Beginner: uncertainty principle?

  • Thread starter frankypoo
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  • #1
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I'm a newcomer to physics (know of any good beginner forums?) and was wondering why the uncertainty principle is true.
 

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  • #2
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Because when you shine light on a particle to see it, you are using at least one quantum. This quantum will affect the particle in a way we can't predict, therfore limiting the knowledge one can have of a particle's position and velocity.
 
  • #3
Integral
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This is one of those things that is the way it is, because that is the way it is. Why is it that way? Don't know. It just is. Physics cannot address why, we can only observe and analize.

Philosophers talk about why, not Physicists.
 
  • #4
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Originally posted by frankypoo
I'm a newcomer to physics (know of any good beginner forums?) and was wondering why the uncertainty principle is true.
first of all, the uncertainty principle isn t always true. knowing when it is true and when it is not goes a long way towards explaining why it is true (when it is)

whenever two observables do not commute, then they obey the heisenberg uncertainty relationship.

when the observables do not commute, they can be shown to be fourier transforms of one another, and fourier transforms exchange narrow signals for wide signals.

therefore, if one signal is wide, the other one has to be narrow, because they do not commute.
 
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  • #5
turin
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Originally posted by lethe
when the observables do not commute, they can be shown to be fourier transforms of one another, ...
Wow, I've never heard that before. Are you over-generalizing for the sake of clarity, or is this absolutely true? Can you give a simple proof?
 
  • #6
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Originally posted by turin
Wow, I've never heard that before. Are you over-generalizing for the sake of clarity, or is this absolutely true? Can you give a simple proof?
I think Lethe really meant two observables which obey canonical commutation relations, i.e. [itex][\hat{A},\hat{B}] = i\hbar[/itex], at least concerning the Fourier transform bit. (But it is true that any two non-commuting observables obey a Heisenberg uncertainty relation, even if their commutator is more complicated than [itex]i\hbar[/itex].)
 
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  • #7
jcsd
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It's called the generalized uncertainty principle and for two observables [itex]q[/itex] and [itex]r[/itex] can be stated as :

[tex]\Delta q \Delta r \geq \frac{1}{2}|\langle[\hat{Q},\hat{R}]\rangle|[/tex]
 
  • #8
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Originally posted by turin
Can you give a simple proof?
sure, i can give a proof. here we go:

suppose i have two self adjoint observables A, and B with continuous spectra satisfying the following commutation relation:
[tex]
[A,B]=i
[/tex]

let a and a' be eigenvalues for A, and b be an eigenvalue for B.

[tex]
<a'|[A,B]|a>=(a'-a)<a'|B|a>=i\delta(a'-a)
[/tex]
but also:
[tex]
<b|B=b<b|
[/tex]
so
[tex]
<b|B|a>=b<b|a>
[/tex]

putting these together yields:
[tex]
\begin{multline*}
<b|B|a>=\int da'<b|a'><a'|B|a>\\
=-i\int da'<b|a'>\delta'(a'-a)=i\frac{d}{da}<b|a>
\end{multline*}
[/tex]
solving this differential equation yields:
[tex]
<b|a>=e^{-iab}
[/tex]
finally, take a generic state in the Hilbert space, [itex]|\psi>[/itex]:

[tex]
<b|\psi>=\int da<b|a><a|\psi>=\int da \ e^{-iab}<a|\psi>
[/tex]

and so they are fourier transforms
 
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  • #9
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Originally posted by Ambitwistor
I think Lethe really meant two observables which obey canonical commutation relations, i.e. [itex][\hat{A},\hat{B}] = i\hbar[/itex], at least concerning the Fourier transform bit
yes, very good. what i said is only true for variables obeying canonical commutation relations.
 
  • #10
turin
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Thanks alot lethe; that was excellent!




Originally posted by lethe
... observables A, and B with continuous spectra ...
Does this condition contain a condition of non-degeneracy? I guess it doesn't matter since you are taking the derivative of the delta function (something I thought you would've frowned upon, though)?



Originally posted by lethe
yes, very good. what i said is only true for variables obeying canonical commutation relations.
This was the main part of my confusion.
 
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  • #11
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Originally posted by turin

Does this condition contain a condition of non-degeneracy? I guess it doesn't matter since you are taking the derivative of the delta function (something I thought you would've frowned upon, though)?
this makes no assumptions about degeneracy.

derivatives of the delta function(al) are perfectly well defined.
 
  • #12
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Back to the beginning post...

IMHO, the best way to understand HUP is to picture the analogy of a piece of music. What HUP says is essentially that certain pairs of properties are like the frequency and time of this music. Since frequency is based on repeated oscillations over time, it makes no sense to talk of the music having an exact frequency at an exact time - there is always a minimal degree of uncertainty.

PF is a good forum, beginner or not. :wink:
 

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