Ising Ferromagnet: Spontaneous symmetry breaking or not?

Question

In explaining/introducing second-order phase transition using Ising system as an example, it is shown via mean-field theory that there are two magnetized phases below the critical temperature. This derivation is done for zero external magnetic field $B=0$ and termed spontaneous symmetry breaking The magnetic field is then called the symmetry breaking field. But, if the symmetry breaking occurs "spontaneously" at zero external field why do we need to call the external magnetic field the symmetry breaking field? I am confused by the terminology.

Answer 1

This is mostly a question of definitions:

Spontaneous symmetry breaking occurs when the underlying laws of a physical system have a symmetry, but the ground state does not. For an Ising system with $B=0$, $$H = \sum_{i,j} J_{ij} s_i s_j$$ we can see explicitly that the energy of a state $\{s_i\}$ is precisely the same as the energy of the state with every spin flipped, $\{-s_i\}$. Nevertheless, the ground state does not have this symmetry - all of the spins are either up or down! There are a lot of subtleties to this idea, especially in how it relates to the limit of infinite system size - I really recommend reading Goldenfeld's Lectures on Phase Transitions and the Renormalization Group to understand this more deeply.

By contrast, with $B \neq 0$, the symmetry is explicitly broken - the Hamiltonian $$H = \sum_{i,j} J_{ij} s_i s_j + B \sum_i s_i$$ does not have the $s\to-s$ symmetry. These are two different ideas.

Answer 2

You don't need a B-field for spontaneous symmetry breaking to occur. In the Ising model, there is a term in the free energy which punishes misalignment of spins. For low enough temperatures (i.e. when the quadratic coefficient changes sign) this term dominates the usual entropy driven term. At that point symmetry is broken and you get non-zero expectation values for the magnetization. The term 'spontaneous symmetry breaking' is coined as there are infinite many minima of the corresponding action, but you choose only one. You can also consider the Ising model with an external field, the critical temperature shall be higher then because the B-field 'helps' the misalignment term I talked about above.

Answer 3

The mean field theory states that the average magnetization is $$\bar{\mu}=\frac{\sum_{\text{states}}\mu e^{\mu\beta B_E}}{\sum_{\text{states}}e^{\mu\beta B_E}}=\frac{M}{N},$$ where $M$ is the total magnetization, $\mu=\pm\mu_0$ the two possible magnetizations (states), $N$ the number of spins and the effective field acting on spins $B_E=B_{\text{external}}+B'$; $B'$ is the molecular field (interaction between spins). There is two importants cases:

• case 1 $\bar{\mu}=0\rightarrow M=0$ no spontaneous magnetization (random behavior), so $T\geq T_c$;
• case 2 $\bar{\mu}=\mu_0\rightarrow M_\infty=N\mu_0$ saturation value, $T\rightarrow T_c$.

We can now evaluate the total magnetization $$M=N\mu_0\tanh{\Big[\frac{\mu_0}{k_BT}(B+B')\Big]},$$ now if the external field is set to zero $B=0$, we get $$\frac{M}{M_\infty}=\tanh{\Big[\frac{\mu_0}{k_BT}B'\Big]}=\frac{z_+-z_-}{z}=\frac{\#\text{spins up}-\#\text{spins down}}{\#\text{spins tot}}.$$ We need to do an assumption (a self consistent one's) that is: $B'\propto M$. If you think that $\Delta E=\pm\mu_0B'=J(z_+-z_-)$ so you get the closed relation for $M$ $$\frac{M}{M_\infty}=\tanh{\Big[\frac{Jz}{k_BT}\frac{M}{M_\infty}\Big]}.$$ The critical temperature became $T_c=\frac{Jz}{k_B}$ and the analysis of the following relation $$X=\tanh{aX},a=\frac{T_c}{T},X=\frac{M_s}{M_\infty}$$ is the last thing to do. If we have $X\simeq0$ so the relation became $X=X$ only for $X=0$ ($M=0$, $a=1$) no magnetization. The other case, $a<1$, $X=0$ and again $M=0$ no spontaneous magnetization $T>T_c$.The last case is $a>1,(T<T_c)$ and $X\neq0\neq M$, net spontaneous magnetization. This mean Spontaneous Symmetry breaking in MFT, the transition between disordered and ordered phases.