# Eigenstates for Hamiltonian

1. Jun 3, 2013

### Denver Dang

1. The problem statement, all variables and given/known data
Assume a Hilbert space with the basis vectors $\left| 1 \right\rangle$, $\left| 2 \right\rangle$ and $\left| 3 \right\rangle$, and a Hamiltonian, which is described by the chosen basis as:
$$H=\hbar J\left( \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{matrix} \right),$$
where $J$ is a constant.

Now, show that the linear combination $\left| {{\psi }^{\left( k \right)}} \right\rangle =\sum\nolimits_{n=1}^{3}{c_{n}^{\left( k \right)}\left| n \right\rangle }$, where $c_{n}^{\left( k \right)}={{e}^{in\cdot 2\pi k/3}}/\sqrt{3}$ with $k = 1,2,3$, is eigenstates for $H$.

2. Relevant equations

3. The attempt at a solution
I've been told that for $k = 1$, as an example, I should get:
$$H\left| {{\psi }^{\left( 1 \right)}} \right\rangle =\hbar J\left( \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{matrix} \right)\left( \begin{matrix} {{e}^{2\pi i/3}} \\ {{e}^{4\pi i/3}} \\ 1 \\ \end{matrix} \right)\frac{1}{\sqrt{3}}$$
And from this you see that it is an actual eigenstate, since I end up with:
$$H\left| {{\psi }^{\left( 1 \right)}} \right\rangle =-\hbar J\left| {{\psi }^{\left( 1 \right)}} \right\rangle$$

My question is, how did the guy, which I got this from, come up with the vector for $\left| {{\psi }^{\left( 1 \right)}} \right\rangle$.
I can't seem to figure that out :/

2. Jun 3, 2013

### TSny

The expression for $\left| {{\psi }^{\left( 1 \right)}} \right\rangle$ just comes from letting $k=1$ in $\left| {{\psi }^{\left( k \right)}} \right\rangle =\sum\nolimits_{n=1}^{3}{c_{n}^{\left( k \right)}\left| n \right\rangle }$

Did you try writing that out explicitly and then interpreting as a column matrix?

3. Jun 3, 2013

### Denver Dang

I just figured it out :)

Thank you.