Download Complete Set of Commuting Operators in Quantum Mechanics and more Study notes Quantum Mechanics in PDF only on Docsity! 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 (nx , ny , nz ) to fully specify a state [ or equivalently ( n , ny , nz ), ( n , nx , nz ), etc where 2 2 2x y zn n n n= + + ]. Just specifying n (or just nx ) is insufficient, since the eigenstates of (or ) are Ĥ xĤ 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: xˆ ˆ[ H , H ] 0= , x yˆ ˆ[ H ,H ] 0= ) then there exists 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: (nx , ny , nz )] are required to uniquely specify a state, then there must exit N commuting operators [example: ] 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. x y z ˆ ˆ ˆ( H , H ,H ) F-24
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