Talk:Hermitian adjoint

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Shouldn't one use inner product notation here rather than bra-ket notation?

I.e. \langle Ax,y\rangle rather than \langle Ax | y\rangle.

PJ.de.Bruin 00:49, 5 Jul 2004 (UTC)

at least one concrete example would be nice! otherwise it seems A*=A^{-1} as in A*A=AA*=I... anyone?

Contents

[edit] Hermiticity & Self-Adjointness: Distinction

The distinction is not made clear. A is self-adjoint if:

\displaystyle\begin{array}{rl} A^* &= A, \text{ and}\\ \text{dom}(A^*) &= \text{dom}(A). \end{array}

Hermiticity does not necessarily guarantee the latter statement.

Dirc 20:52, 14 February 2007 (UTC)

Hermiticity and self-adjoitness are synonymous. (the notation A* = A implicitly implies Dom(A) = Dom(A*), usually.) so doesn't quite make sense to say self-adjointness is not guaranteed by Hermiticity. on the other hand, being symmetric does not imply an operator is self-adjoint/Hermitian in general. Mct mht 23:48, 14 February 2007 (UTC)

[edit] Proof?

Hmm, the article says:

Moreover, \| A^* A \| _{op} = \| A \| _{op}^2

I don't really see how this follows from the properties above it. Can someone provide a simple proof for this?

Never mind, think I found it. How dumb can you be?
A * A is Hermitian, as one can easily show. Therefore
\displaystyle\begin{array}{rl} ||T^* T|| & = \sup_{||x|| = 1} |\langle T^* T x, x \rangle| = \sup_{||x|| = 1} |\langle T x, T x \rangle| \\ & = \sup_{||x|| = 1} ||T x||^2 = \left( \sup_{||x|| = 1} ||T x|| \right)^2 = ||T||^2 \end{array}

[edit] Uses?

This page explains what adjoints are, but just reading this page it isn't clear why it might be useful to define "adjoint". —Ben FrantzDale 01:18, 9 April 2007 (UTC) Sometimes a solution to a problem involving an adjoint converts into a solution to the original problem as for example in the case of some second order differential equations.

I think the intro already gives some useful info:

Adjoints of operators generalize conjugate transposes of square matrices to (possibly) infinite-dimensional situations. If one thinks of operators on a Hilbert space as "generalized complex numbers", then the adjoint of an operator plays the role of the complex conjugate of a complex number.

If this is still too vague, perhaps we could add a line about switching over linear operators from vector spaces to their dual (like with vectors kets to bras), and how about one would like to write down an inner product between, say v' and w, where v = A x for some linear operator A? --CompuChip 14:04, 9 April 2007 (UTC)
How about adding a simple concrete nontrivial example, or two? 140.109.169.94 14:28, 23 October 2007 (UTC)

[edit] Adjoint of a bounded linear operator between normed spaces

Neither I easily found a link from this article to adjoint of general lin. bound. operator nor is it coevered in this article. —Preceding unsigned comment added by 131.111.8.103 (talk) 16:46, 14 February 2008 (UTC)