# Ambiguity with Dirac Notation

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.

• What do you mean by "the action of a covector on a vector" ? Usually, this action is exactly the inner product so there is no ambiguity ! Mar 27 '18 at 22:50
• Indeed, Dirac notation assumes by default that all bases are orthonormal and breaks down otherwise, since the factors of the metric are missing. I remember getting all kinds of contradictions before learning this when I started. Mar 27 '18 at 22:57
• @knzhou That sounds like the basis of a good answer Mar 28 '18 at 3:41

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 $\langle \phi | \psi \rangle = (|\phi\rangle, |\psi\rangle)$.