representation theorem for their duals (see below). Thus, the Dirac bra-ket notation can be adapted to quaternionic Hilbert spaces: $$\mid f{\mathbf{q}}\rangle=\mid f\rangle{\mathbf{q}},\hspace{1cm}\langle f{\mathbf{q}}\mid=\overline{{\mathbf{q}}}\langle f\mid\;,$$ for a right quaternionic Hilbert space, with $\vert f\rangle$ denoting the vector $f$ and $\langle f\vert$ its dual vector.

### Left Quaternionic Hilbert Space

Let $V_{H}^{L}$ be a linear vector space under left multiplication by quaternionic scalars. For $f,g,h\in V_{H}^{L}$ and ${\mathbf{q}}\in H$, the inner product $$\langle\cdot\mid\cdot\rangle:V_{H}^{L}\times V_{H}^{L}\longrightarrow H$$ satisfies the following properties

1.  $\overline{\langle f\mid g\rangle}=\langle g\mid f\rangle$

2.  $\|f\|^{2}=\langle f\mid f\rangle>0$ unless $f=0$, a real norm

3.  $\langle f\mid g+h\rangle=\langle f\mid g\rangle+\langle f\mid h\rangle$

4.  $\langle {\mathbf{q}}f\mid g\rangle={\mathbf{q}}\langle f\mid g\rangle$

5.  $\langle f\mid {\mathbf{q}}g\rangle=\langle f\mid g\rangle\overline{{\mathbf{q}}}$

Again, we shall assume that the space $V_{H}^{L}$ together with $\langle\cdot\mid\cdot\rangle$ is a separable Hilbert space. Also, $$\label{leftcs}
\mid {\mathbf{q}}f\rangle=\mid f\rangle\overline{{\mathbf{q}}},\hspace{1cm}\langle {\mathbf{q}}f\mid={\mathbf{q}}\langle f\mid.$$ Note that, because of our convention for inner products, for a left quaternionic Hilbert space, the bra vector $\langle f\mid$ is to be identified with the vector itself, while the ket vector $\mid f \rangle$ is to be identified with its dual. (There is a natural left multiplication by quaternionic scalars on the dual of a right quaternionic Hilbert space and a similar right multiplication 