This is the same kind of thing as saying that the electric field of a photon is a vector transverse to the propagation. If you have a photon going in the +z direction, it can be linearly polarised in the +x direction, and just by rotating the detector's linear polariser, we can observe $\pm45^\circ$. Working this out tells us that electric fields are vectors, not spinors, not higher order tensors, etc. Spin-1, not spin-half, not spin-2, etc.
I'm confused as to when one uses $\mapsto$ and when one uses $\to$. From what I understand, we use $\to$ when dealing with sets and $\mapsto$ when dealing with elements but I'm not entirely sure.
For example which of the two is used for the following? $$\begin{pmatrix} x \\ y \\ z \end{pmatrix}...
hm well i'd like to include the data of the specific state $\rho$ as well as the space...
since i know that $M: A \rightarrow B$ and $a \in A \mapsto b \in B$ is proper. But I'd like to (it seems) combine both notions into a single notation? since emphasizing the particular element and the spaces is what i wish to do \
i guess "M is defined by $a \in A \mapsto b \in B$" is close enough
@SillyGoose Can you be a bit more specific what the context here is? Your notation seems a bit confused to me and this seems like another case where you've prematurely abstracted too much from what you actually want to express
@naturallyInconsistent I apologise to anyone who wants to help me but can no longer see my question. Enough members of the SE community felt that this important question should be killed because of some petty SE rules, so let it be dead.
@Pygmalion we tend to be bit fussy about what is suitable for the main SE site, but one of the functions of the chat is to discuss questions that aren't suitable for the main site. For example homework questions are banned on the main site but we're quite happy to discuss them in the chat.
If you wanted to discuss your question here I'm sure lots of people would be willing to help.
Note that closing your question is not a personal attack on you. It is just the statement that the question doesn't fit within the scope of the site.
@ACuriousMind The situation is this: consider the conjugation of some tensor product state $\rho$ by $U$, i.e. $U\rho U^\dagger$. I am trying to express in math the idea that $U$ sends one of the tensor factors of $\rho$ to another state in the same tensor factor space.
$\rho \in \bigotimes_i \mathcal{H}_i$ and is of the form $\rho = \rho_1 \otimes \rho_i \otimes \rho_2$ where $\rho_i$ belongs to the $i$th tensor factor and $\rho_1$ and $\rho_2$ belong to the correct tensor product spaces. and $U: \bigotimes_i \mathcal{H}_i \rightarrow \bigotimes_i \mathcal{H}_i$
In that case your $U$ is itself the tensor product of individual $U_i : H_i \to H_i$. If it just holds for one $i$ then write $\otimes_j H_j = H_i \otimes H_\text{rest}$ and state that $U = U_i\otimes U_\text{rest}$, where $U_i : H_i\to H_i$ and $U_\text{rest} : H_\text{rest} \to H_\text{rest}$
@JohnRennie I do not take it personally, but I hate administrative rigour, which I think is the biggest obstacle to progress. I have already made up my mind on the question, so there is no need for discussion here any more. The only loss is for all the people who will not be able to see the question.
While the change has made closing homework questions very effective it is also causing lots of valid questions to be closed as homework. We seem to have a small hard core of users with a rather extreme view of what constitutes homework, and the current three vote rule allows them to close questions regardless of what the rest of us think.
Also, I think in these striking times, where we still have a backlog of close votes to clear, it might be helpful to be a bit fast and loose with the closing.
@JohnRennie Is the problem that you think this hard core is larger than three but smaller than five, or that the three vote threshold means the close review will be completed so fast that not enough other people have a chance to vote to leave open?
because if it's the latter, getting the close reviews to complete faster was the point of lowering the threshold; the flip side of the coin here is the position that the larger threshold allows a small hard core of users who will answer any question no matter how off-topic to answer questions regardless of what the rest of us think ;)
I think you and I both know who the two hardest core close voters are. The two of them seem to vote instantly on any question that offends their sensibilities, then it only takes one more close vote to close the question.
Effectively two people are now dictating the SE content policy.
@JohnRennie The underlying problem is that the number of frequent close reviewers is barely double-digit
you see the same 2, 3, 4 names on close banners all the time because barely anyone else is reviewing questions - and you don't see the ones who are more lenient since "leave open" reviews are invisible from just looking at the question
Yes, and if more of use were active reviewers the problem would be reduced. However not all members are retired ex-physicists with too much free time :-)
Reverting the three vote change seems the only option available.
raising the threshold back to 5 votes will likely lead to the close review queue overflowing again, i.e. lots of reviews aging away for lack of reviewers instead of being decided one way or the other
I mean, I agree with you that the current state is suboptimal, but I'm also afraid that going back to five votes will effectively turn off community moderation because getting to five votes takes so long most low-effort off-topic questions are answered by that point anyways. I don't see an easy way to decide which is worse
The best option would be to increase the power of leave open votes but that would require the SE to act and realistically they're not going to change anything.
Conversely, the 3 vote threshold means it's easier to re-open questions that were over-zealously closed. I don't work review queues, but I do cast reopen votes on questions I encounter organically. And I post comments explaining why I believe a question shouldn't be closed under our homework policy.
Also, there are lots of questions that are technically against the HW policy, but you can see that the OP really wants to ask a conceptual question, but they don't know how to express their question that way. This is especially a problem for people who come from a culture with a heavy emphasis on "teaching to the exam".
They may even say stuff like "I'm not asking for the answer, I want to know how to figure out the answer". But their question still gets closed. Some voters close anything that smells vaguely homework-like, especially if it contains any calculations. :(
Several years ago, we got a steady stream of "homework dump" questions, or as they're known on Math.SE, PSQs: Problem Statement Questions. The OP literally posts the problem statement from a textbook or exercise sheet, and expects us to provide an answer that that they can submit to their teacher.
We still get a few questions like that, but IMHO, we frightened off most of the PSQers, and most OPs these days do try to make an effort. But it can be hard for some of them to comprehend our "conceptual question" policy.
interestingly there is a number of recent papers on the topic :P
there are three camps: 1) partitioning is determined by experimental set up, 2) partitioning is determined whether the hamiltonian is local or not, 3) partitioning is determined whether each subsystem admits pointer states
just to remove one of the assumptions it makes. it is more "foundational" in that it is not so "practical"
it would make decoherence more satisfying to be able to start with just hilbert space and hamiltonian and not assume a partitioning of the hilbert space is the idea
but decoherence indeed doesnt happen in any system
u need this interaction with environment
perhaps u want to start with an abstract hilbert spac and hamiltonian, and derive the conditions for decoherence that is more general than parititioning
it would still b some condition becuz decoherence doesnt happen in any system
How do we identify position of two points if we specify position in one parenthesis . In cartesian we define position of two points separately $A = (x_1 , y_1 , z_1)$ and $B = (x_2 , y_2 , z_2 )$ . It is understandable .
@123 it is part of the definition of the config space. for e.g. when u write x1...x6, it is understood that x1,x2,x3 belong to the first particle and x4,x5,x6 belong to the second
u can also write (x1, y1,z1, x2, y2, z2)
the 1 and 2 make it clear that it's different particles. the x,y,z are the physical space dimensions
@123 the time evolution of the system simply becomes a line in the configuration space. now u can apply variations to this line and formulate the principle of least action
because the current state is just a point. then the time evolution is a line
in the physical space, the time evolution looks complicated. u hav to follow multiple points simultaneously
It means we always known $(x_1 , x_2 , x_3) = (x_1 , y_1 , z_1)$ it is position of first particle. and $(x_4 , x_5 , x_6) = (x_2 , y_2 , z_2)$ is position of second particle. We write all together in compact form $q = x_i$
and what those angles physically describe does not need to be a point in $\mathbb{R}^n$, it might also be the angles some lever in a machine forms with something else or whatever
generalized coordinates are truly generalized: These are values that, in some sense, describe the state of your system. There is no requirement that these coordinates directly relate to the position of anything, or that they have to be valued in $\mathbb{R}^n$
generalized coordinates are truly generalized: These are values that, in some sense, describe the state of your system. There is no requirement that these coordinates directly relate to the position of anything, or that they have to be valued in $\mathbb{R}^n$
so we just represent this particle's position by an angle
other times, we are studying not particles, but rigid bodies. their state is given by the position of center of mass and a vector for their orientation
so u need the usual (x,y,z) in R^3 for the center of mass
@123 well the "state" of a system is whatever you need to be able to say "I know everything I want to know about this system at this instant". For a point particle its usually just its position, for a rigid body it's its position and its orientation, for two metal pipes connected by a joint it might just be the angle between the pipes, etc.
the point of generalized coordinates is that we don't really assume anything specific about the physical meaning of these coordinates
@RyderRude Aah ookay. it means we need minimum coordinates which required to specify the state of system (motion of system) used in generalized coordinate
the minimum is for convenience. nothing is stopping u from representing the rigid body in terms of its million particles each in cartesian co-ordinates
or to represent a particle whose motion is confined to a circle in cartesian co ordinates
but the point of generalised co ordinates is that u can use whatever numbers u want to FULLY specify the current sttae of ur system
this is y it's called generalised. u r not restricted to cartesian
@RyderRude something is stopping you: using too many variables either turns the theory into a gauge theory or means you have to muck around with constraints; at the introductory level the set of generalized coordinates is always minimal
@123 we cannot. this is i say it's for convenience to reject them
@ACuriousMind yes. i mean that nothing is stopping u in principle
minimality isnt part of the definition of generalised co ordinates. but u should always prefer it because there is practically no way to handle a rigid body in terms of its particles
So in generalized coordinate system we can use anything (position, angle, vector , momentum etc..) which represent the state of system . And this should be minimum
you want analogues to position as your coordinates: What distinguishes coordinates from velocities is that if I show you a photo of the system at one instant of time, you'll be able to tell me the coordinates of the system but not its velocities
the values of coordinates make sense for the system at one point in time; the notion of velocity or momentum makes sense only as the derivative of some trajectory through configuration space (configuration space=space of all possible values for the chosen generalized coordinates)
@ACuriousMind It means the procedure of finding velocities , acceleration from position is same in generalized coordinates as we do in cartesian by using derivatives?
it is by the definition of configuration space that you have only positions as co ordinates. we use this space because it's great for Lagrangian mechanics
we do have other spaces like phase space which use both momentum and position as co ordinates
@ACuriousMind Aaah.. i see. It means methodology of finding other parameters from position is same in generalized coordinates. But we use minimum set of coordinate which can specify complete state of system.
Lagrangian mechanics happens on the tangent space of the configuration space, Hamiltonian mechanics happens on the cotangent space. The cotangent space is also called phase space.
@ACuriousMind My point is that configuration space is used to specify position using generalized coordinates. As we do in cartesian. then we create other parameters from that. like velocity , acceleration, vector?
you should not think of "generalized coordinates" as being the same kind of thing as "Cartesian coordinates"
"Cartesian coordinates" are one way of expressing a vector in $\mathbb{R}^n$
"generalized coordinates" is a name for whatever you use to specify the state of your system, and that state doesn't have to be a vector in $\mathbb{R}^n$
@123 suppose u have two particles of position and momentum (x1,p1) and (x2,p2). when u clump these together, (x1,p1,x2,p2), u get phase space. here too, the positions need not be cartesian
@ACuriousMind Means. Let say in cartesian system. The smallest or first object is position of a point. Then from that we can create displacement if it move in space, then from it we create velocity using time derivative, then acceleration by time derivative velocity.
Does the same way we go in generalized coordinates?
@123 I think we need to take a step back here and consider what we're really doing: We want to do physics, not create a nice list of random quantities.
the point here is that "mechanics" comes down to the following problem: You have some set of generalized coordinates $q_i$ that describe what the "position" of the system you're interested in is at any given moment
the problem you want to solve is: How can I predict $q_i(t)$, i.e. where the system will be in the future
the point of both Lagrangian and Hamiltonian mechanics is that this kind of problem has a very general solution, and you don't need to necessarily think about "accelerations" or "displacements" or "forces" etc. to arrive at it
but if it wasn't, then there would be no point to it, right? If all this did was re-express ordinary Newtonian mechanics in weirder terminology, it wouldn't be useful
yes. in cofiguration space, u would only need generalised positions and their first derivatives. the acceleration and forces are replaced by the Euler Lagrangr eqn @123
Newton's formulation focuses on Forces as if they were the real quantities...it definitely hides the focus from trajectories, which are actually measured
@ACuriousMind So what is the drawback of using Newtonian Mechanics? It work well. I have read in one book. Whatever examples we learned in previous books was so easy and solve able . But not all problems can be solved by NM.
as I said: Lagrangian and Hamiltonian mechanics are a step up in abstraction: They allow us to treat a much broader class of "time evolutions" with the same tools, they allow us to reason very generally about the existence of conserved quantities, for instance, and see their relation to symmetry
also they make certain problems that are really ugly in the Newtonian formulation more tractable, e.g. when it comes to constraints (a particle that is constrained to move on a given track, three particles connected to each other through springs, whatever)
in fact one of the first things people usually do is either derive a limited form of Lagrangian mechanics from Newtonian mechanics or derive $F=ma$ from Lagrangian mechanics and some assumptions
@RyderRude funnily enough finding the Wilson-Fisher critical point using the epsilon expansion and linearised beta functions was one of the first examples of explicit diagonalisation that came to mind
but yeah I'd agree that there isn't much linear algebra in the sense of matrix crunching compared to say textbook NRQM
I'm confused as to when one uses $\mapsto$ and when one uses $\to$. From what I understand, we use $\to$ when dealing with sets and $\mapsto$ when dealing with elements but I'm not entirely sure.
For example which of the two is used for the following? $$\begin{pmatrix} x \\ y \\ z \end{pmatrix}...
