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# Heisenbergs uncertainty principle Watch

1. Hello there,
I've got a question here that I'm not sure how to do. Any help would be very appreciated.

"Estimate the minimum uncertainty in the velocity of a particle whose position is known to within 1nm"

I believe there is an equation Dq = h(bar)/(2mDv)
so we plug in for Dq and rearrange for Dv?
thanks
ourkid
2. Yes, it's fairly easy to show that for any operators, the following relationship holds:

where is the root mean square deviation of the operator q, defined as

,

where angle brackets denote the expectation value of the operator. Finally the bit in brackets is the commutator of the operators A and B, i.e.
.

This is quite easy to prove, and I can show you the proof if you like.

Anyway, what you need to do then is to find the commutator of position and velocity. Since you know that linear momentum is p = m*v, it's easier just to work out the commutator of the momentum and the position, since you know the operator for momentum straight away.
So you have (where f is a ghost function)
(using the product rule; then cancel the ghost function on both sides)

Good, so now just plug it into the formula, and use v = p/m. Since mass is a constant, it simply falls out at the beginning of the equation, so you have

Or in summary

When talking about the minimum uncertainty, this is where the equality holds (i.e. the 'equal' part of 'greater than or equal to'). This is the formula you quoted. So now just insert the numbers and you get your answer out. I'm assuming that 'known within' means that the standard deviation is 1 nm. Since you don't know the particle's mass, you'll need to give your answer as a function of mass.
3. Sinuhe, congratulations on overcomplicating a simple question

(Original post by Ourkid)
"Estimate the minimum uncertainty in the velocity of a particle whose position is known to within 1nm"

I believe there is an equation Dq = h(bar)/(2mDv)
so we plug in for Dq and rearrange for Dv?
thanks
ourkid
Yup thats exactly how to do it, the minimum uncertainty exists when the relation has an equals sign. As sinuhe said it will be in terms of m
4. (Original post by EierVonSatan)
Sinuhe, congratulations on overcomplicating a simple question
Thanks, I'm quite pleased with myself now.

But seriously, my last paragraph WAS actually vaguely related to what he asked.

(Although if this is an exam question, surely it isn't enough just to state the formula and plug some numbers in, is it? You need some sort of rationalisation to get the marks, I should think.)
5. (Original post by Sinuhe)
Thanks, I'm quite pleased with myself now.

But seriously, my last paragraph WAS actually vaguely related to what he asked.

(Although if this is an exam question, surely it isn't enough just to state the formula and plug some numbers in, is it? You need some sort of rationalisation to get the marks, I should think.)
Ahh relax im just teasing, and yeah you did answer the question

But honestly if that was an exam question (aimed at first years on a compulsory introductory course) then i imagine it would only be a 1 mark question...the derivation is unlikely to be asked! I would say the limit would be 'solve the schrodinger equation for a particle in a 1D well...'
6. (Original post by EierVonSatan)
Ahh relax im just teasing, and yeah you did answer the question
Yeah, my comeback wasn't very good, was it? Should have asked what you were getting me as a prize or something ... Maybe next time.

But honestly if that was an exam question (aimed at first years on a compulsory introductory course) then i imagine it would only be a 1 mark question...the derivation is unlikely to be asked! I would say the limit would be 'solve the schrodinger equation for a particle in a 1D well...'
Hmm, yes, I guess most first year questions are about the particle in a box, simple harmonic oscillator or particle on a ring problems. But this question did come up in a first year exam a few years ago (I'm now clutching at straws to try to justify my answer really ):
- Show that the operators for location and linear momentum in one dimension are noncommutative, and that the commutator is .
This was worth 5 marks though - very generous for one line of work!

Updated: May 18, 2008
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