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Proof of Heisenberg's Uncertainty Principle 
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#1
Mar805, 09:07 AM

P: 297

Is there any mathematical or physical proof or derivation of Heisenberg's Uncertainty principle out there? Can someone send me a link to one or provide a proof if it isn't too complicated? I know that in quantum mechanics if two operators don't commute then we can't measure both of these simulataneously. Why does this correlation exist?



#2
Mar805, 09:10 AM

P: 4,006



#4
Mar805, 09:14 AM

P: 4,006

Proof of Heisenberg's Uncertainty Principle
Besides, check out the original manuscript in my last entry of my journal
marlon 


#5
Mar805, 09:46 AM

P: 297

I don't really follow this derivation. Why does he have to show that (deltaA')^2=(deltaA')^2?
And how do we know <A'psi,B'psi><B'psi,A'psi>=2Im<A'psi,B'psi> 


#6
Mar805, 08:12 PM

P: 932

http://en.wikipedia.org/wiki/Uncertainty_principle
Go to the section "Generalized uncertainty principle". 


#7
Mar805, 08:55 PM

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The nonmathematized version (the one involving wavemechanics formalism) is in the beginning of Davydov's book,IIRC.Anyway,every book on QM has the proof for the generalized version:
[tex] \Delta \mathcal{A}\cdot\Delta \mathcal{B}\geq\frac{1}{2} \langle[\hat{A},\hat{B}]_{}\rangle _{\psi\rangle} [/tex] Daniel. 


#8
Mar805, 09:05 PM

P: 932

Daniel, why is there a minus sign as the subscript of the commutators? I never understood that.



#9
Mar805, 10:12 PM

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Because it is the commutator...?
[tex] [\hat{A},\hat{B}]_{}=:\hat{A}\hat{B}\hat{B}\hat{A} [/tex] and that's how it's elegantly specified the fact that one speaks about commutators of (linear) operators... Daniel. 


#10
Mar805, 10:24 PM

P: 932

Oh... OK. I thought the commutator automatically implied a negative sign, and we have the anticommutator for the version with the plus sign. Ok never mind about that.
I have another question though. If we reverse the order of the commutator, i.e. [tex][\hat{A},\hat{B}] = [\hat{B}, \hat{A}][/tex], we get a minus sign in the uncertainty. But is that of any significance if the product of the variances is negative as opposed to positive? 


#11
Mar805, 10:28 PM

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If you haven't seen so far,there's a modulus after performing the average of the commutator on the (pure) quantum state [itex] \psi\rangle [/itex]
Daniel. 


#12
Mar805, 10:31 PM

P: 932

Oh yes, of course. Sorry, it's 04:30AM here in England, and my mind's not thinking straight!



#13
Mar905, 11:19 AM

P: 297

If I can understand this, I understand the derivation of the Uncertainty Principle. But I can't make sense of this one line in the original link provided by Marlon. 


#14
Mar905, 11:51 AM

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[tex]\langle \psi\phi \rangle = \langle \phi\psi \rangle^*[/tex] where the * denotes complex conjugation. So [tex]\langle A\psiB \psi \rangle=\langle B\psiA \psi \rangle^*[/tex] For any complex number [itex]z[/itex] we have [itex]zz^*=2i\Im(z)[/itex]. 


#15
Mar905, 11:58 AM

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This is very simple.
[tex] \langle \psi\hat{A}\hat{B}\psi\rangle =:u\in \mathbb{C} [/tex] (1) Then,using the property: [tex] \langle \psi\hat{A}\hat{B}\psi\rangle = \langle \psi\hat{B}\hat{A}\psi\rangle ^{*} [/tex] (2) ,we can write the LHS of the equality u wish to prove as: [tex] uu^{*} [/tex] (3) The RHS of the equality you want to prove is [tex] 2Im \ u  [/tex] (4) Take the generic algebraic for "u" [tex] u=:a+ib [/tex] (5) [tex] \Rightarrow u^{*}=aib [/tex] (6) and then [tex] uu^{*}=2ib=2b=2Im \ u [/tex] (7) q.e.d. Daniel. 


#16
Mar905, 02:19 PM

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#17
Mar905, 02:37 PM

P: 297

Thanks a lot everyone. Much appreciated.



#18
Mar1605, 09:31 AM

P: 1,781

The very simplest answer is this.
If you want to determine the frequency of a signal by counting pulses, your frequency determination gets more accurate the longer you count. In QM, frequency determies energy so the less time you have to count the frequency, the less certain you will be of the exact energy. 


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