# Are broken symmetry states non-stationary?

In his famous paper, More is Different (link), Philip W. Anderson states that in the context of quantum mechanics :

[...] the state of the system, if it is to be stationary, must always have the same symmetry as the laws of motion which govern it.

However, that does not seem to be true in general. The most trivial example I can think of is a Hamiltonian equal to the identity operator (symmetric to any transformation) with respect to which any state is stationary. There are a lot of other examples.

Anderson obviously means something more. In which context does his statement apply?

EDIT : In the paper, Anderson gives the example of the ammonia molecule. Here is a quote to the discussion that follow. I want to know if these precise statements are true and how.

no stationary state of a system (that is, one which does not change in time) has an electric dipole moment. If ammonia starts out from the above unsymmetrical state, it will not stay in it very long. By means of quantum mechanical tunnelling, the nitrogen can leak through the triangle of hydrogens to the other side, turning the pyramid inside out, and, in fact, it can do so very rapidly. A truly stationary state can only be an equal superposition of the unsymmetrical pyramid and its inverse.

• I think "stationary" means "equilibrium" here. Commented Sep 7, 2020 at 16:11
• The identity operator...with respect to which basis in the Hilbert space?? Commented Sep 7, 2020 at 16:50
• @NorbertSchuch for me these terms are synonymous, does it change anything? Commented Sep 7, 2020 at 19:11
• @sintetico it makes no difference, the identity operator is the same with respect to any orthonormal basis Commented Sep 7, 2020 at 19:12
• Great question. I do not have time to read again the article and give an organic reply right now, I'll do it. Let me just point out that Anderson starts considering ammonia, then larger systems. The larger the system, the longer it persists in one broken-symmetry state: "Hydrogenphosphide [..] which is twice as heavy as ammonia, inverts, but at one tenth the ammonia frequency." In the thermodynamic limit (infinite system size) broken-symmetry states do indeed become stationary. To summarize, I think that quote is strictly speaking valid only for finite-size systems. Commented Sep 9, 2020 at 7:33

Consider a generic Hamiltonian $$H$$ and a unitary operator $$U$$. This operator is a symmetry of the system if and only if it commutes with the Hamiltonian $$[H,U]=0$$. I consider only one symmetry operator for simplicity.
If a state $$|\psi\rangle$$ is stationary and non-degenerate, then it is an eigenvector of the Hamiltonian $$H$$, and since $$[H,U]=0$$ it is also an eigenvector of the symmetry operator. That is, the stationary state has the symmetry of the Hamiltonian.
If there are a set of degenerate $$|\psi_i\rangle$$ states instead, in principle the individual states are not necessarily eigenvector of the symmetry operator. This is at the core of the concept of spontaneous symmetry breaking. However, the eigenspace spanned by the combination of the degenerate eigenvectors $$\sum c_i |\psi_i\rangle$$ is invariant under the action of the symmetry operator $$U$$. Let's consider your example. The identity operator $$H=1$$ has only degenerate states (all eigenstates are degenerate). The eigenspace spanned by these degenerate states coincide with the full Hilbert space, which is trivially invariant under the action of the symmetry operator $$U$$.