Is operator that is made from orthonormal operator also orthonormal?

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.

 

I now understand why you used that notation. The answer to your question about is U unitary is "yes" but the reasons you give are not adequate, nor correctly stated.

First, you must distinguish between an operator and a set of vectors. They are not at all the same thing. So what you call G is a set of vectors, and what you call U is an operator that does something to the vectors. So it is wrong (and worse yet, confusing) to write ##U=\{x_2,x_3,x_1\}##

The vectors in G are stated to be orthonormal. The U doesn't arise from any other operator. It is an operator whose values are described on a normal basis, and which is thus defined for the entire space.

Now you must come up with some valid reasons why U is unitary. First, you must define what is meant by unitary, and then you must show that U fulfills those conditions.

In my reply to you I replaced the ##| \Psi_i > \hspace 5px with \hspace 5px x_i##. So if you will just read my answer that way, I think you could start on this problem.

Start reading at the line that says let ## x_1, x_2, x_3 ...##