Quantum Physics Quantum Physics Help Forum  7Likes
May 20th 2019, 04:46 AM

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Originally Posted by neila9876 The most funny event might be that some physicist are still trying to seek the border between classical physics and QM. 
Actually this is a BIG field in Physics. If we can't prove that QM reproduces Classical Physics on the macroscopic scale then QM is wrong. That would be a huge problem for the Standard Model and might wreck it entirely.
Otherwise you seem to be saying that QM is complicated, more complicated than you would like it to be. Fair enough. It is. No one says that Nature has to be simple.
Dan
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May 20th 2019, 07:38 PM

#12  Senior Member
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 pig vs pig
it seems that this topic has to return to the beginning of this thread...

 
May 22nd 2019, 03:58 PM

#13  Senior Member
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 uncertainty vs uncertainty principle
@dragon:
What I read about the uncertainty principle before is a simple and classical one. It tells: position and momentum can not be determined at the same time.
It it ok too?

 
May 22nd 2019, 05:40 PM

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Originally Posted by neila9876 @dragon:
What I read about the uncertainty principle before is a simple and classical one. It tells: position and momentum can not be determined at the same time.
It it ok too? 
The Uncertainty Principle does not apply to Classical Physics. It does hold for QM, but not for SR and GR, which are Classical in behavior.
Specifically, in this case it says that $\displaystyle \Delta x \cdot \Delta p < \dfrac{\hbar}{2}$. (Some prefer just $\displaystyle \hbar$ here. It doesn't really matter a whole lot...different uses use different constants.) $\displaystyle \Delta x$ is the uncertainty in the measured position and $\displaystyle \Delta p$ is the uncertainty in the measured momentum.
(To be precise we have $\displaystyle \Delta P \cdot \Delta Q < \dfrac{\hbar}{2}$ where P and Q are canonical pairs of variables in Classical Physics. So we could be using E and t, for example.)
Dan
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May 22nd 2019, 06:34 PM

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 x vs p
@dragon:
Maybe Chinese English again. haha...
What I say "classical" above does mean "classical physics", it means "a more traditonal method of description". I don't know how to speak it accurately in English...
Put aside E and t, please...just talk about x and p for simplicity sake.
It seems that you are talking both positon and monentum can not be determined at the same time, while what I asked is if one of them can be determined?

 
May 22nd 2019, 08:52 PM

#16  Forum Admin
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Originally Posted by neila9876 @dragon:
Maybe Chinese English again. haha...
What I say "classical" above does mean "classical physics", it means "a more traditonal method of description". I don't know how to speak it accurately in English...
Put aside E and t, please...just talk about x and p for simplicity sake.
It seems that you are talking both positon and monentum can not be determined at the same time, while what I asked is if one of them can be determined? 
A made a brief comment about that in a different thread.
Here's the short version. If we assume that the particle's wavefunction (wave packet is more specific) then the uncertainty in the position can be thought of "width at half maximum" of the curve. As it happens the momentum representation of this wavefunction is a Dirac delta function, which gives $\displaystyle \Delta p = 0$, that is the momentum is specified exactly. (For a bit more of an explanation, see post #2, here.)
In practice neither $\displaystyle \Delta x$ nor $\displaystyle \Delta p$ are determined exactly.
Dan
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May 23rd 2019, 07:37 PM

#17  Senior Member
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 classical vs QM
Of course, if position x and momentum p are both determined at the same time, it will be classical physics.
Look at the simplest wave function for free particle in QM (or the “glass” in this topic):
Ψ = A exp( i(p•x– Et )) omit “h” and vector marks for convenience of observation.
When position x and momentum p are both determined at the same time, it will no longer be a wave function, instead, an oscillation function. It seems that the “wave character” should disappear either.
What interesting is it could be considered as a special situation and the calculation of probability is still applicable. “A” is a constant. ｜A｜² is a constant. It means the probability of appearance of the particle in any space point is the same. In turn, it means the position of the free particle can not be determined.
Contradiction…
Classical physics vs QM, who is wrong?
Or, it demonstrates that a coin has two aspects and classical physics describes the certain aspect while QM describes the uncertain aspect?
How to solve the “seems” contradiction? The fourth (special) spacial dimension is needed to accommodate the uncertainty (probability of appearance).

 
Jun 20th 2019, 07:57 PM

#18  Senior Member
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How to view a photon from the viewpoint of man?
I remember that another guy ****9876 post an article “Analysis of Probability Wave Speed
” last year in the thread “First Bohr Track Radius”, in the attachment. It’s in attachment here too.
……..
When v → C, Lim u = C,
It means photon is the perfect wavicle, in another word: photon is (a section of electromagnetic) wave. It’s not this guy who can tell more…haha
Something interesting is : considering a photon appear in a 3D space point, and it also propagates ahead in the form of (electromagnetic) wave. That means light speed in photon frame is also C.
For other kinds of particles (wavicles), (probability) wave is (probability) wave, particle is particle; for a photon, the probability wave is the electromagnetic wave and is the photon itself.

 
Jun 20th 2019, 08:20 PM

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Originally Posted by neila9876 How to view a photon from the viewpoint of man?
I remember that another guy ****9876 post an article “Analysis of Probability Wave Speed
” last year in the thread “First Bohr Track Radius”, in the attachment. It’s in attachment here too.
……..
When v → C, Lim u = C,
It means photon is the perfect wavicle, in another word: photon is (a section of electromagnetic) wave. It’s not this guy who can tell more…haha
Something interesting is : considering a photon appear in a 3D space point, and it also propagates ahead in the form of (electromagnetic) wave. That means light speed in photon frame is also C.
For other kinds of particles (wavicles), (probability) wave is (probability) wave, particle is particle; for a photon, the probability wave is the electromagnetic wave and is the photon itself. 
Two comments:
1) The speed of light's symbol is "c", not "C".
2) May I be the first to say: What???? Can you please explain what you are trying to say a bit more clearly? I can't follow this at all!
Dan
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Jun 21st 2019, 05:30 PM

#20  Senior Member
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@Dragon:
Sorry, Dan, I can't understand it clearly either. I just feel interesting to think it this way, just as someone plays math game elsewhere...
If you are willing to understand it, you will understand...

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