Inner Product Space - Mathematical Method

Note:The development of concepts need to keep extension to infinite analog in mind

The identity operator and the Kroncker’s Delta¶

Let $V$ be a vector space endorsed with an inner product $\langle\cdot| \cdot \rangle$ of dimension $n$ and $\beta = \{e_\nu\}$ be an orthonormal basis. Then we can compute the inner product between arbitrary $x,y\in V$ as follows:

\begin{align*} \langle x | y \rangle &= \langle e_{\nu} | x \rangle^* \langle e_{\nu} | y\rangle \\ &= \langle x | e_{\nu} \rangle \langle e_{\nu} | y\rangle \\ &= \langle x \lvert |e_{\nu} \rangle \langle e_{\nu}| \rvert y\rangle \end{align*}

(1)

But $\langle x | y \rangle = \langle x | I | y \rangle$ , hence we get that

Help! What is happening here?

Step 1 We first rewrite inner product as a sum of component-component product. In $\mathbb{C}^3$ for instance,

\begin{align*} \langle x | y\rangle &= \sum_{\nu} (x^{\nu})^* y^{\nu} \\ &= \sum_{\nu} \langle e_{\nu} | x \rangle^* \langle e_{\nu} | y \rangle \end{align*}

(3)

Note in the last step we envicted orthonormality of $\{e_\nu\}$ . Adopting Einstein notaiton we get the first equality.

Step 2 We are bascially writing two consecutive inner product $\langle x, \alpha \rangle \langle \beta, y \rangle$ as a bilinear form $\langle x, A y\rangle$ where $A=| \alpha \rangle \langle \beta |$ . Consider this example in $\mathbb{R^3}$ :

\begin{align*} \langle x, \alpha \rangle \langle \beta, y \rangle &= \begin{pmatrix} x^1 & x^2 & x^3 \end{pmatrix} \begin{pmatrix} \alpha^1 \\ \alpha^2 \\ \alpha^3 \end{pmatrix} \begin{pmatrix} \beta^1 & \beta^2 & \beta^3 \end{pmatrix} \begin{pmatrix} y^1 \\ y^2 \\ y^3 \end{pmatrix} \\ &= \begin{pmatrix} x^1 & x^2 & x^3 \end{pmatrix} \begin{pmatrix} \alpha^1 \beta^1 & \cdots & \alpha^1\beta^3 \\ \vdots & & \vdots \\ \alpha^3 \beta^1 & \cdots & \alpha^3\beta^3 \end{pmatrix} \begin{pmatrix} y^1 \\ y^2 \\ y^3 \end{pmatrix} \\ \end{align*}

(4)

In otherword, we are just invoking associative law(which holds in vector space):

\begin{align*} (x\alpha^T) (\beta y^T)&=x(\alpha^T \beta) y^T \end{align*}

(5)

where the outer product $\alpha^T \beta=A$ .

We denote bilinear forms $x^TAy$ in sandwich notations: $\langle x |A|y\rangle$

(Step 3) The only possble opeartor to be sandwitched without changing the result:

\begin{align*} \langle x|Ay\rangle &= \langle x | y\rangle \end{align*}

(6)

is when $A=I$ . We therefore conclude that the outer product $|e_{\nu}\rangle \langle e_{\nu}|$ is the identity. In math notation, this is saying:

For orthonormal basis $\{e_{\nu}\}$ , we have
$\sum_{\nu=1}^n e_{\nu}e_{\nu}^T=I$
(7)

Now we consider representing $|x\rangle$ in $\beta$ :

\begin{align*} |x\rangle &= (|e_{\nu} \rangle \langle e_{\nu}|) |x\rangle \\ &= |e_{\nu}\rangle \langle e_{\nu}|x\rangle \end{align*}

(8)

Now consider taking the inner product with another basis $e_{\mu}$ :

\begin{align*} \langle e_{\mu}|x\rangle &=\langle e_{\mu}|e_{\nu}\rangle \langle e_{\nu}|x\rangle \end{align*}

(9)