TITLE: Is a bra the adjoint of a ket?
QUESTION [15 upvotes]: The instructor in my quantum computation course sometimes uses the equivalence
$$(\left|a\right>)^\dagger\equiv\left<a\right|$$
I understand that this is true for the typical matrix implementation of Dirac's bra-ket notation, and taking $x^\dagger\equiv\bar{x}^T$; but does it follow from the definition of an inner product space and the adjoint? Should I be able to derive this as a fact from definition of an inner product space and the adjoint of linear operators on that space or is it simply a notational convention?

REPLY [3 votes]: Given a complex Hilber space $H$ we can define the complex conjugate of $H$, $\overline{H}$, which is all the same than $H$ except that the scalar multiplication by $z\in \mathbb{C}$ is changed by the conjugate $\overline{z}$. Obviously, there is a conjugate linear isomorphism between $H$ and $\overline{H}$ 
On the other hand, we can define the continuous dual $H^*$ of $H$ like the vector space of all continuous and linear maps from $H$ to $\mathbb{C}$. 
The inner product gives rise to a morphism from $H$ to his dual $H^*$ that is conjugate-linear:
$$
\begin{array}{rcc}
\Phi: H & \longmapsto & H^*\\
v & \longmapsto & \langle v,\cdot \rangle
\end{array}
$$
The Riesz representation theorem tell us that $\Phi$ is an isomorphism (this is only evident in the finite dimensional case). Moreover, it is an isometry respect to the norm.
In conclusion, $H^* \cong\overline{H}$, since the composition of two conjugate-linear maps is a genuine linear map. This conclusion is what justifies the bra-ket notation: For an element $v\in H$ we will write $|v \rangle$ and for $\Phi(v)\in H^*$ we will write $\langle v|$. This way, for $v,w\in H$, you know that $\Phi(v)(w)=<v,w>$, but in the new notation is, directly
$$
\langle v|w \rangle=\langle v,w \rangle
$$
Observe that to $z\cdot|  v \rangle$ we associate $\langle v |\cdot\overline{z}$. The scalar is at the right because is an habit of physicist. 
A linear map between two complex Hilbert spaces $f: H_1\mapsto H_2$ gives rise to other linear map
$$
\overline{f}: \overline{H_1} \longmapsto \overline{H_2}
$$
Let us identify $\overline{H_i}$ with $H_i^*$ and call $\Phi_i$ to the conjugate linear isomorphism from $H_i$ to $H_i^*$ (for example, $\Phi_1(v)=<v,\cdot >$). We will take $\overline{f}=\Phi_2\circ f \circ \Phi_1^{-1}$.
We are going to study this in the bra-ket notation. Suppose $|v\rangle \in H_1$ such that $f(|v\rangle)=|w \rangle$, what is $\overline{f}(\langle v |)$? Obvisouly, is $\langle w |$. But let $M$ be the matrix of $f$ respect to the basis $\{|e_i\rangle\} \subset H_1$, $\{|g_j\rangle\} \subset H_2$. What is the marix of $\overline{f}$ respect to $\{\langle e_i|\} \subset H_1^*$, $\{\langle g_j|\} \subset H_2$?
If we take, say for example, $\langle e_1|$ and follow the compositions that produce $\overline{f}$ we obtain
$$
\overline{f}(\langle e_1|)=\langle \sum_j m_{j1}g_j|= \sum_j \langle g_j|\cdot \overline{m}_{j1}
$$
But wait a moment. Physicist has the habit that represent $|v\rangle$ like a column vector, and $\langle v|$ like a row vector; and, in the latter case, apply the matrix of linear maps multiplying by the right. Putting all together we obtain that the matrix of $\overline{f}$ is
$$
M^{\dagger}:=\overline{M}^T
$$
 which is calle the Hermitian conjugate of $M$.
And we can write that to $M|v\rangle$ we associate
$$
\langle v | M^{\dagger}
$$