Interpretation of the uncertainty principle

In summary, the uncertainty principle states that it is impossible to know both the position and momentum of a particle with arbitrary precision. This principle applies to any non-commuting self-adjoint operators associated with classical observables. "Knowing" in this context means determining the observables with maximum precision. The principle states that the accuracy of the apparatus can be made as small as desired, but the standard deviations of position and momentum will always obey Heisenberg's formula. This is because the theory of measurement in quantum mechanics is statistical and relies on the concepts of compatible and non-compatible observables. The deviations from the average values of position and momentum are represented by \Delta x and \Delta p_x, respectively. This can be seen in the
  • #1
broegger
257
0
I have a fundamental question regarding the uncertainty principle (position and momentum). This principle states that it is impossible simultaneously to know both position and momentum with arbitrarily good precision - even in principle.

Now, what does "know" mean? No one can prevent you from doing a simultaneous measurement on position and momentum, and the values you get must be the position and momentum at the time of measurement (within the accuracy of the apparatus, which - in principle - can be made as small as you like). The position and momentum becomes well-defined when measured.

By my understanding the only thing that the uncertainty principle says is that if we perform identical experiments and we get a small spread in x-values we will get a large spread in p-values according to [tex] \Delta x \Delta p_x \geq \bar{h} [/tex]. [tex]\Delta x[/tex] should be interpreted as the standard deviation in the measured x-values and [tex]\Delta p_x[/tex] as the standard deviation in the measured p-values.
 
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  • #2
broegger said:
I have a fundamental question regarding the uncertainty principle (position and momentum).


Excellent.

broegger said:
This principle states that it is impossible simultaneously to know both position and momentum with arbitrarily good precision - even in principle.

That's the simplified version due to Werner Heisenberg.The uncertainty principle is much more general and applies to any non commuting self adjoint operators associated by quantization to classical (Hamiltonian) observables.

broegger said:
Now, what does "know" mean?

Knowing means determining the observables with arbitrary (hence even maximum) precision.That is to say that for observables A and B,[tex] \Delta A [/tex] and [tex] \Delta B [/tex] can be 0,either both,or only one (irrelevant which).


broegger said:
No one can prevent you from doing a simultaneous measurement on position and momentum, and the values you get must be the position and momentum at the time of measurement (within the accuracy of the apparatus, which - in principle - can be made as small as you like). The position and momentum becomes well-defined when measured.

Now comes the beautiful part of QM.Since,for an observable A,u define [tex] \Delta A =: A-<A>[/tex],which in self adjoined operator context reads [tex] \Delta A\hat =: A\hat-<A\hat>_{I\psi>}[/tex],and the commutation relationship between the self adjoint operators [tex] x\hat \mbox{and} p_{x}\hat [/tex] tells u that their commutator is proportional to the unit operator,then the principle stated above steps in and tells u that whichever the quantum state [tex]{I\psi>}[/tex] pure or mixed,u determine the averages of the operators [tex] x\hat \mbox{and} p_{x}\hat [/tex] u have no chance of finding [tex] \Delta x \mbox{or/and} \Delta p_{x} [/tex] nil.
So yes,u can make the accuracy of the apparatus as small as u like,but when measuring simultaneously position on an axis and momentum on the same axis always the standard deviations will obbey Heisenberg formula.

broegger said:
By my understanding the only thing that the uncertainty principle says is that if we perform identical experiments

Depends on what u mean by "identical" experiments.To me,there are no "identical" experiments,therefore the theory of measurement (in QM especially) is merely statistical.

broegger said:
and we get a small spread in x-values we will get a large spread in p-values according to [tex] \Delta x \Delta p_x \geq \bar{h} [/tex]. [tex]\Delta x[/tex] should be interpreted as the standard deviation in the measured x-values and [tex]\Delta p_x[/tex] as the standard deviation in the measured p-values.

The theory of measurement in QM is very important and relies on the basic concepts of compatible and noncompatible observables.I would advise u to read upon these matters...


DANIEL-CRISTIAN CIOBOTU.
 
  • #3
broegger said:
No one can prevent you from doing a simultaneous measurement on position and momentum, and the values you get must be the position and momentum at the time of measurement (within the accuracy of the apparatus, which - in principle - can be made as small as you like). The position and momentum becomes well-defined when measured.
Nature certainly can! In order for you to measure position "as small as you like" you will have to use light, or electrons (as in an electron microscoe), or whatever, with a very small wavelength. The smaller the wavelength, the greater the energy contained in a "quantum"- and whatever you are measuring with changes the momentum of the object you are measuring. The point of the uncertainty principle is that there ARE pairs of quantities that CANNOT be measured simultaneously- position and momentum are one such pair.
(In terms of quantum mechanics, only quantities whose corresponding operators COMMUTE can be measured simultaneously- and most operators do not commute.)
 
  • #4
What are [tex]\Delta x[/tex] and [tex]\Delta p_x[/tex] exactly? They are not deviations from the "true values" of [tex]x[/tex] and [tex]p_x[/tex] (like classical uncertainties) since position and momentum are in general not well-defined...?

Let's say we have a stream of electrons accelerated from rest through a constant potential and directed through a single slit and recorded on a screen. If the slit is very narrow the x-positions of the electrons are well-defined when going through, but the x-momentum has a huge spread corresponding to a wide pattern on the screen. Is this not the uncertainty principle in action?

Please check out this thread https://www.physicsforums.com/showthread.php?t=48579 (especially post #9), which I think contradicts some of the things you are saying.

I'm not saying you're wrong (you probably aren't). I just want a better understanding of this.
 
  • #5
broegger said:
What are [tex]\Delta x[/tex] and [tex]\Delta p_x[/tex] exactly? They are not deviations from the "true values" of [tex]x[/tex] and [tex]p_x[/tex] (like classical uncertainties) since position and momentum are in general not well-defined...?

They are the deviations from the average values,as stated above:[tex] \Delta \hat{A} =: \hat{A} - <\hat{A}>_{|\psi>} \hat{1}[/tex].

broegger said:
Let's say we have a stream of electrons accelerated from rest through a constant potential and directed through a single slit and recorded on a screen. If the slit is very narrow the x-positions of the electrons are well-defined when going through, but the x-momentum has a huge spread corresponding to a wide pattern on the screen. Is this not the uncertainty principle in action?

Yes it is,it's basically what Feynman says in his book.

broegger said:
Please check out this thread https://www.physicsforums.com/showthread.php?t=48579 (especially post #9), which I think contradicts some of the things you are saying.

In what sense??

broegger said:
I'm not saying you're wrong (you probably aren't). I just want a better understanding of this.

The "classical" slit experiment is the perfect example of an experimental proof that Heisenberg's principle doesn't just hold,but it actually rules the bahavior of matter at microscopical level.
 

1. What is the uncertainty principle?

The uncertainty principle is a fundamental concept in quantum mechanics that states that it is impossible to know certain pairs of physical properties of a particle, such as its position and momentum, with absolute precision at the same time.

2. How does the uncertainty principle relate to the behavior of particles?

The uncertainty principle tells us that at the quantum level, particles do not behave in predictable ways like classical objects. Instead, they exist in a state of probability and have a range of possible outcomes for their properties.

3. Can the uncertainty principle be violated or overcome?

No, the uncertainty principle is a fundamental aspect of quantum mechanics and cannot be violated or overcome. It is a natural limit to our ability to measure and understand the behavior of particles.

4. What are the implications of the uncertainty principle?

The uncertainty principle has major implications for our understanding of the physical world, as it challenges our intuitive understanding of cause and effect and the concept of determinism. It also has practical applications in fields such as quantum computing and cryptography.

5. How does the uncertainty principle impact our daily lives?

The uncertainty principle may seem like a concept that only applies to the microscopic world, but it has implications for our daily lives. For example, it plays a role in the development of new technologies and materials, and it has also been used to explain phenomena such as the stability of atoms and the behavior of electrons in chemical reactions.

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