# A slightly trickier system: the Heisenberg model

We can now apply the transfer matrix method to a more complicated system, the so called Heisenberg model. This model is identical to the Ising one with the exception that this time the spin variables can assume vectorial values instead of being discrete scalars. In other words the degrees of freedom of the system will be ${\displaystyle N}$ vectors ${\displaystyle {\vec {S}}_{i}}$, each residing on a site of the lattice and with unitary module, namely ${\displaystyle |{\vec {S}}_{i}|=1}$. If there is no external field[1], i.e. ${\displaystyle H=0}$, then the reduced Hamiltonian of a one-dimensional Heisenberg model is:

${\displaystyle -\beta {\mathcal {H}}=K\sum _{i=1}{\vec {S}}_{i}\cdot {\vec {S}}_{i+1}}$
and with periodic boundary conditions (${\textstyle {\vec {S}}_{N+1}={\vec {S}}_{1}}$):
${\displaystyle Z_{N}=\sum _{\lbrace {\vec {S}}_{i}\rbrace }e^{K\sum _{i=1}^{N}{\vec {S}}_{i}\cdot {\vec {S}}_{i+1}}=\sum _{\lbrace {\vec {S}}_{i}\rbrace }e^{K{\vec {S}}_{1}\cdot {\vec {S}}_{2}}e^{K{\vec {S}}_{2}\cdot {\vec {S}}_{3}}\cdots e^{K{\vec {S}}_{N}\cdot {\vec {S}}_{1}}=\operatorname {Tr} {\boldsymbol {T}}^{N}}$
where the transfer matrix ${\displaystyle {\boldsymbol {T}}}$ is such that:
${\displaystyle \langle {\vec {S}}_{i}|{\boldsymbol {T}}|{\vec {S}}_{i+1}\rangle =e^{K{\vec {S}}_{i}\cdot {\vec {S}}_{i+1}}}$
Also in this case the transfer matrix can be written in the form:
${\displaystyle {\boldsymbol {T}}=\sum _{i}\lambda _{i}|t_{i}\rangle \langle t_{i}|}$
so that, for example:
${\displaystyle \langle {\vec {S}}_{1}|{\boldsymbol {T}}|{\vec {S}}_{2}\rangle =e^{K{\vec {S}}_{1}\cdot {\vec {S}}_{2}}=\sum _{i}\lambda _{i}\langle {\vec {S}}_{1}|t_{i}\rangle \langle t_{i}|{\vec {S}}_{2}\rangle =\sum _{i}\lambda _{i}f_{i}({\vec {S}}_{1})f_{i}^{*}({\vec {S}}_{2})}$
where ${\displaystyle f_{i}({\vec {S}}_{j})=\langle {\vec {S}}_{j}|t_{i}\rangle }$ is in general a complex function.

Now, in order to determine the eigenvalues ${\displaystyle \lambda _{i}}$ we must use the fact (which we will not prove) that plane waves ${\displaystyle e^{-i{\vec {r}}\cdot {\vec {k}}}}$ can be decomposed in spherical harmonics ${\displaystyle Y_{\ell m}}$ in this way:

${\displaystyle e^{-i{\vec {r}}\cdot {\vec {k}}}=4\pi \sum _{\ell =0}^{\infty }\sum _{m=-\ell }^{\ell }i^{\ell }j_{\ell }(|{\vec {k}}||{\vec {r}}|)Y_{\ell m}^{*}({\hat {k}})Y_{\ell m}({\hat {r}})}$
where ${\displaystyle j_{\ell }}$ are the spherical Bessel functions:
${\displaystyle j_{\ell }(xy)=-{\frac {i^{\ell }}{2}}\int _{0}^{\pi }\sin \theta e^{ixy\cos \theta }P_{\ell }(\cos \theta )d\theta }$
and ${\displaystyle P_{\ell }(\cos \theta )}$ is the ${\displaystyle \ell }$-th order Legendre polynomial. In this case, setting ${\displaystyle {\vec {r}}={\vec {S}}_{1}}$ and ${\displaystyle {\vec {k}}={\vec {S}}_{2}}$ we get:
${\displaystyle e^{k{\vec {S}}_{1}\cdot {\vec {S}}_{2}}=4\pi \sum _{\ell =0}^{\infty }\sum _{m=-\ell }^{\ell }i^{\ell }j_{\ell }(-iK)Y_{\ell m}^{*}({\vec {S}}_{1})Y_{\ell m}({\vec {S}}_{2})}$
This suggests that the eigenvalues have the form:
${\displaystyle \lambda _{\ell m}(K)=4\pi i^{\ell }j_{\ell }(-iK)}$
(which actually doesn't depend on ${\displaystyle m}$). Therefore, the partition function is:
${\displaystyle Z_{N}=\operatorname {Tr} {\boldsymbol {T}}^{N}=\sum _{\ell =0}^{\infty }\sum _{m=-\ell }^{\ell }\lambda _{\ell m}(K)^{N}=\sum _{\ell =0}^{\infty }(2\ell +1)\lambda _{\ell }^{N}(K)}$

And from now on, from the explicit computation of the eigenvalues (which we are not going to do) we can determine all the thermodynamics of the system.
1. We consider this particular case because otherwise we can't apply the transfer matrix method.