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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.

user140255
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    I think "stationary" means "equilibrium" here. – Norbert Schuch Sep 07 '20 at 16:11
  • The identity operator...with respect to which basis in the Hilbert space?? – sintetico Sep 07 '20 at 16:50
  • @NorbertSchuch for me these terms are synonymous, does it change anything? – user140255 Sep 07 '20 at 19:11
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    @sintetico it makes no difference, the identity operator is the same with respect to any orthonormal basis – user140255 Sep 07 '20 at 19:12
  • Sure. The Gibbs (=thermal equilibrium) state does not break symmetries. (Well, it has long range order, which to some people is synonymous to symmetry breaking ... ) – Norbert Schuch Sep 07 '20 at 19:13
  • I see, but in the paper he explicitly refers to stationary as being non-evolving – user140255 Sep 07 '20 at 19:19
  • Sorry, you are right, I said something stupid. – sintetico Sep 07 '20 at 22:15
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    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. – fra_pero Sep 09 '20 at 07:33

1 Answers1

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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 $U$.

In short, if a stationary state is non degenerate, it is invariant under the symmetries of the Hamiltonian. If there are some stationary states which are degenerate, their linear combinations, as a whole (eigenspace) are invariant under the symmetries of the Hamiltonian. Broken symmetry ground states belong to this category, they are stationary and degenerate.

sintetico
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  • Thank you for your answer. However, it is not quite what I need. Anderson stresses this affirmation as fundamental and it does seem like he would make a mistake on that. I added a quote from the paper in my question to guide answers. – user140255 Sep 09 '20 at 02:20
  • Of course Anderson did not make any mistake! This is not what I wrote. – sintetico Sep 09 '20 at 09:29
  • Anderson does not say that you can choose an invariant state in the degenerate eigenspace of an hamiltonian, he explicitly says that no broken symmetry state can be stationary, which contradicts what you say. – user140255 Sep 09 '20 at 12:53
  • I will clarify with the next edit. Maybe my wording was not very clear, but my argument is not in contradiction with Anderson 's statement – sintetico Sep 09 '20 at 14:59