There is, therefore, a fixed probability per unit time of the system experiencing the effect of one of the decay operators. You can think of the Lindblad equation as averaging over many processes with this constant per-unit-time probability of collapse. When you average over that kind of process you always get exponential decay.
We also specialize in semantics, a bit of politics, gossip, some fanning the flames of site drama, and a relatively large portion of making fun of @Danu.
@TanMath : there's been a lot of chat about game playing. Obviously one wouldn't expect this chatroom to be 100% physics, but sometimes it's been like 0% physics.
@JohnDuffield Yes. I would point out that last time I asked politely that the gaming conversation move to a private room the participants of that conversation did so almost immediately with no problems.
This is a rather common form in nature because it arises from anything where you have an energy which goes as the square of the difference of two canonical operators.
For example, if my system is a harmonic oscillator and it couples to other oscillators via a spring force, my coupling energy is $(x - x_\text{bath})^2$.
Here $x$ is the position of my system and $x_\text{bath}$ is the position of a mode of the environment.
@Jimself : re your carping, I'm not wrong about the speed of light. Nor was Einstein. Nor is Shapiro or Koks or Wright. Instead you are. Now try to understand it instead of clinging to popscience version of relativity. If need be ask a question and I'll answer it with robust references to Einstein and the evidence and to other authors and papers.
A "Lindblad" equation is an equation which models a system which is connected to an environment.
The Lindblad equation does not say anything about the internal structure of that environment.
It only contains a representation of how the environment affects the system under consideration.
Instead of the statement you just made, I'd say that many system-environment pairs are such that if you were to model the system with a Lindblad equation, you could find the collapse operators by the method I wrote there.
Can anyone explain why the pure tensors can be thought of as a cone? It's mentioned in my online lecture notes but exactly from a lecture I wasn't at, so I don't know what they mean.
@0celo7 Well, I'll be going to the UK and my parents will be in a different country, so I need my passport to get there, and whenever I want to travel outside. There's also a bunch of legal stuff I have to sort out which can only be done within the UK, hence the need for other documentation. I guess since you're staying in the US, this isn't a problem for you.
@DanielSank No, I do like Python, that's what I use in prototyping. It's just that it wasn't really designed with mathematics in mind, like Matlab or Julia, and your comments about Matlab being bad ("hurting yourself with [it]") were not really fair especially considering that the competition is still currently behind. That said, Matlab is not open which is the main reason I dropped it (but I dropped Windows for the same reason, so I'm a bit more uptight about this than most).
@JamalS I understand the passport, but AFAIK I've never needed my birth certificate and I'm certainly not going to take it and risk it being stolen or lost.
The universe is in a long, slow decline to darkness
Sorry folks but it's true, read the article!
My question
Using either your imagination and soft-science or actual hard-science, can you suggest ways that the human race and/or other sentient life-forms stops this eventual decline.
Specifical...
@skillpatrol I don't think that is an okay thing to have
I mean, double account are only allowed as long as they don't interact, and I don't use their votes to "boost" my weight, i.e. doubly vote on the same things. Using a second account to circumvent the close vote limit seems quite the same to me, as it is obviously intended that no non-mod should have more than 24 close votes a day.
That doesn't make any sense - if I think a question should be closed and I have votes left, why should I wait to cast them? (There even is a meta.SE post discussing this, it says vote to close as soon as you can - you can always retract if the question is improved)
@skillpatrol AFAIK, they have unlimited votes, but those are binding, and they are encouraged to only use them in clear-cut cases or when it is evident that the community review can't keep up with the amount of closeworthy questions.
@skillpatrol: You have to work on your reading skill: 0celo7 was trying to provide a reason why Danu might've though DanielSank knows something about what Danu does in Amsterdam - and came up with the link that weed is available at both places. That's all that has happened here.
At @0celo7's stage in life he has to start carefully thinking about what it means to be on his own; and there is going to be temptation out there to do stuff that he hasn't done before.
Pre Edit 4:
Pointing out that thread's popularity is a direct result of programming community interest in said subject, and that someone likely stands to become rich and/or famous by solving the problem none of the answers, save my own, came even close to resolving.
Long salutation desc...
I did read more often, that one can not measure the angular momentum exactly in 2 different directions, but can exactly measure the angular momentum in one direction an the absolute value of the angular momentum (or the square)
@DominicMichaelis What do you mean with "not linear"? It is a sum of a product of linear operators, and thus still a linear operator (the sum of a product of matrices is just a matrix).
@ACuriousMind if I have some function and meaure the square of the total angular momentum, and takes the function times 2, the square of the total angular momentum should be taken by 2^2
The states of quantum mechanics are rays in Hilbert spaces. This means that multplying a state (whether it is represented by a function or something else) with any complex number does nothing
So if you multiply a state by 2, the angular momentum of the state stays the same.
@ACuriousMind the square of the angular momentum is never negative, but every linear function which takes a positive value, must take negative values either?
@DominicMichaelis I don't understand what you say. Yes, the squared operator never has negative eigenvalues, it is positive-definite. But it is an operator, a matrix, it doesn't "take values".
@DominicMichaelis Because $L^2(f)$ is just the composition of applying $L$ twice $f\mapsto L(f) \mapsto L(L(f))$, and the composition of two linear functions is linear! (It's linear algebra: The product of two matrices is a matrix, and all matrices represent linear functions on the space)
It can't be the pointwise product because a general Hilbert space has no notion of such a thing! I very much advise to not always think of wavefunctions first, but of general vectors.
@ACuriousMind oki then for sure it is clear that $L^2$ is linear, but i don't see that it is the norm of the angular momentum, because that should be the angular momentum of the angular momentum, shouldn't it ?
@DominicMichaelis The squared angular momentum is the operator $L_x^2 + L_y^2 + L_z^2$. For system with definite angular momentum it has the same eigenvalue for every vector, and writing that value as $l(l+1)$, $l$ is what we commonly call "the angular momentum (quantum number" of a quantum syste,
Don't try to think of angular momentum as an ordinary vector as in classical mechanics, that is an analogy that does more harm than good, imo
Whenever you see a power on an operator, it means the n-fold composition with itself. (That's also why you won't see non-integer exponents on operators)
I don't like the calculations in the notes, he does the easy stuff very very explicit (like simplifying expressions and stuff like that) but every interesting point like changing limits etc or some approximations and why they are allowed
I was wondering if I could get some quick help on a conceptual issue I have with motional EMF? The equation has a minus sign for a conductor moving in a uniform magnetic field: EMF = -Blv where l is the length of the conductor and v is the velocity. Given a magnetic field into the page, and a conductor moving to the right, I would expect positive charges to collect at the top and negative charges to collect at the bottom, making the nodal voltage at the top more positive compared to the bottom.
However, EMF = -Blv implies that the top is more negative?
I drew it out because words suck at relaying images.
@TanMath My specialty is uh... I guess... working in the lab. I dunno I do a lot of things: electronics, programming, circuit design, fabrication, chip layout, physics of qubits (i.e. simple quantum mechanics problems), signal processing, and a lot of going between theorist language and practical lab language.
I do have Massoud's email but I don't think I should give it out if it's not public. I'm happy to forward a message to him. You can find my email via my profile page.
@0celo7 I don't know what "integrate as t goes to infinity" means. What are you integrating over, and is the limit taken outside or inside the integral?
If reasonably smooth means that the function has some notion of "width" and goes itself to zero at infinity, then I'd say yes, the integral will also go to zero.
@0celo7 No, I just wanted to rephrase your question in nicer terms. Integrating something multiplied with e^itx is just Fourier transforming, no?
But yes, it is true. E.g. the Fourier transform of a Schwartz function is a Schwartz function.
There's also the Riemann-Lebesgue lemma stating that the transform of L^1 functions vanishes at infinty, which is even stronger.
I think the Riemann-Lebesgue lemma is the best rigorous formulation of the "fast oscillations cancel" that physicists like to use.
@0celo7: Are you on mobile? I thought you arrived?
The universe is in a long, slow decline to darkness
Sorry folks but it's true, read the article!
My question
Using either your imagination and soft-science or actual hard-science, can you suggest ways that the human race and/or other sentient life-forms stops this eventual decline.
Specifical...
Pre Edit 4:
Pointing out that thread's popularity is a direct result of programming community interest in said subject, and that someone likely stands to become rich and/or famous by solving the problem none of the answers, save my own, came even close to resolving.
Long salutation desc...
