# Uncertainty principle free particle

• semc
In summary, for a free particle, the uncertainty relation can be written as (Delta lambda) (Delta x) >= (Delta lambda)^2 / 4pi, using [(Delta p)(Delta x)>=h/4pi] and h/p=lambda, and relating Delta lambda to p and Delta p. This can be derived by ignoring terms of order Delta lambda ^2 and using the trick of rationalizing a complex denominator.
semc
Show that for a free particle the uncertainty relation can be written as (Delta lambda) (Delta x) >= lambda^2 / 4pi

Firstly I am sorry not writing this in latex but I gave up after trying to write it in latex for half an hour. Would be great if anyone can show me how to do it.

Using [(Delta p)(Delta x)>=h/4pi ] and h/p=lambda, I got (Delta x)>=(Delta lambda)/4pi. Multiplying throughout by (Delta lambda) I got (Delta lambda) (Delta x) >= (Delta lambda)^2 / 4pi. So how do I proceed from here?

Thanks

Using [(Delta p)(Delta x)>=h/4pi ] and h/p=lambda, I got (Delta x)>=(Delta lambda)/4pi.

How did you get that?

Firstly, to type in latex you bracket your code as follows:
(tex) \Delta x = \frac{\partial A}{\partial B} (/tex)( with () --> [] ).

In the advanced editor you should also see a latex reference button (Sigma).

Note that the closing tex bracket uses a forward slash while latex code uses backslash as an escape char.

I have problems with the preview of latex code as it uses cached images but the actual post will be OK (if my latex code is OK).

As to your question I think you need to relate Delta lambda to p and Delta p. Assuming the "Delta's" are small I think one can use differential rules (but absolute value the results since the Delta error is a magnitude):

Given
$$\lambda = \frac{h}{p}$$
or
$$p = \frac{h}{\lambda}$$
hence
$$\Delta \lambda = \Delta\left(\frac{h}{p}\right) = \frac{h \Delta p}{p^2}$$
and
$$\Delta p= \Delta\left(\frac{h}{\lambda}\right) = \frac{h \Delta \lambda}{\lambda^2}$$

I think that's the key to the problem.

If you need a bit more rigor begin with:
$$\Delta p = \left| \frac{h}{\lambda} - \frac{h}{\lambda + \Delta \lambda} \right|$$

combine the fractions and then multiply top and bottom by $$\lambda - \Delta \lambda$$ and ignore $$O(\Delta \lambda^2)$$ terms.

Thanks so much for the help on the latex coding!

I thought h/p=lambda so
$$\Delta \lambda = \frac{h}{\Delta p}$$
Is this wrong?

I expanded
$$\Delta p = \left| \frac{h}{\lambda} - \frac{h}{\lambda + \Delta \lambda} \right|$$
but I got
$$\frac{h \Delta \lambda}{\lambda^2 + \lambda \Delta \lambda}$$
So do I ignore the $$\lambda \Delta \lambda$$ to get the $$\Delta p$$? Why can we ignore that term?

Last edited:
semc said:
Thanks so much for the help on the latex coding!

I thought h/p=lambda so
$$\Delta \lambda = \frac{h}{\Delta p}$$
Is this wrong?
Yep.
If you divide by a small change you get a big change and you shouldn't.
[edit:] Note that with differentials if you rather write:
$$h = \lambda p$$
then you get:
$$0 = d\lambda p + \lambda dp$$
so
$$dp = -\frac{pd\lambda}{\lambda}$$
[end edit]

I expanded
$$\Delta p = \left| \frac{h}{\lambda} - \frac{h}{\lambda + \Delta \lambda} \right|$$
but I got
$$\frac{h \Delta \lambda}{\lambda^2 + \lambda \Delta \lambda}$$
So do I ignore the $$\lambda \Delta \lambda$$ to get the $$\Delta p$$? Why can we ignore that term?

Here's the trick:
$$\frac{h \Delta \lambda}{\lambda(\lambda+ \Delta \lambda)}\cdot \frac{\lambda-\Delta \lambda}{\lambda - \Delta \lambda} =\frac{h (\lambda\Delta\lambda -\Delta \lambda^2}{\lambda(\lambda^2 - \Delta \lambda^2)}$$
Ignore order Delta lambda ^2 terms and you have:
$$=\frac{h \lambda\Delta\lambda }{\lambda^3}= \frac{h\Delta\lambda}{\lambda^2}$$

This is the same "trick" used in deriving the derivative of 1/x or "rationalizing" a complex denominator.

I got it thanks alot

## 1. What is the uncertainty principle in relation to a free particle?

The uncertainty principle states that it is impossible to accurately know both the position and momentum of a particle at the same time. This means that the more precisely we know the position of a particle, the less we know about its momentum, and vice versa.

## 2. How does the uncertainty principle affect the behavior of a free particle?

The uncertainty principle has a significant impact on the behavior of a free particle. It means that the particle's exact position and momentum can never be determined with complete accuracy, and its trajectory cannot be predicted with certainty. This leads to the probabilistic nature of quantum mechanics, where the behavior of particles can only be described in terms of probabilities.

## 3. Is the uncertainty principle limited to free particles?

No, the uncertainty principle applies to all particles, whether they are free or bound. It is a fundamental principle of quantum mechanics that is applicable to all particles at the atomic and subatomic level.

## 4. Can the uncertainty principle be violated or overcome?

No, the uncertainty principle is a fundamental law of nature and cannot be violated or overcome. It is a consequence of the wave-particle duality of quantum particles and is a fundamental limit to our ability to measure and understand the behavior of particles.

## 5. How does the uncertainty principle impact our understanding of the physical world?

The uncertainty principle challenges our traditional understanding of the physical world, which is based on classical mechanics and determinism. It suggests that at the subatomic level, the behavior of particles is inherently random and unpredictable. This has profound implications for our understanding of the nature of reality and the limitations of our scientific knowledge.

• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
222
• Introductory Physics Homework Help
Replies
21
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
779
• Introductory Physics Homework Help
Replies
23
Views
433
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
15
Views
1K
• Quantum Physics
Replies
2
Views
589
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
938