# Integral Representation of Hermitian Conjugation

My Quantum Physics textbook asserts that, $$⟨u│u⟩=\int_{-∞}^{∞}\left|u\right|^{2}dx$$ Where the term, $$⟨u│u⟩$$ denotes the product of u and its Hermitian conjugate.

What I am confused about is where did this "representation" come from? And how can it be derived mathematically? I don't see how this integral "representation" makes any mathematical sense.

• Can you please specify what "data types" we're talking about here? Presumably, $u$ is meant to be a function over $\Bbb R$. However, in order for "the product of u and its Hermitian conjugate" to make sense, $u$ must presumably be a vector (or perhaps a sequence of coefficients). – Ben Grossmann Jan 23 at 6:38
• Yes, "u" is a vector. The Textbook also states that "u", is a vector with an infinite amount of entries corresponding to a value out-putted by the "u" within the integral, given a real input. – Matrix001 Jan 23 at 6:46
• If you tell me that "$u$ is a vector with an infinite amount of entries, each "entry" corresponds to a value out-putted by the $u$... given a real input", then the only interpretation I can reasonably make is that $u$ is a function $u$ is a function defined over $\Bbb R$. If that is the case, then there is no way to take "the product of $u$ and its Hermtian conjugate". – Ben Grossmann Jan 23 at 6:51

In quantum mechanics, the expression $$\langle u | v \rangle$$ is used to denote the inner-product of two vectors $$u$$ and $$v$$ (or, as a physicist might insist, $$|u\rangle$$ and $$|v\rangle$$). What exactly this inner product is depends on the context.
When $$u,v$$ are vectors $$u = (u_1,\dots,u_n)$$ and $$v = (v_1,\dots v_n)$$ over $$\Bbb C$$, their inner-product is defined by $$\langle u|v\rangle = \sum_{k=1}^n u_k^* v_k.$$ When $$u,v$$ are functions $$u,v: \Bbb R \to \Bbb C$$, their inner-product is defined by $$\langle u|v \rangle = \int_{-\infty}^\infty u^*(x) v(x)\,dx.$$ So, what exactly $$\langle u|v \rangle$$ means depends on how it is defined for the given context. In all cases though, this function has the properties that define an inner-product.
$$⟨u│u⟩=\int_{-∞}^{∞}u(x) \overline{u(x)}dx$$, hence $$⟨u│u⟩=\int_{-∞}^{∞}\left|u(x)\right|^{2}dx$$, since for a complex number $$z$$ we have $$z \overline{z}=|z|^2.$$
• Yes, but were does,$$⟨u│u⟩=\int_{-∞}^{∞}u^{*}u\ dx$$ come from? – Matrix001 Jan 23 at 6:46
• $u^*= \overline{u}.$ – Fred Jan 23 at 8:41