My undergraduate classes in Q.M describes the adjoint of a linear operator purely as a mathematical formality. At this point, I'd like a deeper and heuristic understanding of it.

My questions are based on page 24 and 28 of the text An introduction to quantum computing by P. Kaye.

Page 24 outlines the definition of the dual of a Hilbert space $\mathcal{H}^{*}$

Definition: Let $\mathcal{H}$ be a Hilbert space. The Hilbert space $\mathcal{H}^{*}$ is defined as the set of linear maps $\mathcal{H} \rightarrow \mathbb{C}$. We denote elements of $\mathcal{H}^{*}$ by $\langle \mathcal{X} |$, where the action of $\langle \mathcal{X} |$ is $\langle \mathcal{X} | : | \psi \rangle \mapsto \langle \mathcal{X} | \psi \rangle \in \mathbb{C}$.

Indeed, the Hilbert space $\mathcal{H}^{*}$ is just the space of all linear function - each of which with a corresponding row vector form or dual vector - that maps a vector from $\mathcal{H}$ to $\mathbb{C}$ by contracting a bra and a ket, or, if you're a mathematician, a row vector with a column vector.

On page 28, Each map in $\mathcal{H}^{*}$ corresponds to some vector $\langle{\phi'} | $. The adjoint of the operator $T$, denoted $T^{\dagger}$, is defined as the linear map that sends $\langle {\phi} |$ to $\langle {\phi'} |$, where $\langle \phi | (T |\psi \rangle) = \langle \phi ' |\psi \rangle$.

First, by definition, any linear operator on $\mathcal{H}^{*}$ maps dual vectors in $\mathcal{H}^{*}$ to $\mathbb{C}$ so this appears to contradicts the statement made by the author that the adjoint $T^{\dagger}$ maps elements in $\mathcal{H}$ to $\mathcal{H}$.

Second, if the aforementioned is true, how does this validates the claim that $\langle \phi | (T |\psi \rangle) = \langle \phi ' |\psi \rangle$?

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    $\begingroup$ To me, the adjoint story is MUCH more understandable in normal, everyday linear algebra notation instead of bra-ket notation. $\endgroup$ – Randall Apr 21 at 3:16
  • $\begingroup$ Try reading this page en.wikipedia.org/wiki/Transpose_of_a_linear_map $\endgroup$ – IAmNoOne Apr 21 at 3:38
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    $\begingroup$ "By definition, any linear operator on $\mathcal{H}^*$ maps dual vectors in $\mathcal{H}^*$ to $\mathbb C$" Absolutely not. The elements of $\mathcal{H}^*$ are linear maps $\mathcal{H}\rightarrow\mathbb C$. Linear operators on $\mathcal{H}^*$, by definition, are linear maps $\mathcal{H}^*\rightarrow\mathcal{H}^*$. $\endgroup$ – eyeballfrog Apr 21 at 4:38
  • $\begingroup$ @eyeballfrog I may have overlooked this. Would you like to expand on this as an answer? I will accept your contribution and give it an upvote if you do. $\endgroup$ – Mathematicing Apr 21 at 8:53

I'll try to help clear things up.

Let $f$ be a continuous linear map $\mathcal{H}\rightarrow \mathbb C$, and $\mathcal{H}^*$ be the set of all such maps. The Riesz representation theorem shows that for each such $f$, there is a unique element $\left|\phi\right>$ of $\mathcal{H}$ such that $f(\left|\psi\right>) = \left<\phi\right.\left|\psi\right>$ for all $\left|\psi\right> \in \mathcal{H}$. We then make a small abuse of notation and identify $f$ with $\left<\phi\right|$. The Riesz theorem also shows that $\mathcal{H}^*$ is itself a vector space, and that the map $\mathcal{H}\rightarrow \mathcal{H}^*$ defined by $\left|\phi\right> = \left<\phi\right|$ is antilinear (it preserves addition and replaces scalar multiplication by scalar multiplication by the complex conjugate).

Now suppose $T:\mathcal{H}\rightarrow\mathcal{H}$ is a linear operator on $\mathcal{H}$. It can be shown that there is another linear operator $T^\dagger:\mathcal{H}\rightarrow\mathcal{H}$ such that $\left<\phi\right|(T\left|\psi\right>) = \left<T^\dagger\phi\right|(\left|\psi\right>)$ for all $\left|\psi\right> \in \mathcal{H}$ and $\left<\phi\right| \in \mathcal{H}^*$, and that this satisfies $T^{\dagger\dagger} = T$ and $(AB)^\dagger = B^\dagger A^\dagger$ (Exercise: prove this). We then use this to define a corresponding linear operator on $\mathcal{H}^*\rightarrow \mathcal{H}^*$, which we will write in postfix notation: $\left<\phi\right|T^* = \left<T^\dagger\phi\right|$.

So why did we do all this? The point is that, with this notation, for every linear operator $T$ on $\mathcal{H}$ and for all for all $\left|\psi\right> \in \mathcal{H}$ and $\left<\phi\right| \in \mathcal{H}^*$, we have $\left<\phi\right|(T\left|\psi\right>) = (\left<\phi\right|T^*)(\left|\psi\right>)$. This means we can drop the parentheses and the $^*$ and have an unambiguous symbol: $\left<\phi\right|T\left|\psi\right>$. Whether we interpret this as $T$ acting on $\left|\psi\right>$ (making it equal to $\left<\phi|T\psi\right>$) or $T^*$ acting on $\left<\phi\right|$ (making it equal to $\left<T^\dagger\phi|\psi\right>$) doesn't matter--we get the same result regardless. And this even works with compound operators: $$ \left<\phi\right|AB\left|\psi\right> = \left<\phi\right|A\left|B\psi\right> = \left<A^\dagger\phi\right|B\left|\psi\right> = \left<\phi|AB\psi\right> = \left< A^\dagger\phi|B\psi\right> = \left< B^\dagger A^\dagger\phi|\psi\right> = \left< (AB)^\dagger \phi|\psi\right>. $$ This completes the analogy to matrix notation in finite-dimensional vector spaces, where we write the inner product of two vectors $(x,y)$ as $x^Ty$, and for a linear operator $A$, we can unambiguously write $x^TAy$ to represent both $(x,Ay)$ and $(A^Tx,y)$.


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