
#1
Dec711, 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=(Ai)(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=(Ai)(A+i)^1 so (U)(x) = (Ai)((A+i)^1)x (U acting on an x) Then (Ux,y)= {INTEGRAL}(Ai)((A+i)^1)x y(conjugate) dx (1) = {INTEGRAL}x(Ai)((A+i)^1)(both conjugate)y(all three conjugate) dx (2) =(x,U*y) and so deduce (U*)(y) = (A+i)((Ai)^1)y and so the adjoint of U is U*=(A+i)(Ai)^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 (Ai)(A+i)^1 which is conjugated and then also the whole of (AI)((A+i)^1)y which is also conjugated. Sorry if my notation is confusing, if unsure just ask. Thanks for your help in advance! 



#2
Dec711, 04:38 PM

HW Helper
P: 6,189

Hi Ad123q!
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]? 



#3
Dec711, 06:06 PM

P: 120

[(Ai)(A+i)^{1}]* = [(A+i)^{1}]*(Ai)* = [(A+i)*]^{1}(Ai)*
=(A*  i)^{1}(A*+i) = (A  i)^{1}(A+i) Now for instance multiply this with the original operator (A  i)^{1}(A+i)(Ai)(A+i)^{1} Note that A+i and Ai commute hence you get the result. Similiarly for the other way around 


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