Unitary Transformation

  • #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 ]
 

Answers and Replies

  • #2
dextercioby
Science Advisor
Homework Helper
Insights Author
13,023
576
What number is positive and has unit modulus ?
 
  • #3
<P|¯UU |P> is positive and ¯UU =1
 
  • #4
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
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 ?
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].
 
  • #5
dextercioby
Science Advisor
Homework Helper
Insights Author
13,023
576
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.
 
  • #6
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
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:

Related Threads on Unitary Transformation

  • Last Post
Replies
1
Views
209
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
3
Views
3K
Replies
8
Views
7K
Replies
1
Views
752
Replies
3
Views
878
Replies
9
Views
1K
Top