Smallest possible uncertainty in the positon of an electron

[bina0001]

Homework Statement

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}

Homework Equations

\Delta x \Delta p=\frac{h}{4\pi}

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

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}<br /> =\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 can't get rid off.

And are the assumtions i am making valid?

 

[kuruman]

You may find this wikipedia reference useful

http://en.wikipedia.org/wiki/Compton_wavelength

 

[bina0001]

ah ok thanks tahnks, that's a bit helpful.
But i am still a bit unclear. That mean i don't 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?

 

[gabbagabbahey]

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

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=____?

 

[gabbagabbahey]

\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 can't get rid off.

And are the assumtions i am making valid?

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.

 

[JayKo]

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

 

[gabbagabbahey]

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

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

 

[JayKo]

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

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

 

[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 uncertainty in E of a particle can be m_{0}c^{2}, because anything beyond that it would have enough energy to form another particle.

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?

 

[bina0001]

anyone?

 

[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? :)