Justification of Bra-Ket notation in Quantum Mechanics

Given a complex Hilbert 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 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, \(H^* \cong\overline{H}\), 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 the corresponding \(\Phi(v)\in H^*\) we will write \(\langle v|\). This way, for \(v,w\in H\), you know that \(\Phi(v)(w)=\), but in the new notation is, directly

\[ \langle v||w \rangle=\langle v,w \rangle \]

Observe that to \(z\cdot| v \rangle\) we associate \(\langle v |\cdot\overline{z}\). 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

\[ \overline{f}: \overline{H_1} \longmapsto \overline{H_2} \]

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, \(\Phi_1(v)=\)). 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. Physicists have a habit of repressentig \(|v\rangle\) 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 called the Hermitian conjugate or conjugate transpose of \(M\).

And we can write that to \(M|v\rangle\) we associate

\[ \langle v | M^{\dagger}. \]

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