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If $[A,B]=0$ then there is a unitary transformation $U_{A,B}$ that diagonalises both $A$ and $B$ simultaneously. This transformation depends on the pair of commuting operators $(A,B)$, so that for a different pair there could be a different unitary. Assume that all the eigenvalues of $H$ have multiplicity 1. Then there exists a unique (up to a phase factor) ...


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If $H$ commutes with $A_1$, then it will indeed share an eigenbasis with it. Your mistake is in supposing that it will share the same eigenbasis with both $A_i$s. Examples are easy to provide: On the trivial side, if $H=E_0\mathbf 1$ is trivial, then it shares an eigenbasis with $A_1=x$ and it shares an eigenbasis $A_2=p$, but it cannot share an ...



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