Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

So I was trying to think through the statement that the uncertainty principle can characterize metallic bonding. I know that the uncertainty principle is: $\Delta p \Delta x = \frac{\hbar}{2}$.

And that in metals I heard that metallic bonding can be thought of as the spreading out of the wave function in space. I think that this makes sense because the electrons are no longer bound to just one atom.

I can see that if the space that the electron could be found ($\Delta x$) increased, say by a factor of 10, then the momentum range ($\Delta p$) will decrease by a factor of 10. However I don't see how this says anything about the bonding energy. My understanding of $\Delta p$ is that is the range of momentum values around a measured value. So wouldn't increasing $\Delta x$ just make the range around the momentum (and energy) tighter instead of lower. I thought that in order to make a stable bond we needed a lower energy.

I was wondering if I was thinking about this the wrong way.

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.