Are symmetries in the equation necessarily symmetries in the corresponding solution(s)?

Question

I wonder whether the symmetries in the equations (such as the heat equation, the wave equation, the Schrödinger equation, Maxwell equations) are reflected into their solution(s). I.e., assuming that for a particular equation, a particular initial and boundary conditions, if the equation has, say, a rotational, translational as well as a Lorentz invariance, can I always expect the solution (assuming it's unique) to preserve these symmetries?

In electro and magnetostatics, when the equation to be solved had a rotational symmetry around, say, the z-axis, then we assumed that the solution had no dependence on the azimuthal angle coordinate. So I think that in that case, by looking at the solution, I can guess that the equation has rotational invariance around an axis and that the boundary condition also have that symmetry. Are there examples where such a supposition regarding the form of the solution fail? In other words, can a solution to an equation not contain a symmetry that the underlying equation has? If so, how can one explain it?

Answer 1

Let me concentrate for a moment in the geodesic equation, since it is an ordinary differential equation and easier to describe. The set of all solutions of the geodesic equation is equivalent to the space of initial conditions of the theory, since an initial coordinate plus an initial momentum completely determine the geodesic and vice versa. So instead of talking about the set of solutions we can talk about the space of initial conditions in other words the phase space of the theory.

Starting from a specific solution which corresponds to a specific point in phase space, we have two cases.

1) The initial point is invariant under a group of symmetries of the Hamiltonian.

2) The initial point is invariant only under a subgroup of the group of symmetries of the Hamiltonian.

The first case corresponds to spontaneously unbroken symmetry of the classical system, while the second to a spontaneously broken symmetry of the system. Spontaneous symmetry breaking is not just a quantum phenomenon, it has a classical counterpart. Moreover, in classical theory, spontaneous symmetry breakdown can occur in finite number of degrees of freedom in contrast to quantum theory where it requires an infinite number of degrees of freedom.

In the following answer, I tried to explain how the above phenomenon is realized in the geodesic equation of a particle in $3$-space on one hand and a particle moving on a sphere on the other hand.

In the first case (free particle in $3D$) an initial condition starting from the origin and having a momentum along the $z$-axis. Both the position and momentum are invariant under the group of rotations about the $z$- axis (a $U(1)$ group). In this case this $U(1)$ symmetry group is not spontaneously broken. In the sphere case, if we choose to start from the south pole then the position is invariant under the group of rotations about the $z$ axis but the momentum is not because it must be tangent to the sphere. Thus this is the case of a spontaneously broken symmetry.

What happens now is that in the first case all points of the geodesic trajectory are invariant under the rotation about the $z$-axis, while in the second case, except for the south and the north pole, the classical trajectory is not invariant under the rotation about the $z$-axis. In the sphere case the rotation group maps a solution (great circle) into another solution.

Now, for the equations expressed as partial differential equations; basically the principle is the same, but the initial conditions are the field configurations on a Cauchy surface. The space of initial conditions is infinite dimensional in this case, but in many cases we can show that it has the properties of a phase space (symplectic or Poisson). In these cases also a a partially invariant initail solution corresponds to spontaneous symmetry breaking, which causes the configurations of almost all times to be non-invariant under the symmetry.

Answer 2

In reading an article about misconceptions in QM , I found the following by chance:

If a one-dimensional potential has reflection symmetry about any point, then the probability density functions associated with each energy eigenstate will also posses that symmetry. This fact is so reasonable (and so strongly stressed by many textbooks) that many students slip into believing that it holds for other symmetries, such as the rotational symmetry of the Coulomb problem, where it does not. I have often seen this misconception submerge and propagate unseen until the student takes a course in solid state physics, where it emerges to make the Bloch theorem appear trivial.

So the answer to my question is "No."

I do not understand the reference of the word I put in bold. What does that word refer to?

Anyway, I tried to understand Bloch theorem. Apparently, in a solid the Hamiltonian describing Bloch electrons has a lack of translational invariance (the translational symmetry is broken), except for very particular translations: the ones by a Bravais lattice vector $\vec R$. So the Hamiltonian can be thought to have a translational symmetry invariance when the translation is a vector multiple of $\vec R$. If the answer to the question is a yes, then the wavefunction of the Bloch electron should also have that symmetry. However, according to the following, it doesn't:

The Bloch wavefunction is $\psi _{\vec k} (\vec r)=e^{i\vec k \cdot \vec r}u_{\vec k} (\vec r)$ where $u_{\vec k} (\vec r)$ shares the same invariance with respect to Bravais lattice vector translations than the Hamiltonian, namely $u_{\vec k} (\vec r) = u_{\vec k} (\vec r + \vec R)$. A quick calculation reveals that $\psi _{\vec k} (\vec r + \vec R)=\psi _{\vec k}(\vec r)e^{i\vec k \cdot \vec R} \neq \psi _{\vec k}(\vec r)$. Hence the wavefunction of the Bloch electrons lacks the symmetry of translations by a Bravais lattice vector that the Hamiltonian possesses.

Answer 3

1. Perhaps a simple counterexample is instructive: The EOM $$\ddot{\bf r}~=~0$$ of a free point particle in Newtonian mechanics is invariant under a Galilean boost
   $$\dot{\bf r}\quad \longrightarrow \quad\dot{\bf r}+{\bf v},$$ but a solution/trajectory is not.

2. Counterexamples exist also in field theory. If you like this question you may also enjoy reading this Phys.SE post.