Bra-ket notation

Explanation: It is a notation used in Quantum Mechanics... bla bla bla...

Justification:
Given a complex Hilbert space $H$ we can define the complex conjugate of $H$ , $\overset{―}{H}$ , which is all the same than $H$ except that the scalar multiplication by $z \in C$ is changed by the conjugate $\overset{―}{z}$ . Obviously, there is a conjugate linear isomorphism between $H$ and $\overset{―}{H}$ .
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 $C$ .

The inner product gives rise to a morphism from $H$ to his dual $H^{*}$ that is conjugate-linear:

\begin{array}{rcc} Φ : H & ⟼ & H^{*} \\ v & ⟼ & ⟨ v, \cdot ⟩ \end{array}

The Riesz representation theorem tell us that $Φ$ is indeed an isomorphism (this is only evident in the finite dimensional case! See dual vector space). Moreover, it is an isometry respect to the norm.
In conclusion, $H^{*} ≅ \overset{―}{H}$ , 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 $| v ⟩$ and for the corresponding $Φ (v) \in H^{*}$ we will write $⟨ v |$ . This way, for $v, w \in H$ , you know that $Φ (v) (w) =< v, w >$ , but in the new notation is, directly

⟨ v | | w ⟩ = ⟨ v, w ⟩

Observe that to $z \cdot | v ⟩$ we associate $⟨ v | \cdot \overset{―}{z}$ . I put the scalar here at the right side 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

\overset{―}{f} : \overset{―}{H_{1}} ⟼ \overset{―}{H_{2}}

Let us identify $\overset{―}{H_{i}}$ with $H_{i}^{*}$ and call $Φ_{i}$ to the conjugate linear isomorphism from $H_{i}$ to $H_{i}^{*}$ (for example, $Φ_{1} (v) =< v, \cdot >$ ). We will take $\overset{―}{f} = Φ_{2} \circ f \circ Φ_{1}^{- 1}$ .

We are going to study this in the bra-ket notation. Suppose $| v ⟩ \in H_{1}$ such that $f (| v ⟩) = | w ⟩$ , what is $\overset{―}{f} (⟨ v |)$ ? Obvisouly, is $⟨ w |$ . But let $M$ be the matrix of $f$ respect to the basis ${| e_{i} ⟩} \subset H_{1}$ , ${| g_{j} ⟩} \subset H_{2}$ . What is the marix of $\overset{―}{f}$ respect to ${⟨ e_{i} |} \subset H_{1}^{*}$ , ${⟨ g_{j} |} \subset H_{2}$ ?

If we take, say for example, $⟨ e_{1} |$ and follow the compositions that produce $\overset{―}{f}$ we obtain

\overset{―}{f} (⟨ e_{1} |) = ⟨ \sum_{j} m_{j 1} g_{j} | = \sum_{j} ⟨ g_{j} | \cdot {\overset{―}{m}}_{j 1}

But wait a moment. Physicists have a habit of repressentig $| v ⟩$ like a column vector, and $⟨ v |$ 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 $\overset{―}{f}$ is

M^{†} := {\overset{―}{M}}^{t}

which is called the Hermitian conjugate or conjugate transpose of $M$ .
And we can write that to $M | v ⟩$ we associate

⟨ v | M^{†} .