Uncertainty principle free particle

[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

 

[hikaru1221]

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

How did you get that?

 

[jambaugh]

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.

 

[semc]

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
<br /> \Delta p = \left| \frac{h}{\lambda} - \frac{h}{\lambda + \Delta \lambda} \right| <br />
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?

 

[jambaugh]

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
<br /> \Delta p = \left| \frac{h}{\lambda} - \frac{h}{\lambda + \Delta \lambda} \right| <br />
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.

 

[semc]

I got it thanks alot