# Spontaneous Symmetry Breaking

In Spontaneous symmetry breaking we have got that, a field $$\phi= \pm \sqrt{\frac{-m^2}{\lambda}}.$$ Now in order to get the unstable minima we need to guess the mass $m^2 <0$. But can mass be negative? Why are we guessing so?

One more question is that, how this lead us to the solution of nonlinear equation?

• I don't understand your second question, can you say more? – innisfree Mar 28 '14 at 20:41
• Yes, but real scalar fields ($\phi\in\mathbb{R}$) or complex scalar fields ($\phi\in\mathbb{C}$)? ;-) – Wouter Mar 28 '14 at 21:36
• Yeah, I'm not too sure about the solitons, but isn't the idea of spontaneous symmetry breaking for a real scalar field due to the fact that you have a Lagrangian density of the form $\mathcal{L} = \partial_{\mu}\phi\partial^{\mu}\phi \mp m^2\phi^2 - \lambda\phi^4$ (which is the most general form you can write down that is Lorentzinvariant and $\mathbb{Z}_2$ invariant, and yields a renormalizable theory). Now, the minus sign corresponds to a potential $V_-(x)=m^2\phi^2+\lambda\phi^4$, which only has one extremum, a minimum at $\phi=0$. [to be continued...] – Wouter Mar 28 '14 at 21:46
• [...continued] The plus sign corresponds to three extrema, a local maximum at the origin and two global minima at $\phi = \pm\sqrt{\frac{m^2}{2\lambda}}$. One of these two physically equivalent groundstates is chosen, which constitutes the spontaneous breaking of symmetry. But perhaps soliton theory isn't about this, I can't seem to recall much about this. Then again, I'm not sure we've really treated solitons in-depth in our course... – Wouter Mar 28 '14 at 21:50
• Yeah, I think @Wouter is right. You should think of the $m^2$ as the coupling between fields and not the mass (before symmetry breaking). After symmetry breaking, it turns out that this coupling has turned into a mass term for the fields, and the mass is positive. Regarding your second question: I think it is way too broad. Whole books have been written about solitons, and you will need to acquire some mathematical skills. For instance, you will need to have at least an intuitive understanding of homotopy theory. – Hunter Mar 28 '14 at 23:31

The mass can be supposed negative without much effort in the Ginzburg-Landau model:

$$F=\int dr\left[\dfrac{\beta}{2}\left(\left|\Psi\right|^{2}-\dfrac{\alpha}{\beta}\right)^{2}+\dfrac{1}{2m}\left|-\mathbf{i}\hslash\nabla\Psi\right|^{2}\right]$$

with $\alpha\propto\left(T-T_{c}\right)$ negative for temperatures $T$ below the critical temperature $T_{c}$. You can call $\alpha$ a mass if you like, but it is a parameter of the model in fac, like $\beta$, ....

The minimum of the free energy $F$ above is $F=0$ given for

$$\Psi_{\infty}^{2}=\dfrac{\alpha}{\beta}\Rightarrow\Psi_{\infty}=\pm\sqrt{\dfrac{\alpha}{\beta}}$$

for positive $\alpha$ (so I choose $\alpha\propto\left|T-T_{c}\right|$ in fact). It corresponds to the space-independent field configuration for $x\rightarrow\infty$, such that $\nabla \Psi_{\infty} =0$. We nevertheless realise that there are two field configurations, $\pm\Psi_{\infty}$.

We next suppose a position dependent 1D problem, and real field for simplicity (it doesn't alter the main conclusion below, but it changes the exact solutions). We ask ourself: is there a space dependent solution tending to $\pm\Psi_{\infty}$ when $x\rightarrow \pm\infty$ ? Well, two possibilities:

• $\Psi\left(x\rightarrow+\infty\right)=\Psi\left(x\rightarrow-\infty\right)=\Psi_{\infty}$ or $\Psi\left(x\rightarrow+\infty\right)=\Psi\left(x\rightarrow-\infty\right)=-\Psi_{\infty}$ in which case there is no space-dependent solution

• $\Psi\left(x\rightarrow+\infty\right)=-\Psi\left(x\rightarrow-\infty\right)=\Psi_{\infty}$ in which case there is a space-dependent solution (called an instanton, or a soliton) which connects the two non-trivial minima of the model

We now find the space-dependent solution. The equation of motion we should verify is

$$\dfrac{\delta F}{\delta\Psi}=0\Rightarrow\xi^{2}\dfrac{d^{2}\Psi}{dx^{2}}+2\Psi\left(1-\dfrac{\alpha}{\beta}\Psi^{2}\right)=0$$

where we use that the field $\Psi$ is real (otherwise the factor 2 in front of the second term disappears, this is the simplifying hypothesis to find a simple instanton) and where $\xi=\hbar/\sqrt{2m\alpha}$ is the characteristic length of the problem (in the theory of superconductivity, it is called the coherence length). You see this is a non-linear equation, so solutions are complicated (when they are known). Here, you can nevertheless easily show that

$$\Psi\left(x\right)=\pm\sqrt{\dfrac{\alpha}{\beta}}\tanh\dfrac{x-x_{0}}{\xi}=\pm\Psi_{\infty}\tanh\dfrac{x-x_{0}}{\xi}$$

is the solution we were looking at. You can choose $x_{0}=0$ for the initial condition. This solution smoothly connects the two non-trivial minima $\pm\Psi_{\infty}$ as required.

The local energy (the integrand of $F$) for this solution has a bump around $x_{0}$ of characteristic size $\xi$.