# Heisenberg Uncertainty in wavelength and position

1. Jul 20, 2015

### Feynman.12

1. The problem statement, all variables and given/known data

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

3. The attempt at a solution

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

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

2. Jul 20, 2015

### Orodruin

Staff Emeritus
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.)

3. Jul 20, 2015

### Feynman.12

Here is my working

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4. Jul 20, 2015

### Orodruin

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

5. Jul 20, 2015

### Feynman.12

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 $\Delta p = h/\Delta \lambda$

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6. Jul 20, 2015

### Orodruin

Staff Emeritus
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 $p$.

7. Jul 20, 2015

### BvU

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...)