Finding the Adjoint of an Operator on a Hilbert Space

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SUMMARY

The discussion focuses on finding the adjoint of an operator \( u \) defined on a Hilbert space \( H \), where \( u(x) = \langle b, x \rangle a \). The key equation discussed is \( \langle u x, y \rangle = \langle x, u y \rangle \), which leads to the conclusion that \( u \) is an antilinear operator due to the inner product's linearity in its first argument. The participants clarify that if the inner product is defined as antilinear in the first argument, the definition of the adjoint must be adjusted accordingly.

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Homework Statement


Let H be a Hilbert space, with a in H and b in H.

Let u be an operator on H with u(x)=<b,x>a

Find the adjoint of u.

Thanks!

Homework Equations


<ux,y>=<x,u*y>


The Attempt at a Solution


<ux,y>=<<b,x>a,y> = <b,x><a,y> = <b<a,y>,x>

So it seems that all I need to do is flip that last inner product round. But that will introduce a complex conjugate, stopping me from getting the equation to the right form (maybe?).
 
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Is <<b,x>a,y> = <b,x><a,y> really correct?
 
The inner product is linear in its first argument, so I think so.
 
pyf said:
The inner product is linear in its first argument, so I think so.

Ok, I'm used to the convention where it's linear in the second argument and antilinear in the first. Are you sure?
 
I see. Well, this is the form that we've been given. Any ideas?
 
pyf said:
I see. Well, this is the form that we've been given. Any ideas?

Check your definitions again. I have that for a constant c, <x,cy>=c <x,y> and <cx,y>=c* <x,y>. If you take that sort of an inner product the problem works out nicely. If you take the opposite convention then u(cx)=<b,cx>a=(c*)(<b,x>a)=(c*)u(x). That means u ISN'T a linear operator. It's antilinear. The definition of 'adjoint' for antilinear operators needs to be modified. If you REALLY have a backwards convention for the inner product and a forwards convention for the definition of adjoint, then that's messed up.
 

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