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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?

Qmechanic
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ric.san
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  • Note that $|x\rangle$ is not an element of $\mathscr{H}$; in particular, the position states do not form a complete orthonormal basis (e.g. $\langle x|x\rangle$ is not well-defined). – Tobias Fünke Jul 09 '21 at 10:08
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    Additionally, the inner product you've given is only applicable for $\mathscr{H}=L^2(\mathbb{R})$ functions, no? – Tobias Fünke Jul 09 '21 at 10:14
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    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? – Emilio Pisanty Jul 09 '21 at 10:26
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    @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. – Tobias Fünke Jul 09 '21 at 10:28
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    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. – Cosmas Zachos Jul 09 '21 at 10:57
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    Related, probably useful: Bra-Ket Notation. – Emilio Pisanty Jul 09 '21 at 12:04

1 Answers1

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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.

Wihtedeka
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