# On Dirac notation: inner product vs base representation

The Dirac notation $$\langle a | b \rangle$$ seems somewhat ambiguous.

1. On one hand, it can be seen as inner product of elements $$a(x)$$ and $$b(x)$$ of the Hilbert space $$\scr H$$, namely: $$\langle a | b \rangle ={\displaystyle}\int_\mathbb R a^*(x) \ b(x) \ dx.\tag{1}$$

2. On the other hand, it's the evaluation of $$b$$ at its $$a$$th component, with respect to a particular orthonormal base for $$\scr H$$.

• discrete case. $$\langle n | b\rangle = b_n, \tag{2d}$$ where $$\displaystyle \sum_n |n\rangle\langle n | = \mathbb I\$$ and $$\langle n | m \rangle = \delta_{nm}.$$

• continuous case. $$\langle x | b\rangle = b(x), \tag{2c}$$ where $$\displaystyle \int_\mathbb R |x\rangle\langle x | = \mathbb I\$$ and $$\langle x | x' \rangle = \delta(x- x').$$

The obvious conclusion is that you are free to see $$\langle a | b \rangle$$ in both ways, that is, 1. and 2. are equivalent.

But it can't be! For instance:

$$b(x) \stackrel{(2c)}{=} \langle x | b\rangle \stackrel{(1)}{\neq} \displaystyle \int _\mathbb R x b(x) dx.\tag{3}$$

So? How can I choose the right way to see it a priori?

• Additionally, the inner product you've given is only applicable for $\mathscr{H}=L^2(\mathbb{R})$ functions, no? Commented Jul 9, 2021 at 10:14
• Your assertions 1 and 2 are indeed equivalent. You will get more useful answers if you explain in more detail why you think they are incompatible. What do you mean by "$b(x) \neq \displaystyle \int _\mathbb R x b(x) dx$", and why do you think those two should be equal? Commented Jul 9, 2021 at 10:26
• @EmilioPisanty I think OP is saying that $\langle x|b\rangle = \displaystyle \int _\mathbb R\mathrm{d}x\, x\, b(x)$ from the perspective of a scalar product. Commented Jul 9, 2021 at 10:28
• Your (3) is nonsense: Dirac takes special care in his book to remind you the wave function of $|x\rangle$ is a delta function, not x. Commented Jul 9, 2021 at 10:57
• Related, probably useful: Bra-Ket Notation. Commented Jul 9, 2021 at 12:04

Given $$\mathbb{I} = \int |x\rangle\langle x| \,\mathrm{d}x$$ You can rewrite the scalarproduct $$\langle x | b\rangle = b(x)$$ by inserting a 'one' in the middle \begin{align*} \langle x | b \rangle &= \langle x | \mathbb{I} | b\rangle = \int \langle x | x'\rangle \langle x'| b\rangle \,\mathrm{d}x' \\ &= \int \delta(x-x')b(x') \,\mathrm{d}x' \\ &= b(x) \end{align*}
• I think the mistake in OP's equation (3) is simply that it is not an inner product between two $L^2$ functions; so equation (1) is not applicable. Commented Jul 9, 2021 at 11:12