# Question on the inner product of wavefunctions [closed]

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 $$(\Psi,\Psi)$$?

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

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: $$\left\langle \int_{\mathbb{R}} \psi(x) |x\rangle \, dx \ | \int_{\mathbb{R}} \psi(x') |x' \rangle \, dx' \right\rangle = \int_{\mathbb{R}} \psi(x') \left\langle \int_{\mathbb{R}} \psi(x) | x \rangle \, dx \ | x' \right\rangle dx' =$$ $$\int_{\mathbb{R}} \int_{\mathbb{R}} \psi(x') \psi^{*}(x) \langle x | x\rangle' dx' dx = \langle \Psi | \Psi \rangle = \int_{\mathbb{R}} \int_{\mathbb{R}} \psi^{*}(x') \psi(x) \delta(x-x') \, dx' \, dx = \int_{\mathbb{R}} |\psi(x)|^2 \, dx$$