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I have a system of equations: \begin{cases} f\left(x_{1}\right)+f\left(x_{2}\right)+P=0\\ \\ g\left(x_{1}\right)+g\left(x_{2}\right)=0 \end{cases}

where $f$ and $g$ are some functions, $P$ is a parameter that I want to vary, and $\boldsymbol{x}=\left[x_{1}, x_{2}\right]$ is the unknown. Let's say that for $P<P^{*}$ the solution of my equation is a stable $$ \underline{\boldsymbol{x}}=\left[\begin{array}{c}\alpha\left(P\right) \\\alpha\left(P\right) \end{array}\right] $$ while for $P>P^{*}$ the solution $\underline{\boldsymbol{x}}$ becomes unstable, and I also got two new stable solutions $$ \begin{align} \overline{\boldsymbol{x}}=\left[\begin{array}{c}\beta\left(P\right) \\\gamma\left(P\right) \end{array}\right]\qquad\text{and}\qquad \overline{\overline{\boldsymbol{x}}}=\left[\begin{array}{c}\gamma\left(P\right)\\\beta\left(P\right)\end{array}\right] \end{align} $$ where $\alpha\left(\cdot\right)$, $\beta\left(\cdot\right)$ and $\gamma\left(\cdot\right)$ are functions that depend on $f\left(\cdot\right)$ and $g\left(\cdot\right)$.

Clearly the system of equations is always (i.e. $\forall P$) symmetric under exchange of $x_{1}$ and $x_{2}$, but the solutions (i.e. $\overline{\boldsymbol{x}}$ and $\overline{\overline{\boldsymbol{x}}}$) are not for $P>P^{*}$. Moreover, for $P>P^{*}$, in numerical simulations the solution converge to one between $\overline{\boldsymbol{x}}$ and $\overline{\overline{\boldsymbol{x}}}$. In other terms, I think this is a pitchfork bifurcation. Is this an example of spontaneous symmetry-breaking?

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  • $\begingroup$ SSB occurs when the vacuum in a QFT isn't invariant under a symmetry transformation. you describe solutions that aren't "invariant" - but solutions to EOM usually "break" a symmetry. throw a ball in a particular direction, you've "broken" rotation etc but of course the dynamics are still symmetric $\endgroup$ – innisfree May 18 '15 at 20:25
  • $\begingroup$ that said, the set of all the solutions should be as symmetric as the original equations $\endgroup$ – innisfree May 18 '15 at 20:36
  • $\begingroup$ There are other kinds of spontaneous symmetry breaking. Every crystal is an example of SSB, as is every balanced needle that falls to one side. $\endgroup$ – user27118 May 19 '15 at 0:46
  • $\begingroup$ According to en.wikipedia.org/wiki/Symmetry_breaking, there are two kinds of symmetry-breaking. The first is when symmetry is broken explicitly, e.g. the Zeeman effect. In this case you apply an external magnetic field that breaks the SO(4) symmetry of the hydrogen atom. The field is aligned along one axis, so the hamiltonian and the equations of motions are not symmetric anymore under rotations. The second case is when you don't have any term that explicitly breaks the symmetry of the equations, but you get solutions that are not invariant, without any external source of asymmetry. $\endgroup$ – user2983638 May 19 '15 at 8:23
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    $\begingroup$ So it seems to me that the case I described in my post is an example of the second case, since I don't have any asymmetric term in my equations. I mean, when P>P* and I get different solutions, the equations are still symmetric under exchange of x_1 and x_2. So I think this should be a SSB w.r.t. the symmetry group S_2 of permutations en.wikipedia.org/wiki/Symmetric_group $\endgroup$ – user2983638 May 19 '15 at 8:29
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Spontaneous symmetry breaking requires more than two distinct solutions to a system of equations, because there are often physical considerations not captured in the equations that determine the solution or combination of solutions chosen.

For example, take the 1D elastic collision between two objects of equal mass, one stationary and one initially moving at speed $v_0$. Momentum and energy are conserved, so after the collision

$$ v_1^2 + v_2^2 - v_0^2 = 0 $$

$$ v_1 + v_2 - v_0 = 0 $$

(These aren't exactly what you posted, but the concept is the same.) There are two solutions, one in which the first mass becomes stationary and the second mass carries on with the same speed as the original, and another in which the second mass remains stationary and the first mass moves through it uninterrupted, which we discard on physical grounds. This is not spontaneous symmetry breaking. Your equations could be an instance of spontaneous symmetry breaking, for example if the solutions of $\mathbf{x}$ describe degenerate configurations in the free energy of some system - but we would have to know what they refer to.

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  • $\begingroup$ Thanks user27118. Well, my equations describe possible solutions of the membrane potentials in a neural network when you increase anesthetics over a critical concentration. There shouldn't be any reason to discard one of them. Moreover, I suspect that during simulations the system chooses one of these solutions due to numerical approximations, that act as a source of noise (which is not captured by the equations, as you correctly said). $\endgroup$ – user2983638 May 19 '15 at 8:10
  • $\begingroup$ Moreover, for P<P* I got a stable solution, while for P>P* two stable ones, plus $\underline{x}$, which is unstable but still a valid solution. So to me it looks like the situation shown here en.wikipedia.org/wiki/Spontaneous_symmetry_breaking#/media/… $\endgroup$ – user2983638 May 19 '15 at 8:37
  • $\begingroup$ That sounds like SSB of the regular physical type to me. $\endgroup$ – user27118 May 19 '15 at 14:50
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Well, in his book "Symmetries in Science" (see attached picture), B. Gruber describes SSB without mentioning the ground state at all. His words describe exactly what I found in my system, so I conclude this is indeed a SSB, even if I don't have an energy function.

enter image description here

It is maybe a slightly more general definition of SSB compared to that of Hamiltonian systems.

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