Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
I removed the wikipedia link because my question is specific to the Ising system and not to symmetry breaking in general – Sankaran Jan 29 '13 at 20:07
I haven't come across the term 'symmetry breaking field' when learning about this. But perhaps it signifies the effective magnetic field $H_{eff} = H + ZJM$ where $H$ would be zero in the case of spontaneous symmetry breaking, $Z$ is the number of nearest neighbours, $J$ the strength of the interaction and $M$ the magnetisation. So maybe the term 'symmetry breaking field' relates to $H_{eff} = ZJM_{T<T_c}$ where $M_{T<T_c}$ is the magnetisation of the system below the critical temperature. – Wouter Jan 29 '13 at 20:10
up vote 4 down vote accepted

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.

share|cite|improve this answer
I thought so but wanted to confirm the terminology. Thanks for a simple and elegant confirmation. – Sankaran Apr 30 '13 at 17:21

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.