Heisenberg Uncertainty in wavelength and position

[Feynman.12]

Homework Statement

Show that $\Delta\lambda\Delta\ x>lamdba^2/4*pi$

The Attempt at a Solution

When I substitute de Broglie's p=h/lambda I get the equation of

$\frac {\Delta\x}{\Delta\lambda} > 1/(4*pi )$

 

[Orodruin]

You need to show more of your work. How are we supposed to tell where you went wrong if you only provide us with your final result? (I anyway have a pretty good idea of where you have gone wrong, but I want to see exactly what you did first.)

 

[Feynman.12]

You need to show more of your work. How are we supposed to tell where you went wrong if you only provide us with your final result? (I anyway have a pretty good idea of where you have gone wrong, but I want to see exactly what you did first.)

Here is my working

 

[Orodruin]

The first row is not correct. $p = h/\lambda$ does not imply $\Delta p = h/\Delta \lambda$. What is $d\lambda/dp$?

 

[Feynman.12]

The first row is not correct. $p = h/\lambda$ does not imply $\Delta p = h/\Delta \lambda$. What is $d\lambda/dp$?

I have attached a worked solution in which I came to the right answer so I believe it must be right. However, I still don't understand intuitively why $p = h/\lambda$ does not imply $\Delta p = h/\Delta \lambda$

 

[Orodruin]

I have attached a worked solution in which I came to the right answer so I believe it must be right. However, I still don't understand intuitively why $p = h/\lambda$ does not imply $\Delta p = h/\Delta \lambda$

Because it is a matter of how a small change in $\lambda$ changes $p$. If you used your formula, a small change in $\lambda$ would give a huge change in $p$.

 

[BvU]

I have attached a worked solution in which I came to the right answer so I believe it must be right. However, I still don't understand intuitively why $p = h/\lambda$ does not imply $\Delta p = h/\Delta \lambda$

If $\ \ y = {1\over x}\ \$ then surely $\ \ {dy\over dx} = -{1\over x^2}\ \ \Rightarrow \ \ dy = -{dx\over x^2}\$ . Change d to $\Delta$ and voila !

(Sorry for barging in, Oro...)