Register to reply

Wave function at high symmetry point

by jianglai
Tags: function, point, symmetry, wave
Share this thread:
jianglai
#1
May2-12, 03:35 PM
P: 2
How to prove that wave function at [itex]\Gamma[/itex] point can always be a real function? I know it is not true for general k point, but for [itex]\Gamma[/itex] and other high symmetry point like X, is there a simple proof?

Thanks!
Phys.Org News Partner Physics news on Phys.org
'Squid skin' metamaterials project yields vivid color display
Scientists control surface tension to manipulate liquid metals (w/ Video)
Simulation method identifies materials for better batteries
DrDu
#2
May3-12, 02:13 AM
Sci Advisor
P: 3,630
This is hard to prove as it is wrong in general. E.g. once spin orbit coupling cannot be neglected, the orbitals have to be chosen complex.
sam_bell
#3
May3-12, 11:24 PM
P: 67
Hmm. If we ignore spin-orbit then this seems easy. Note that the complex conjugate of the Bloch wave at gamma is also a solution of the Schrodinger equation. That means the real and imaginary parts are separately solutions. A similar argument should work at other high-symmetry points if -k = k + K where K is a reciprocal lattice vector.

DrDu
#4
May4-12, 02:05 AM
Sci Advisor
P: 3,630
Wave function at high symmetry point

A clean discussion involves the assumption and discussion of time reversal symmetry. If there are no spin orbit coupling effects, time reversal will be represented by complex conjugation and the single particle wavefunctions in a periodic potential can always be chosen real as then E(k)=E(-k) so that instead of the solutions [itex]\psi_k(x)=u_k(x)\exp(ikx)[/itex] and [itex]\psi_{-k}=(\psi_k(x))^*[/itex] real valued combinations can be chosen. For k=0, only one real function will be obtained.
If spin orbit coupling is taken into account, time reversal is no longer just complex conjugation so that it does not always guarantee real valuedness. This is known as Kramers degeneracy.
sam_bell
#5
May4-12, 03:39 AM
P: 67
Quote Quote by DrDu View Post
A clean discussion involves the assumption and discussion of time reversal symmetry. If there are no spin orbit coupling effects, time reversal will be represented by complex conjugation and the single particle wavefunctions in a periodic potential can always be chosen real as then E(k)=E(-k) so that instead of the solutions [itex]\psi_k(x)=u_k(x)\exp(ikx)[/itex] and [itex]\psi_{-k}=(\psi_k(x))^*[/itex] real valued combinations can be chosen. For k=0, only one real function will be obtained.
If spin orbit coupling is taken into account, time reversal is no longer just complex conjugation so that it does not always guarantee real valuedness. This is known as Kramers degeneracy.
That's right, but I think we want to keep our wave-functions as Bloch waves. In other words, we're really aking where in k-space we can choose the periodic function u_k(r) to be real. I guess you could do what you said for all k if you wanted to work with stationary boundary conditions (in opposition to the conventional Born-von Karmen).
DrDu
#6
May4-12, 04:13 AM
Sci Advisor
P: 3,630
Quote Quote by sam_bell View Post
That's right, but I think we want to keep our wave-functions as Bloch waves. In other words, we're really aking where in k-space we can choose the periodic function u_k(r) to be real. I guess you could do what you said for all k if you wanted to work with stationary boundary conditions (in opposition to the conventional Born-von Karmen).
I think it also works with Born- von Karman boundary conditions. So basically the only reason why we have to use complex u_k is because we insist on complex exp(ikx) instead of sin(ikx) or cos(ikx).
jianglai
#7
May10-12, 12:09 PM
P: 2
Thank you both for the reply! I think I get a sense of it now. Without spin-orbital coupling, for any [itex]k[/itex], [itex] \psi_{nk}(r) [/itex] and [itex]\psi_{nk}(r)^* = \psi_{-nk}(r) [/itex] are degenerate (in [itex]H[/itex]). But Bloch state are simultaneous eigenstates for both [itex]H[/itex] and translation [itex] T_R [/itex], and only at [itex] -k = k + G [/itex] are [itex]\psi_{nk}(r) [/itex] and [itex]\psi_{-nk}(r)[/itex] degenerate in [itex]T_R[/itex] as well, which means we can take a linear combination of them and get rid of the imaginary part. For a general [itex]k[/itex] however, [itex]\psi_{nk}(r)+\psi_{-nk}(r)[/itex] would be a real-valued eiginstate of [itex]H[/itex] that's not a Bloch state.
DrDu
#8
May10-12, 02:54 PM
Sci Advisor
P: 3,630
Couldn't have formulated it better!


Register to reply

Related Discussions
High symmetry point in GaAs superlattice Advanced Physics Homework 0
High symmetry points for a GaAs superlattice Atomic, Solid State, Comp. Physics 1
High symmetry points Atomic, Solid State, Comp. Physics 3
High symmetry points Atomic, Solid State, Comp. Physics 1