PageF-23 - a set of orthonormal simultaneous eigenstates of...

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F-23 Notice that in 1D problems, like the 1D infinite well or the 1D SHO, we only needed one number (n) to uniquely specify an eigenstate. This state label is called a quantum number or q-number and it is always in a 1-to-1 correspondence with the eigenvalues of an observable operator. But in 3D problems, we need 3 quantum numbers (n x , n y , n z ) to fully specify a state [ or equivalently ( n , n y , n z ), ( n , n x , n z ), etc where 2 2 2 x y z n n n n = + + ]. Just specifying n (or just n x ) is insufficient, since the eigenstates of ˆ H (or x ˆ H ) are degenerate . In cases with degeneracy, more than one quantum number is required to specify a state, and the other quantum numbers are associated with other operators that must commute with the first. If two operators commute (examples: x ˆ ˆ [H,H ] 0 = , x y ˆ ˆ [H ,H ] 0 = ) then there exists
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Unformatted text preview: a set of orthonormal simultaneous eigenstates of both operators. We proved this for the case of operators with non-degenerate states, but it is also true when there are degeneracies. (We will show below that when operators do not commute, it is impossible to find simultaneous eigenstates.) Claim: If N quantum numbers [example: (n x , n y , n z )] are required to uniquely specify a state, then there must exit N commuting operators [example: x y z ˆ ˆ ˆ (H ,H ,H ) ] whose simultaneous eigenstates are non-degenerate and whose N eigenvalues provide the quantum numbers that uniquely label the state. Such a set of operators is called a complete set of commuting operators (CSCO). We will give a proof later, when we talk about matrix formulation of QM....
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