Ambiguity with Dirac Notation

Question

When using Dirac notation, how do we distinguish the difference between the inner product and the action of a covector on a vector?

As I understand it, $\vert x \rangle$ is an element of say a Hilbert space $\mathcal{H}$ and when we write $\langle x \vert$ we denote the corresponding linear map in $\mathcal{H}^*$ which takes a vector in $\mathcal{H}$ to $\mathbb{C}$.

Here's where my confusion lies: $\langle x \vert y \rangle$ is supposed to denote the "inner product" defined by the metric/orthonormality relation $\langle e_i \vert e_j \rangle = \delta _{ij}$, where $e_i$ denotes the $i^{ith}$ basis vector. But how do we distinguish this from the action of a covector $\langle x \vert$ on a vector $\vert y \rangle$?

Moreover, it seems like this notation only works in the case where the basis of $\mathcal{H}^*$ is defined such that $e^i(e_j) = \delta^{i}_j$. If we had chosen an arbitrary metric $\langle e_i \vert e_j \rangle = g_{ij}$ then this notation wouldn't make sense anymore.

Answer 1

The point of Dirac notation is that there is no need to distinguish them. There is a canonical isomorphism between vectors and covectors given by the inner product, so that if we denote the inner product by $(\cdot,\cdot)$, given $|\phi\rangle$ and $|\psi\rangle$ we define $\langle \phi |$ by $\langle \phi | \psi \rangle = (|\phi\rangle, |\psi\rangle)$.

The notation is independent of the basis. What does depend on the basis is the relation between the components: the components of a covector are the complex conjugates of the components of the vector only if the basis is orthonormal; otherwise you have to lower/raise indices with the metric.