Scalar product - Aetherwisp

The scalar product, also called dot product, is an operation between two vectors that produces a Scalar, hence the name. If the vectors are complex, it is also called the Hermitian product. It is denoted in a variety of ways: $(v,w)$ , $v\cdot w$ or $\braket{ v | w }$ (in braket notation).

In finite-dimensional vector spaces, such as $\mathbb{R}^{N}$ or $\mathbb{C}^{N}$ , the scalar product is defined as a sum of products of the vector components. Let $v,w$ be two vectors in the space, then the scalar product between the two is

(v,w)\equiv\sum\limits_{i=0}^{N}v_{i}^{*}w_{i}

where $v_{i}$ and $w_{i}$ are the $i$ -th component of the vectors and the $*$ symbol denotes the complex conjugate. If $v_{i}$ is real, it can safely be removed, as $v_{i}^{*}=v_{i}$ for any real. It can also be expressed geometrically as

(v,w)\equiv\lVert v \rVert \lVert w \rVert \cos\theta

where $\theta$ is the angle between the two vectors and $\lVert \cdot \rVert$ refers to the norm of a vector.

The scalar product of an element with itself satisfies all of the properties of a norm and is called the norm induced by the scalar product, and denoted as usual: $(v,v)=\lVert v \rVert$ .

The scalar product can also be defined on infinite-dimensional spaces, such as Hilbert spaces or L^p spaces, by converting sums to integrals. Then taking two functions $f(x)$ and $g(x)$ in such a space, the scalar product between them is

(f,g)\equiv\int_{-\infty}^{+\infty}f^{*}(x)g(x)\ dx,\qquad( f, f)=\int_{-\infty}^{+\infty}|f(x)|^{2}\ dx

Properties#

The scalar product satisfies the following properties:

It is linear in the right-hand member[^1]. Given $\alpha,\beta \in \mathbb{C}$ , $( v, \alpha w + \beta z)=\alpha ( v, w)+\beta ( v, z)$ .
It is antilinear in the left-hand member, $( \alpha v + \beta w, z)=\alpha^{\ast} ( v, w)+\beta^{\ast} ( v, z)$ . Over reals, it is linear in both members (also said to be bilinear).
It is distributive, $( v, w+u)=( v, w)+( v, z)$ .
Commutation leads to the conjugate: $( v, w)=\overline{( w, v)}$ . Over reals, it is commutative.
The product of an element with itself is always nonnegative, $( v, v)\geq0$ .
The product of an element with itself is zero only if and only if $v$ is zero, $( v, v)=0\Leftrightarrow v=0$ .
It satisfies the Cauchy-Schwarz inequality, $(v,w)\leq \lVert v \rVert\lVert w \rVert$ .
It is a continuous operation, i.e. given a sequence $\{x_{n}\}$ converging to $x\in \mathcal{H}$ where $\mathcal{H}$ is a Hilbert space, it holds that for every $y\in \mathcal{H}$ we have $\lim\limits_{n \rightarrow \infty}(y,x_{n})=(y,x)$ .

In basis expression#

Consider the vectors $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ in some vector space $V$ equipped with a Basis $\{ \mathbf{e}_{1},\ldots,\mathbf{e}_{n} \}$ . They can be expressed as a Linear combination of basis vectors as $\mathbf{v}_{1}=\sum_{i=1}^{n}\alpha_{i}\mathbf{e}_{i}$ and $\mathbf{v}_{2}=\sum_{j=1}^{n}\beta_{j}\mathbf{e}_{j}$ , using suitable values $\alpha_{i}$ and $\beta_{i}$ . The scalar product between these two is

\begin{align} \mathbf{v}_{1}\cdot \mathbf{v}_{2}&=\left( \sum_{i=1}^{n} \alpha_{i}\mathbf{e}_{i} \right)\cdot\left( \sum_{j=1}^{n} \beta_{j}\mathbf{e}_{j} \right)=\sum_{i=1}^{n} \alpha_{i}\mathbf{e}_{i}\cdot\left( \sum_{j=1}^{n} \beta_{j}\mathbf{e}_{j} \right)= \\ &=\sum_{i=1}^{n} \alpha_{i}\sum_{j=1}^{n} \beta_{j}(\mathbf{e}_{i}\cdot \mathbf{e}_{j})=\sum_{i=1}^{n}\sum_{j=1}^{n} \alpha_{i}\beta_{j}\delta_{ij}= \\ &=\sum_{i=1}^{n} \alpha_{i}\underbrace{ (\beta_{1}\delta_{i1}+\ldots+\beta_{n}\delta_{in}) }_{ \beta_{i}\delta_{ii} }=\sum_{i=1}^{n} \alpha_{i}\beta_{i} \end{align}

using the Kronecker delta.

Su matrici#

Il prodotto scalare può anche essere applicato su matrici. Prendiamo $\hat{A},\hat{B}\in M_{2}(\mathbb{C})$ matrici $2\times2$ . Considero la traccia della matrice $Tr(\hat{A})=\sum\limits_{i=1}^{n}A_{ii}$ . E' importante notare che la traccia è unica e non cambia in base alla rappresentazione della matrice (ossia la base rispetto al quale è espressa):

Tr\hat{A}=\sum\limits_{i=1}^{n}\langle\psi_{i}|\hat{A}\psi_{i}\rangle\;\forall\;\text{base o.n.}\{|\psi_{i}\rangle\}_{i=1}^{n}\in\mathbb{C}^{n}

La matrice $\hat{U}=[\langle \psi_{k}|\phi_{i}\rangle]$ è unitaria, ossia $\hat{U}\hat{U}^{\dagger}=\hat{\mathbb{1}}=\hat{U}^{\dagger}\hat{U}$ .

Prendendo $|\psi\rangle,|\phi\rangle\in\mathbb{C}^{N}$ . Il prodotto scalare possiamo scriverlo come

\langle \psi|\phi\rangle=\sum\limits_{i=1}^{n}\psi^{\ast}_{i}\phi_{i}=\sum\limits_{i=1}^{n}\langle \psi|\chi_{i}\rangle\langle \chi_{i}|\phi\rangle=\ldots$$usando una base ortonormale $\{|\chi_{i}\rangle\}^{n}_{i=1}\in\mathbb{C}^{N}$ da cui, usando la completezza dei [[Proiettore|proiettori]]

\ldots=\langle \psi|\underbrace{\left( \sum\limits_{i=i}^{n}\hat{P}{\chi{i}}\right)}\limits_{\hat{\mathbb{1}}}|\phi\rangle

Possiamo generalizzare a $|\psi\rangle,|\phi\rangle\in L^{2}(\mathbb{R},dx)$, ricordando che $\psi^{\ast}_{i}=\overline{\langle \chi_{i}|\psi\rangle}$,

\langle \psi|\phi\rangle=\int_{\mathbb{R}}\psi^{\ast}(x)\phi(x)dx=\int_{\mathbb{R}}\langle \psi|x\rangle \langle x|\phi\rangle=\ldots

ma possiamo descrivere $\phi(x)$ come

\phi(x)=\int \delta(\bar{x}-x)\phi(\bar{x})d\bar{x}

Si ha inoltre

\int_{\mathbb{R}}|x\rangle\langle x|dx=\hat{\mathbb{1}}

dove $|x\rangle\langle x|$ è uno *pseudoproiettore* di una base di pseudovettori. Allora tornando indietro

\ldots=\langle \psi|\left( \int_{\mathbb{R}}|x\rangle\langle x| dx\right)|\phi\rangle

[^1]: Being on the right-hand member specifically is actually a convention. Using the right-hand side is typical in physics. In mathematics, the left-hand side is more commonly used. The same applies to antilinearity, with mathematicians usually place on the right-hand side instead.