3
$\begingroup$

The Ising model is defined with the Hamiltonian:

$$ H = -\sum_{<i,j>}S_i^z\cdot S_j^z $$

What is the difference between quantum version and classical version? My intuition is that the classical version is equal to quantum version in any dimension and any lattice.

If we add a transverse field, I think there will be difference, the Hamiltonian is:

$$ H = -\sum_{<i,j>}S_i^z\cdot S_j^z - h\sum_i S_i^x $$ My intuition is that the quantum and classical will be different, because in quantum mechanics $[S^x,S^z]\neq0$. However I still don't know the difference in detail.

$\endgroup$
5
$\begingroup$

You are correct that for $h=0$ the quantum Ising model reduces to the classical model. Assuming a 2D square lattice this model has been solved exactly by Onsager. It undergoes a phase transition at a certain critical temperature which is signaled by the order parameter $M_2 = \left( \frac{1}{N} \sum_i S_i^z \right)^2$. Below this temperatur the system breaks the symmetry of the disordered phase and becomes ferromagnetic. The universality class (basically characteristic scaling exponents) of this transition is given by the 2D Ising class.

For $h > 0$ the model becomes non-trivial in a quantum mechanical sense because of the commutation relation you mentioned. However, the much of the physics remain. The system again undergoes a phase transition with the order parameter $M_2 = \langle \left(\frac{1}{N} \sum_i S_i^z \right)^2 \rangle$. This refers now to a quantum mechanical expectation value in the canonical ensemble at some temperature T. The critical temperature depends on the value of $h$, and the universality class along this line $T_c(h)$ is equivalent to the previous case of the 2D Ising class. Something new happens when the line reaches $T \rightarrow 0$, where the line terminates in a quantum critical point (QCP) with a new universtality class. In this model, the universality class of the QCP is again related to the classical Ising model, but in 3 dimensions. The imaginary time axis in the path integral formalism becomes an additional dimension and therefore allows for a 2+1 quantum to classical mapping.

$\endgroup$

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.