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? $\endgroup$
    – Jakob
    Jul 9 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 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. $\endgroup$
    – Jakob
    Jul 9 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 at 10:57
  • 1
    $\begingroup$ Related, probably useful: Bra-Ket Notation. $\endgroup$ Jul 9 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*}

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. $\endgroup$
    – Jakob
    Jul 9 at 11:12
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    $\begingroup$ @Jakob his mistake is using the wrong wave function! $\endgroup$ Jul 9 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 at 11:21

The problem with Dirac's braket notation is that it is used without care and physics community is somehow OK with that. For a brief and concise explanation of the Dirac's braket notation please look at my previous answer (points 2, 8, 11, 12, 26, 31 and 32 in particular).

1- $|\alpha\rangle$ is a ket representing a state of a system.

2- $\langle x|\alpha\rangle$ is the wavefunction representing the state $|\alpha\rangle$ in (continuous) position space and can be written as $$\psi_{\alpha}\left ( {x} \right )=\langle {x}|\alpha\rangle$$

3- The scalar product $\langle \beta|\alpha\rangle$ can be written as $$\begin{align*} \langle \beta|\alpha\rangle&=\int\mathrm{d}{x}\langle \beta|{x}\rangle\langle {x}|\alpha\rangle \\&=\int\mathrm{d}{x}\psi_{\beta}^*\left ( {x} \right ) \psi_{\alpha} \left ( {x} \right ) \end{align*}$$ where

$$\int \mathrm{d}{x}|{x}\rangle\langle{x}|=\mathbb{I}$$

  • $\begingroup$ Which careless use of the bra-ket notation, deviating from your presentation, are you claiming the physics community tolerates? What's your basis for that? $\endgroup$
    – J.G.
    Jul 9 at 15:12
  • $\begingroup$ Maybe I should have said scientific community instead, including quantum chemists, too. I cannot point to a particular example right now but I have seen in many places that people call a $|\psi\rangle$ wave function. $\endgroup$ Jul 12 at 6:10

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