In Solid State Physics by Ashcroft and Mermin, we see in chapter 12 (figure 12.8) that there are open orbits for electrons(or holes) in k-space. But there is no reason there to explain why there should be open orbits (or the constant energy surfaces stick in some directions to each other).

Can anyone help in understanding the physical basis for this assumption?

  • $\begingroup$ Contemplate the opposite: how can you require that there be no constant energy surfaces that cross unit cell boundaries? The physical basis for the assumption is that, in fact, such open orbits are extremely common and help map out the Fermi surfaces. $\endgroup$ – Jon Custer Nov 4 '16 at 14:47
  • $\begingroup$ @JonCuster: sorry that I can't grasp the meaning of what you want to say... if explain more, it would be appreciable... thanks alot $\endgroup$ – P.A.M Nov 4 '16 at 14:52
  • $\begingroup$ If you have free electrons that can move between unit cells, than there will be constant energy surfaces that connect one unit cell to another. These surfaces will have some relation to the underlying crystal symmetry, and thus may result in open orbits for a properly oriented field with respect to the crystal. Again, the hard thing to do is try to reason how there could not be constant energy surfaces connecting unit cells. (This is a powerful tool often in physics - if you can't understand the reason why, try to construct the reason why not). $\endgroup$ – Jon Custer Nov 4 '16 at 15:06
  • $\begingroup$ @JonCuster: yes.thanks for your help in reasoning and mention the free electron case... now my question is that: is there a physical basis for those open orbits to happen? what happens ? why should electron not move in in circles and move in somewhat straight way in these orbits when we apply a magnetic field? $\endgroup$ – P.A.M Nov 4 '16 at 15:12
  • $\begingroup$ @JonCuster - Maybe you should clarify that by unit cells you actually mean unit cells in reciprocal k-space. $\endgroup$ – freecharly Nov 4 '16 at 15:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.