Alphanumeric, don't be ridiculous. I have not been e-mailing people sporadically. The doctor I was talking about is a facebook friend.
But you obviously think your work good enough to be evaluated by such a person.
I've 'tinkered' in a few things, come up with minor results which I haven't read anywhere, but I wouldn't email them to a friend with a PhD in that area for their comments. I might, during a conversation at the pub, discuss the area with them but even then the discussion wouldn't go down to the level of "Here's a print out of my work, can you read it".
I don't know where you get ''often'' from. I said in this specific example that it refers to time. You said it didn't, remember?
I said that often 0, as an index, can refer to time. In the sense that a 4 vector is of the form $$v = v^{a}\frac{\partial}{\partial x^{a}} = v^{0}\frac{\partial}{\partial t} + v^{1}\frac{\partial}{\partial x} + v^{2}\frac{\partial}{\partial y} + v^{3}\frac{\partial}{\partial z}$$. That is
often used in relativity. If you'd read much relativity (not wordy explanations, actual relativity) you'd know that my use of the word 'often' is entirely valid.
Thanks, I'm glad I got your okay to do so....
What would you know I have read? Or what I have taken in?
I gauge it from your posts, just as people gauge my level of understanding from my posts and everyone else does for everyone else's posts. If I couldn't add up and didn't know what a function was few people would think I'd spent much time reading mathematics. You wouldn't have to know
precisely what I had or hadn't read, you'd be able to tell that if I haven't heard of one of the most fundamental concepts in mathematics then I can't have gotten far in my reading of it. Or if I couldn't add up then I am unlikely to be able to do more advanced stuff.
You can't add vectors. You can't even divide by 2 or take square roots properly!! And you expect us to think you're familiar with more advanced stuff?
Trying to get to know the works. I know near enough nothing about electromagnetism.
You just like to talk about vorticity and about how you've come up with a result yourself?
I do, I link to papers all the time you disingenuous prick.
You link to wordy essays. The rare occasion you do link to something with an ArXiv or journal reference you don't show any understanding of it.
I speak of any closed system with an energy. In the case of the wave function encoding the information of that equation, the energy vanishes.
They encode all the relevent information about the equation, which is really the whole point.
Clearly my explanation went over your head. If you had a little more experience with basic differential equations and operators you'd have understood why there isn't a unique bijection between the wave function and the Hamiltonian. This isn't even quantum mechanics, its analysis 101.
For instance, consider the particle trapped in a box of constant potential. You end up with a system of the form $$H = \frac{\partial^{2}}{\partial x^{2}} + k$$. If you solve for the wave-function the equation $$H \psi(x) = 0$$ then you have
TWO linearly independent solutions and thus infinitely many possible solutions, ie $$\psi(x) = A\psi_{1}(x) + B \psi_{2}(x)$$ (where the individual solutions happen to be sin and cos). Given
just the Hamiltonian you cannot reconstruct the wave function uniquely. Basically the equation $$H\psi(x) = 0$$ is a second order differential equation, which means that you're going to need initial conditions to uniquely determine the wave function. That's something true of any kind of differential system, you have to have initial/boundary conditions in order to extract out a specific solution from the (typically) infinite choices you have.
Going the other way, if you give me a wave function then the question "What is the Hamiltonian for this wave function which satisfies $$H\psi = 0$$?" is akin to asking "Give me the
unique second order differential operator for which this wave function is a zero eigenvalue eigenfunction.". There's infinitely many such operators, trivially so since if H satisfies that condition then so does kH.
So your "They uniquely encode one another" isn't a matter of arguing physics, this is
fundamental mathematics. The issue of uniqueness of solutions to equations is one 'even physicists' should know.
As I said, I think I understand the equation better than you, afterall, you've had three maybe more concepts about it yourself.
Applying the Schrodinger equation and actually understanding what the states in the Hilbert space can be isn't 'my concept', its simply an application of definitions. The 0
cannot literally be 'time' since the Hilbert space the equation is defined upon doesn't include that.
I've explained many times and the fact you still don't get it demonstrates that, despite your claims, you don't understand even the simplest thing about Hilbert spaces.
It is a problem of pure general relativity. It acts as a solution of pure relativity. You were wrong, end of.
Are you hoping if you keep saying it it'll make it so?
No, but I think I knew a lot more about a Hilbert space than what you gave me credit for, or the average joe.
And yet you maintain the 0 is literally time.
It wasn't dependance. That's the whole point.
The 0 means there's no time dependence. I never said it means there's time dependence.
and..
You're repeating youself.
When you can grasp something the first time you're told it then maybe I'll stop having to repeat myself.
Hissy fit? I am annoyed when I am lynched. I do know what a Hilbert space is. I know all the basics and preliminaries required to understand a Hilbert Space.
Evidence says otherwise.
Alphanumeric seems to be under the impression I believe the Hamiltonian is time-dependant. As he rightfully states, time is not a variable in the Hamiltonian, so how does it arise in the equations?
$$\hat{H}\psi= i \hbar \partial_t \psi$$
No, you failed to understand what I said, which is because you don't understand the Schrodinger equation. Though its possible to formulate a time dependent Hamiltonian (time dependent operators are in the Heisenberg picture, compared to the Schrodinger one), I explained that the Schrodinger equation relates the time dependency
of the wave function to the state obtained by acting the Hamiltonian on the wave function. If the Hamiltonian acts on the wave function to give zero then it means the wave function has no time dependency. This is nothing but a trivial application of the Schrodinger equation and yet you need to have it explained
again.
I guess in some remedial sense, you could say $$\psi$$ does not depend on time. But in every sense of the above work, it does refer to time, or the absence of it.
No, its not some 'remedial sense', if something satisfies $$\partial_{t}\psi = 0$$ then it has no explicit time dependency. Remember what you were saying about agreed upon definitions? Twisting them to suit your purpose does you no favours.
Whilst I had not explicitely believed an dependance of time with the operator, it seems it can still be allowed using a Dirac Picture, where not only the wave function is time dependant but also the operator:
http://arxiv.org/PS_cache/gr-qc/pdf/0312/0312063v1.pdf
I will just assume you have never heard of it AN, or if you have, must have escaped your notice when trying to trash the thread.
Actually I covered the different 'pictures' in my 3rd year in the 'Principles of Quantum Mechanics' course. And I love how you complain when I supposedly make assumptions about what you have or haven't read yet you have no problem assuming what I have or haven't read.
You seem to be under the misapprehension that you're treading in areas which are advanced and stretching the bounds of what I've done myself. Like I said, the different pictures and time dependency of operators and states is in one of the opening sections of one of the first course in QM I took... *counts quickly in head* .....
6 years ago.
If you hadn't just jumped ahead and decided to copy/paste big words from papers you don't understand and instead you'd actually learnt quantum mechanics step by step you'd have known about the different pictures too. Instead you take a pot shot at me for the supposedly silly notion that operators can be time independent (ironically I never actually referred to such a concept, you misunderstood me), only for your Googling to turn it up a little while later.
I know that threads like this push
your Googling skills to the max and you often come across concepts a few days
after they've come up and then you realise they are common place but some of us don't have that issue. Some of us have this thing called
memory where we can recall almost instantly things we've previously learnt, it makes understanding and discussion so much easier when you have more bits of the puzzle in your head and don't have to use Google every 5 seconds. Try it some time.
In the future, I advise you to read up on things like black holes, before you go about saying their singular regions are not dense points. You obviously know nothing on the works of Penrose and Hawking singularity theorems, for if you had, you would never have made such an assinine comment.
So you've gone from not knowing what 'covariant' means and not being able to add vectors to having a sufficient grasp of the work of Penrose and Hawking that you can tell other people "Do some reading!"?! And you seriously think anyone is going to swallow that load of crap? You claim you're modest and not trying to be a scientist but you want people to think you're well read, ie about Penrose, Hawking, quantum mechanics, Hilbert spaces etc, and you're willing to tell other people "Do your homework first!" when its
painfully clear you haven't yourself.
Its comments like that which make me feel entirely justified in being short with you. You don't know a damn thing about these areas of physics, other than what you can parrot back off Wikipedia, and yet you want people to think you do and you act as if you do. No one buys it. What modicum of scientific merit this thread
might have had goes right out the window when you come up with things like that because
no one believes you understand those things. If you're going to be a hypocritical hack then I'll treat you as such.
