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How did Heisenberg derive his famous principle? 
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#1
Mar1912, 03:14 AM

P: 84

Where
How did he come up with that? 


#2
Mar1912, 06:57 AM

P: 615

If I'm not mistaken Heisenberg himself didn't 'derive' the principle, he thought experimented it and got [itex]\Delta x \ \Delta p \approx \hbar[/itex]
It was later on that other people found general formulas for uncertanties related with any operators; [itex]\sigma _A \sigma _B \ge \frac{[A,B]}{2i}[/itex] And since the commutator (the [A,B] part) for position and momentum is simply [itex]i\ \hbar[/itex] we get [itex]\sigma_x \sigma_p \ge \frac{i\ \hbar}{2 i} = \frac{\hbar}{2}[/itex] 


#3
Mar1912, 07:21 AM

Sci Advisor
P: 1,676

The mathematical inequality actually follows from Fourier analysis. It's application as a physical result  to be interpreted as physical uncertainties of measurable quantities  is generally taken to be an axiom of the theory.



#4
Mar1912, 08:14 AM

P: 1,414

How did Heisenberg derive his famous principle?



#5
Mar1912, 08:17 AM

Mentor
P: 11,602

This has been known since probably sometime in the 1800s, so when physicists started to think of describing particles using waves, and superpositions of them, it was only a matter of time before someone applied this general uncertainty principle to QM. 


#6
Mar1912, 08:48 AM

P: 1,414

I don't really understand the basis for Heisenberg's matrix mechanics, and I don't really understand the basis for Schroedinger's wave equatiion. But those are topics for other threads. 


#7
Mar1912, 09:04 AM

Sci Advisor
P: 1,676




#8
Mar1912, 11:10 AM

P: 1,414




#9
Mar1912, 05:03 PM

Mentor
P: 11,602

Schrödinger came up with his equation by making an analogy between mechanics and optics, which has two distinct "pictures"; geometrical or ray optics on one hand, and wave optics on the other. He reasoned that classical mechanics of particles might be analogous to ray optics, and invented a new mechanics which is analogous to wave optics: http://www.physicsforums.com/showthr...069#post418069 The early development of modern quantum mechanics is based on guesses like this. Schrödinger, Heisenberg and others were groping around, trying to come up with something that works. Only gradually did these bits and pieces come together into what you see in today's textbooks. 


#10
Mar1912, 06:02 PM

Sci Advisor
P: 1,676

Huh. I need to refresh my history. I recall reading that Schroedinger and Co. had some inklings that matter had wavelike properties, and that's what directed him to "look" for a wave equation. Were there any other experiments being done at the time that might have provided hints of waveparticle duality?



#11
Mar1912, 06:07 PM

P: 297

Hey, a quick question.
Isn't the relation dependent on the method we are using to find the variables (like x and v of electron? Like on the microscope? Like for eg if I had superman eyes (i hope this assumption is not an infraction.I am trying to make a point :) )then I would be able to see the electron move,see exactly where it is and also calculate its exact velocity.No problem would occur. 


#12
Mar2012, 03:35 AM

P: 615

No matter what your measuring device you will never be able to tell its position and momentum exactly. Take a dirac delta function (or a wave with one really big spike somewhere), you can tell where it is but you cannot tell it's frequency since it's just one spike. Similarly, take a uniform sine wave, you can tell it's frequency but you cannot tell where it is. This is more the nature of the xp uncertainty principle. If, in the position basis we have a delta function (this just means we know its position exactly) we cannot gain any useful information about its momentum (we cannot tell it's frequency). If we have a delta function in the momentum basis (we know it's momentum exactly) we have a specific frequency (just like the uniform sine wave) and we cannot gain any information about its position. We can get interesting little 'wave packets' where we can have some information about the position and some about the momentum, the gaussian wavepacket gives us the minimum uncertainty [itex]\sigma_x \sigma_p = \frac{\hbar}{2}[/itex] Also, going back to your superhuman eyes example, for these super human eyes to be able to give precice information about where the electron is, you need to bombard it with short and shorter wavelength photons which have higher and higher energies. If you bombard it with long wavelengthed photons, you won't disturb it much but you can essentially only know if the electron is in an area of size proportional to the wavelength of the photon. 


#13
Mar2012, 04:35 AM

P: 297

Oh.Now i get it. The way they usually talk about in books, led me to think it was an apparatus dependent error. Thanks a lot for clearing that up :) 


#14
Mar2012, 05:05 AM

P: 615




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