# Homework Help: Smallest possible uncertainty in the positon of an electron

1. Sep 11, 2009

### bina0001

1. The problem statement, all variables and given/known data
Show that the smallest possible uncertainty in the position of an electron whose speed is given by $$\beta=v/c$$ is
$$\Delta x_{min}=\frac{h}{4\pi m_{0}c}(1-\beta^{2})^{1/2}$$

2. Relevant equations
$$\Delta x \Delta p=\frac{h}{4\pi}$$

$$p=mv= \frac{m_{0}}{\sqrt{1-\beta^{2}}}v$$

3. The attempt at a solution
so from the momentum equation, i multiply in c:

$$p=\frac{m_{0}c}{\sqrt{1-\beta^{2}}}\frac{v}{c} =\frac{m_{0}c}{\sqrt{1-\beta^{2}}}\beta$$

then i diffrenciate with respect to $$\beta$$ to get this:

$$\frac{dp}{d\beta}=\frac{m_{0}c}{\sqrt{1-\beta^{2}}}(1+\frac{\beta^{2}}{1-\beta^{2}})$$

So:

$$\Delta p=\frac{m_{0}c}{\sqrt{1-\beta^{2}}}(1+\frac{\beta^{2}}{1-\beta^{2}})\Delta\beta$$

So then i assumed that for minimum $$\Delta x$$ we need maximum $$\Delta p$$ and thus maximum $$\Delta \beta$$, which will give us $$\Delta \beta=1$$ (because max value of v=c)

But then when i sub this into the uncertainty equation i still have the $$(1+\frac{\beta^{2}}{1-\beta^{2}})$$ term i cant get rid off.

And are the assumtions i am making valid?

2. Sep 11, 2009

### kuruman

3. Sep 11, 2009

### bina0001

ah ok thanks tahnks, thats a bit helpful.
But i am still a bit unclear. That mean i dont have to do any rigourous maths like diffrenciations?

and also i am not clear on the part they said, "when $$\Delta p$$ exceeds $$mc$$, the uncertainty in energy is greater than $$mc^{2}$$" Why?

4. Sep 11, 2009

### gabbagabbahey

In SR, the relationship between energy and momentum is:

$$E=\sqrt{(pc)^2+(m_0c^2)^2}$$

So, $\Delta E \approx \left\vert\frac{\partial E}{\partial p}\right\vert\Delta p=$____?

5. Sep 11, 2009

### gabbagabbahey

You can simplify this further:

$$1+\frac{\beta^{2}}{1-\beta^{2}}=\frac{(1-\beta^2)+\beta^{2}}{1-\beta^{2}}=\frac{1}{1-\beta^{2}})$$

So using this method, the maximum $\Delta p$ would be $m_{0}c(1-\beta^{2})^{-3/2}$, but this is greater than $m_{0}c(1-\beta^{2})^{-1/2}$ by a factor of $(1-\beta^{2})^{-1}$ and the energy argument provides tighter restrictions on $\Delta p$.

6. Sep 11, 2009

### JayKo

hi gabbagabbahey, sidetrack, what is your signature all about?

7. Sep 11, 2009

### gabbagabbahey

It's just a line from the (hilarious IMO) TV series "Better off Ted".

8. Sep 11, 2009

### JayKo

oh i see, quite a new series, wasn't air in my country though. thanks.

9. Sep 11, 2009

### bina0001

sorry to be a bit daft, but still cat really get to the end.

Let me summarize wat i understand so far first.

The maximum possible uncertanty in E of a partical can be $$m_{0}c^{2}$$, because anything beyond that it would have enough energy to form another partical.

and from

$$E=\sqrt{(pc)^{2}+(m_{0}c^{2})^{2}}$$

we get

$$\frac{\partial E}{\partial p}=\frac{pc^{2}}{\sqrt{(pc)^{2}+(m_{0}c^{2})^{2}}}$$

So the uncertainty relation with be:

$$\Delta E=\frac{pc^{2}}{\sqrt{(pc)^{2}+(m_{0}c^{2})^{2}}} \Delta p$$

so set $$\Delta E=m_{0}c^2$$ and $$p=(1-\beta^{2})^{(-1/2)}m_{0}c \beta$$

But i end up getting

$$\Delta p= \frac{m_{0}c}{\beta}}$$

am i doing anything wrong along the way?

Last edited: Sep 12, 2009
10. Sep 12, 2009

### bina0001

anyone?

11. Sep 15, 2009

### mikefish

hi bina, I think you are close to the answear in the very first post...
just cancel the term $$(1+\frac{\beta^{2}}{1-\beta^{2}})$$, since b should be regarded as a given value. it is only dv we need to deal with. however, dv is c*db right? :)

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