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For Hamiltonian $\operatorname H$ and lattice translation operator $\operatorname T$, if

$$\operatorname H\psi=E\psi, \qquad \operatorname T\psi=e^{ik\cdot R}\psi,$$

and

$$\operatorname H\phi=E\phi, \qquad\operatorname T\phi=e^{ik\cdot R}\phi,$$

then $\psi=\phi$?

Here $R$ is lattice vector, $\psi$ can be $\psi_k$ and $\phi$ can be $\psi_{k+K}$, and $K$ of reciprocal lattice.

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2 Answers 2

First express the discrete translation symmetry via Bloch's theorem. So the starting point is a Bloch wave

$$ \psi_{\bf k}({\bf r})~=~e^{i{\bf k}\cdot{\bf r}}u_{\bf k}({\bf r}) $$

for some given crystal momentum ${\bf k}$ modulo an arbitrary reciprocal lattice vector. Or to remove redundancy in the dual lattice description, let ${\bf k}$ belong to the first Brillouin zone. Still there is no reason for $u_{\bf k}({\bf r})$ to be non-degenerate without further information about the system.

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Does this mean $\psi=\phi$? –  richard Mar 11 '13 at 2:02
1  
No. We are talking about solutions to linear equations (if we ignore normalization of the wave function), so the solution space is a vector space. Therefore even if we additionally assume that the solution space is non-degenerate, i.e. one-dimensional, then two solutions may still differ by an overall phase factor (when we put back the normalization condition). –  Qmechanic Mar 13 '13 at 17:42

I am trying hard to think of the reasons why not, so I will try to prove, that we can actually state that $\psi = \phi$.

So now assume as you said, that:

$$ \operatorname H \psi_k = E_k \psi_k $$ and $$ \operatorname H \phi_{k+K} = E_{k+K} \psi_{k+K} $$

So now, if we take the difference between the two and we assume that $\operatorname H$ is the same in both cases, $$ \operatorname H \left( psi_k - \phi_{k+K} \right) = E_k \psi_k - E_{k+K} \phi_{k+K} = E' \left( \psi_k - \phi_{k+K} \right) $$ Where I assumed, that any linear combination of two eigenfunctions will yield an eigenfunction. Now we can rearrange:

$$ \left(E_k - E'\right) \psi_k = \phi_{k+K} \left( E' - E_{k+K} \right) $$

Now we can invoke the orthogonality condition, by taking an inner product with $\psi_k$:

$$ E_k - E' = (E' - E_{k+K}) \left<\psi_k \right| \left. \phi_{k+K} \right> $$

Now if the eigen functions are different, then it means, that $E_k = E'$. If we take an inner product with $\phi_{k+K}$ instead, then we similarly deduce that $E_{k+K} = E' \implies E_k = E_{k+K}$. Now we can start thinking whether the equal eigenvalue condition means degeneracy, or being the eigenvalues being the same.

Suppose the eigenvalues $\psi$ and $\phi$ are not the same, then they must be degenerate. However, I can not think of a way make use of $\operatorname T$ operator in order to get ride of the degeneracy possibility between $\psi$ and $\phi$.

This did not answer your question, but maybe it will give you some thoughts.

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Thanks gns-ank. –  richard Mar 11 '13 at 0:00

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