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Wave function at high symmetry point 
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#1
May212, 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! 


#2
May312, 02:13 AM

Sci Advisor
P: 3,595

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.



#3
May312, 11:24 PM

P: 67

Hmm. If we ignore spinorbit 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 highsymmetry points if k = k + K where K is a reciprocal lattice vector.



#4
May412, 02:05 AM

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P: 3,595

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. 


#5
May412, 03:39 AM

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#6
May412, 04:13 AM

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P: 3,595




#7
May1012, 12:09 PM

P: 2

Thank you both for the reply! I think I get a sense of it now. Without spinorbital 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 realvalued eiginstate of [itex]H[/itex] that's not a Bloch state.



#8
May1012, 02:54 PM

Sci Advisor
P: 3,595

Couldn't have formulated it better!



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