2nd law is simple: There's nothing I can do with that knowledge to decrease entropy (in contrast to the demon in the gas, who actively can use the location information of gas particles to do so). My default position is that all knowledge is of this kind unless you can explain a specific mechanism by which it can be used to violate the 2nd law.
you seem, for some reason, to have the opposite default position, but I haven't really heard any argument for why that should be so
all the arguments about "this measurement has to produce entropy" are only necessary because everyone agrees that the demon's actions reduce entropy, so in order to solve the paradox, people try to argue that its measurements have to increase entropy
but this does not show that, in general, measurements have to increase entropy or something like that
when there's no actions that reduce entropy, there's no paradox to begin with
@MoreAnonymous and what exactly do you think that section does to say anything about my mug?
or the brain state idea, for that matter
the section very specifically explains Szilard's argument that I also made above: Because the demon decreases entropy, the rest of its action - the measurement - must increase entropy if the 2nd law is to hold. If we're not in a situation with a demon (i.e. a specific process that can use specific information to decrease entropy), this just doesn't apply
The joint eigenvectors of H,L^2 and L_z, to which the corresponding wave equation is of the form $\Phi_{n,l,m}(\vec r)$. These are also called bound states and the spectrum is discrete. If that's the case, why in the partial wave analysis in scattering theory, we consider these same wave functions , when there we deal with scattering states, and states characterizing free particles ?
@imbAF you need to look more closely - partial waves usually have no "n" and the spherical Bessel functions that appear there have a different dependence on $r$ - namely $\mathrm{e}^{\pm\mathrm{i}kx}$ instead of the bound $\mathrm{e}^{-kr}$.
When we consider a scattering event, we have the incident particle ray, the scattered one, the transmitted one, which is a superposition of the scattered ray/particles and the incident ray/particles, and this is called a stationary scattering state. My question is, under which conditions is the state that of a stationary scattering state and under which just a scattering state? Stationary means that it
corresponds to one of the eigenstates of the hamiltonian of a particle but it has also a phase with time dependency. But why we say that it is a stationary one?
yes but why do we claim that the superposition of the incident particle ray and that of the scattered particles, is a stationary scattering state, and not a scattering state
a scattering state is a stationary scattering state if it is stationary, there's no trick or subtlety here
@imbAF I don't know what you are referring to and I suspect that you have misunderstand whatever text or lecture you're following. We don't just call some random state "stationary scattering state", we call it that because it is both a scattering state and stationary
it's completely common in quantum mechanics that until you measure the particle and find out what happens it will be in a superposition of both states
this is not qualitatively different at all from the double-slit, Stern-Gerlach or any of the other experiments where we treat the outcome as a superposition until a measurement is performed
cuz I don't know eigenstates of what operator those can be
Nono
@ACuriousMind I don't mean this
But I will have to make things more clear
We have incoming particles. The state of an incident particle is described by wavefunction of a plane wave. because before interacting with the potential the incident particles are free particles, and their states are represented via wavefunctions of plane waves
The state of an incident particle can be an eigenstate of some operator (maybe H,or R or P) or it can be a state that is a superposition of eigenstates, for example a wave paket
Now, after interacting with the potential the particle can either scatter or not. Which means if it scatters, it's characterized by a scattering state, to which a wave function corresponds
or if it doesn't scatter nothing changes about it's state
and we agreed that the state of the incident particle is an eigenstate of linear momentum. So naturally would it be wrong to say that the state which is a superposition of the scattered state and that of the incident one, it's an eigenstate of the linear momentum?
@imbAF sure, that would be wrong (in 1d it's either "pass through" or "reflect" and the reflected state clearly has the negative momentum of the incoming state
@MoreAnonymous no, that is not what I'm saying at all
Ah, it's whatever when I had soundbits of this topic on my lecture
I just find it difficult to understand, how a scattering state, which is already a combination of two different types of states, which also can be superpositions (i.e incident state as a superposition of eigenstates of linear momentum) correspond to the eigenstates of the hamiltonian
@MoreAnonymous my point is that determining e.g. the position of a macroscopic mug is a very different kind of measurement/operation than the microscopic measurements Maxwell's demons need to perform to defeat the 2nd law. If the measurement increases entropy then it will do so regardless of what you choose to do with its result.
But this whole demon argument only shows that the kinds of measurements that the demon would need to perform increase entropy, not that all measurements increase entropy.
so: yes, determining a molecules position in a gas, sure, probably increases entropy. Looking at stuff: probably doesn't. Knowing about some sort of partial brain state: Who knows?
@imbAF there is nothing difficult about this: Think about plane waves: $\mathrm{e}^{\mathrm{i}kx}$ and $\mathrm{e}^{-\mathrm{i}kx}$ have momentum $k$ and $-k$, respectively, but their sum is still an eigenstate of the free Hamiltonian $p^2$ with eigenvalue $k^2$
there isn't some deep physical insight here or anything, it's just a consequence of saying the scattering is elastic: energy needs to be conserved, so whatever the result of the scattering is, it will still have the same energy as the incident state
well the only case I know when I sum eigenstates of an operator, in this case of the Hamiltonian, is when i try to express an arbitrary state as a linear combination of it's eigenstates
and we refer to that as a linear superposition
but then
is this arbitrary state, an eigenstate of the hamiltonian?
And again, you seem to want me to produce a completely coherent theory of classical measurement here, but that's not really what I'm trying to do here
I'm just pointing out there's a missing piece in your argument
you're the one who claims that knowing information like a brain state allows one to violate the 2nd law, you're the one who needs to argue why this is true
@MoreAnonymous maybe the number of photons is the relevant difference! Maybe the point is that measurements that lead to classical paradoxes are effectively forbidden by quantum effects (measurements change the state!)
I mean that there is something different here seems a tad obvious: If you just blow up the Maxwell demon to human size and posit doing the same with a bunch of elastic balls in a gravity-less chamber, obviously no one thinks the human hitting balls above a certain velocity with a bat violates the 2nd law
we haven't even begun talking about the consciousness part of the question because I think the problem is actually a misunderstanding of Maxwell's demon :P
@ACuriousMind I dislike this resolution even though i concede it's valid
@ACuriousMind so the system is in thermal equilibrium here . If one keeps blocking the only the slow moving molecules I think it will violate the 2nd law since the momentum imparted would cause some heat? (Causing the door to go out of equilibrium)
@ACuriousMind wait! measurements do change the state but the average is agreed upon
@ACuriousMind If we consider the case of Time independent degenerate Pertubation theory, for the perturbated energy eigenvalue of the perturbed hamiltonian one can write: $E_{n,d}=E_n^{(0)}+\lambda E_{n,d}^{(1)}+\lambda^2 E_{n,d}^{(2)}$ If the index n represents the eigenvalue we are considering, the superscript (i) the order of the correction term, then what can I say about what the $d$ index represents? The degeneracy ?
@imbAF I'm not sure what you mean by "representing the degeneracy". It's just an index for the linearly independent states with unperturbed energy $E_n$. I.e. if you have two-fold degeneracy, there's two independent states $\psi_{n,1},\psi_{n,2}$ for every energy $E_n$, and so you potentially get two different perturbed energies $E_{n,1}, E_{n,2}$ for these.
@ACuriousMind "representing degeneracy" if we have no degeneracy then the index d is not needed. Which means if an eigenvalue E_n is g_n times degenerated, then it means that we have g_n first order energy correction terms
Well depending where $J=1/2k$ goes up to, it's the $\rho_J$ rep and $u_1,u_2$ are the vector coeffs of the defining rep and you think of the products $u_1^k u_2^r$ as basis vectors of the vector space
but you can do this trick with every kind of representation, it's not anything special for SL(2,C)
for any representation space $V$ of any group, the symmetric and antisymmetric tensors of rank $k$ on $V$ also form a representation of that group (not necessarily irreducible)
it's often the case that you can get all possible irreps as one of these tensor representations
@ACuriousMind Is this what gives rise to the singlet and triplet reps when considering $\psi^{[\alpha} \psi^{\beta]}$ and $\psi^{(\alpha} \psi^{\beta)}$ or something
I've found a wonderful book that seems to answer all my questions and builds everything up explicitly with the relation to physics in mind. It's 'Group Theory and Quantum Mechanics' by Van der Waerden.
how would one describe what is going here: $\langle n^{(0)}|n^{(1)}\rangle$
Inner product between the unperturbed eigenstate and the first order eigenstate correction term of the perturbed eigenstate,to which the unperturbed eigenstate in the inner product is the 0-th term?
@Slereah i have no clue, i'm just reading a proof for what must surely be a classical non-relativistic EM method, and they mentioned fixing the gauge with lorentz gauge condition to resolve degrees of freedom vs variables
and i'm trying to understand the what/why/how of what they're doing, preferably without having to learn what a four vector is etc
i mean i want to learn that stuff someday, but was hoping not yet. challenge one thing at a time yknow
and i can only find one reference for gauge fixing which doesn't mention spacetime four vectors or QM
you can do EM in a way that doesn't make its relativistic nature obvious, but really at its heart it is a relativistic theory - it is no accident that the speed of light as a central object in relativity is a function of the electric and magnetic permeability of the vacuum
and the Lorenz gauge in particular is very much related to relativity because it is invariant under the Lorentz (no relation to Lorenz) transformations of special relativity
but gauge theory is very much a fundamental component of modern physics - all the four fundamental forces are described by a gauge theory in some fashion (though let's not get into how this applies to gravity)
well i'm trying to read about gauge fixing, i like this idea of configuration spaces and unphysical axes. but as soon as i get into it they mention four vectors
sorry **gauge fixing
ah i see, thats pretty cool actually
essentially i was wondering if there was an appropriate "level" at which you can learn gauge fixing before you're up to relativity
