How to find eigenvectors of ugly transfer matrices in stat mech

[paco_uk]

Homework Statement

I'm trying to use the transfer matrix method in statistical mechanics but I'm struggling with the algebra so I'd like to know if there is a simpler way to find the eigenvalues and eigenvectors of a matrix.

For example, studying the lattice gas model produces the transfer matrix:
[tex]
T = \left( \begin{array}{cc}
1 & e^{\beta \mu /2} \\
e^{\beta \mu /2} & e^{\beta ( J - \mu)} \end{array} \right)
[/tex]

Homework Equations

It's not too hard to find the eigenvalues, although they don't look very nice:
[tex]
\lambda = \frac{1 + e^{\beta (J + \mu)}}{2} \pm \sqrt{4e^{\beta \mu} + (1-e^{\beta(J+\mu)})^2}
[/tex]

The Attempt at a Solution

When it comes to the eigenvectors it seems like a hopeless case:
[tex]
\left( \begin{array}{cc}
1 & e^{\beta \mu /2} \\
e^{\beta \mu /2} & e^{\beta ( J - \mu)} \end{array} \right) \left( \begin{array}{c}
a \\
b \end{array} \right) = \lambda
\left( \begin{array}{c}
a \\
b \end{array} \right)
[/tex]

The algebra involved here seems unmanageable, especially as I'm supposed to be able to do this under exam conditions.

I'm supposed to be able to show that:
[tex]
\langle n_i \rangle = \frac{1}{1+e^{-2 \theta}}
[/tex]
where
[tex]
\sinh (\theta ) = \exp(\frac{1}{2} \beta J) \sinh (\frac{1}{2} \beta [J + \mu])
[/tex]

I think that [tex] \langle n_i \rangle [/tex] is given by:
[tex]
\langle n_i \rangle = \langle 0 | \mathbf{C} | 0 \rangle
[/tex]
where [tex] | 0 \rangle [/tex] is the eigenvector corresponding to the largest eigenvalue and:
[tex]
C = \left( \begin{array}{cc}
0 & 0 \\
0 & 1 \end{array} \right)
[/tex]

In fact the examiners report for this question suggests that it is not even necessary to find the eigenvalues of T which is why I am wondering if there is some clever way to spot the eigenvectors without going through all the algebra?

A similar question on the Potts model said something about guessing the eigenvectors from the symmetry of the matrix but I wouldn't know how to start.

 

[Mute]

My best guess at the short cut would be to realize that the lattice gas hamiltonian can be mapped to the Ising model hamiltonian. Hence, if you know the eigenvectors of the Ising model transfer matrix you can find the eigenvectors of the lattice gas matrix.

The mapping between the models is that if [itex]n_i[/itex] is the occupation of site i in the lattice gas model, then the relation [itex]s_i = 2n_i - 1[/itex] gives you the spin [itex]s_i[/i] at site i. You can thus map the lattice gas to the ising model and use all the known results for the Ising model to solve the problem.

If that's not how they want you to do it, then perhaps you could post the entire problem for us?

 

[paco_uk]

Thanks, that's a useful thing to know.

In fact, I came across their solution to this specific question today (that always seems to happen the day after I post a question here). It seems to be more of a neat trick that works here rather than being a generally applicable principal like mapping between models as you suggest. I'll post it anyway in case it's useful to anyone.

Writing [itex] x=e^{\beta \mu /2} [/itex] and [itex] y=e^{\beta J} [/itex], the transfer matrix becomes:
[tex]
T = \left( \begin{array}{cc}
1 & x \\
x & x^2 y \end{array} \right)
[/tex]
Then writing [itex] |0 \rangle = (u,v)^T [/itex], we have
[tex]
\begin{eqnarray}u+xv = \lambda_0 u \\
xu + x^2 y v = \lambda_0
\end{eqnarray}
[/tex]
Eliminating [itex]\lambda_0[/itex] and rearranging gives:
[tex]
0=\left( \frac{u}{v} \right)^2 +\left( xy -\frac{1}{x} \right) \left(\frac{u}{v} \right) -1
[/tex]
with solution
[tex]
\left(\frac{u}{v} \right) = -1/2 \left( xy - \frac{1}{x} \right) \pm \sqrt{\left( xy - \frac{1}{x} \right)^2 +4}
[/tex]

Now the trick is to let [itex] ( xy - 1/x) =2 \sinh{\theta} [/itex] which leads to the solution:
[tex]
\left(\frac{u}{v} \right) = e^{\pm \theta}
[/tex]
where I guess the choice of sign determines which of the eigenvectors is being considered.

With C defined as in my original post, the matrix element is just [itex]u^2[/itex]. Along with the normalisation condition [itex]u^2+v^2=1[/itex], this gives the result:
[tex]
\langle n_i \rangle = \frac{1}{1+e^{-2 \theta}}
[/tex]

and the definition of [itex]\sinh \theta[/itex] matches that in the OP.