What is the Bethe Approximation for a One-Dimensional Ising Model?

[MathematicalPhysicist]

Homework Statement

We suppose that the lattice has coordination number $q$ and now retain as variables a central spin and its shell of nearest neighbors. The remainder of the lattice is assumed to
act on the nearest-neighbor shell through an effective exchange field which we
will calculate self-consistently. The energy of the central cluster can be written
as
$$ H_c = -J\sigma_0 \sum_{j=1}^q \sigma_j -h\sigma_0-h'\sum_{j=1}^q \sigma_j$$

The fluctuating field acting on the peripheral spins $\sigma_1,\ldots , \sigma_4$
has been replaced by an effective field $h'$, just as we previously replaced the interaction of $\sigma_0$ with its first neighbor shell by a mean energy.
The partition function of the cluster is given by:

$$Z_c = \sum_{\sigma_j=\pm 1} e^{-\beta H_c} = e^{\beta h} \bigg( 2\cosh [\beta(J+h')]\bigg)^q + e^{-\beta h} \bigg(2\cosh [ \beta (J-h')]\bigg)^q$$

Homework Equations

The Attempt at a Solution

I don't see how to do this calculation of $Z_c$, I need somehow to separate between $\sigma_j=1$ and $\sigma_j=-1$, and what with $\sigma_0$?

 

[MathematicalPhysicist]

And to tell you the truth this notation as a mathematician seems absurd to me, the index $j$ is inside the sums of spins in the argument of the exponential, and you are summing over $\sigma_j$ outside it?!

That's look like an awful abuse of notation.

 

[TSny]

In $Z_c$ the summation $\sum\limits_{\sigma_j=\pm 1}$ means $\sum\limits_{\sigma_0=\pm 1} \sum\limits_{\sigma_1=\pm 1} \sum\limits_{\sigma_2=\pm 1}...$

It might be a good idea to write out explicitly the case where there are only 2 nearest neighbors ($q$ = 2). Begin by writing out explicitly $H_c$ with no summation symbols.

 

[Nadav Outmezguine]

Ok, Let's first work out the notation.
The partition function can be thought of as merely an normalization of some distribution function, as such, we always must sum over all possible configurations.
The lattice in the above example is comprised of $N$ sites, each has the value $\pm 1$. I t is convenient to define a set $\sigma$ that has $N$ elements corresponding with eace site. Note that there are $2^N$ different sets that can described a physical configuration,

Our Hamiltonian is a "functional" of the "field" $\sigma$, by that I mean that the functional associates an energy value for each configuration. More formally, the Hamiltonian is a map from the space of all possible configuration to the space of all possible energy levels $H:\{\sigma\}\to \{E\}$, where I just introduced the notation $\{\sigma\}$ for the space of all possible configuration (which is a set of $2^N$ sets of size $N$) , and similarly for the space of all possible energy levels.

Now, as mentioned above, in order to get the right partition function we must sum over all possible configurations and thus our partition function is always defined as,
$$Z=\sum_{\{\sigma\}}e^{-\beta H[\sigma]}$$.
The simplest example would be to show the above for the case of two sites,
$$Z_2=\sum_{\{\sigma\}}e^{-\beta H[\sigma]}=e^{-\beta H[\{+,+\}]}+e^{-\beta H[\{-,+\}]}+e^{-\beta H[\{+,-\}]}+e^{-\beta H[\{-,-\}]}$$.

We shall next try to actually calculate the above partition function.
If I understand correctly, we are looking at the site $\sigma_0$ and it's $q$ neighbours.
For a clearer treatment I will define
$$ H_c[\sigma_0,\sigma]=-h \sigma_0-\left(J \sigma_0+h'\right)\sum_{j=1}^q\sigma_i.$$
The partition function will be
$$Z_c=\sum_{\sigma_0=\pm1}\sum_{\{\sigma\}}e^{-\beta H_c[\sigma_0,\sigma]}\equiv\sum_{\sigma_0=\pm1}Z_0[\sigma_0].$$
From the above lines we find,
$$Z_0[\sigma_0]=\sum_{\{\sigma\}}e^{-\beta H_c[\sigma_0,\sigma]}=e^{\beta h \sigma_0}\sum_{\{\sigma\}}\exp\left\{\beta\left(J \sigma_0+h'\right)\sum_{j=1}^q\sigma_i\right\}.$$
Since the exponent of a sum is just the multiplication of exponents, we can rewrite the partition function as,
$$Z_0[\sigma_0]=e^{\beta h \sigma_0}\sum_{\{\sigma\}}\prod_{j=1}^q\exp\left\{\beta\left(J \sigma_0+h'\right)\sigma_i\right\}.$$
Now come the tricky part, since each term in this multiplication is independent of the other terms we can replace the product and the sum in the following way,
$$Z_0[\sigma_0]=e^{\beta h \sigma_0}\prod_{j=1}^q\sum_{\sigma_j=\pm 1}\exp\left\{\beta\left(J \sigma_0+h'\right)\sigma_i\right\}=e^{\beta h \sigma_0}\prod_{j=1}^q2\cosh\left\{\beta\left(J \sigma_0+h'\right)\right\}.$$
We see now that we are multiplying $q$ identical terms, and so we get the nice (almost) final answer,
$$Z_0[\sigma_0]=e^{\beta h \sigma_0}\cosh^q\left[\beta\left(J \sigma_0+h'\right)\right].$$
Now we just need to sum over $\sigma_0=\pm1$ to find that,
$$Z_c=e^{\beta h}\cosh^q\left[\beta\left(J+h'\right)\right]+e^{-\beta h}\cosh^q\left[\beta\left(J-h'\right)\right]$$

 

[TSny]

Looks very good to me.

 

[MathematicalPhysicist]

Thanks Nadav!

But shouldn't we have it:
$$
Z_0[\sigma_0]=e^{\beta h \sigma_0}\prod_{i=1}^q\sum_{\sigma_i=\pm 1}\exp\left\{\beta\left(J \sigma_0+h'\right)\sigma_i\right\}=e^{\beta h \sigma_0}\prod_{j=1}^q2\cosh\left\{\beta\left(J \sigma_0+h'\right)\right\}.
$$

As in your LHS you should be summing over $\sigma_i$ and not $\sigma_j$ as you wrote, in which case I can see how you get the $2\cosh(\ldots )$.

BTW, if both of you are interested there's another question I posted here: https://www.physicsforums.com/threads/one-dimensional-ising-model-in-bethe-approximation.930327/