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I having difficulties in understanding "transfer matrix" in the paper Metastability in the two-dimensional Ising model.

They consider a periodic $N \times \infty$ lattice with the energy

$$ E = -J \sum_{nn} \sigma \sigma - H \sum \sigma $$ for spins $\sigma = \pm 1$, where the first summation occurs over nearest neighbours.

Now they say, "<...>

The associated $2^N \times 2^N$ symmetric transfer matrix $L$ is defined as follows. for two column configurations $\vert \mu \rangle = (\sigma_1, \cdots, \sigma_n)$ and $\vert \mu' \rangle = (\sigma_1', \cdots, \sigma_n')$

$$ \langle \mu \vert L \vert \mu' \rangle = \exp \bigg\{{ \frac{\nu}{2} \sum_{i=1}^{N} \sigma_i \sigma_{i+1} + \sigma_i' \sigma_{i+1}' + \frac{1}{2} h \sum{i=1}^{N} (\sigma_i + \sigma_i') + \nu \sum_{i=1}^N \sigma_i \sigma_i' \bigg\}} $$

where $\nu = J/T$ and $h = H/T$ <...>".

I cannot understand this definition, let alone reconcile it with the one I am used to.

First of all, the column configuration $\mu$ has $N$ components, while $L$ is a $2^N \times 2^N$ matrix, so I cannot make even basic sense of the left-hand side, what operation it represents.

For a say one-dimensional lattice model with nearest-neighbours interaction $U_{ij} = U(\sigma_i, \sigma_j)$ the transfer matrix $V$ is a $2^N \times 2^N$ matrix and has components

$$ V_{ij} = -\exp \big\{ -\frac{U_{ij}}{kT} \big\} $$

What does the first notation mean? Hopefully I will then see how it relates to the latter definition, thanks

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  • $\begingroup$ First write out the answer explicitly for N=2, then write out the answer explicitly for N=3, then you will be able to see the answer to your question for general N. $\endgroup$
    – hft
    Jun 7, 2022 at 16:25
  • $\begingroup$ @hft, I must be blind, I cannot see the LHS meaning it even for N=3. The configurations have 3 elements, and the matrix has 2^3 = 8 elements, how can I perform $a A a$ is $a$ is a vector with 3 components and $A$ is a $8 \times 8$ matrix? But I see your point, I will write the answer explicitly using the LHS and see if it then makes sense to me thanks $\endgroup$
    – Smerdjakov
    Jun 7, 2022 at 16:38
  • $\begingroup$ For the N=3 case there are eight states: |000>; |001>; |010>; |011>; |100>; |101>; |110>; |111>. $\endgroup$
    – hft
    Jun 7, 2022 at 18:33
  • $\begingroup$ I must be explsining myself really poorly. For $N=3$ I see well there are $2^3 = 8$ states, and can choose two from those, use them on the RHS. But what does the LHS stand for? Does not $\langle \mu \vert L \rangle $ stand for vector matrix multiplication? Each of the 8 configurations has 3 components, and $L$ has 8 rows, how can be multiplied? $\endgroup$
    – Smerdjakov
    Jun 7, 2022 at 19:14
  • $\begingroup$ Each of the eight states I wrote down above corresponds to one row of the L matrix (or one row of the E matrix). For example $L_{11} = <000|L|000>$, ..., $L_{88} = <111|L|111>$. If I get a chance I will write up an answer. $\endgroup$
    – hft
    Jun 7, 2022 at 19:23

2 Answers 2

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There are $2^N$ possible spin configurations $|\mu\rangle= (\sigma_1,\sigma_2,\ldots, \sigma_N)$, when $\sigma_i=\pm1$. So although the sum on the RHS is only over $i=1,\ldots N$ there are $2^N$ by $2^N$ possible expressions that can be evaluated to give a matrix entry on the LHS. The matrix $L$ defined by the array of numerical entries $\langle \mu|L|\mu'\rangle$ is therefore $2^N$-by-$2^N$. Indeed, you will need to sum over all $2^N$ possible spin configurations at each intermediate level, and that is exactly what the trace of powers of the $2^N$-by-$2^N$ matrix $L$ is doing..

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  • $\begingroup$ thanks a lot, all you say makes sense to me. $2^N$ choices for $\mu$, same number for $\mu'$, get a $2 ^N \times 2^N$ matrix that is clear. What is totally obscure to me is how can $\langle \mu \vert L \vert \mu ' \rangle$ represent a matrix entry, as a matter of fact what does that notation represent at all, given that the vector has $N$ components and the matrix has $2^N$ rows. Does not $\langle a \vert B \vert c \rangle$ represent vector matrix multiplication $a B$, and the result multiplied by $c$? $\endgroup$
    – Smerdjakov
    Jun 7, 2022 at 17:28
  • $\begingroup$ I understand, I think!, the RHS, now if I were told the LHS has nothing to do with vector matrix multiplication and is just a notational shortcut to express the RHS then all would make sense (the entry $ij$ is given by feeding configurations $i$ and $j$ to the RHS), but that "ket bra" notation is confusing me I wonder if it does somehow express some vector matrix multiplication at all $\endgroup$
    – Smerdjakov
    Jun 7, 2022 at 17:31
  • $\begingroup$ Yes $\langle \mu|L|\mu'\rangle$ is just notation for the matrix element $L_{\mu\mu'}$. There is no Dirac notation needed. $\endgroup$
    – mike stone
    Jun 8, 2022 at 13:04
  • $\begingroup$ I got it many thanks much appreciated $\endgroup$
    – Smerdjakov
    Jun 8, 2022 at 14:19
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I having difficulties in understanding "transfer matrix" in the paper [Metastability in the two-dimensional Ising model][1].

They consider a periodic $N \times \infty$ lattice with the energy

$$ E = -J \sum_{nn} \sigma \sigma - H \sum \sigma $$ for spins $\sigma = \pm 1$, where the first summation occurs over nearest neighbours.

Consider, for example, a periodic 1d lattice of $N=3$ spins. There are eight possible "states," which I choose to number as below:

  1. $|--->$
  2. $|--+>$
  3. $|-+->$
  4. $|-++>$
  5. $|+-->$
  6. $|+-+>$
  7. $|++->$
  8. $|+++>$

where, I my notation |xyz> means the spin on site one is x, site two is y, and site three is z. For example, |+-+> means the spin is up on site 1, down on site 2, and up on site 3. For example, |--+> means the spin is down on site 1, down on site 2, and up on site 3.

The energy of these states are:

  1. $-3J + 3H$
  2. $J+H$
  3. $J+H$
  4. $J-H$
  5. $J+H$
  6. $J-H$
  7. $J-H$
  8. $-3J-3H$

I could write the energy as a 8x8 matrix: $$ \begin{bmatrix} -3J+3H & 0& 0& 0& 0& 0& 0& 0 \\0 & J+H & 0& 0& 0& 0& 0& 0 \\0 & 0 & J+H & 0& 0& 0& 0& 0 \\0 & 0 & 0 & J-H & 0& 0& 0& 0 \\0& 0& 0& 0& J+H& 0& 0& 0 \\0& 0& 0& 0& 0& J-H& 0& 0 \\0& 0& 0& 0& 0& 0& J-H& 0 \\0& 0& 0& 0& 0& 0& 0& -3J-3H \end{bmatrix} $$

In this case I would write the state $|--->$ as: $$ \begin{bmatrix} 1 \\0 \\0 \\0 \\0 \\0 \\0 \\0 \end{bmatrix} $$

In this case I would write the state $|--+>$ as: $$ \begin{bmatrix} 0 \\1 \\0 \\0 \\0 \\0 \\0 \\0 \end{bmatrix} $$

And so on.

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  • $\begingroup$ Thanks that makes sense. Is this a "standard" approach / notation? At the end of the day, the object $\vert \mu \rangle$ is somehow abused. It refers both to a 3-dimensional vector (e.g. 001) as well as the 8-dimensional encoding (e.g. 01000000), am I the only one finding this confusing?? $\endgroup$
    – Smerdjakov
    Jun 8, 2022 at 7:36
  • $\begingroup$ It's common to take this approach in "exact diagonalization" techniques. In quantum mechanics the $|\mu>$ would denote a "state" of the system, and there are many different ways to represent such a state. This choice of matrix representation just amounts to choosing one explicit/concrete basis for doing calculations. Similar to how you can choose to working position space or in momentum space, all bases are fine, just some are more convenient than others sometimes. $\endgroup$
    – hft
    Jun 8, 2022 at 15:54

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