Phy622 Mathematical Methods 3

01-13

Hermitian Inner Product

$\text{IP} : \mathbb{V}\times\mathbb{V}\to \mathbb F$

$\text{IP}(|u\rangle, |v\rangle) = \langle u|v \rangle$

st.

$\langle v|u+w\rangle = \langle v|u \rangle + \langle v|w \rangle$ $\langle v|\alpha u\rangle = \alpha \langle v|u \rangle$ $\langle v|u\rangle = \langle u|v \rangle^*$ $\langle v|v\rangle \ge 0 \text{ and } 0 \iff |v\rangle = 0$

Hence, a natural norm is defined.

$||v|| = \sqrt{\langle v|v \rangle}$

Prove that this is a norm.

Triangle inequality:

$\sqrt{\langle u+v\rangle}\le \sqrt{\langle u\rangle} + \sqrt{\langle v\rangle}$

Cauchy Schwartz inequality says that $\langle u |v\rangle^2 \le \langle u\rangle + \langle v\rangle$

Assume that Cauchy Schwartz inequality is true,

Hence, $||u+v||^2 = + + <u\|v> + <v\|u>$ $(\|\|u\|\| + \|\|v\|\|)^2 = + + 2\sqrt{<u\|u><v\|v>}$

So,

$(||u|| + ||v||)^2 - ||u+v||^2 = 2\sqrt{} - 2 Re <u\|v> \ge 2 <u\|v> (1 - \cos(\theta)) \ge 0$ $\implies \triangle \text{ inequality}$

Although the IP defines a norm, a norm may not give an IP.

Proof:

Obvoiusly, on an IP space, $||u+v||^2 + ||u-v||^2 = 2(||u||^2+||v||^2)$

Counter Example:

Norm: Max norm over $\mathbb V = \mathbb C^n$

Let $|u\rangle = (1, 0, 0…)^T$ Let $|v\rangle = (-1, 3, 0…)^T$

$||u||=1, ||v||=3, ||u+v|| = 3, ||u-v|| = 3$ But above equality does not hold. Hence, not all norms give an IP.

Properties of IP Spaces

$\forall |x\rangle \in V, \langle x|0\rangle = 0$. From the fact that $\langle u| v+(-v)\rangle = \langle u|v\rangle - \langle u|v\rangle = 0$
$\forall |u\rangle, \langle u|v\rangle = 0 \iff |v\rangle = |0\rangle$. If from thm 1. Only if by contradiction. Let $\exist |v\rangle\in \mathbb{V}$. Then $\langle e|v\rangle =0, \forall|e\rangle$. Hence, it must lie outside the space.

Proof of the Cauchy Schwartz Inequality

Thm: $|\langle u| v\rangle|^2\le \langle u\rangle\langle v\rangle$

Proof:

If $\langle u| v\rangle = 0\implies ||u+v||^2 = ||u-v||^2$

If either is $=0$, it is trivial.

If neither is null vector, Let $|w\rangle = |v\rangle - \frac{\langle u|v\rangle}{\langle u|u\rangle} |u\rangle$

Clearly, $\langle u|w\rangle = 0$ Also, $|v\rangle = |w\rangle + \frac{\langle u|v\rangle}{\langle u|u\rangle} |u\rangle$ $\langle v\rangle = \langle w\rangle + \frac{\langle u|v\rangle\langle v|u\rangle}{\langle u\rangle}$

Hence,

$\langle v\rangle \ge \frac{|\langle u|v\rangle|^2}{\langle u\rangle}$

QED

Equality $\implies \langle u|w\rangle =0$ $\implies |v\rangle = \frac{\langle u|v\rangle}{\langle u/u\rangle} |u\rangle$ i.e. $|u\rangle \propto |v\rangle$

$\frac{\langle u | v\rangle}{\sqrt{\langle u\rangle\langle v\rangle}} = \cos(\theta) e^{i\phi},\ \theta \in [0, \pi/2],\ \phi\in[0, \pi]$