# Uncertainty principle free particle

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?

jambaugh
Gold Member
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:
jambaugh
Gold Member
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