# Why is there no kinetic term in the Hamiltonian of the Ising model?

I am used to the Hamiltonian formalism in the context of (quantum) field theory, where as far as I can remember it always has the form of a kinetic term + a potential term. For me the absence of kinetic terms means a theory without dynamics. In Wikipedia the Hamiltonian of the Ising Model reads:

$$H(\sigma) = \sum_{i,j} J_{ij} \sigma_i \sigma_j + \sum_j h_j \sigma_j\,, \tag{1}$$

where the first term corresponds to interactions and where the sum runs over nearest neighbors, so I suppose $$i \neq j$$. The second term corresponds to an external potential.

Why is there no kinetic term? Can the system evolve in time, e.g. by seeing the 2d Ising Model as a (Euclidean) (1+1)d model? How should I picture this Hamiltonian as a system in my head?

• Somehow I think that the (1+1)d suggestion is not correct, and that instead I should see the temperature as the inverse time. Is that true? That still does not explain the absence of kinetic term I suppose.
– Pxx
May 9, 2021 at 10:19
• Related: math.stackexchange.com/questions/124706/… (moreover: "inverse temperature=time" is the Wick rotation, see my answer below). May 9, 2021 at 11:55

I always thought the Ising model as the simplest model for magnetism in which spins (atoms, electrons or particles with spin to be precise) are in a static lattice, and the model only considers the spin-spin interaction.

It's static in the sense that they are in a fixed lattice, but there is dynamics since spins do evolve in time.

If you which to study something else that just a fixed lattice, then search for itinerant magnetism. The Hubbard model is one of the simplest models in which spins aren't just glued and can jump between the lattice spaces.

• No I am happy with the Ising model, but how can I see this time dependence? More explicitly, how can I compute the spin $\sigma_i$ after a time $t$ given an initial configuration $S$ that specifies all the initial spins? On the wikipedia page they talk about the correlators $\langle \sigma_i \sigma_j \rangle$, but that's not time-dependent is it?
– Pxx
May 9, 2021 at 11:13
• You have to solve the Schrödinger equation, using the Ising hamiltonian. Instead of a continuous wavefunction like when solving a particle in a box, here the functions are spinors. Each $\sigma$ is a linear combination of up and down eigenfunctions, so the full solution will be of the type: $\sigma_i(t) = A_i(t) \left | \uparrow \right \rangle + B_i(t) \left |\downarrow \right \rangle$, for each $i$. In these representation, the dynamics is in determining the functions $A_i$ and $B_i$. May 10, 2021 at 0:42
• @FrancisLecuak you are talking about a quantum model, while the OP is about the classical Ising model, which does not have an intrinsic dynamics (or a Schroedinger equation). I try to address these points in my answer below.
May 10, 2021 at 8:54

Just a complement to the other answers, as they do not seem to touch on this issue.

People have of course been studying the dynamics of Ising models for a long time, but they usually (always?) use stochastic dynamics rather than a Hamiltonian evolution. That is, they consider a Markov chain on the set of configurations having the Gibbs measure as stationary distribution. There are many choices. For instance, in order to describe a non-conservative dynamics (evolving magnetization), Glauber dynamics is often used, while a conservative dynamics (fixed magnetization) can be studied using Kawasaki dynamics. See, for instance, these notes (Sections 6.2 and 6.4, respectively) or this book (Chapter 8) for more information.

With such models, one can then study all sorts of dynamical aspects: the approach to equilibrium, metastability, etc.

An underlying dynamics is implicit in the thermodynamic treatment. If there were no dynamics, there couldn't be an approach to equilibrium or spontaneous changes in the system. In brief, the system would not get thermodynamic equilibrium, which is the prerequisite for thermodynamics.

However, in the classical case, kinetic energy plays a separate role from potential energy, and provided the system is Hamiltonian$$^*$$, one can deal with the potential energy term separately. To stress an important point, since the kinetic energy term in the Hamiltonian of a classical system contributes with an additive analytical term in the thermodynamic limit, it does not influence the possibility of phase transitions.

Just to add an explicit indication to the OP, I recall that there is an explicit version of the Ising model equipped with deterministic dynamics: Creutz, M. (1986). Deterministic Ising dynamics. Annals of physics, 167(1), 62-72.

$$^*$$ Of course, it is possible to introduce stochastic dynamics, as indicated by @YvanVelenik, and there is quite a large literature about that. However, the introduction of stochastic dynamics is not necessary to answer the original question.

You probably have in mind a dynamical "Hamiltonian system", which is typically defined over a phase space (typically position-momentum). In this case it makes sense to talk about potential and kinetic energy.

Here you are doing equilibrium statistical mechanics. The $$H$$ of the Ising model is not the full Hamiltonian of the underlying dynamical system. In fact, at this basic level, the Ising model is not defined over a "phase space", but rather over a "configuration space" that is the space of all spin configurations. There is no explicit time (this is equilibrium thermodynamics/statistical mechanics), and there are no canonically conjugate variables.

From an abstract point of view (useful in optimization problems), the function $$H$$ of the Ising model may be called "cost of the configuration".

If you really want to introduce time, then you have to perform a Wick rotation (but by doing this you end up with a quantum system, see also this). In this case you will be able to "see" a discrete time derivative in the Hamiltonian, which can be used to define a sort of kinetic energy. However, this would not solve your conceptual problem, as the Wick rotation is just a formal way to map quantum systems into thermodynamic systems.

In other words: there is no way to compute the $$i$$-spin after a time $$t$$ given an initial configuration that specifies all the initial spins within the framework of equilibrium statistical mechanics. At equilibrium the temporal dynamics of each spin is ruled by thermal fluctuations, which description is beyond the formulation you are referring to in your question.

Edit: here I am not saying that every statistical mechanical model does not have a kinetic term. However, are some statistical mechanical models that are not defined on a phase space, but rather on a configuration space.

I also completely agree with the nice answer of @Qmechanic. You can add to $$H$$ a "kinetic" term $$\sum_i^N \sigma_i ^2/2 = N/2$$, which is just an irrelevant constant! However, this does not answer to the more "philosophical" part of OP's question. I believe that, in the end, the answer is: not all statistical mechanics models stem from an underlying microscopic Hamiltonian dynamical model (despite they have a so-called "Hamiltonian"). After all, the degrees of freedom appearing in the irrelevant kinetic term $$N/2$$ are still the $$\sigma_i$$, and not some canonically conjugate variables.

• I see, thanks for your answer. It would be nice if the person who put you a -1 could explain its downvote. I have one follow-up question: in principal you could imagine the same system (nearest-neighbors interaction) but in non-thermal equilibrium, right? So you start with a configuration that does not minimize $H$ and let it evolve until it reaches equilibrium.
– Pxx
May 9, 2021 at 12:09
• I have no idea, probably it has been down voted because of a misunderstanding (you can upvote it if you find it useful): I am NOT saying that EVERY statistical mechanics model has no kinetic energy. I am referring to the Ising one. The Ising model has no kinetic energy, it accounts for the spin-spin or spin-external field interactions. May 9, 2021 at 12:33
• For you second question: yes, you can do that. It's the idea beyond the Monte Carlo simulation of the Ising model. Note, however, that it is a "fake" dynamical evolution which is used to sample configurations from the Gibbs ensemble (typically by using the Metropolis algorithm)! It is not an evolution coming from a dynamical Hamiltonian. May 9, 2021 at 12:33

In classical statistical mechanics (as opposed to quantum stat-phys), the dynamical (kinetic energy) part of the Hamiltonian decouples from the interaction (potential) part, since both momentum and position are independent. The same is true for classical spin systems: their thermodynamics is independent from the dynamics, which has to be added by hand.

Depending on the choice of dynamics (an example of which is Glauder dynamics), the dynamics exponents at the thermal phase transition changes, see the review by Hohenberg and Halperin.

Of course, classical spin models are microscopically coming from a corresponding quantum model. However, it is well known that thermal phase transitions of quantum systems are described by static classical models (here meaning described by the classical Ising model without dynamics). How to find the classical dynamics of a quantum spin model at a thermal transition is still quite an open question, since computing the dynamics of a quantum system at a phase transition is very hard.

For what its worth, recall that a kinetic term in the Hamiltonian for a rigid body in classical mechanics is $$H=\sum_{i=1}^3\frac{L_i^2}{2I_i}$$.

Or more abstractly: A kinetic term of the form $$H=\sum_{i=1}^n\frac{p_i^2}{2m_i}$$.

In that sense a kinetic term for the Ising model would be $$\propto \sum_i\sigma^2_i$$, which is just a constant that can be dropped, since the spin is $$\sigma_i\in\{\pm 1\}$$.

• Thanks for your answer! I am not sure I follow, the first term in $(1)$ is of the form $\sigma_i \sigma_j$ with $i \neq j$ and not of the form $\sigma_i \sigma_i$... Could you explain in a bit more detail how the analogy with rigid bodies work?
– Pxx
May 9, 2021 at 11:15
• I updated the answer. May 9, 2021 at 11:37
• I disagree with the answer. The thermodynamics of classical spin models is always defined without dynamics, independent of the constraint (take for instance a Blum-Cappel model, where $\sigma=-1,0,1$. The true reason is that the dynamics and static part of a classical model decouple in the partition function (as is easier to see for a collection of particles interacting with some potential).