God and Quantum Mechanics

[lionking]

*Sigh*

What I am pointing out is your reaction to people who do know what they are talking about when they point out fundamental errors of yours. Reactions like telling sol invictus he knows no physics.

Oh, and use a spellchecker.

 

[ben m]

And the integral ben... i am
surprised you do not know it. The integral expresses the domain (or the point of collapse of some wave [latex]\psi[/latex]). The definition of the integral is the collapse of the wave function itself. You just showed a notation of the same context, but one which doesn't take into respect time.

a) Collapse onto what eigenvector space? Position? Velocity? The squeezing operator?

b) If I am wrong about some point of notation, surely there's some reliable source to back you up. Please go back to your quantum physics textbook where you learned this---you're a student, right?---and give a full citation, including title, author, and page, which will convince me that your function "is the collapse" and takes "into respect time".

 

[Singularitarian]

a) Collapse onto what eigenvector space? Position? Velocity? The squeezing operator?

b) If I am wrong about some point of notation, surely there's some reliable source to back you up. Please go back to your quantum physics textbook where you learned this---you're a student, right?---and give a full citation, including title, author, and page, which will convince me that your function "is the collapse" and takes "into respect time".

This is a joke right? No one here has heard of the absolute square value of the wave function?

Link: http://en.wikipedia.org/wiki/Wave_function

''It is commonly applied as a property of particles relating to their wave-particle duality, where it is denoted ψ(position,time) and where | ψ | 2 is equal to the chance of finding the subject at a certain time and position.[1] ''

I have simply given the absolute sqaure an integral in respect with the time's domain. This is perfectly logical mathematics. The entire expression yields statistically a positive value of 1. Hence, the collapse.

 

[Singularitarian]

*Sigh*

What I am pointing out is your reaction to people who do know what they are talking about when they point out fundamental errors of yours. Reactions like telling sol invictus he knows no physics.

Oh, and use a spellchecker.

I could piss myself reading this **** man! Funniest **** i've seen all day.

 

[lionking]

I could piss myself reading this **** man! Funniest **** i've seen all day.

Childish response.

I am only following this thread in the so far fruitless hope that you will admit that you do not know what you think you know about physics. If I were agruing with professionals like you are and getting monumentally trashed, I would slink back into my hole.

But keep it up, it's funny as hell.

 

[Singularitarian]

I don't care if i sound childish. I'm sick of trying to be adult when i have people telling me i don't know what i am talking about. I mean, i can't believe for instance, a forum supposidly having its share of scientists haven't heard of the asbolute sqaure of the wave function. I just find it fascinating, as to your sheepish behaviour.

 

[ben m]

This is a joke right? No one here has heard of the absolute square value of the wave function?

Link: http://en.wikipedia.org/wiki/Wave_function

''It is commonly applied as a property of particles relating to their wave-particle duality, where it is denoted ψ(position,time) and where | ψ | 2 is equal to the chance of finding the subject at a certain time and position.[1] ''

I have simply given the absolute sqaure an integral in respect with the time's domain. This is perfectly logical mathematics. The entire expression yields statistically a positive value of 1. Hence, the collapse.

The integral [latex]\int \mid \psi(\vec{x},t) \mid^2 d^3\vector{x}[/latex] (more universally [latex]\langle \psi \mid \psi \rangle[/latex]) is indeed equal to 1. This has, however, nothing to do with collapse; it's a statement of the conservation of probability. It is true whether or not you make an observation of any sort.

The integral [latex]\int \mid \psi(\vec{x},t) \mid^2 d^3\vector{x} dt[/latex] has units of time, but at least it's time-translation invariant. It is not generally equal to one. It also does not appear to have anything to do with collapse. I am not aware of any place this integral is used.

The integral [latex]\int \mid\psi(\vec{x},t)\mid^2 dt[/latex] has units of time/length^3. It is not generally equal to one and depends on both space and time. It also does not appear to have anything to do with collapse. I am not aware of any place this integral is used.

Your definition for "collapse" seems to be different than the mainstream one.

 

[Singularitarian]

The integral [latex]\int \mid \psi(\vec{x},t) \mid^2 d^3\vector{x}[/latex] (more universally [latex]\langle \psi \mid \psi \rangle[/latex]) is indeed equal to 1. This has, however, nothing to do with collapse; it's a statement of the conservation of probability. It is true whether or not you make an observation of any sort.

The integral [latex]\int \mid \psi(\vec{x},t) \mid^2 d^3\vector{x} dt[/latex] has units of time, but at least it's time-translation invariant. It is not generally equal to one. It also does not appear to have anything to do with collapse. I am not aware of any place this integral is used.

The integral [latex]\int \mid\psi(\vec{x},t)\mid^2 dt[/latex] has units of time/length^3. It is not generally equal to one and depends on both space and time. It also does not appear to have anything to do with collapse. I am not aware of any place this integral is used.

Your definition for "collapse" seems to be different than the mainstream one.

Anyone who has worked on statistical wave mechanics knows that the probability density is what defines the collapse of the wave function, the point at which you might find a particle in a given state, whatever state you chose.

You can't allay this one by stating what you did. I could have done that, but i'm not referring to actual eigenstates such as a position or an energy, but instead i gave the mechanism in which such states are observable, again with the probability density equation i provided. And its certainly not different from the mainstream one. In Doctor Cramers Transactional Interpretation has the collapse of the wave function defined by the absolute square of the wave function <E(t_1)|O(t_2)>, where E stands for an echo wave, and O is for an offer wave. The statistical mathematics are actually also used in the Copenhagen interpretation as a basis rule. So you are simply wrong

 

[Mashuna]

Childish response.

I am only following this thread in the so far fruitless hope that you will admit that you do not know what you think you know about physics. If I were agruing with professionals like you are and getting monumentally trashed, I would slink back into my hole.

But keep it up, it's funny as hell.

It's kind of like watching Sideshow Bob stepping on a rake, then turning to step on another, ad infinitum.

 

[Singularitarian]

It's kind of like watching Sideshow Bob stepping on a rake, then turning to step on another, ad infinitum.

How can anyone (hence yourself) support that paragraph, when evidently these ''professionals'' (supposidly) crushing my every move here, or something along those lines, don't even know about the absolute square of the wave function, and being a [probability density] for collapse equations. I mean, both of them where seriously like, ''what does that mean?''

This statistical mathematics is taught very early on in University. I can't even contemplate why they never recognized it.

 

[lionking]

How can anyone (hence yourself) support that paragraph, when evidently these ''professionals'' (supposidly) crushing my every move here, or something along those lines, don't even know about the absolute square of the wave function, and being a [probability density] for collapse equations. I mean, both of them where seriously like, ''what does that mean?''

This statistical mathematics is taught very early on in University. I can't even contemplate why they never recognized it.

Spellcheck is that abc thing in the top right corner.

 

[Mashuna]

How can anyone (hence yourself) support that paragraph, when evidently these ''professionals'' (supposidly) crushing my every move here, or something along those lines, don't even know about the absolute square of the wave function, and being a [probability density] for collapse equations. I mean, both of them where seriously like, ''what does that mean?''

This statistical mathematics is taught very early on in University. I can't even contemplate why they never recognized it.

Turn, <boing>, SPLAT!

 

[BNRT]

Only difference is, is that it cannot be applied to one observer, such as an atom. It requires all inertial observers, including us.

Now I'm confused. According to you, does the theory of relativity need a human observer, all human observers, or, since you're talking about atoms, everything in the universe as observers?

 

[GeeMack]

Yes he did, but his theory initiates awareness, the knowledge of time passing more slowely for observers. Why doesn't anyone understand this makes his theory observer-dependant... its quite simple really, inavoidable. Telling me it is not observer-dependant, is like telling me a Mars Bar has no chocolate.

Usually when you have to ask why nobody understands something that you're trying to express, the answer seems to come down to one of these two possibilities. First, what you're saying can't be understood. It's nonsense. Second, you're incapable of communicating effectively. You aren't making your point clear enough to be understood.

So maybe you should reassess your position, since there seems to be a very good possibility that what you're saying is crap. And if, after reassessing your position, you still feel it's not nonsense, then you probably should reassess your method of trying to communicate it. Because if your position is sound and you can't get anyone to understand, then in all likelihood, your ability to communicate is crap.

 

[Upchurch]

I don't care if i sound childish. I'm sick of trying to be adult when i have people telling me i don't know what i am talking about.

That's because you don't know what you are talking about. Experiments with orbiting atomic clocks have experienced time dilation with no observers present. Gravitational lensing has been recorded without anyone out near the sun to watch it happen. Relativistic effects are not observer dependent.

That Einstein used them in his thought experiments does not indicate that they are crucial to the effects occurring. I challenge you to find something Einstein wrote that indicates otherwise.

 

[ben m]

Anyone who has worked on statistical wave mechanics knows that the probability density is what defines the collapse of the wave function, the point at which you might find a particle in a given state, whatever state you chose.

You didn't give a probability density. You gave the integral of a probability density. The probability integral---presuming you meant the one which equates to 1---is the one thing you know about a wavefunction without observing it. It's always 1.

If you had said "| \Psi |^2 can only be observed by collapsing \Psi" I would have shrugged and said 'close enough'. But that's not what you said.

 

[Foster Zygote]

I could piss myself reading this **** man! Funniest **** i've seen all day.

I had a similar reaction when I saw you address a science fiction plot device as though it was a real physics theory.

 

[Singularitarian]

Now I'm confused. According to you, does the theory of relativity need a human observer, all human observers, or, since you're talking about atoms, everything in the universe as observers?

Does it require observers, more specifically us?

I want to say ''yes'', but it's a reserved ''yes'' because of the steps required to think this way.

Take for analogy, a theory like relativity. It has to do with noticing differential times for all inertial observers. Since we classify as an inertial observer, the theory of special relativity includes us then as being an observer-dependant model. It not magical or anything, its practically the form of the theory, and how it demonstrates the effects of time dilation and gravitational dilation.

If you removed our role in varifying the experiments set by relativity, then relativity only works for atomic observers. To show we are no different to the experience of that, Einsteins demonstrated the twin paradox, making us essentially within the same mix of principles.

 

[Singularitarian]

You didn't give a probability density. You gave the integral of a probability density. The probability integral---presuming you meant the one which equates to 1---is the one thing you know about a wavefunction without observing it. It's always 1.

If you had said "| \Psi |^2 can only be observed by collapsing \Psi" I would have shrugged and said 'close enough'. But that's not what you said.

It was still right.

 

[Singularitarian]

Usually when you have to ask why nobody understands something that you're trying to express, the answer seems to come down to one of these two possibilities. First, what you're saying can't be understood. It's nonsense. Second, you're incapable of communicating effectively. You aren't making your point clear enough to be understood.

So maybe you should reassess your position, since there seems to be a very good possibility that what you're saying is crap. And if, after reassessing your position, you still feel it's not nonsense, then you probably should reassess your method of trying to communicate it. Because if your position is sound and you can't get anyone to understand, then in all likelihood, your ability to communicate is crap.

I don't even know how to respond to this slugde of words.