# Symmetry of E(k) in the first BZ

1. Oct 1, 2012

### hokhani

Why the curve E(k) in the first brillouin zoon is symmetric? For example why in the first BZ of a one-dimensional lattice we have E(k)=E(-k)?

2. Oct 1, 2012

### PhysTech

This is true in general, not just for 1-D. By Bloch's theorem the solutions of your Schrödinger equation will be Bloch states given by

$\psi_{\textbf{k}}(\textbf{r}) = \exp(i\textbf{k}.\textbf{r})u_{\textbf{k}}(\textbf{r})$

We can note a property of the Bloch states when $\textbf{k} \rightarrow -\textbf{k}$. When we take the Hermitian conjugate of the above equation we get

$\psi_{\textbf{k}}^*(\textbf{r}) = \exp(-i\textbf{k}.\textbf{r})u_{\textbf{k}}^*(\textbf{r})$

Changing $\textbf{k} \rightarrow -\textbf{k}$ in the above equation we get

$\psi_{-\textbf{k}}^*(\textbf{r}) = \exp(i\textbf{k}.\textbf{r})u_{-\textbf{k}}^*(\textbf{r})$

Comparing this with the very first equation it can be seen that

$\psi_{-\textbf{k}}^*(\textbf{r}) = \psi_{\textbf{k}}(\textbf{r})$

$u_{-\textbf{k}}^*(\textbf{r}) = u_{\textbf{k}}(\textbf{r})$

The Bloch states must obviously satisfy Schrödinger's equation

$H \psi_{\textbf{k}}(\textbf{r}) = \left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right] \psi_{\textbf{k}}(\textbf{r}) = E_{\textbf{k}} \psi_{\textbf{k}}(\textbf{r})$

Now, we know that the Hamiltonian is Hermitian (i.e. $H = H^\dagger$). Therefore taking the Hermitian conjugate of the above equation we get:

$H^\dagger \psi_{\textbf{k}}^*(\textbf{r}) = \left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right] \psi_{\textbf{k}}^*(\textbf{r}) = E_{\textbf{k}} \psi_{\textbf{k}}^*(\textbf{r})$

$E_{\textbf{k}}$ stays the same since eigenvalues of a Hermitian operator are real. Now, once again changing $\textbf{k} \rightarrow -\textbf{k}$ we get

$\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right] \psi_{-\textbf{k}}^*(\textbf{r}) = E_{-\textbf{k}} \psi_{-\textbf{k}}^*(\textbf{r})$

Using the property of Bloch states shown above the $\psi_{-\textbf{k}}^*(\textbf{r})$ can be replaced to give

$\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right] \psi_{\textbf{k}}(\textbf{r}) = E_{-\textbf{k}} \psi_{\textbf{k}}(\textbf{r})$

Comparing this with the original Schrödinger equation you can see that

$E_{-\textbf{k}} = E_{\textbf{k}}$

3. Oct 3, 2012

### hokhani

Thank you PhysTech.

4. Oct 3, 2012

### DrDu

E(k)=E(-k) is a consequence of time reversal invariance.

5. Oct 4, 2012

### hokhani

Could you please explain some more?

6. Oct 4, 2012

### DrDu

This is basically what phystech tried to explain. Ignoring spin, the time reversal operation consist in taking t->-t and in taking the complex conjugate. Taking the complex conjugate is often the same as taking the hermitian adjoint as phystech implied, but not always.
In going through the proof by phystech you can convince yourself that it is necessary to take the complex conjugate and not the hermitian adjoint, although both operations coincide for the hamiltonian considered.
E.g. when a magnetic field is present, the Hamiltonian is still hermitian but not equal to it's complex conjugate as
$H=\frac{1}{2m} (\mathbf{p}+ie\mathbf{A})^2 +V(\mathrm{r})$
In position representation, p will be represented by an anti-symmetric matrix while A is a symmetric matrix so that their relative sign changes on taking the complex conjugate.
Similar effects can be observed when spin orbit coupling is present. In fact there are some materials, e.g. ferromagnets, where E(k)=E(-k) is violated as time reversal symmetry is spontaneously broken.