Physics Help Forum The Meaning of the Uncertainty Principle

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Jan 24th 2016, 01:42 AM   #1
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The Meaning of the Uncertainty Principle

I have started this thread because the phrase "uncertainty in momentum" was used in another thread.

 Another thing about this that students will need to learn is what happens when a photon whose momentum has some uncertainty to it.
Need help with: Absorption of electrons/photons

Point for discussion.

What does this mean?

Does it mean that the (quantum) object does not have an exactly defined momentum?

Or does it mean that there is an exact number, just that we can never know it?

The UP arises naturally in the pure mathematics of the Schrodinger Wave equation or Dirac's relativistic version, as a result of non commutativity of operators.

So mathematically the quoted uncertainty means that the calculated momentum doesn't satisfy the equation perfectly. i.e. the equation is not a perfect model.

Physically we can interpret this as noting that we get slightly different results (solutions) depending upon whether we calculate the momentum first or the the position first. This slight difference is, of course, the uncertaintly.

 Jan 24th 2016, 08:09 AM #2 Senior Member     Join Date: Apr 2008 Location: Bedford, England Posts: 668 Just to get the ball rolling. OK, I admit that I am speculating way outside my knowledge here. Could it imply that the momentum and position are not independent properties? You are actually looking at a single property MomentumPosition, they only make sense together, try to look at them independently and you get a seemingly incongruous result. __________________ You have GOT to Laugh !
Jan 25th 2016, 07:40 AM   #3
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 Originally Posted by studiot I have started this thread because the phrase "uncertainty in momentum" was used in another thread. Need help with: Absorption of electrons/photons Point for discussion. What does this mean? Does it mean that the (quantum) object does not have an exactly defined momentum?
Just as there is a wave function for position, i.e. Psi(x, t), there is also a wave function for momentum, i.e. Psi(p, t). If this function is not a Dirac delta function then it means that measurement of momentum can yield more than one possible result. The same thing holds for all quantum observables. But until an observable is measured it doesn't even have a value.

 Jan 25th 2016, 01:24 PM #4 Senior Member     Join Date: Apr 2008 Location: Bedford, England Posts: 668 I have always had a problem with the Copenhagen interpretation because of the vagueness of the term "observed" In my own mind I view the term "observed" as indicating an interaction with the wider universe (rather than an interaction with a conscious being). The uncollapsed wave-functions seem to be fairly delicate, requiring the system to be carefully isolated from wider interactions if they are to be maintained. I understand that almost macroscopic objects have been induced into superposed states (buckyballs, viruses, even recently a tiny tuning fork), However as the system becomes more complex, a superposed quantum state seems to become harder to achieve. Would I be correct in suspecting that more complex systems have more sharply defined (more tightly constrained) wave-functions? __________________ You have GOT to Laugh ! Last edited by MBW; Jan 26th 2016 at 04:59 AM.
Jan 25th 2016, 02:56 PM   #5
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 Just as there is a wave function for position, i.e. Psi(x, t), there is also a wave function for momentum, i.e. Psi(p, t).
As I understand it, there is only one wave function.
Momentum is derived by applying the momentum operator as a premultiplier.
Yes, MBW, momentum and position are linked since one is the Fourier transform of the other.
So once the momentum is obtained the position follows as a FT.
So alternatively you can apply a position operator to obtain a position and then transform this to momentum.

But you cannot perform both operations together.

This is my point.

Jan 25th 2016, 08:50 PM   #6
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 Originally Posted by studiot As I understand it, there is only one wave function.
The terminology can be tricky. Normally if you see the term wave function without a qualifier then it refers to the position representation of the quantum state $\displaystyle |\Psi(x, y, z, t)>$. Otherwise one speaks of the wave function in the q representation where q is an observable such as position, momentum, energy, etc.

 Originally Posted by studiot Momentum is derived by applying the momentum operator as a premultiplier.
In quantum mechanics there are various terms which refer to momentum:

(1) Momentum operator, P.
(2) Eigenvalues of momentum operator, p.
(3) Expectation of momentum, <p>.

Momentum eigenvalues are what are measured in the lab. To find these eigenvalues of P apply it to the wave function and the following will result

P$\displaystyle \Psi$ = p$\displaystyle \Psi$

As you an see one simply applies P to $\displaystyle \Psi$ and will result multiplied by a constant. That constant is the momentum eigenvalue.

 Originally Posted by studiot Yes, MBW, momentum and position are linked since one is the Fourier transform of the other.
It's the momentum representation and position representation of the wave function which are Fourier transform pairs, not momentum itself.

 Jan 27th 2016, 09:16 AM #7 Senior Member     Join Date: Apr 2008 Location: Bedford, England Posts: 668 What is Uncertain? Position and Momentum are the generally quoted pairing that exhibit this issue, are there other pairings (or perhaps more complex arrangements) where this issue applies? __________________ You have GOT to Laugh !
Jan 27th 2016, 12:05 PM   #8
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 Originally Posted by MBW Position and Momentum are the generally quoted pairing that exhibit this issue, are there other pairings (or perhaps more complex arrangements) where this issue applies?
Yes. However I can't think of any off hand though. First of all please keep in mind that whenever you see the term momentum with linear mechanical momentum. It always means conjugate momentum

But there is an example from EM which might interest you. If there is a charged particle moving in a magnetic field then the conjugate momentum is given by p = mv + qA/c. So you can see that the momentum conjugate to r is not conjugate to mv.

Let me contact a friend of mine who teaches this stuff and ask him and I'll get right back to you with a more detailed answer.

Last edited by Pmb; Jan 27th 2016 at 07:45 PM.

Jan 27th 2016, 12:27 PM   #9
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 Originally Posted by MBW Position and Momentum are the generally quoted pairing that exhibit this issue, are there other pairings (or perhaps more complex arrangements) where this issue applies?
Well that one's easy.

The UP states that the product of momentum and position < = a constant times Planck's constant.

So any pair with the same dimensions (ie LMT^-1 times L that is L^2MT^-1) as Planck's constant will do.

A common alternative is energy and time.

 Jan 27th 2016, 02:29 PM #10 Senior Member     Join Date: Apr 2008 Location: Bedford, England Posts: 668 Leading Questions? My previous post was leading up toward asking if UP always involves a time verses space combination, or to put it another way, is UP fundamentally an issue related to space-time combinations? __________________ You have GOT to Laugh !

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