1
$\begingroup$

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?

$\endgroup$
6
  • 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
3
$\begingroup$

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.

$\endgroup$
3
  • 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
  • 2
    $\begingroup$ @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
0
$\begingroup$

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

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

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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