Fixed boundaries in 1D Ising model What are the differences for solving the one dimensional Ising model for fixed boundaries using the transfer matrix, compared with periodic boundaries?
this picture show the solution for periodic boundaries
 A: Your question is fairly broad, so I'm going to give a fairly general answer. I have no intention of discussing the details of a solution to this problem, as that would be "off topic" for this site. Hopefully, you just need a pointer or two in the right direction.
You can find more details in a few places, including this answer to a previous question, 
and these online lecture notes. 
In the notation of your source reference, for energy
$$
\mathcal{H} = -J\sum_{\langle i,j\rangle} s_i s_j - h\sum_i s_i
$$
the transfer matrix is
\begin{align*}
T_{s,s'} &= \exp\{\beta J ss' + \tfrac{1}{2}\beta h(s+s')\}
\\
\text{or}\qquad
\mathbb{T} &= \begin{pmatrix} 
e^{\beta(J+h)} & e^{-\beta J} \\ e^{-\beta J} & e^{\beta(J-h)} \end{pmatrix} .
\end{align*}
We can think of $s$ and $s'$ as the indices of the $2\times 2$  matrix, taking two values $\pm1$, or, if you like $\uparrow$ and $\downarrow$. It is associated with the bond between successive spins. To fit this picture, only half of the external field acting on $s$, and half of the field on $s'$, is included. The other half of the field on $s$ appears in the transfer matrix for the previous bond; the other half of the field on $s'$ appears in the transfer matrix for the next bond. This symmetric form works well if we have periodic boundary conditions. The expression for the partition function 
$$
Z = \sum_\alpha \big(\mathbb{T}^N\big)_{\alpha\alpha} = \text{Tr}\, \mathbb{T}^N
$$
involves summing over the spin values $s_2,s_3,\ldots s_N$ (the $N-1$ matrix multiplications involved in calculating $\mathbb{T}^N$), then identifying $s_{N+1}$ with $s_1$ and summing over those values, $\sum_\alpha$ as well, to end up with the trace.
This simple form leads to an elegant solution in terms of 
$\lambda_1$ and $\lambda_2$, the two eigenvalues of $\mathbb{T}$.
You are asking about the differences for a chain of spins with fixed boundaries. The main difference is that we will not be taking the trace. Nonetheless, the matrix product has the same interpretation. Consider the matrix element
$$
\big(\mathbb{T}^N\big)_{\alpha\beta}
$$
Again, spins $s_2,s_3,\ldots s_N$ have been summed over. We have fixed $s_1=\alpha$. The value $\beta$ is the fixed orientation of an extra spin, $s_{N+1}$, which we do not identify with $s_1$. So, this looks like the partition function of an $N+1$-spin system, of which the end spins are fixed, and the $N-1$ interior spins are summed over the values $\pm1$.
It is not perfect, because only half the field on $s_1$, and half the field on $s_{N+1}$ has been included. Physically, we should either count the full effect of the field on these spins, or decide that the field does not apply to these spins. Either way, we can correct the expression by multiplying $(\mathbb{T}^N)_{\alpha\beta}$ by some constant involving $\beta h$ and the two values $s_1=\alpha$ and $s_{N+1}=\beta$. But in the end, because these spins are fixed, this decision will only affect our choice of the "zero of energy". Provided we bear this in mind, it will be satisfactory to write
$$
Z_{\alpha\beta} = (\mathbb{T}^N)_{\alpha\beta} .
$$
When you come to solve this, it will still be useful to compute
the eigenvalues of $\mathbb{T}$, namely $\lambda_1$, $\lambda_2$,
and the corresponding eigenvectors $\vec{p}_1$ and $\vec{p}_2$.
Moreover, if you
construct the $2\times 2$ matrix $\mathbb{P}$ whose columns are 
$\vec{p}_1$ and $\vec{p}_2$,
you can diagonalize $\mathbb{T}$:
$$
\tilde{\mathbb{P}}\mathbb{T}\mathbb{P} =
\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}
$$
where $\tilde{\mathbb{P}}$ is the inverse (i.e. the transpose in this case) of $\mathbb{P}$. Inserting 
unity in the form $\mathbb{P}\tilde{\mathbb{P}}$ 
in between every pair of $\mathbb{T}$ matrices in our equation for $Z_{\alpha\beta}$ will simplify it dramatically.
This should all be familiar from the periodic boundary case.
However, again, you must pay attention to the "end effects",
since we are not performing a trace.
The other main difference is that we may be interested in the magnetization as a function of position in the chain. Because of the fixed spins at the ends, the properties of the chain will not be uniform. Consider the $n^\text{th}$ spin $s_n$. The probability that it takes a value $\gamma$ will be
$$
\frac{(\mathbb{T}^n)_{\alpha\gamma}(\mathbb{T}^{N-n})_{\gamma\beta}}{(\mathbb{T}^N)_{\alpha\beta}}
$$
where we don't sum over $s_n$.
If we want the average magnetization for $s_n$,
then we do want to sum over $s_n$,
and we should include a factor $+1$ when $s_n=+1$,
and $-1$ when $s_n=-1$. In terms of matrices this can be written
$$
\frac{(\mathbb{T}^n\mathbb{D}\mathbb{T}^{N-n})_{\alpha\beta}}{(\mathbb{T}^N)_{\alpha\beta}} \qquad\text{where}\qquad
\mathbb{D}=\begin{pmatrix} +1 & 0 \\ 0 & -1 \end{pmatrix} .
$$
Again, I'm not going to go any further with the solution, that's up to you. But I hope that this gives sufficient general background.
