Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Unitary Transformation

  1. Nov 2, 2011 #1
    In Dirac’s text the equation ¯UUα=α¯UU is well proven . Next it is said that since ¯UU commutes with all linear operators so it must be a number . Further since ¯UU and its complex conjugate are same so ¯UU is a real number . Also Dirac mentions that 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 ? [¯U being the complex conjugate of U ]
     
  2. jcsd
  3. Nov 2, 2011 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    What number is positive and has unit modulus ?
     
  4. Nov 3, 2011 #3
    <P|¯UU |P> is positive and ¯UU =1
     
  5. Nov 3, 2011 #4

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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],
    [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].
     
  6. Nov 3, 2011 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    A thorough argument replacing Dirac's heuristic derivation goes like this:

    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.
     
  7. Nov 3, 2011 #6

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
    Last edited: Nov 3, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Unitary Transformation
  1. Unitary transformation (Replies: 5)

  2. Unitary transforms (Replies: 3)

Loading...