Uncertainty principle free particle

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  • #1
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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
 

Answers and Replies

  • #2
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Using [(Delta p)(Delta x)>=h/4pi ] and h/p=lambda, I got (Delta x)>=(Delta lambda)/4pi.

How did you get that?
 
  • #3
jambaugh
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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
[tex] \lambda = \frac{h}{p}[/tex]
or
[tex] p = \frac{h}{\lambda}[/tex]
hence
[tex]\Delta \lambda = \Delta\left(\frac{h}{p}\right) = \frac{h \Delta p}{p^2}[/tex]
and
[tex]\Delta p= \Delta\left(\frac{h}{\lambda}\right) = \frac{h \Delta \lambda}{\lambda^2}[/tex]

I think that's the key to the problem.

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

combine the fractions and then multiply top and bottom by [tex]\lambda - \Delta \lambda[/tex] and ignore [tex]O(\Delta \lambda^2)[/tex] terms.
 
  • #4
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Thanks so much for the help on the latex coding!

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

I expanded
[tex]
\Delta p = \left| \frac{h}{\lambda} - \frac{h}{\lambda + \Delta \lambda} \right|
[/tex]
but I got
[tex] \frac{h \Delta \lambda}{\lambda^2 + \lambda \Delta \lambda} [/tex]
So do I ignore the [tex]\lambda \Delta \lambda[/tex] to get the [tex]\Delta p[/tex]? Why can we ignore that term?
 
Last edited:
  • #5
jambaugh
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Thanks so much for the help on the latex coding!

I thought h/p=lambda so
[tex]\Delta \lambda = \frac{h}{\Delta p} [/tex]
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:
[tex]h = \lambda p[/tex]
then you get:
[tex] 0 = d\lambda p + \lambda dp[/tex]
so
[tex]dp = -\frac{pd\lambda}{\lambda}[/tex]
[end edit]

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

Here's the trick:
[tex] \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)}[/tex]
Ignore order Delta lambda ^2 terms and you have:
[tex]=\frac{h \lambda\Delta\lambda }{\lambda^3}= \frac{h\Delta\lambda}{\lambda^2}[/tex]

This is the same "trick" used in deriving the derivative of 1/x or "rationalizing" a complex denominator.
 
  • #6
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I got it thanks alot
 

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