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Proof of Heisenberg's Uncertainty Principle

by Ed Quanta
Tags: heisenberg, principle, proof, uncertainty
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Ed Quanta
#1
Mar8-05, 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?
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marlon
#2
Mar8-05, 09:10 AM
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http://www.cbloom.com/physics/heisenberg.html

marlon
marlon
#3
Mar8-05, 09:11 AM
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Quote Quote by Ed Quanta
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?

Because Mother Nature has chosen to behave like this

marlon

marlon
#4
Mar8-05, 09:14 AM
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Proof of Heisenberg's Uncertainty Principle

Besides, check out the original manuscript in my last entry of my journal

marlon
Ed Quanta
#5
Mar8-05, 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>|=2|Im<A'psi,B'psi>|
masudr
#6
Mar8-05, 08:12 PM
P: 932
http://en.wikipedia.org/wiki/Uncertainty_principle

Go to the section "Generalized uncertainty principle".
dextercioby
#7
Mar8-05, 08:55 PM
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The nonmathematized version (the one involving wave-mechanics 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.
masudr
#8
Mar8-05, 09:05 PM
P: 932
Daniel, why is there a minus sign as the subscript of the commutators? I never understood that.
dextercioby
#9
Mar8-05, 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.
masudr
#10
Mar8-05, 10:24 PM
P: 932
Oh... OK. I thought the commutator automatically implied a negative sign, and we have the anti-commutator 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?
dextercioby
#11
Mar8-05, 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.
masudr
#12
Mar8-05, 10:31 PM
P: 932
Oh yes, of course. Sorry, it's 04:30AM here in England, and my mind's not thinking straight!
Ed Quanta
#13
Mar9-05, 11:19 AM
P: 297
Quote Quote by Ed Quanta

And how do we know |<A'psi,B'psi>-<B'psi,A'psi>|=2|Im<A'psi,B'psi>|?

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.
Galileo
#14
Mar9-05, 11:51 AM
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Quote Quote by Ed Quanta
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.
Generally:

[tex]\langle \psi|\phi \rangle = \langle \phi|\psi \rangle^*[/tex]
where the * denotes complex conjugation.

So

[tex]\langle A\psi|B \psi \rangle=\langle B\psi|A \psi \rangle^*[/tex]

For any complex number [itex]z[/itex] we have [itex]z-z^*=2i\Im(z)[/itex].
dextercioby
#15
Mar9-05, 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] |u-u^{*}| [/tex] (3)

The RHS of the equality you want to prove is

[tex] 2|Im \ u | [/tex] (4)


Take the generic algebraic for "u"

[tex] u=:a+ib [/tex] (5) [tex] \Rightarrow u^{*}=a-ib [/tex] (6)

and then

[tex] |u-u^{*}|=|2ib|=2|b|=2|Im \ u| [/tex] (7)

q.e.d.

Daniel.
Tom Mattson
#16
Mar9-05, 02:19 PM
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Quote Quote by Ed Quanta
I don't really follow this derivation. Why does he have to show that (deltaA')^2=(deltaA')^2?
He's not showing that. He's showing that [itex]( \Delta A )^2=< \psi , A'^2 \psi >[/itex]. The fact that he ended up with [itex]( \Delta A )^2=( \Delta A )^2[/itex] simply means that he completed the proof.
Ed Quanta
#17
Mar9-05, 02:37 PM
P: 297
Thanks a lot everyone. Much appreciated.
Antiphon
#18
Mar16-05, 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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