We know first order phase transitions, like second order transitions, also have order parameters. They too can be dealt with Landau theory.

What kind of temperature dependence of the order parameter (a discontinuous step function?) is needed to explain a discontinuous phase transition?

How is that functional dependence achieved through Landau's approach to phase transition?

What is the physical way of understanding such discontinuous change within the arena of Landau theory? Is it some kind of tunnelling that happens between the minima of free energy?

  • $\begingroup$ Landau's theory deals with phase transitions where the symmetry of the ground state changes, which is why it needs an order parameter. A first-order transition does not require any change of symmetry (otherwise, there could be no critical point where a first-order transition turns into a smooth crossover), and thus does not need an order parameter. In such a case, you cannot use Landau's theory to describe a first-order transition. An exception is the class of systems close to a critical point with a small symmetry-breaking perturbation. This is the situation discussed by @knzhou. $\endgroup$ Jun 10 '18 at 10:36

Yes, Landau theory can explain discontinuous phase transitions. As a toy example, consider a free energy of the form $$F(\phi) = a(T) \phi - b(T) \phi^2 + c(T) \phi^4$$ where $a(T)$ changes sign at $T = T_0$, and both $b$ and $c$ are positive about $T_0$. At $T = T_0$, the two minima of the free energy have the same value. For higher and lower temperatures, which minimum is lower changes.

This is a first order phase transition, since the equilibrium order parameter $\phi$ changes discontinuously. Note that the higher minimum is still metastable. Physically, thermal fluctuations allow $\phi$ to find the global minimum. Landau theory does not account for such fluctuations, it just postulates they happen.

Such phase transitions are generic when the theory does not have $\mathbb{Z}_2$ symmetry, since the linear term is allowed. If we do have $\mathbb{Z}_2$ symmetry but no other symmetries, then a second order phase transition is instead generic, taking place for $$F(\phi) = - b(T) \phi^2 + c(T)\phi^4$$ when $b(T)$ changes sign, assuming $c(T)$ positive.

  • $\begingroup$ You said, "as the sign of $a$ changes, the minimum changes". Can we say therefore the dependence of the order parameter on temperature difference $(T-T_0)$ a step function? You said, "Physically, thermal fluctuations allow $\phi$ to find the global minimum" I believe that the change of minimum is abrupt. Is it via tunneling or just hopping over the barrier? Thanks! $\endgroup$ Jun 9 '18 at 12:18
  • 2
    $\begingroup$ Yes, the order parameter changes with a step function. How the change happens depends on the specific physics of the situation. For example, if we are talking about the vacuum of a QFT, it would be via quantum tunneling. For the freezing of water into ice, it could be by bubble nucleation. $\endgroup$
    – knzhou
    Jun 9 '18 at 12:44
  • $\begingroup$ A clarification: How could you say, in the first case, that the minima are symmetrically situated on the $\phi$-axis about $\phi=0$ (one at $-\phi_0$ and another at $+\phi_0$)? $\endgroup$ Jun 28 '19 at 17:26
  • $\begingroup$ @mithusengupta123 This follows if the theory has $\phi \to -\phi$ symmetry. $\endgroup$
    – knzhou
    Jun 28 '19 at 17:42
  • $\begingroup$ The first free energy (first eq.) doesn't have that symmetry. $\endgroup$ Jun 28 '19 at 17:46

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.