# Wave function at high symmetry point

1. May 2, 2012

### jianglai

How to prove that wave function at $\Gamma$ point can always be a real function? I know it is not true for general k point, but for $\Gamma$ and other high symmetry point like X, is there a simple proof?

Thanks!

2. May 3, 2012

### DrDu

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. May 3, 2012

### sam_bell

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.

4. May 4, 2012

### DrDu

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 $\psi_k(x)=u_k(x)\exp(ikx)$ and $\psi_{-k}=(\psi_k(x))^*$ 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. May 4, 2012

### sam_bell

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).

6. May 4, 2012

### DrDu

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).

7. May 10, 2012

### jianglai

Thank you both for the reply! I think I get a sense of it now. Without spin-orbital coupling, for any $k$, $\psi_{nk}(r)$ and $\psi_{nk}(r)^* = \psi_{-nk}(r)$ are degenerate (in $H$). But Bloch state are simultaneous eigenstates for both $H$ and translation $T_R$, and only at $-k = k + G$ are $\psi_{nk}(r)$ and $\psi_{-nk}(r)$ degenerate in $T_R$ as well, which means we can take a linear combination of them and get rid of the imaginary part. For a general $k$ however, $\psi_{nk}(r)+\psi_{-nk}(r)$ would be a real-valued eiginstate of $H$ that's not a Bloch state.

Last edited: May 10, 2012
8. May 10, 2012

### DrDu

Couldn't have formulated it better!