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If the position state $|x\rangle$ is complete and orthonormal I understand that $$\langle x | y \rangle = \delta(x-y).\tag1$$

However, why exactly $$\langle x' | x \rangle = \delta(x'-x)\tag2$$

How do we know that $x'$ is orthogonal to $x $?

here, they use eq. (2), but we don't know anything about $x'$.

And the explications here and here seems to always assume the statement above. Is it only something with the notation that I don't understand or is it something else?

Qmechanic
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InvisibleParticle
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3 Answers3

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First,

$$ \langle x|p\rangle = e^{ipx} $$

as it is a plain wave. Now, with the identity

$$ \int\frac{dp}{2\pi}|p\rangle\langle p| = 1 $$ you can get $$ \langle x|x'\rangle = \int \frac{dp}{2\pi}\langle x|p\rangle\langle p|x'\rangle = \int \frac{dp}{2\pi}e^{ip(x-x')} = \delta(x-x') $$

Vicky
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Well in the momentum basis ($\hbar=1$):

$$ \langle x'|x \rangle \propto \int{e^{-ipx'}e^{ipx} dp} = \int e^{ip(x-x')} dp\propto \delta(x-x') $$

JEB
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$|x>$ is a state with definite position. It describe a state of a particle whose position is definitely at point x. Now in QM, the notion of inner products gives you how two states is connected with each other. Saying that the particle is definitely at point $x'$ is of course completely different in saying that the particle is at point x. That is why, that the inner product of $<x|x'> = 0 $ in the case that $x' \neq x$ and $<x|x'> = 1 $ if $x = x'$, under the assumption that the kets here are normalized states. In general, two states that are orthogonal means that they are completely different with each other.

King Meruem
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    $\langle x|x' \rangle$ is not 1 when $x = x'$ but a Dirac delta – Vicky Feb 09 '23 at 16:23
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    Omg i forgot that it is Dirac delta, i thought that is Kronecker delta. It is because for continues cases Dirac delta is use and for discreet cases Kronecker delta is used. But the point in doing that is, note that inner product <x|x'> can be written in an integral form from negative to positive infinity. Note that Dirac delta has a property that integral of Dirac delta from negative to positive infinity is equal to 1. Hence, by defining <x|x'> to be like this, then one can ensure the orthogonality condition between position states. – King Meruem Feb 09 '23 at 23:07
  • You're mixing spaces. The inner product of two position states can never be 1 but a Dirac delta. That's done to ensure $|\psi(t)\rangle = \int dx |x\rangle\langle x|\psi(t)\rangle = \int dx |x\rangle \psi(t, x)$. And that is how you relate wavefunction $\psi(t, x)$ to state $|\psi(t) \rangle$. This way, if $|\psi(t)\rangle \equiv |x'\rangle$, for the previous formula that relates wf to state to be coherent, one must impose $\langle x|x' \rangle = \delta(x-x')$ (not 1). – Vicky Feb 11 '23 at 11:33