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Is operator that is made from orthonormal operator also orthonormal?
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[QUOTE="brmath, post: 4514716, member: 486151"] I am a little confused about the statement of the problem. First G cannot be both an operator and a basis. A space with a basis of 3 elements is 3-dimensional and isomorphic to $$R^3$$ It is still a Hilbert space, but we would ordinarily call it a vector space. Second, I don't understand the notation $$| \Psi _1 >$$ So this may not be the right question but I'm going to assume you are asking this: Let $$x_1,x_2,x_3$$ be a basis for the space H and let U be an operator that maps$$ x_1 \rightarrow x_2; x_2 \rightarrow x_3; x_3 \rightarrow x_1$$ Is U a unitary operator? The definition of unitary operator is a bounded operator U such that $$(1)...UU^* = U^*U = I \hspace 5px where \hspace 5px U^*$$ is the adjoint of U. ( Note I said "the" adjoint, implying that it is unique. It's worth checking that out.) If this is your problem you need to find an adjoint operator for U. Since U has only 3 operations, you should be able to find its adjoint by inspection. Then see if your proposed adjoint fits with (1). If this is the wrong problem, please write again and clarify what you mean. [/QUOTE]
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Is operator that is made from orthonormal operator also orthonormal?
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