Rotational invariance and degeneracy (quantum mechanics)

AI Thread Summary
A Hamiltonian that is invariant under all rotations implies that its eigenstates are also eigenstates of the angular momentum operator L². This leads to a degeneracy of 2l+1 for these eigenstates, where l represents the angular momentum quantum number. The discussion highlights the use of both generic Hilbert space and 3D real Hilbert space for calculations. The original poster initially struggled with the problem but later confirmed they found a solution. The key takeaway is the relationship between rotational invariance and the properties of eigenstates in quantum mechanics.
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Homework Statement




Show that if a Hamiltonian H is invariant under all rotations, then the eigenstates of H are also eigenstates of L^{2} and they have a degeneracy of 2l+1.


Homework Equations



The professor told us to recall that

J: \vec{L}=(L_x,L_y,L_z)

L_z|l,m\rangle=m|l,m\rangle

L_\pm=L_x\pm iL_y

L_\pm|l,m\rangle= \hbar\sqrt{l(l+1)-m(m\pm1)} |l,m\pm 1\rangle


The Attempt at a Solution



I have been reading as much materials as I can, but I still have no clue at all on how to solve it at this moment. Can anyone help? Thank you so much!
 
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I solved it!

Thank goodness! I have solved it now.

One can do the calculations in either generic Hilbert space for general rotations or in 3D real Hilbert space.
 
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