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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.

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  • $\begingroup$ 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). $\endgroup$ – Ben Grossmann Jan 23 at 6:38
  • $\begingroup$ 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. $\endgroup$ – Matrix001 Jan 23 at 6:46
  • $\begingroup$ 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". $\endgroup$ – Ben Grossmann Jan 23 at 6:51
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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.

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  • $\begingroup$ @daw actually the QM convention is reversed, but thanks for pointing that out. $\endgroup$ – Ben Grossmann Jan 23 at 14:21
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$⟨u│u⟩=\int_{-∞}^{∞}u(x) \overline{u(x)}dx$, hence $⟨u│u⟩=\int_{-∞}^{∞}\left|u(x)\right|^{2}dx$, since for a complex number $z$ we have $z \overline{z}=|z|^2.$

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  • $\begingroup$ Yes, but were does,$$⟨u│u⟩=\int_{-∞}^{∞}u^{*}u\ dx$$ come from? $\endgroup$ – Matrix001 Jan 23 at 6:46
  • $\begingroup$ $u^*= \overline{u}.$ $\endgroup$ – Fred Jan 23 at 8:41
  • $\begingroup$ Yes, the asterix denotes the complex conjugate of "u". $\endgroup$ – Matrix001 Jan 23 at 10:23

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