It is common knowledge in Physics that eigenstates share the symmetries of the Hamiltonian.(adsbygoogle = window.adsbygoogle || []).push({});

And it is trivial to show that this is true for the eigenspaces. Let g be an element of a symmetry group of Hamiltonian H, [tex]M_g[/tex] its representation, [tex]\left| \phi \right>[/tex] an eigenvector and [tex]\lambda[/tex] the corresponding eigenvalue. If g is an element from a symmetry group of H it is given that:

[tex]M_gHM_{g^{-1}}\left|\phi\right>=\lambda\left|\phi\right>[/tex]

Thus

[tex]HM_{g^{-1}}\left|\phi\right> = \lambda M_{g^{-1}}\left|\phi\right>[/tex]

So we see that the eigenspace for [tex]\lambda[/tex] is closed under the operations of [tex]M_g[/tex]

Is there some theorem, that I can decompose this eigenspace into eigenstates of M or how do I proceed from here? This is important for Bloch functions and crystallography. I read something about Schur's lemma being involved.

So to phrase a proper question:

Is every Eigenfunction of a Hamiltonian,invariant up to a scale factor of unity magnitude under the operation of the Hamiltonian's symmetry groups? And if this is so how do I show this?

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# Symmetry of eigenstates

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