- **Atomic, Solid State, Comp. Physics**
(*http://www.physicsforums.com/forumdisplay.php?f=64*)

- - **Why only first brillouin zone?**
(*http://www.physicsforums.com/showthread.php?t=165966*)

why only first brillouin zone?I realise this a fundamental flaw in my understanding but I can't seem to get my head around it.
I understand that bragg reflection occurs at the zone boundary and so electron wavevectors are diffracted back into the 1st brillouin zone, but what is the justification for only considering the 1st brillouin zone, for example, when considering energy gaps? This is also connected with my earlier question about why Umklapp processes are required, where large k is transmitted back into 1st brillouin zone by G: K1+K2=K3+G |

The reason why it is only necessary to consider the first Brillouin zone lies in the fact that the wave function in a crystal is written as a function with the periodicity of the lattice times a plane wave, so that in one dimension:
[tex] \psi_{nk}(x) = e^{i k r} u_{nk}(x) [/tex] The plane wave is unique only up to a reciprocal lattice vector. This is seen by considering the reciprocal lattice vectors which in 1D are given by [tex]\mathbf{G}=\frac{2n\pi}{a}[/tex]: [tex] e^{i G_{n1} r}= e^{i G_{n2} r}, [/tex] for all values of n1 and n2. The interval is often chosen to be -pi/a to pi/a to get the band structure around a symmetric interval. |

Ok, thanks for that
I understand Bloch waves. So are you saying that a bloch wave travelling through the crystal is the same in all brillouin zones so consideration of which one is arbitary? So what about the electrons which remain within the 1st brillouin zone as their wavevector K is large enough to experience diffraction (laue condition: delta K=G & at zone boundary min kin for diffraction = (delta k/2) = G/2) ? I think i understand. The pure isotropic crystal has translational symetry, any position has same properties as that position translated by the lattice vectors. So every brillouin zone is equivalent and choice between any 1st zone, or any 2nd zone is arbitrary. So electrons which are bound behave in the same way in every brillouin zone, as do electrons passing through the crystal. so may only consider 1st brilluoin zone. is that the reasoning or am I missing something? Sorry to go on about it! |

All times are GMT -5. The time now is 01:43 AM. |

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.

© 2014 Physics Forums