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?

  • 1
    $\begingroup$ Additionally, the inner product you've given is only applicable for $\mathscr{H}=L^2(\mathbb{R})$ functions, no? $\endgroup$ Jul 9, 2021 at 10:14
  • 1
    $\begingroup$ 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? $\endgroup$ Jul 9, 2021 at 10:26
  • 2
    $\begingroup$ @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. $\endgroup$ Jul 9, 2021 at 10:28
  • 5
    $\begingroup$ 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. $\endgroup$ Jul 9, 2021 at 10:57
  • 1
    $\begingroup$ Related, probably useful: Bra-Ket Notation. $\endgroup$ Jul 9, 2021 at 12:04

1 Answer 1


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*}

This explains how to get the right result, but I'm honestly not sure where you made a mistake.

  • 2
    $\begingroup$ 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. $\endgroup$ Jul 9, 2021 at 11:12
  • 2
    $\begingroup$ @Jakob his mistake is using the wrong wave function! $\endgroup$ Jul 9, 2021 at 11:17
  • $\begingroup$ Yes, these are two ways of seeing the same problem I guess. Anyway, thank you all $\endgroup$
    – ric.san
    Jul 9, 2021 at 11:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.