How Are Eigenstates Determined for a Given Hamiltonian Matrix?

[Denver Dang]

Homework Statement

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}<br /> 0 &amp; 1 &amp; 1 \\<br /> 1 &amp; 0 &amp; 1 \\<br /> 1 &amp; 1 &amp; 0 \\<br /> \end{matrix} \right),<br />
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.

Homework Equations

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}<br /> 0 &amp; 1 &amp; 1 \\<br /> 1 &amp; 0 &amp; 1 \\<br /> 1 &amp; 1 &amp; 0 \\<br /> \end{matrix} \right)\left( \begin{matrix}<br /> {{e}^{2\pi i/3}} \\<br /> {{e}^{4\pi i/3}} \\<br /> 1 \\<br /> \end{matrix} \right)\frac{1}{\sqrt{3}}<br />
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 :/Thanks in advance.

 

[TSny]

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 :/Thanks in advance.

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?

 

[Denver Dang]

I just figured it out :)

Thank you.