Math, asked by navi993500, 7 months ago

unitary operator drfinition and condition.
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Answers

Answered by sidhupari

Answer:

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Step-by-step explanation:

unitary operator is a bounded linear operator

U : H → H

on a Hilbert space H that satisfies U*U = UU* = I, where U* is the adjoint of U, and I : H → H is the identity operator. The weaker condition U*U = I defines an isometry. The other condition, UU* = I, defines a coisometry.

Answered by alenronaldo63514

Answer:

Definition 1. A unitary operator is a bounded linear operator U : H → H on a Hilbert space H that satisfies U*U = UU* = I, where U* is the adjoint of U, and I : H → H is the identity operator. The weaker condition U*U = I defines an isometry. The other condition, UU* = I, defines a coisometry.

Step-by-step explanation:

s \|^2 - \overline{\lambda} \langle U(\lambda x), U(x) \rangle - \lambda \langle U(x), U(\lambda x) \rangle \\

&= |\lambda|^2 \| x \|^2 + \| \lambda x \|^2 - \overline{\lambda} \langle \lambda x, x \rangle - \lambda \langle x, \lambda x \rangle \\

&= 0

\end{align}

Analogously you obtain

\| U(x+y)-(Ux+Uy)\| = 0.

Properties

The spectrum of a unitary operator U lies on the unit circle. That is, for any complex number λ in the spectrum, one has |λ| = 1. This can be seen as a consequence of the spectral theorem for normal operators. By the theorem, U is unitarily equivalent to multiplication by a Borel-measurable f on L2(μ), for some finite measure space (X, μ). Now UU* = I implies |f(x)|2 = 1, μ-a.e. This shows that the essential range of f, therefore the spectrum of U, lies on the unit circle.

A linear map is unitary if it is surjective and isometric. (Use Polarization identity to show the only if part.)

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unitary operator drfinition and condition. #mathematical physics it's urgent please. best ans will be marked as brainliest wrong will be reported​

Answers

unitary operator drfinition and condition.
#mathematical physics
it's urgent please.
best ans will be marked as brainliest
wrong will be reported