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I want to make sure whether I do understand the transverse Ising model correctly or not. The classical Ising model describes the interaction between spins in a grid and the state of spins can be either +1 or -1. The transverse-field Ising model which is the quantum version of the classical Ising model. The possible outcomes of a single spin $\sigma_i$ can either be +1 and -1 therefore it can be represented as a Hamiltonian of Z-axis. And now if we want to present the coupling between two spins (based on the definition of the Ising model, see below)

$$H=-\sum_{ij}J_{ij}\sigma_{i}^{z}\sigma_{j}^{z}-\sum_{i}h_{i}\sigma_{i}^{z},$$

then we can present it, namely $\sigma_{i}^{z} \otimes \sigma_{j}^{z}$ in a matrix form as the following:

$$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 &0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 &0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 &0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$$

And this is a product of two Pauli Z-matrix. Do I understand the transverse Ising model correct?

And why do we need unit matrixes here (based my university lecture, $\sigma_{i}^{z}:=I\otimes\ldots\otimes I \otimes\sigma_{z} \otimes I\otimes\ldots\otimes I$, where $\sigma_{z}$ is in the i-th position of the tensor product. I think the $\sigma_{z}$ is a single spin, isn't it?)? Are unit matrixes necessary? Thank you for your answer!

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Your Hamiltonian is not transverse.

In your Hamiltonian, your magnetic field monomial $-\sum_{i}h_{i}\sigma_{i}^{z}$ is in the same direction as (longitudinal with) your interaction terms $-\sum_{ij}J_{ij}\sigma_{i}^{z}\sigma_{j}^{z}$ - i.e. both are in the $Z$ direction. This is a longitudinal model and not a transverse model.

A transverse-field model would be something like:

$$H=-\sum_{ij}J_{ij}\sigma_{i}^{z}\sigma_{j}^{z}-\sum_{i}h_{i}\sigma_{i}^{x}.$$

The interactions are $\sigma^z_i\sigma^z_j$ (Pauli Z's) while magnetic field is $\sigma^x_i$ (Paul X).


The identity matrices are implied in the Hamiltonian

Regarding your matrices, the middle matrix is an antiferromagnetic interaction matrix acting on two qubits. But, the leftmost and rightmost matrices are both just $I_2=I\otimes I$.

Usually by convention these identity matrices are implied in the Hamiltonian. For example if you have $L$ qubits in your lattice, your full Hamiltonian matrix would be of dimension $2^L\times 2^L$, but that's too too big to write, and it's only a two-body Ising model (or 2-local Hamiltonian). For such local Hamiltonians we implicitly have the identity matrices to avoid having to write the entirety of the Hamiltonian.

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  • $\begingroup$ Do you have any reference regarding longitudinal model? $\endgroup$ Commented Mar 5, 2023 at 16:01

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