# Zero-Point Energy, Relativity

1. Nov 12, 2008

### JoshuaR

1. The problem statement, all variables and given/known data
Consider an electron confined in a region of nuclear dimensions (about 5fm). Find its minimum possible KE in MeV. Treat as one-dimensional. Use relativistic relation between E and p.

2. Relevant equations
KE = p2/(2a) = $$\hbar$$2/(2ma2)
p = h/$$\lambda$$
E = hf
E2 = (mc2)2 + (pc)2
E = mc2 + KE

3. The attempt at a solution
First attempt: $$KE = p^2/(2m) = (pc)^2/(2mc^2) = (E^2-(mc^2)^2)/(2mc^2) Thus, K2mc^2=E^2-(mc^2)^2 = K^2 + 2Kmc^2 + (mc^2)^2.$$However, here everything cancels out and I get:
K^2 = 0, which can't be right. (Answer in book is 40 MeV)

Another approach: $$E^2=(pc)^2+(mc^2)^2=(mc^2)^2 + 2mc^2K + K^2$$(setting the invariant energy squared equal to the energy squared where E = K + mc^2

This leads to a quadratic equation: $$K = -2mc^2 \pm$$ $$\sqrt{(mc^2)^2+(pc)^2}$$
OR
K = -2mc^2 $$\pm$$ $$\sqrt{(mc^2)^2+(hc/\lambda)^2}$$

When I plug in those values, I don't get 40 MeV. I feel like I'm missing something very simple and fundamental. This shouldn't be a hard problem. Any help would be appreciated, thanks.

2. Nov 12, 2008

### weejee

KE=p^2/2m is only valid for p<<mc

You should only use E = sqrt[(mc2)2 + (pc)2] and therefore,
KE = E - mc^2 = sqrt[(mc2)2 + (pc)2] - mc^2.

3. Nov 12, 2008

### JoshuaR

That was indeed my guess on the second approach. If I take just that:
KE = $$sqrt[(mc^2)^2 + (pc)^2] - mc^2$$, I still don't get there. Using the formula $$p = h/\lambda$$

If I do as I did on the "second approach" then setting energies equal, I also have trouble. I'm a bit puzzled because I am sure I am missing something simple, and this is just straightforward.

4. Nov 12, 2008

### borgwal

You should transform the statement that the electron is confined to small region in space to a statement about the uncertainty in position.

Then, you use the uncertainty relation to find a minimum for the expectation value of p^2.

That's the value you substitute in your relativistic relation between E and p, and you extract K from that, as you already correctly noted.

5. Nov 12, 2008

### weejee

You get 40Mev if you use p=hbar/lambda.

6. Nov 12, 2008

### borgwal

Yes...and the correct explanation for that is the one I posted.

7. Nov 12, 2008

### JoshuaR

Mmkay, I think I've tried that.

I have $$\delta p\delta x = \hbar/2.$$

Thus, when I use the KE form from before:$$KE = sqrt[(mc^2)^2 + (\hbar c/\delta x*2)^2] - mc^2$$

I must be doing incorrect math... sigh.

Last edited: Nov 12, 2008
8. Nov 12, 2008

### JoshuaR

Gah!

Where's the math mistake?

$$-.511MeV + sqrt[.511^2 + (6.58E-22*3E8/(2*5E-15))^2)$$

9. Nov 12, 2008

### weejee

What you wrote down is the minimum uncertainty condition, which is only satisfied for special cases such as the gaussian wave packet.

Normally people use $$\delta p\delta x \sim \hbar$$.

Actually, since it is an order of magnitude estimation, you can't really say it's wrong even if you use values like 2*hbar or hbar/2.

In my opinion, any value between 10MeV and 100Mev is OK for the answer.

10. Nov 12, 2008

### borgwal

\delta x is not equal to 5fm, the better estimate is 2.5fm: it's the uncertainty in the position, and we can assume the expectation value of the position of the electron corresponds to the center of the region.

Apart from that, I agree with weejee: an order of magnitude estimate will have to do: the actual physics is too complicated to get the "real" answer.

11. Nov 12, 2008

### JoshuaR

I see. It all makes so much more sense now, lol. Just take away that darn factor...

New question on this. I figured I should use minimum uncertainty for $$\delta x$$ since we were confining it to a box of size 5E-15m. Do you suppose my problem here was assuming the box was $$\delta x$$? After all, $$\delta x$$ is usually only one side of the wave packet (akin to radius, not diameter). Thus, if I have a region of some size, then $$\delta x$$ is half the width of the region. Then when I plug in for minimum uncertainty, I get 2*0.5*width...

Thanks for the help.

12. Nov 12, 2008

### JoshuaR

Right, borgwal, I think that was the problem. Thanks again!

13. Nov 12, 2008

### borgwal

Welcome!

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