0
$\begingroup$

we came across a peculiarity when calculating the partition function of $N$ spins $s_i=\pm1$ with Hamiltonian $$H=-J\sum_{i=1}^Ns_is_{i+1}$$ where we impose periodic boundary conditions such that $s_{N+1}=s_1$.
Now we can calculate the partition function as $$Z=\sum_{spins}e^{-\beta H[s]}=\sum_{s_1=\pm1}...\sum_{s_N=\pm1}\prod_{i=1}^Ne^{\beta Js_is_{i+1}}=\prod_{i=1}^N(\sum_{s_i=\pm1}e^{\beta Js_is_{i+1}})$$
The sum over $s_i=\pm1$ on the RHS can be performed without specifying the value of $s_{i+1}$ since $$\sum_{s_i=\pm1}e^{\beta Js_is_{i+1}}=e^{\beta Js_{i+1}}+e^{-\beta Js_{i+1}}=2\cosh(\beta J)$$ for $s_{i+1}=\pm1$. This leaves us with $$Z=(2\cosh(\beta J))^N$$ But we could have used a different technique to calculate the partition function:
Define $$h_1(s)=\cfrac{1+s}{2} \quad h_2(s)=\cfrac{1-s}{2}$$ such that $$\sum_{s=\pm1}h_\tau(s)=1$$ $$h_\tau(s)h_\rho(s)=\delta_{\tau\rho} h_\rho(s)$$ With that we can expand any function $f(s_i,s_{i+1})$ in terms of $h_\tau(s_i)$ and $h_\rho(s_{i+1})$. In particular:$$k_{i}=:e^{\beta Js_is_{i+1}}=T_{\tau\rho}h_\tau(s_i)h_\rho(s_{i+1})$$ Using the properties from above, we can see that $$\sum_{s_{i+1}=\pm1}k_{i}k_{i+1}=\sum_{s_{i+1}=\pm1}T_{\tau\rho}T_{\mu\nu}h_\tau(s_i)h_\rho(s_{i+1})h_\mu(s_{i+1})h_\nu(s_{i+2})=h_\tau(s_i)T_{\tau\rho}T_{\rho\nu}h_\nu(s_{i+2})$$ Since $$Z=\sum_{s_1=\pm1}...\sum_{s_N=\pm1}\prod_{i=1}^Nk_i$$ we get $$Z=\sum_{s_1=\pm1}h_\tau(s_1)[T...T]_{\tau\rho}h_\rho(s_{N+1})=Tr(T^N)$$ The transfer Matrix $T$ has the form $$T=\begin{bmatrix} e^{\beta J} & e^{-\beta J} \\ e^{-\beta J} & e^{\beta J} \end{bmatrix}$$ it is symmetric and can therefore be diagonalized with orthogonal matrices which gives us for the partition function $$Z=Tr(\mathcal O^\top \hat{T}^N \mathcal O)=\sum_{i}\lambda_i^N$$ where we used the cyclic propertry of the trace and the $\lambda_i$ are the Eigenvalues of $T$ The result of this calculation gives $$Z=(2\cosh(\beta J))^N+(2\sinh(\beta J))^N=(2\cosh(\beta J))^N[1+(\tanh(\beta J))^N]$$ And this is where the problem arises: The two results are different. For large $N$ we obtain our old result since $|\tanh(x)|<1$.
But for any finite $N$ there is an additional contribution of $(2\sinh(\beta J))^N$
Did we somehow overcount in our second method or did the first method not capture the system correctly?
Furthermore, what is the physical significance of "small" eigenvalues? I heard in a lecture that if one unique, largest eigenvalue of a transfer matrix exists, then the corresponding eigenvector is the state of thermal equilibrium. This makes sense to me since one could consider a normalized transfer matrix (in a sense of $\lambda_{max}=1$) and repeatedly apply it to some initial state. But what do the other eigenvectors represent? Which information is lost when we neglect the small eigenvalues in our partition sum?
I would be very happy if someone could demystify this issue.

$\endgroup$
  • 2
    $\begingroup$ In your first treatment, you seem to be summing over $s_i$ without taking into account the interaction with the previous spin $s_{i-1}$. $\endgroup$ – LonelyProf Feb 1 at 19:49
  • $\begingroup$ each pair of neighbouring spins is summed over once in my first sum. Introducing the interactions of $s_i$ and $s_{i-1}$ would just double count those pairs. $\endgroup$ – eapovo Feb 1 at 19:52
  • 1
    $\begingroup$ What I mean is, I don't think that your rearrangement of sums and product, in your first method, is correct for periodic boundaries. Maybe you could write out the left side, and the right side, of your first equation for $Z$, explicitly for a three-spin system with periodic boundaries (8 terms). $\endgroup$ – LonelyProf Feb 1 at 20:49
  • $\begingroup$ Thank you, I tried that and it became clear to me that the rearrangement is invalid for periodic boundaries. $\endgroup$ – eapovo Feb 1 at 21:07
0
$\begingroup$

I think the first way in which you do it has a flaw. Suppose that the single term did depend on both $s_i$ and $s_{i+1}$, then $$ \sum_{s_1=\pm 1}\cdots\sum_{s_N=\pm 1}\prod_{i=1}^N f(s_i,s_{i+1}) = \prod_{i=1}^N\left(\sum_{s_i=\pm 1}f(s_i,s_{i+1})\right)\,, $$ wouldn't make sense because all $s_{i+1}$ would survive even after the sum has been performed. If you had a function of $s_i$ only, then such an identity would hold $$ \sum_{s_1=\pm 1}\cdots\sum_{s_N=\pm 1}\prod_{i=1}^N f(s_i) = \prod_{i=1}^N\left(\sum_{s_i=\pm 1}f(s_i)\right)\,. $$ However I can't figure out why you still get a correct result in the $N\to\infty$ limit.

$\endgroup$
  • 1
    $\begingroup$ Roughly speaking, the approach of the OP will work for free boundaries (summing progressively over each spin, from one end to the other), and gives the quoted result. And when $N\rightarrow\infty$, the boundary conditions become unimportant. $\endgroup$ – LonelyProf Feb 1 at 20:54
  • $\begingroup$ Thank you, I must agree that the expression does not make sense when there is dependency on $s_{i+1}$, I guess it was just too tempting that in both cases the result is the same. $\endgroup$ – eapovo Feb 1 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.