When taking the inner product of a wavefunction $\Psi$ with itself, denoting the inner product as $(\Psi,\Psi)$, since $$\Psi(x)=\int \psi(x)\vec{x}dx$$ letting $$\overline\Psi(x')=\int \overline\psi(x')\vec{x'} dx'$$ would $$(\Psi,\Psi)= \int \overline\psi(x')\psi(x)\delta(x'-x)dx'dx$$ or would the $dx'$ not be included? If I am wholly wrong, how would I go about representing $(\Psi,\Psi)$?

  • 3
    $\begingroup$ Could you please explain your integral representation of the time-dependent state? I have never seen such an expression. $\endgroup$
    – NDewolf
    Oct 5, 2020 at 17:41
  • 2
    $\begingroup$ I don't think any of these expressions are correct. The inner product between two wave functions is not a double integral. $\endgroup$
    – Charlie
    Oct 5, 2020 at 17:42
  • $\begingroup$ Sorry Just realized that is wrong $\endgroup$ Oct 5, 2020 at 17:42
  • $\begingroup$ @Charlie can you explain where I am wrong and how? $\endgroup$ Oct 5, 2020 at 17:43
  • 2
    $\begingroup$ @NDewolf He probably meant something like: $$| \psi \rangle = \int_{\mathbb{R}} \langle x | \psi \rangle |x\rangle \, dx$$ but in different notation. $\endgroup$ Oct 5, 2020 at 17:43

1 Answer 1


I don't know how exactly you went from $(1)$ and $(2)$ to $(3)$. But, here is a nice trick. In our Hilbert Space $\mathbb{V}$, the eigenbasis $|x\rangle$ satisfies: $$\hat{I} = \int_{\mathbb{R}} |x\rangle \langle x| \, dx$$ where $\hat{I}$ is the identity operator. Thus, $$\langle \Psi | \Psi \rangle = \langle \Psi | \hat{I} | \Psi\rangle = \langle \Psi| \int_{\mathbb{R}} | x \rangle \langle x | \, dx \ |\Psi\rangle = \int_{\mathbb{R}} \langle \Psi | x \rangle \langle x|\Psi\rangle \, dx = \int_{\mathbb{R}} \psi^{*}(x) \psi(x) \, dx \Longrightarrow$$ $$\langle \Psi | \Psi \rangle = \int_{\mathbb{R}} |\psi(x)|^2 \, dx$$ Note that this is the same as: $$\langle \Psi | \Psi \rangle = \int_{\mathbb{R}} \int_{\mathbb{R}} \psi^{*}(x') \psi(x) \delta(x-x') \, dx' \, dx $$ but it is completely unnecessary to do that. You could have also arrived at the latter result through the following route: $$\left\langle \int_{\mathbb{R}} \psi(x) |x\rangle \, dx \ | \int_{\mathbb{R}} \psi(x') |x' \rangle \, dx' \right\rangle$$ By using the $-$ linearity in the second argument and anti-linearity in the first $-$ properties of the inner product, we get: $$\left\langle \int_{\mathbb{R}} \psi(x) |x\rangle \, dx \ | \int_{\mathbb{R}} \psi(x') |x' \rangle \, dx' \right\rangle = \int_{\mathbb{R}} \psi(x') \left\langle \int_{\mathbb{R}} \psi(x) | x \rangle \, dx \ | x' \right\rangle dx' = $$ $$\int_{\mathbb{R}} \int_{\mathbb{R}} \psi(x') \psi^{*}(x) \langle x | x\rangle' dx' dx = \langle \Psi | \Psi \rangle = \int_{\mathbb{R}} \int_{\mathbb{R}} \psi^{*}(x') \psi(x) \delta(x-x') \, dx' \, dx = \int_{\mathbb{R}} |\psi(x)|^2 \, dx$$


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