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- Thread starter bikashkanungo
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In summary: To see that this implies that U is a normed space isomorphism, let y be an arbitrary element of H, and let x be an element of H such that \langle x,y\rangle=\|y\|. The x that does the job is then U^*y, since then we get \|y\|=\langle x,y\rangle=\langle U^*y,y\rangle=\|U^*y\| (because U^* is an isometry), so that \|y\|=\|U^*y\|. This shows that U^*y=U^*U^*Uy=U^*U^*U^*Uy=U^*U^*y

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What number is positive and has unit modulus ?

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<P|¯UU |P> is positive and ¯UU =1

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Before this, you concluded that [itex]U^*U[/itex] is a real number times the identity operator. So what you're saying here is that if r is a real number and [itex]\langle P|rI|P\rangle=\langle P|P\rangle[/itex] for all [itex]|P\rangle[/itex], then r=1. The left-hand side is obviously equal to [itex]r\langle P|P\rangle[/itex]. So for all [itex]|P\rangle[/itex],bikashkanungo said:for any ket |P> , <P|¯UU |P> is positive and equal to <P|P> , so ¯UU can be taken as equal to 1 . How does the last equation is concluded ?

[tex]r\langle P|P\rangle=\langle P|P\rangle.[/tex] All you need to know to conclude that r=1 is that there's a [itex]|P\rangle[/itex] such that [itex]\langle P|P\rangle\neq 0[/itex].

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Let U be a linear operator acting on a separable Hilbert space [itex] \mathcal{H}[/itex] subject to the condition

[tex] \forall \psi\in D(U) \subset \mathcal{H}, \, \langle U\psi,U\psi\rangle = \langle \psi,\psi\rangle [/tex]

It follows that U is bounded, hence continuous and can be extended through continuity to all vectors in the Hilbert space. If it's bounded and defined everywhere, it admits an unique adjoint, so that the isometry condition becomes

[tex] \langle \psi, \left(U^{\dagger}U - \hat{1}\right)\psi \rangle = 0 [/tex]

It follows that [itex] \left(U^{\dagger}U - \hat{1}\right)\psi \in \mathcal{H}^{\text{orthogonal}} \Rightarrow \left(U^{\dagger}U - \hat{1}\right)\psi = 0 [/itex]

The operator in the brackets is forced then to be the 0 operator, since [itex]\psi [/itex]is arbitrary. Then

[tex] U^{\dagger}U = \hat{1} [/tex].

The statement [itex] UU^{\dagger} = \hat{1} [/itex] follows simply from the fact that, because U is bounded, its double adjoint is equal to U.

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Another approach: One possible definition of a unitary operator is: U is said to be unitary if it's a normed space isomorphism (a linear bijective isometry) from H onto H. A linear isometry is obviously bounded. We can prove that a unitary operator defined this way satisfies [itex]\langle Ux,Uy\rangle[/itex] for all x,y. This implies that [itex]U^*U=1[/itex]. We can also prove that the set B(H) of bounded linear operators satisfy the definition of a C*-algebra. The norm of U then follows immediately from the C*-identity [itex]\|A\|^2=\|A^*A\|[/itex], which is satisfied by all members of B(H).

D'oh, for a moment I thought that this thread was about determining the norm of a unitary operator, but it's about proving that U*U=1. In this approach, the definition includes the condition [itex]\|Ux\|=\|x\|[/itex] for all x. This clearly implies that [itex]\langle Ux,Ux\rangle=\langle x,x\rangle[/itex] for all x. Now if we let y,z be arbitrary, and insert stuff like y+z and y-z into that result, we will see (after a little algebra) that [itex]\langle Ux,Uy\rangle=\langle x,y\rangle[/itex] for all x,y.

D'oh, for a moment I thought that this thread was about determining the norm of a unitary operator, but it's about proving that U*U=1. In this approach, the definition includes the condition [itex]\|Ux\|=\|x\|[/itex] for all x. This clearly implies that [itex]\langle Ux,Ux\rangle=\langle x,x\rangle[/itex] for all x. Now if we let y,z be arbitrary, and insert stuff like y+z and y-z into that result, we will see (after a little algebra) that [itex]\langle Ux,Uy\rangle=\langle x,y\rangle[/itex] for all x,y.

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A unitary transformation is a mathematical operation that preserves the inner product of vectors. In other words, it maintains the length and angle between vectors, and therefore, the overall structure of a vector space. It is an important concept in quantum mechanics, where it represents a change of basis in a vector space.

Dirac's text, "The Principles of Quantum Mechanics," is a foundational work in quantum mechanics. In this text, Dirac presents his notation, including the use of kets and bras, to represent vectors and dual vectors in a vector space. He also discusses the importance of unitary transformations in quantum mechanics.

In Dirac's notation, ¯U represents the Hermitian conjugate of U, which is a mathematical operation that involves taking the transpose and complex conjugate of a matrix. The equation ¯UU = 1 means that the product of a unitary matrix U and its Hermitian conjugate is equal to the identity matrix, 1. This is an important property of unitary matrices, and it allows for the inverse of a unitary matrix to be easily calculated.

To prove ¯UU = 1 in Dirac's notation, you would need to show that the product of each element of the matrix U and its corresponding element in the Hermitian conjugate matrix ¯U is equal to 1. This can be done by using the properties of the Hermitian conjugate, such as taking the transpose and complex conjugate of matrices, and the properties of unitary matrices, such as the fact that U^-1 = ¯U.

The importance of proving ¯UU = 1 in Dirac's notation lies in its application to quantum mechanics. Unitary matrices play a crucial role in quantum mechanics, as they represent transformations that preserve the structure of a vector space. Proving ¯UU = 1 in Dirac's notation shows that unitary transformations are reversible, which is a fundamental property in quantum mechanics. It also allows for the calculation of the inverse of a unitary matrix, which is necessary in many quantum mechanical calculations.

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