Register to reply

Proving Cayley Transform operator is unitary

by Ad123q
Tags: cayley, operator, proving, transform, unitary
Share this thread:
Ad123q
#1
Dec7-11, 03:07 PM
P: 19
Hi,

Was wondering if anyone could give me a hand.

I need to prove that the Cayley Transform operator given by U=(A-i)(A+i)^-1 is UNITARY, ie that UU*=U*U=I where U* is the adjoint of U (I am given also that A=A* in the set of bounded operators over a Hilbert space H).

My solution so far, is this correct?

U=(A-i)(A+i)^-1 so

(U)(x) = (A-i)((A+i)^-1)x (U acting on an x)

Then (Ux,y)= {INTEGRAL}(A-i)((A+i)^-1)x y(conjugate) dx (1)

= {INTEGRAL}x(A-i)((A+i)^-1)(both conjugate)y(all three conjugate) dx (2)

=(x,U*y)

and so deduce (U*)(y) = (A+i)((A-i)^-1)y

and so the adjoint of U is U*=(A+i)(A-i)^-1

It can then be checked that UU*=U*U=I

As you can see my main query is the mechanism of finding the adjoint of U for the given U.

For clarity in step (1) it is just the y which is conjugated, and in step (2) it is (A-i)(A+i)^-1 which is conjugated and then also the whole of (A-I)((A+i)^-1)y which is also conjugated. Sorry if my notation is confusing, if unsure just ask.

Thanks for your help in advance!
Phys.Org News Partner Science news on Phys.org
'Smart material' chin strap harvests energy from chewing
King Richard III died painfully on battlefield
Capturing ancient Maya sites from both a rat's and a 'bat's eye view'
I like Serena
#2
Dec7-11, 04:38 PM
HW Helper
I like Serena's Avatar
P: 6,189
Hi Ad123q!


Quote Quote by Ad123q View Post
Hi,

Was wondering if anyone could give me a hand.

I need to prove that the Cayley Transform operator given by U=(A-i)(A+i)^-1 is UNITARY, ie that UU*=U*U=I where U* is the adjoint of U (I am given also that A=A* in the set of bounded operators over a Hilbert space H).

My solution so far, is this correct?

U=(A-i)(A+i)^-1 so

(U)(x) = (A-i)((A+i)^-1)x (U acting on an x)

Then (Ux,y)= {INTEGRAL}(A-i)((A+i)^-1)x y(conjugate) dx (1)

= {INTEGRAL}x(A-i)((A+i)^-1)(both conjugate)y(all three conjugate) dx (2)

=(x,U*y)

and so deduce (U*)(y) = (A+i)((A-i)^-1)y

and so the adjoint of U is U*=(A+i)(A-i)^-1

It can then be checked that UU*=U*U=I
How do you conclude this from your expression for U*?

Btw, instead of using the integral, can't you simply use the properties of the adjoint operator?
That is, [itex](AB)^*=B^*A^*[/itex] and [itex](A^{-1})^*=(A^*)^{-1}[/itex]?


Quote Quote by Ad123q View Post
As you can see my main query is the mechanism of finding the adjoint of U for the given U.

For clarity in step (1) it is just the y which is conjugated, and in step (2) it is (A-i)(A+i)^-1 which is conjugated and then also the whole of (A-I)((A+i)^-1)y which is also conjugated. Sorry if my notation is confusing, if unsure just ask.

Thanks for your help in advance!
Sina
#3
Dec7-11, 06:06 PM
P: 120
[(A-i)(A+i)-1]* = [(A+i)-1]*(A-i)* = [(A+i)*]-1(A-i)*
=(A* - i)-1(A*+i) = (A - i)-1(A+i)

Now for instance multiply this with the original operator

(A - i)-1(A+i)(A-i)(A+i)-1

Note that A+i and A-i commute hence you get the result. Similiarly for the other way around


Register to reply

Related Discussions
Is the unitary operator unique? Quantum Physics 2
Cayley-Hamilton theorem for Operator Differential Equations 1
Unitary operator Calculus & Beyond Homework 1
Is it true that unitary transform in QM corresponds to canonical transform Quantum Physics 1
Do you know what this particular unitary operator is? Quantum Physics 7