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I'm trying to show/convince myself the following statement is correct (I haven't been able to find any similar posts):

"There is no reversible quantum operation that transforms any input state to a state orthogonal to it."

I can see how this could be true based on the operation being unitary and that you could likely find some input state that isn't transformed into an orthogonal state. Is there a simple way to show/prove this? My unsuccesful approach has been:

Let $U$ be some unitary transformation/operation, and $|x\rangle$ some input state decomposed as $$|x\rangle=\lambda_1|0\rangle + \lambda_2|1\rangle + \lambda_1|2\rangle + \cdots + \lambda_n|n\rangle$$ where $$\lambda_1 \lambda_1+\lambda_2 \lambda_2+\cdots+\lambda_n\lambda_n=1,$$ and

$$U|x\rangle = \langle x|U^\dagger$$

I'm trying to show that there exists an input state $x$ such that $\langle x|U^\dagger|x\rangle = 0$ isn't true.

I've tried to make use of the properties of unitary matrices but haven't had much luck.

Any assistance or suggestions on alternative approaches on how to show this would be greatly appreciated.

It might be worth mentioning that this fact depends on the use of a

complexHilbert space for quantum mechanics. In a purely real Hilbert space, it is not true; for instance, in R^2 (the Euclidean plane), a rotation of 90 degrees around the origin rotates all real vectors into an orthogonal vector, more or less by definition.If you expand the space to C^2, though, this same matix (no considered as a complex 2x2 matrix) does indeed have two eigenvectors, with eigenvalues +i and -i, and there are many vectors that are transformed into a non-orthogonal vector by this matrix. – Alan Geller – 2019-10-03T19:24:55.937